Mercurial > hg > mpdl-xml-content
annotate texts/XML/archimedes/it/pacio_summ2_X02_it_1494.xml @ 10:d7b79f6537bb
Version vom 2009-02-14
| author | Klaus Thoden <kthoden@mpiwg-berlin.mpg.de> |
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| date | Thu, 02 May 2013 11:08:12 +0200 |
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1 <?xml version="1.0"?> |
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2 <archimedes> |
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3 <info> |
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4 <author>Pacioli, Luca</author> |
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5 <title>Tractatus geometrie. Summa de Arithmetica et Geometria, Proportioni et Proportionalita Pars II</title> |
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6 <date>1494</date> |
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7 <place>Venice</place> |
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8 <translator></translator> |
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9 <lang>it</lang> |
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10 <cvs_file>pacio_summa2_X02_it_1494.xml</cvs_file> |
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11 <locator></locator> |
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12 </info> |
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13 <pb></pb> |
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14 <p class="folio"> |
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15 folio 1r |
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16 </p> |
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17 <p class="main"> |
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18 <lb></lb> |
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19 </p> |
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20 <p class="runhead"> |
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21 Distinctio prima. Capitulum primum. 1 |
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22 </p> |
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23 <p class="main"> |
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24 <lb></lb> |
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25 </p> |
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26 <p class="head"> |
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27 Tractatus Geometrie. Pars secunda principalis huius operis et primo eius divisio. |
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28 </p> |
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29 <p class="main"> |
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30 <lb></lb> |
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31 Ora col nome di Jesu. Segue la seconda parte principale dela presente<lb></lb> |
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32 opera. In la quale (commo in principio promettemmo) se tracta dela quantitá conti-<lb></lb> |
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33 nua: cioé geometria quanto ala pratica se aspecti: e anco la theorica<lb></lb> |
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34 de tutte le operationi sempre con degni fondamenti de philosofi chiari e aperti per<lb></lb> |
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35 litterati e vulgari commo nel processo vederasse. E questa tutta chia-<lb></lb> |
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36 maremo tractato. E divideremola in .8. altri parti partiali a reverentia<lb></lb> |
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37 dele .8. beatitudine. E ciascuna sia decta distinctione: e quelle poi subsequenter<lb></lb> |
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38 distingueremo in capituli. Nela prima voglio dimostrare comme le<lb></lb> |
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39 figure quadrate e triangolari sonno da essere recate a quantitá di bracci qua-<lb></lb> |
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40 dri o de altra mesura. Nela seconda voglio mostrare quando un<lb></lb> |
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41 ponto è dato fuor o dentro d’ uno triangolo e da quello si muovi una li-<lb></lb> |
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42 nea a sapere la sua quantita. Nela terza del modo de trovare l’ area delle<lb></lb> |
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43 superficie di .4. lati col modo de misurare le figure di piú di .4. faccie. Nela quarta del modo di mesu-<lb></lb> |
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44 rare li cerchi: e le superficie in monte. Nela quinta del modo de dividere le superficie in parti. Nela sex-<lb></lb> |
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45 ta el modo di trovare l’ area corporale deli corpi. Nela septima del modo de misurare col viso, cioé col<lb></lb> |
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46 vedere. Nela octava e ultima: alcuno caso trattaremo de geometria belli e gentili e così faremo<lb></lb> |
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47 fine lasciando lo mendare del superfluo o mancamento a chi leggi et cetera.<lb></lb> |
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48 <lb></lb> |
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49 </p> |
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50 <p class="head"> |
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51 Divisio et continentia prime distinctionis. |
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52 </p> |
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53 <p class="main"> |
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54 <lb></lb> |
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55 Rettamente volendo tractare è di bisogno accioché particularmente sia trovato quello che<lb></lb> |
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56 desideri: dividere questa distinctione in .8. capituli. Deli quali il primo conterá certe diffinitioni.<lb></lb> |
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57 El .2o. certe demostrationi e conclosioni del primo de Euclide. El .3o. certe conclusioni e demo-<lb></lb> |
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58 strationi del .2o. de Euclide. Quarto certe dimostrationi e conclusioni del sexto de Eu-<lb></lb> |
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59 clide. Quinto in che modo se usa a misurare secondo lo strumento fiorentino secondo el quale in questo tra-<lb></lb> |
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60 ctato al piú ci reggemo. Sexto comme se misurano le figure quadrate. Septimo commo se misura-<lb></lb> |
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61 no li triangoli de ciscuna sorte. Octavo e ultimo comme si truovano li chatetti: over perpendicu<lb></lb> |
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62 lari deli triangoli e comme secondo un modo vulgare s’ usa in sul terreno a misurare. E così veniamo<lb></lb> |
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63 ala prima parte. De quinque circa quam principaliter pratica geometrica versatur et ad bonum<lb></lb> |
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64 agrimensorem spectant. Et de eo pro declaratione cum principiis per sé notis. Capitulum primum prime <lb></lb> |
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65 distinctionis. |
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66 </p> |
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67 <p class="main"> |
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68 Cinque cose sonno necessarie a sapere a chi vuole essere perfectamente pratico nell’ arte di<lb></lb> |
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69 geometria. Delle quali la prima è puncto. Seconda linea. Terza angolo. Quarta superfi-<lb></lb> |
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70 cie. Quinta e ultima corpo. E peroché noi seguitiamo per la magior parte. L. Pisano Io inten-<lb></lb> |
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71 do de chiarire che quando si porrá alcuna proposta senza autore quella sia di detto. L. E quan-<lb></lb> |
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72 do d’ altri sia qui sará l’ autorità aducta. Adunche diffinendo diremo in questo modo. Puncto è quello che<lb></lb> |
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73 non á parte. La linea è una lunghezza senza ampieza quasi una via imaginata: della quale li termini sonno<lb></lb> |
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74 .2. puncti. E sonno di doi maniere linee. Una è detta linea recta. L’ altra è detta linea curva. Li-<lb></lb> |
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75 nea recta è quella che da un ponto a un altro è menata diritta. Linea curva è quella che fa arco. L’ an-<lb></lb> |
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76 golo è il toccamento di doe linee. E possonsi comporre gli angoli de linee recte e di linee curve. E l’ an-<lb></lb> |
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77 golo de linee recte si dice Angolo rettilineo. E l’ angolo di due linee curve si dice curvilineo.<lb></lb> |
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78 L’ angolo de rette linee puó essere in .3. modi. De’ quali uno è quello che è facto dalla squadra e chiamase<lb></lb> |
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79 angolo retto. Un altro se dice angolo obtuso. E questo è quello che è magiore che ’l retto. E un altro se<lb></lb> |
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80 dice angolo acuto: e questo è quello che è minore che ’l retto. Quando una linea retta stará sopra una<lb></lb> |
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81 linea retta e gli due angoli sieno infra loro iguali ciascuno di quegli angoli se dice angolo retto. E la li-<lb></lb> |
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82 nea che sta sopra l’ altra se dice chatetto: over perpendiculare. El termine è fine dela cosa. La figu-<lb></lb> |
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83 ra è quella che sotto a uno o piú termini è constituita. La figura di rette linee: è quella che è circundata<lb></lb> |
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84 da linee rette. La superficie è quella che á lunghezza e larghezza: della quale li termini sonno le linee. La<lb></lb> |
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85 figura di .3. lati è quella che da .3. linee rette è fatta. La figura quadrilatera è quella che è fatta da .4. li-<lb></lb> |
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86 nee rette. La figura multilatera è quella che è fatta da molte linee. Cerchio è una figura piana<lb></lb> |
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87 contenuta da una sola linea che è nominata circunferentia over periferia. Dentro ala qual linea è uno pon-<lb></lb> |
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88 to detto centro di cerchio dal quale tutte le linee che sonno menate alla circunferentia son-<lb></lb> |
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89 no eguali fra loro. Diametro di cerchio è una linea recta che passa sopra il centro: e da ciascun la-<lb></lb> |
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90 to tocca la circunferentia e divide il cerchio in due parti equali. Semicirculo, cioé mezo cerchio è<lb></lb> |
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91 una figura piana contenuta dal diametro del cerchio e dala mitá dela circunferentia. Portione di cer-<lb></lb> |
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92 chio è una figura contenuta d’ una linea retta e dela parte dela circunferentia magiore over minore<lb></lb> |
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93 del semicirculo. Settore di cerchio è una figura piana contenuta da doi linee recte producte<lb></lb> |
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94 dal centro ala perifera: e compreso da quella l’ archo: cioé parte d’ essa perifera. Le linee equedistanti sonno<lb></lb> |
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95 <lb></lb> |
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96 <pb></pb></p> |
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97 <p class="folio"> |
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98 folio1v |
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99 </p> |
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100 <p class="main"> |
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101 <lb></lb> |
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102 </p> |
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103 <p class="runhead"> |
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104 Distinctio prima. Capitulum secundum. |
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105 </p> |
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106 <p class="main"> |
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107 <lb></lb> |
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108 quelle che in una medesima superficie collocate e infinitamente menate da ciascuno lato mai<lb></lb> |
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109 si toccano. E sonno dette linee equidistanti: quelle che menate sopra quelle una linea recta fa-<lb></lb> |
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110 rá li .2. angoli dentro iguali a .2. angoli retti. E l’ angolo di fuora del’ uno sia iguali al’ angolo<lb></lb> |
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111 dentro del’ altra. El quale è aposto a quello di fuora. Comme sieno .2. linee equidistanti .ab. e .cd.<lb></lb> |
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112 Sopra le quali passi la linea recta .ez. segando le linee sopra i ponti .fg. Dico che quando l’ an-<lb></lb> |
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113 golo .bfz. con l’ angolo .fgd. sonno iquali a .2. angoli retti allora le ditte linee sonno equidistan-<lb></lb> |
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114 ti. Overo quando l’ angolo .efb. di fuora è iguali al’ angolo .fgd. Over l’ angolo .efa. di fuora<lb></lb> |
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115 sia iguali al’ angolo .fgc. dentro. Allora le ditte linee sonno equidistanti commo chiaro appa-<lb></lb> |
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116 re. Li corpi sonno di molte maniere comme colonne: cassi: Pozzi: Arche: Piramidi: e altre<lb></lb> |
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117 figure secondo la loro diversitá. Le quali figure nella sexta distinctione apertamente sieno mo-<lb></lb> |
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118 stre. Aduncha a questa prima parte over capitulo faren fine.<lb></lb> |
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119 <lb></lb> |
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120 Substantia omnium conclusionum libri primi Euclidis brevissima. Capitulum secun-<lb></lb> |
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121 dum. prime distinctionis.<lb></lb> |
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122 <lb></lb> |
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123 Senza el tractato de Euclide male si puó fare per coloro che vogliono <lb></lb> |
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124 misurare<lb></lb> |
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125 over attendere ad alcuna sottilitá. E peró quelle demostrationi: diffinitioni: con-<lb></lb> |
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126 clusioni che io vederó necessarie quelle porró solamente. Ponendo il testo. Impero-<lb></lb> |
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127 chè aprovato è per tutti li geometri e non ha bisogno le sue cose con demostratio-<lb></lb> |
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128 ni fare chiare. E in questa parte diremo certe cose nel suo primo libro dette: niente di meno bi-<lb></lb> |
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129 sognano consentire queste cose che sonno ditte .5. petitioni, cioé che: Da un ponto a un altro una li-<lb></lb> |
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130 nea si puó menare diritta. Seconda: che sopra uno centro si puó fare uno cerchio di quan-<lb></lb> |
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131 to spatio vuoi. Terza: che tutti gli angoli retti in fra loro sonno iguali. Quarta: quando una<lb></lb> |
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132 linea retta caderá sopra .2. linee rette e gli .2. angoli da una parte presi e sieno minori di .2. an-<lb></lb> |
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133 goli retti, quelle doi linee senza dubio menate in quella parte si congiogneranno. La quinta:<lb></lb> |
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134 doi linee rette non inchiudano superficie. Le sequenti se dicano conceptioni, cioé:<lb></lb> |
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135 E quelle cose che a una medesima cosa sonno iguali, infra loro sonno iguali. E, se<lb></lb> |
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136 le cose iguali sonno multiplicate o partite per cose iguali, l’ avenimento sia iguali. E,<lb></lb> |
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137 se dele cose iguali si trae over s’ agiongne cose iguali, lo rimanente: over l’ agiognimen-<lb></lb> |
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138 to sia iguali. E, se dele cose iguali s’ agiugne over trae le cose non iguali, el rimanente: over lo<lb></lb> |
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139 agiognimento sia non iguiali. Ogni tutto è magiore che la sua parte. Se una cosa si pone<lb></lb> |
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140 sopra a un’ altra e non avanza e non n’ é avanzata, siano intra loro iguali. Prima conclosione.<lb></lb> |
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141 Sia data una linea retta terminata della quale voglio si facia uno triangolo equi-<lb></lb> |
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142 latero e ciascun suo lato sia la detta linea: cioé che per gli altri .2. lati sia quanto la<lb></lb> |
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143 detta linea. .2. |
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144 </p> |
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145 <p class="main"> |
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146 Sia dato un ponto ad alcuna linea recta: dallo quale si debba menare una linea iguali ala<lb></lb> |
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147 ditta linea. .3. |
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148 </p> |
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149 <p class="main"> |
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150 Sieno proposte .2. linee e sia de bisogno dela magiore torne una parte iguali ala minore. .4. |
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151 </p> |
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152 <p class="main"> |
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153 D’ ogni .2. triangoli deli quali li .2. lati del’ uno sonno iguali a .2. lati del’ altro e gli .2. an-<lb></lb> |
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154 goli di quelli triangoli contenuti da’ detti .2. lati iguali fieno infra loro iguali, cioé l’ uno a<lb></lb> |
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155 l’ altro. Dico che gli altri .2. angoli del’ uno triangolo fieno iguali agli altri .2. angoli<lb></lb> |
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156 del’ altro triangolo: collocati in medesimo luogo. E la basa del’ uno alla basa del’ al-<lb></lb> |
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157 tro. E tutto il triangolo a tutto el triangolo. .5. |
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158 </p> |
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159 <p class="main"> |
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160 D’ ogni triangolo de .2. lati iguali gli angoli che sonno sopra la basa sonno iguali. .6. |
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161 </p> |
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162 <p class="main"> |
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163 Ancora se .2. angoli d’ alcuno triangolo sonno iguali li lati che fanno quegli angoli son-<lb></lb> |
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164 no iguali infra loro. .7. |
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165 </p> |
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166 <p class="main"> |
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167 Se da .2. ponti terminanti alcuna linea .2. linee usciranno: e congiongniasi a uno ponto<lb></lb> |
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168 e da quelli medesimi ponti altre .2. linee iguali ale prime ciascuna alla sua conterminale a una<lb></lb> |
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169 che in altro ponto concorrino in quella medesima parte è impossbile. .8. |
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170 </p> |
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171 <p class="main"> |
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172 D’ ogni .2. triangoli deli quali e .2. lati dell’ uno sonno iguali a .2. lati del’ altro. E la basa<lb></lb> |
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173 del’ uno ala basa del’ altro sia iguali. Dico e .2. angoli che contengono e .2. lati igual del’ uno:<lb></lb> |
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174 sonno iguali a .2. angoli che contengono e .2. lati iguali del altro. .9. |
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175 </p> |
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176 <p class="main"> |
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177 Sia dato uno angolo lo quale voglio dividere per mezo, cioé per igual parti. .10. |
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178 </p> |
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179 <p class="main"> |
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180 Sia data una linea retta la quale voglio dividere in .2. parti iguali. .11. |
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181 </p> |
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182 <p class="main"> |
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183 Sia data una linea retta nella quale sia uno ponto alo quale bisogni menare una per-<lb></lb> |
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184 pendiculare. .12. |
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185 </p> |
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186 <p class="main"> |
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187 Sia dato un ponto fuore della linea data: dal quale bisogni menare una perpendiculare<lb></lb> |
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188 infino alla linea data. .13.<lb></lb> |
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189 <lb></lb> |
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190 <pb></pb></p> |
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191 <p class="folio"> |
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192 folio 2r |
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193 </p> |
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194 <p class="main"> |
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195 <lb></lb> |
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196 </p> |
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197 <p class="runhead"> |
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198 Distinctio prima. Capitulum secundum. 2 |
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199 </p> |
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200 <p class="main"> |
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201 <lb></lb> |
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202 Ogni linea retta che sta sopra una linea retta li .2. angoli fatti da quello sonno retti o<lb></lb> |
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203 sonno iguali a .2. angoli retti sempre. .14. |
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204 </p> |
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205 <p class="main"> |
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206 Se doi linee dal ponto d’ una linea escono in diverse parti. E gli .2. angoli fatti intorn’ a quel-<lb></lb> |
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207 le sieno retti, over iguali a .2. retti, quelle doi linee che escono congionte fieno e una sola linea. <lb></lb> |
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208 .15. |
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209 </p> |
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210 <p class="main"> |
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211 D’ ogni .2. linee rette segandose infra loro: gli angoli contraposti sonno iguali. E per questo<lb></lb> |
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212 è manifesto li .4. angoli fatti da quelle sonno iguali a .4. angoli retti. .16. |
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213 </p> |
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214 <p class="main"> |
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215 Se ciascuno deli lati d’ un triangolo si mena diritto fará l’ angolo de fuora magiore che<lb></lb> |
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216 ciascuno degli angoli dentro oposti a quello angolo. Cioé comme se si menerá del triangolo<lb></lb> |
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217 .abc. il lato .ab. infino al .d. Dico l’ angolo .cbd. essere magiore che l’ angolo .cab. over dell’ angolo .acb.<lb></lb> |
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218 Li .2. angoli d’ ogni triangolo sonno minori di .2. angoli retti: commo tolli qual .17.<lb></lb> |
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219 voli .2. angoli del triangolo .abc. Dico che fienno minori di .2. angoli retti. .18. |
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220 </p> |
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221 <p class="main"> |
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222 El magiore angolo d’ ogni triangolo è sempre opposito al piú longo lato di quel triangolo. Comme<lb></lb> |
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223 sia l’ angolo .bac. del triangolo .acb. magiore di ciascuno degli altri angoli. Dico il lato .bc., ch’ é oppo-<lb></lb> |
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224 sto a quello angolo sia magiore di ciascuno degli altri lati del triangol detto. .19. |
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225 </p> |
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226 <p class="main"> |
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227 El magiore lato d’ ogni triangolo è sempre opposto al magiore angolo del detto trian-<lb></lb> |
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228 golo. E questo per la passata se manifesta. .20. |
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229 </p> |
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230 <p class="main"> |
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231 D’ ogni triangolo li .2. lati sempre fieno magiori che ’l terzo lato. Comme sia il triangolo .abc.<lb></lb> |
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232 Dico che lo lato .ab. e .ac. agionti insiemi sonno magiori delo lato .bc. E così degli altri. .21. |
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233 </p> |
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234 <p class="main"> |
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235 Se da .2. ponti terminali d’ uno triangolo doi linee usciranno e congionghise infra ’l triango-<lb></lb> |
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236 lo (cioé dentro al triangolo) quelle doi linee sonno piú brievi che gli doi lati del triangolo che<lb></lb> |
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237 da detti ponti si muovano. E conterranno magiore angolo. Comme sia il triangolo .abc. e dagli<lb></lb> |
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238 ponti .b. e .c. si menino le linee .bd. e .cd. che si congionghino nel ponto .d. dico che le .2. dette linee<lb></lb> |
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239 sonno minori dela linea .ab. e .ac. E che l’ angolo .d. fatto dale .2. linee: è magiore che l’ angolo<lb></lb> |
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240 .bac. del detto triangolo. .22. |
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241 </p> |
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242 <p class="main"> |
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243 Sienno proposte .3. linee rette. De le quali le. 2. quali voi sieno magiori del’ altra. E volsi<lb></lb> |
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244 dele .3. ditte linee rette constituire un triangolo: faremo in questo modo. Sienno .3. linee<lb></lb> |
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245 date rette .a.b.c. e agiontone insieme .2. qual voi: sienno magiori che l’ altra (che altramen-<lb></lb> |
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246 te el triangolo non si potrebbe conporre per la .20a.) e voglio dele .3. dette linee conporre uno triango-<lb></lb> |
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247 lo. Piglieró una linea retta che sia .de. Ala quale non pongo fine determinato: e dala parte de<lb></lb> |
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248 .d. piglio .df. iguali alla linea .a. comme insegna la .3a. E poi tolgo .fg. iguali ala linea .b. e il .gh.<lb></lb> |
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249 tolgo iguali ala linea .c. E fatto il ponto .f. centro scriveró uno cerchio secondo la quantitá .fd. E sia<lb></lb> |
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250 cerchio .kd. E ancora il ponto .g. faró centro di cerchio e discriveró un cerchio secondo la quantitá<lb></lb> |
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251 .gh. e sia cerchio .hk. li quali .2. cerchi si segheranno in .2. ponti de’ quali l’ uno è il ponto .k. E produ-<lb></lb> |
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252 reró la linea .kf. e .kg. e sia fatto el triangolo .fkg. constituto di .3. linee iguali alle .3. linee da-<lb></lb> |
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253 te: imperoché .fk. sia iguali ala linea .a. e .gk. sia iguali ala linea .c. e .fg. è iguali ala linea .b. e co-<lb></lb> |
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254 sí è il proposito. .23. |
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255 </p> |
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256 <p class="main"> |
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257 Sia dato una linea retta sopra la quale nel termine d’ essa voglio fare uno angolo<lb></lb> |
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258 iguali a uno angolo dato. sia dato la linea .fe. e nel termine .f. voglio fare uno<lb></lb> |
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259 angolo iguali al’ angolo contenuto dale linee .a.b. dato: agiongneró alo detto ango-<lb></lb> |
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260 lo la basa .c. e s’ ará uno triangolo .abc. E ala linea .fe. agiongneró la linea .fd. dala parte del .f. in<lb></lb> |
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261 modo che sia una colla linea .fe. e porró .fd. iguali al lato .a. E dela linea .fe. ne piglieró il ponto<lb></lb> |
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262 .g. e sia la linea .fg. iguali al lato .b. e la linea .gh. porró iguali ala linea over basa .c. e sopra il<lb></lb> |
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263 ponto .f. faró centro e descriveró uno cerchio .dk. secondo la quantitá .fd. E similmente sopra il pon-<lb></lb> |
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264 to .g. porró il pie’ dele sexte immobile: e farró il cerchio .hk. secondo la quantitá .gh. li quali .2. cer-<lb></lb> |
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265 chi s’ intersegano nel ponto .k. Onde menerai la linea .fk. e la linea .gk. e haremo le .2. linee.<lb></lb> |
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266 .kf. e .fg. del triangolo .kfg. iguali a’ .2. lati .a. e .b. del triangolo .abc. e la basa .gk. iguali ala ba-<lb></lb> |
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267 sa .c. comme si pose. Aduncha l’ angolo .f. è iguali al’ angolo dato: che è il proposito. .24. |
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268 </p> |
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269 <p class="main"> |
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270 Ogni .2. triangoli de’ quali .2. lati dell’ uno ae due lati dell’ altro sonno iguali. E l’ an-<lb></lb> |
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271 golo che è fatto dae due lati dell’ uno triangolo sia magiore che l’ altro angolo<lb></lb> |
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272 fatto dae .2. lati iguali dell’ altro triangolo. Dico la basa di quello triangolo che<lb></lb> |
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273 á magiore angolo sia magiore che la basa del’ altro triangolo che á minore an-<lb></lb> |
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274 golo. Comme sienno .2. triangoli .abc. e .def. E sia i lati .ab. e .ac. del triangolo .abc. iguali a doe<lb></lb> |
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275 lati .de. e .df. del triangolo .def. E l’ angolo .a. sia magiore dell’ angolo .d. dico che la<lb></lb> |
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276 basa .bc. sia magiore dela basa .ef. E questo è il proposito. .25. |
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277 </p> |
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278 <p class="main"> |
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279 E cosí per averso dico che stando li .2. triangoli in detta forma ma sia la basa .bc. ma-<lb></lb> |
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280 giore dela basa .ef. dico l’ angolo .a. essere magiore dello angolo .d. .26.<lb></lb> |
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281 <lb></lb> |
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282 <pb></pb></p> |
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283 <p class="folio"> |
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284 folio 2v |
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285 </p> |
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286 <p class="main"> |
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287 <lb></lb> |
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288 </p> |
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289 <p class="runhead"> |
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290 Distinctio prima. Capitulum secundum. |
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291 </p> |
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292 <p class="main"> |
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293 <lb></lb> |
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294 D’ ogni .2. triangoli de’ quali e .2. angoli del’ uno ai doi angoli del’ altro ciascuno al<lb></lb> |
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295 suo relativo sia iguali. E la basa del’ uno sia iguali ala basa del’ altro sará ciascu-<lb></lb> |
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296 no de’ .2. lati del’ uno iguali ae .2. lati del’ altro ciascuno al suo respiciente e l’ ango-<lb></lb> |
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297 lo opposto ala basa del’ uno è iguali al’ angolo opposto ala basa del’ altro. E tut-<lb></lb> |
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298 to il triangolo sia iguali a tutto el triangolo. Comme sia e .2. angoli .b. e .c. del triangolo .abc. iguali<lb></lb> |
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299 ae .2. angoli .e.f. del triangolo .def. E la basa .bc. sia iguali ala basa .ef., dico l’ angolo .a. essere<lb></lb> |
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300 iguali al’ angolo .de. E gli doi lati .ab. e .ac. del triangolo .abc. essere iguali ae .2. lati .de. e .df. del<lb></lb> |
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301 triangolo .def. e tutto il triangolo .abc. sia iguali a tutto il triangolo .def. .27. |
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302 </p> |
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303 <p class="main"> |
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304 Se una linea retta caderá sopra .2. linee recte e gli .2. angoli coalterni fra loro fieno<lb></lb> |
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305 iguali: quelle .2. linee certamente fieno equedistanti. Comme sia la linea .ab. retta che<lb></lb> |
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306 caggia sopra le linee .cd. e .ef. e seghi le dette linee ne’ ponti .g. e .h. E sia l’ angolo .dgh.<lb></lb> |
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307 iguali al’ angolo .chg., dico le linee .cd. e .ef. essere equedistanti. .28. |
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308 </p> |
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309 <p class="main"> |
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310 Se una linea caderá sopra .2. linee e sia l’ angolo di fuora iguali al’ angolo opposto<lb></lb> |
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311 dentro. Dico le doe linee essere equedistanti. Over quando e .2. angoli di fuora da<lb></lb> |
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312 una parte over e .2. angoli dentro da una parte fieno iguali a .2. angoli retti dico<lb></lb> |
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313 quelle .2. linee ancora essere equedistanti. Comme sia caduta la linea .ab. sopra la<lb></lb> |
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314 linea .cd. e sopra la linea .ef. e sia l’ angolo .agd. di fuora iguali al’ angolo .ghf. dentro. Allora di-<lb></lb> |
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315 co le doe linee .cd. e .ef. sonno equedistanti. Over presi e .2. angoli .agc. e .bhe. di fuori e sieno<lb></lb> |
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316 iguali a .2. angoli retti, dico ancora le .2. linee, cioé .cd. e .ef. essere equedistanti. .29. |
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317 </p> |
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318 <p class="main"> |
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319 Se a .2. linee equedistanti caderá una linea, fieno .2. angoli coalterni iguali. Over e .2. an-<lb></lb> |
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320 goli intrinseci da una parte iguali a .2. retti fieno. Over e .2. angoli di fuora fieno iguali a .2.<lb></lb> |
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321 retti. E questo chiaro appare per le .2. passate. .30. |
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322 </p> |
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323 <p class="main"> |
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324 Se sieno .2. o piú linee a una linea equedistanti, dico tutte infra loro sonno equedistanti.<lb></lb> |
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325 Comme sia la linea .ab. e .cd. equedistanti ala linea .ef., dico che .ab. e .cd. sonno infra loro eque-<lb></lb> |
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326 distanti. .31. |
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327 </p> |
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328 <p class="main"> |
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329 Sia dato un ponto fuori d’ una linea dal quale bisogni menare una linea equedistan-<lb></lb> |
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330 te a quella linea data. Nota che s’ intende che ’l ponto sia dato fuori della linea: quan-<lb></lb> |
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331 do menato la linea da ogni lato quanto voi non passerá sopra quello ponto. Sia<lb></lb> |
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332 adunca il ponto .a. dato fuori della linea .bc., dal quale è bisogno menare la linea equedistan-<lb></lb> |
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333 te ala linea .bc. Meneró la linea .ac. comme viene. E constitueró uno angolo .cae. iguali al’ ango-<lb></lb> |
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334 lo .bca. sia adonca .ae. equedistante al .bc. che è il proposito: perché sonno coalterni .32. |
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335 </p> |
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336 <p class="main"> |
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337 Se si mena el lato d’ alcuno triangolo per lo dritto sia l’ angolo di fuora iguali ad a-<lb></lb> |
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338 mendoi gli angoli a quello opposti. E tutti .3. gli angoli d’ uno triangolo son iguali<lb></lb> |
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339 a .2. retti angoli. Comme sia il triangolo .abc. del quale il lato .bc. si meni infino al<lb></lb> |
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340 .d. dico l’ angolo .c. di fuora essere iguali ad amendoi insiemi gli angoli .a. e .b. dentro.<lb></lb> |
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341 E che, agionto insiemi tutti .3. gli angoli di quel triangolo, cioé l’ angolo .a. e l’ angolo .b. e l’ an-<lb></lb> |
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342 golo .c. fieno quanto .2. angoli retti. .33. |
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343 </p> |
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344 <p class="main"> |
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345 Se alle sommitá di .2. linee equedistanti e iguali .2. linee sonno congionte, elle fieno igua-<lb></lb> |
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346 li e ancora equedistanti. Comme sia la linea .ab. equedistante e iguali ala linea .dc.<lb></lb> |
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347 dico che, menato la linea .ac. e .bd. fieno iguali e equedistanti: cioé che la linea .ac.<lb></lb> |
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348 sia iguale e equedistante ala linea .bd. che si manifesta. .34. |
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349 </p> |
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350 <p class="main"> |
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351 Ogni superficie d’ equedistanti lati le linee e gli angoli ex aversi collocati di quelle<lb></lb> |
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352 superficie sonno iguali. E il diametro la divide per mezzo. Comme sia la superficie de<lb></lb> |
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353 equedistanti lati .abcd., cioé che .ac. sia equedistante al .db. e .dc. sia equedistante<lb></lb> |
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354 al .ab. Dico che .db. sia iguali al .ac. e .cd. sia iguali al .ab. e l’ angolo .a. sia iguali al’ an-<lb></lb> |
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355 golo .d. e l’ angolo .c. al’ angolo .b. E il diametro .da. dividerá la detta superficie per .2. iguali parti,<lb></lb> |
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356 é questo chiaro e manifesto. .35. |
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357 </p> |
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358 <p class="main"> |
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359 Tutte le superficie d’ equedistanti lati sopra una basa e in medesime linee constitu-<lb></lb> |
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360 te sonno infra loro iguali. Comme sieno .2. linee equedistanti .ab. e .cd. infra le quali si fa-<lb></lb> |
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361 cia la superficie .acfe. d’ equedistanti lati sopra la basa .ce. e ancora, sopra la medesi-<lb></lb> |
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362 ma basa, si faccia la superficie .gc. e .be. d’ equedistanti lati, dico le .2. pprima dette superficie essere iguali.<lb></lb> |
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363 Tutti e pararelli in base iguali e in medesime linee constituti sonno .36.<lb></lb> |
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364 iguali. El paralello è una superficie di .4. lati equedistanti. Adonca sieno .2. superficie<lb></lb> |
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365 paralellograme: cioé. 2. paralelli .abcd. e .efgh. d’ equedistanti lati e habino le base .cd. e<lb></lb> |
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366 .gh. iguali. E sienno infra doe linee medesime: cioé equedistanti. Dico il paralello<lb></lb> |
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367 .abcd. essere iguali al paralello .efgh. .37.<lb></lb> |
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368 <lb></lb> |
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369 <pb></pb></p> |
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370 <p class="folio"> |
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371 folio 3r |
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372 </p> |
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373 <p class="main"> |
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374 <lb></lb> |
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375 </p> |
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376 <p class="runhead"> |
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377 Distinctio prima. Capitulum secundum. 3 |
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378 </p> |
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379 <p class="main"> |
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380 <lb></lb> |
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381 Tutti i triangoli fatti con una medesima basa e infra doi medesime linee equedistanti sonno in-<lb></lb> |
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382 fra loro iguali. Comme sienno .2. triangoli .abc. e .dbc. fatti con una basa .bc. e infra doi linee medesi-<lb></lb> |
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383 me: cioé equedistanti che sonno .ac. e .bf., dico e detti .2. triangoli essere iguali infra loro. .38. |
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384 </p> |
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385 <p class="main"> |
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386 Se i triangoli sopra le base iguali caderanno infra doi linee equedistanti, saranno infra loro igua-<lb></lb> |
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387 li. Comme sienno .2. triangoli .abc. e .def. fatti sopra le base .bc. e .ef. iguali e sienno infra le linee .ag. e<lb></lb> |
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388 .bh. equedistanti. Dico che li sonno iguali infra loro: cioé il triangolo .abc. è iguali al triangolo .def. .39. |
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389 </p> |
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390 <p class="main"> |
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391 Tutti i triangoli iguali, se haranno una medesima basa. E verso una parte e fienno fat-<lb></lb> |
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392 ti infra .2. linee equedistanti. Comme sienno .2. triangoli .abc. e .dbc. fatti sopra la basa .bc. e<lb></lb> |
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393 sienno infra loro iguali e verso una parte. Dico che li sonno infra doi linee equedistanti: que-<lb></lb> |
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394 sta è conversa alla .37a. E nota che di questa e dela passata che se alcuna linea retta<lb></lb> |
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395 segherá e .2. lati d’ alcuno triangolo per igual parte: ella sará al terzo lato equedistante. Comme sia il trian-<lb></lb> |
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396 golo .abc. del quale e .2. lati .ab. e .bc. sienno segati dala linea .de. per igual parte: cioé .ab. nel ponto .d. e<lb></lb> |
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397 .bc. nel ponto .e., dico la linea .de. sará equedistante ala linea .ac. .40. |
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398 </p> |
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399 <p class="main"> |
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400 Se .2. triangoli iguali sopra le base iguali d’ una medesima linea: cioé che menando<lb></lb> |
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401 l’ una basa in verso l’ altra sia con quella una medesima linea verso una parte, dico essere in-<lb></lb> |
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402 fra doi linee equedistanti e detti triangoli. Sienno .2. triangoli .abc.dfe. fatti iguali e son-<lb></lb> |
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403 no sopra .2. base iguali che sonno .bc. e .fe. d’ una medesima linea .be. e verso una parte.<lb></lb> |
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404 Dico i ditt[i] triangoli essere infra doi linee equedistanti che sia il proposito. .41. |
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405 </p> |
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406 <p class="main"> |
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407 Se uno paralello e uno triangolo fienno fatti in medesime base e in equedistanti linee. El<lb></lb> |
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408 paralello sia doppio al triangolo. Comme sia il paralello .abcd. e il triangolo .edb. cia-<lb></lb> |
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409 scuno sopra la basa .bd. e sienno fatti infra le linee .af. e .bh. che sonno equedistanti. Dico<lb></lb> |
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410 el detto paralello essere doppio al detto triangolo. Similmente si puó provare che<lb></lb> |
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411 si ’l paralello e il triangolo sonno nelle base iguali e nelle linee equedistanti, fienno fatti in questo mo-<lb></lb> |
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412 do che ’l paralello sará doppio al triangolo. E, benché Euclide non la ponesse, facilmente si ma-<lb></lb> |
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413 nifesta. .42. |
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414 </p> |
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415 <p class="main"> |
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416 Io voglio desegnare una superficie d’ equedistanti lati che habbia e .2. angoli contra-<lb></lb> |
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417 posti iguali a uno angolo dato e la detta superficie sia iguali a uno triangolo dato.<lb></lb> |
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418 Comme sia lo dato angolo .a. e lo asegnato triangolo sia .bcd. voglio fare una superfi-<lb></lb> |
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419 cie de equedistanti lati iguali al triangolo dato dela quale e .2. angoli contraposti sienno iguali a l’ an-<lb></lb> |
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420 golo .a. dato. Meneró la linea .bf. dal ponto .b. equedistanti ala linea .cd. e divido la basa .cd.<lb></lb> |
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421 nel ponto .e. per lo mezzo. E meneró la linea .be. e dal ponto .b. meneró .bf. equedistante ala linea<lb></lb> |
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422 .cd. commo è ditto. E sia per la .38a. el triangolo .bed. iguali al triangolo .bec. Perché il triangolo<lb></lb> |
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423 .bed. è mittá del triangolo .bcd. Constitueró sopra il ponto .e. l’ angolo .deg. iguali al’ angolo .a. e<lb></lb> |
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424 faró il paralello .gedf. iguale al detto triangolo: imperoché per la passata egli è doppio al trian-<lb></lb> |
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425 golo .bed. adonca habiamo constituto uno paralello .gedf. iguali al triangolo .bcd. e gli .2. an-<lb></lb> |
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426 goli contraposti: cioé l’ angolo .ged. e l’ angolo. gfd. ciascuno è iguali al’ angolo .a. che è il pro-<lb></lb> |
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427 posito. .43. |
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428 </p> |
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429 <p class="main"> |
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430 E suplimenti d’ ogni paralello che sonno fatti dal diametro sonno infra loro iguali.<lb></lb> |
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431 Comme sia il paralello .abcd. nel quale si faccia il diametro .bc. E menisi .ef. equedistante<lb></lb> |
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432 al’ uno e l’ altro lato .ab. e .cd. la quale segherá il diametro nel ponto .k. Alo quale .gkh.<lb></lb> |
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433 equedistanti al’ uno e l’ altro lato .ca. e .bd. E sia il paralello .abcd. diviso in .4. paralelli de’ quali<lb></lb> |
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434 .2. eckh. e .gkbf. sonno detti stare intorno al diametro, gli altri: cioé .aegk. e .kfhd. sonno detti<lb></lb> |
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435 supplementi; questi .2. supplementi dico che sonno infra loro iguali. .44. |
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436 </p> |
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437 <p class="main"> |
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438 Sia proposta una linea ala quale sia de bisogno desegnare una superficie d’ equedistanti la-<lb></lb> |
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439 ti: ala quale sienno gli angoli contraposti iguali al’ angolo dato e la detta superficie sia<lb></lb> |
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440 iguali al triangolo asegnato. Desegnare una superficie d’ equedistanti lati a una li-<lb></lb> |
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441 nea over sopra una linea e di quella linea fare uno lato ala detta superficie. Sia adon-<lb></lb> |
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442 ca la data linea .ab. e il dato angolo .c. e il dato triangolo .def. Sopra ala linea .ab. voglio de-<lb></lb> |
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443 segnare una superficie d’ equedistanti lati per tal modo che la linea .ab. sia uno lato dela detta superficie.<lb></lb> |
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444 Dela quale superficie e .2. angoli contraposti sienno iguali al’ angolo .c. E quella tutta sia iguali al trian-<lb></lb> |
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445 golo .def. Ora é da questa ala .42a. imperoché qui è dato uno lato della superficie, cioé la linea .ab. e a quel-<lb></lb> |
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446 la non è dato alcuno. E questo adonca volendo fare ala linea .ab., agiongo secondo la rettitudine de-<lb></lb> |
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447 la linea data .ag. che pongo iguali ala linea .ef. che è la basa del dato triangolo. Sopra la<lb></lb> |
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448 quale constituiró uno triangolo iguale e equilatero al triangolo dato in questo modo. Faró l’ angolo<lb></lb> |
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449 .agk. iguali al’ angolo .e. e l’ angolo .gak. iguali al’ angolo .f. per la .23a. E perché la linea .ga. è posta<lb></lb> |
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450 iguali ala linea .ef. sará per la .26a. il triangolo .gka. iguale e equilatero al triangolo .efd.; divideró<lb></lb> |
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451 <lb></lb> |
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452 <pb></pb></p> |
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453 <p class="folio"> |
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454 folio 3v |
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455 </p> |
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456 <p class="main"> |
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457 <lb></lb> |
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458 </p> |
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459 <p class="runhead"> |
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460 Distinctio prima. Capitulum tertium. |
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461 </p> |
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462 <p class="main"> |
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463 <lb></lb> |
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464 adonca .ga. per igual parte nel ponto .h. e meneró .hk. E meneró dal ponto .k. la linea .kn. equedistan-<lb></lb> |
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465 te ala linea .gb., sará adonca per la .38a. el triangolo .ahk. iguali al triangolo .khg. Alora sopra il<lb></lb> |
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466 ponto .a. faró l’ angolo .gal. iguali al’ angolo dato: cioé al’ angolo .c. e compiuto, sopra la basa .ah. infra le<lb></lb> |
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467 linee .gb. e .mn. equedistanti faró la superficie d’ equedistanti lati .lmah. la quale per la .41a. sia doppia al trian-<lb></lb> |
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468 golo .kha. Onde ella è iguali a tutto il triangolo .kga. Per la qual cosa sará iguali al triangolo<lb></lb> |
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469 .def. proposto. Meneró adonca .bn. equedistante ala linea .al. e produceró il diametro .na. El quale me-<lb></lb> |
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470 neró infino che concorrerá nel ponto .o. e compiuto faró una superficie d’ equedistanti lati .mnoq. E me-<lb></lb> |
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471 neró la linea .la. infino che concorrerá con la linea .qo. che sia la linea .lap. Sará adonca per la pas-<lb></lb> |
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472 sata .abpq. iguali al supplemento .lmha. Per la qual cosa e al triangolo .def. E perché per la<lb></lb> |
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473 .15a. l’ angolo .lah. è iguali al’ angolo .bap. e peró l’ angolo .bap. è iguali al’ angolo .c. E peró è fatto sopra<lb></lb> |
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474 la data linea .ab. la superficie d’ equedistanti lati .abpq. iguali al dato triangolo .def. dela quale l’ uno e l’ al-<lb></lb> |
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475 tro angolo contraposti sonno iguali al’ angolo .c. E quelli .2. angoli sonno l’ angolo .a. e l’ angolo .q., cioé<lb></lb> |
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476 dico che ciascuno deli angoli .a. e .q. sonno iguali al’ angolo .c. dato. E così habiamo il proposito. .45. |
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477 </p> |
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478 <p class="main"> |
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479 Io voglio d’ una linea data comporre un quadrato. Comme sia data la linea .ab. dela qua-<lb></lb> |
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480 le voglio descrivere uno quadrato: dali ponti .a. e .b. meno le linee .ac. e .bd. perpendicu-<lb></lb> |
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481 lari ala linea .ab. che fienno equedistanti per la .28a. e fienno .ac. e .bd. E farolle iguali ala<lb></lb> |
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482 linea .ab. E menise la linea .cd., sia quella iguale e equedistante ala linea .ab. per la .33a. E<lb></lb> |
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483 perché l’ uno e l’ altro angolo è retto: cioé l’ angolo .a. e .b. per l’ ultima parte dela .29a. sará .c. e .d. retto.<lb></lb> |
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484 Adonca per la diffinitione de’ quadrati .abcd. é quadrato ch’ é il proposito. .46. |
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485 </p> |
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486 <p class="main"> |
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487 In ogni triangolo rettangolo el quadrato del lato opposto al’ angolo retto è iguale<lb></lb> |
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488 ae .2. quadrati che dagli altri doi lati sonno constituti. Comme sia il triangolo .abc. Del<lb></lb> |
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489 quale l’ angolo .a. sia retto. Dico che ’l quadrato delo lato .bc. è iguali ali .2. quadrati fatti<lb></lb> |
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490 delo lato .ab. e delo lato .ac. che è il proposito. .47. |
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491 </p> |
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492 <p class="main"> |
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493 Se ’l quadrato d’ uno lato d’ un triangolo é quanto li quadrati degli altri doi lati, quello angolo è ret-<lb></lb> |
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494 to. Questa è opposta ala passata. Havendo con brevitá veduto il primo de Euclide. Adonca<lb></lb> |
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495 alcune conclusioni del secondo diremo e peró al secondo capitolo faciendo fine direm del terzo.<lb></lb> |
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496 Declaratio e demostratio secundi libri Euclidis. Capitulum tertium prime distinctionis.<lb></lb> |
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497 Ogni paralello d’ angoli retti è detto contenersi in sule doi linee che fanno gli ango-<lb></lb> |
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498 li retti. El paralello (commo é detto) è una superficie d’ equedistanti lati. El paralello di<lb></lb> |
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499 retti angoli è una superficie havente tutti gli angoli retti ed é fatto del produtto d’ u-<lb></lb> |
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500 no de’ suoi lati contenente l’ angolo retto nel’ altro lato. E peró è detto contenersi in<lb></lb> |
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501 su quelli lati. |
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502 </p> |
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503 <p class="main"> |
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504 Quelle superficie che sonno intorno al diametro d’ ogni spatio di paralello sonno detti<lb></lb> |
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505 paralelli che stanno intorno al diametro. De’ quali ciascuno con gli .2. supplementi si no-<lb></lb> |
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506 mina gnomone. Sia dicemmo nela .43a. quali erano li supplementi e quali paralelli che<lb></lb> |
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507 stanno intorno al diametro. Dico adonca che sia uno paralello .abcd. del quale il diametro .ad.<lb></lb> |
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508 sia diviso da doi linee .ef. e .gh. menate equedistanti a’ lati opposti del detto paralello segantesi<lb></lb> |
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509 sopra il diametro .ad. nel ponto .k. e fanno .4. paralelli i quali sonno li paralelli .agek.kfhd.<lb></lb> |
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510 .kehc.kgfb. Deli quali paralelli .agek. e .khfd., perché il diametro del gran paralello gli se-<lb></lb> |
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511 ga per mezzo, si dicono paralelli che stanno intorno al diametro. Gli altri che ’l diametro non<lb></lb> |
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512 gli sega sonno detti supplementi che amendoi e supplementi agionti con l’ uno o con l’ altro de’ paralelli<lb></lb> |
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513 che stanno intorno al diametro se dici Gnomone. Commo se dimostra da lato.<lb></lb> |
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514 Sienno proposti .2. qual voi quadrati. Dove al’ uno sia de bisogno agiongnere uno gnomo-<lb></lb> |
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515 ne iguali al’ altro quadrato. Commo sienno proposti .2., cioé .ab. e .cd. E sia proposto di fa-<lb></lb> |
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516 re un gnomone intorno al quadrato .ab. iguale al quadrato cd. Per la qual cosa fare, me-<lb></lb> |
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517 nise un lato del quadrato .ab. infino ala equalitá delo lato del quadrato .cd. continuo e deritto. E sia<lb></lb> |
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518 .fe. El quale .fe. sia iguali a uno lato del quadrato .cd. E dal ponto .e. si meni una linea retta al .a.<lb></lb> |
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519 e sia il triangolo .afe. ortogonio imperoché l’ angolo .f. è retto. E il quadrato .cd. è commo el quadrato<lb></lb> |
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520 del .ef. E il quadrato .fa. è iguali al quadrato .ab. Adonca il quadrato .ae. è iguali al quadrato .ab. e .cd.<lb></lb> |
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521 E perché .efa. è triangolo, li lati .ef. e .fa. sonno magiori che lo lato .ae. per la .20a. del primo. Ma lo<lb></lb> |
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522 lato .fa. è iguali alo lato .fb. per la ragione dela quadratura: cioé perché è lato del quadrato .ab. Adon-<lb></lb> |
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523 ca .ef. e .fb. sonno magiori delo lato .ae. Adonca .eb. è magiore che .ae. Piglise adonca del .be. lo<lb></lb> |
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524 eguale del .ea. nel ponto .o. in tal modo che ’l .bo. sia quanto .ea., adonca il quadrato .bo. è iguali ali detti .2.<lb></lb> |
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525 quadrati, per la qual cosa, sopra la linea .bo. si constituisca un quadrato che sia quadrato .bong. El qua-<lb></lb> |
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526 le quadrato agiongne al quadrato .ab. quello gnomone: cioé il paralello .ingh. e il paralello<lb></lb> |
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527 .ahfo. E peró quello gnomone è iguale al quadrato .cd. che è il proposito.<lb></lb> |
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528 <lb></lb> |
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529 <pb></pb></p> |
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530 <p class="folio"> |
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531 folio 4r |
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532 </p> |
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533 <p class="main"> |
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534 <lb></lb> |
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535 </p> |
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536 <p class="runhead"> |
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537 Distinctio prima. Capitulum quartum. 4 |
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538 </p> |
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539 <p class="main"> |
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540 <lb></lb> |
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541 Denanze in questo, nella parte principale de arithmetica, dicemmo dele .11. conclusioni del secon-<lb></lb> |
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542 do de Euclide exemplificando abastanza, peró qui non le replico: conciossiaché li sienno a tuo<lb></lb> |
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543 piacere. Nondimeno qui repigliaró le .4. ultime per esserci al proposito. .11. |
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544 </p> |
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545 <p class="main"> |
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546 Sia data una linea che s’ abbia a dividere in questo modo che quello ch’ é fatto di tut-<lb></lb> |
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547 ta la linea per la menor parte sia iguale al quadrato dela parte magiore. Dicise fatto<lb></lb> |
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548 d’ una linea in un’ altra: quella superficie che è composta: over contenuta dale .2. linee con<lb></lb> |
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549 gli angoli retti: cioé commo la compositione del paralello rettangolo. Sia adonca la<lb></lb> |
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550 data linea .ab. la quale voglio dividere in tal modo che quello ch’ é fatto del .ab. nella menore<lb></lb> |
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551 parte sia iguale al quadrato dela magiore parte. Scriveró il quadrato di tutta la linea .ab.<lb></lb> |
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552 che sia .abcd. E il lato .bd. divideró per igual parti nel ponto .e. e produceró .ae. e faró .ebf. in modo<lb></lb> |
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553 che .ef. sia iguale del .ae. E del .bf. descriveró il quadrato .bfgh. dico .ab. essere divisa com-<lb></lb> |
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554 mo voi: cioé .ah. e .hb., cioé che, multiplicato .ab. in .ah., è iguale al quadrato .hbfg. Adonca<lb></lb> |
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555 .ab. è divisa che l’ una parte é .ah. e l’ altra .hb. che è il proposito. E nota che non bisogna afa-<lb></lb> |
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556 tigarsi in volere dividere in quello modo uno numero ratiocinato perché è impossibile commo per la<lb></lb> |
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557 .29a. del .6o. e anche uno incidente dela .16a. del .9o. si manifesta. .12. |
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558 </p> |
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559 <p class="main"> |
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560 Nelli triangoli che hano uno angolo obtuso tanto è il quadrato delo lato ch’ é sotto-<lb></lb> |
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561 posto al’ angolo obtuso piú che .2. quadrati degli altri doi lati che fanno quello angolo<lb></lb> |
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562 obtuso: quanto è .2. volte quello ch’ é fatto d’ uno di quelli lati che tengono l’ angolo nello<lb></lb> |
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563 agiongnimento a quello lato dove cade la perpendiculare. Commo sia il triangolo .abc.<lb></lb> |
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564 havente l’ angolo .a. obtuso. E dal ponto .c. si meni la perpendiculare ala linea .ba. che, per necessitá ca-<lb></lb> |
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565 derá fuore del triangolo .abc. E, se non cadesse fuore, l’ angolo .a. sarebbe retto: over menor ch’ el<lb></lb> |
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566 retto che l’ uno e l’ altro è impossibile. Sia adonca .cd. perpendiculare sopra la linea .ab. e produce-<lb></lb> |
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567 ró la linea .ab. infino al .d. Dico che ’l quadrato del lato .bc. che è sottoposto al’ angolo obtuso è tan-<lb></lb> |
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568 to magiore de’ .2. quadrati dele .2. linee .ab. e .ac. contenenti quello angolo: quanto è quello del .ba. in .ad.<lb></lb> |
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569 doi volte che chiaro appare per la figura passata. E nota che una linea si dici potere tan-<lb></lb> |
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570 to quanto è il quadrato constituto dala detta linea. .13. |
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571 </p> |
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572 <p class="main"> |
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573 D’ ogni triangolo oxigonio tanto puol meno il lato ch’ é opposto al’ angolo acuto de-<lb></lb> |
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574 gli altri .2. lati: quanto è il doppio del quadrato del lato dove cade la perpendiculare ala<lb></lb> |
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575 distantia delo angolo acuto. Quello che è qui posto del’ angolo acuto del triangolo<lb></lb> |
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576 oxigonio ha la veritá d’ ogni angolo acuto di ciascuno triangolo, o voli ortogo-<lb></lb> |
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577 nio: over ampligonio: over oxigonio. Commo sia il triangolo .abc. havente l’ angolo .c. acuto. On-<lb></lb> |
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578 de si meni la perpendiculare dal’ angolo .b., over dal’ angolo .a., in sula faccia delo .ac., over del .bc. E<lb></lb> |
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579 sempre la perpendiculare caderá intra il triangolo. E, se fosse il triangolo che havesse uno angolo ret-<lb></lb> |
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580 to, muovisi la perpendiculare da quello angolo retto: conciosiacosaché ogni triangolo á .2. ango-<lb></lb> |
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581 li acuti. E, se fosse triangolo ampligonio, muovisi dal’ angolo ampligonio. Adonca meneró la perpen-<lb></lb> |
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582 diculare .ad. in sula faccia .bc., dico che ’l quadrato .ab. (che è opposto al’ angolo .c. acuto) é tanto<lb></lb> |
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583 meno che’ .2. quadrati di .2. linee .ac. e .bc., quanto è il doppio di quello ch’ é facto del .bc. in .dc. E<lb></lb> |
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584 questo è il proposito. .14. |
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585 </p> |
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586 <p class="main"> |
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587 Scrivase uno quadrato iguale a uno triangolo dato. Commo sia dato uno triangolo<lb></lb> |
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588 .a. al quale vogliamo scrivere uno quadrato iguale: e farassi in questo modo. Scriverasse<lb></lb> |
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589 una superficie d’ equedistanti lati e di retti angoli iguale al triangolo dato secondo la do-<lb></lb> |
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590 ctrina dela .42a. del primo. E sia la superficie .bcde. Dela quale, se i lati fienno iguali, haremo quello<lb></lb> |
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591 che cerchiamo: imperoché la ditta superficie sará quadrata. Ma, se i lati sonno non iguali, alora agion-<lb></lb> |
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592 gneró il magiore al menore e sia .cf. iguale a uno de’ lati menori e .bc. sia uno de’ magiori, adon-<lb></lb> |
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593 ca .bf. sará iguali a’ .2. lati della detta superficie, cioé a uno magiore e a uno menore. Dapoi la li-<lb></lb> |
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594 nea .bf. divideró per igual parti nel ponto .g. e faró .g. centro. E sopra la linea .bf., secondo la quantitá dela<lb></lb> |
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595 linea .bg., descriveró uno mezzo cerchio .bhf. e il lato .ec. produceró insino a tanto che segherá<lb></lb> |
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596 la circunferentia nel ponto .h. Dico che ’l quadrato .choi. è iguale al triangolo dato: che è il propo-<lb></lb> |
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597 sito. E nota che per questo modo si truova il lato tetragonico di ciascuna figura de linee rette,<lb></lb> |
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598 commo se sia fatta. Imperoché ogni tale figura risolvaremo in triangoli e a ciascuno di quelli<lb></lb> |
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599 triangoli trovaremo el lato tetragonico secondo la doctrina di questa. E trovaremo, per la pe-<lb></lb> |
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600 nultima del primo, una linea che possa per tuti li lati tetragonichi trovati. E questo basti et cetera.<lb></lb> |
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601 Basta questo quanto al terzo capitolo e del quarto diremo. |
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602 </p> |
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603 <p class="main"> |
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604 Demostratio omnium conclusionum sucincte sexti libri Euclidis. Capitulum quartum. Prime definitioni.<lb></lb> |
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605 Le superficie simili sonno quelle dele quali gli angoli del’ una sonno iguali agli angoli de-<lb></lb> |
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606 l’ altra e i lati che contengono quelli lati sonno in medesima proportione. Comme sienno .2.<lb></lb> |
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607 <lb></lb> |
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608 <pb></pb></p> |
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609 <p class="folio"> |
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610 folio 4v |
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611 </p> |
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612 <p class="main"> |
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613 <lb></lb> |
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614 </p> |
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615 <p class="runhead"> |
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616 Distinctio prima. Capitulum quartum. |
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617 </p> |
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618 <p class="main"> |
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619 <lb></lb> |
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620 triangoli .abc.def. e sia l’ angolo .e. del triangolo .def. iguali al’ angolo .b. del triangolo .abc. e l’ an-<lb></lb> |
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621 golo .a. iguali al’ angolo .d. e l’ angolo .c. iguale al’ angolo .f. e la proportione del .ab. al .de. commo<lb></lb> |
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622 .ac. al .df. e il .bc. al .ef. allora e fienno simili. |
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623 </p> |
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624 <p class="main"> |
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625 Le superficie de lati mutui sonno quelle nele quali e lati sonno nela proportionalitá non<lb></lb> |
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626 continua retransitive. Commo sienno .2. quadrilateri .abc.def. e la proportione del .ab.<lb></lb> |
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627 de lato primo al .de. de lato secondo sará commo la proportione del .ef. de lato secondo<lb></lb> |
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628 alo lato .bc. lato del primo. Alora queli .2. quadrilateri sonno di lati mutui over mu-<lb></lb> |
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629 tachefia che così se dicano. |
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630 </p> |
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631 <p class="main"> |
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632 La linea se dici esser divisa secondo la propotione havente il mezo et due extremi: quando quel-<lb></lb> |
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633 la medesima proportione è de tutta la linea ala magiore parte commo la magiore parte ala minore. <lb></lb> |
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634 Conclusio prima. |
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635 </p> |
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636 <p class="main"> |
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637 Se fienno .2. superficie di rette linee .e. d’ equedistanti lati over de’ triangoli, e sia una<lb></lb> |
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638 medesima alteza la loro: tanto è la proportione del’ una al’ altra quanto la basa del’ u-<lb></lb> |
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639 na ala basa del’ altra. Commo sienno doi paralelli .abc. et .def. d’ iguale alteza, di-<lb></lb> |
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640 co la loro proportione é commo .bc. al .ef. e simile de’ triangoli. Commo sienno .2. triango-<lb></lb> |
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641 li .abc. e .def. dove, menate le linee perpendiculari .ag. e .dh. che sienno iguali: dico tal propor-<lb></lb> |
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642 tione è l’ una al’ altra commo .bc. al .ef. .2.<lb></lb> |
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643 <lb></lb> |
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644 Se una linea retta segherá .2. lati d’ uno triangolo e sia equedistante al terzo lato. Di-<lb></lb> |
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645 co che quella linea sega queli lati proportionalmente. E similmente, per averso, se quel-<lb></lb> |
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646 la linea sega queli lati proportionalmente, ela sará equedistante al terzo lato. Com-<lb></lb> |
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647 mo sia il triangolo .abc. e la linea .de. seghi .2. lati del triangolo, cioé .ab. e .ac. ne’<lb></lb> |
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648 ponti .d. e .e. e sia equedistante al lato .bc. Dico che tale proportione è .ad. al .bd. commo è .ae. al .ec.<lb></lb> |
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649 E, quando tale proportione é del .ad. al .db. commo .ae. al .ec., allora la linea .de. sia equedistante<lb></lb> |
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650 ala linea .bc. .3.<lb></lb> |
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651 <lb></lb> |
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652 Se d’ alcuno degli angoli d’ alcuno triangolo una linea retta si meni infino ala ba-<lb></lb> |
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653 sa in modo che la divida quello angolo per .2. parti iguali, dico tal proportione è dele<lb></lb> |
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654 parti dela basa commo è del’ uno lato al’ altro. E quando una retta divide uno an-<lb></lb> |
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655 golo in modo che le parti dela basa sonno in proportione commo l’ uno de’ .2. altri lati del tri-<lb></lb> |
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656 angolo al’ altro lato: allora quel angolo è diviso in .2. parti iguali. Commo sia el triangolo .abc. del qua-<lb></lb> |
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657 le l’ angolo .a. sia diviso in .2. parti iguali dala linea .ad., dico che tale proportione è del .bd. al .dc.<lb></lb> |
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658 commo .ba. al .ac. E così per averso. .4. |
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659 </p> |
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660 <p class="main"> |
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661 D’ ogni .2. triangoli de’ quali gli angoli del’ uno agli angoli del’ altro sonno iguali, e lati de’ dit-<lb></lb> |
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662 ti triangoli che contengono e simili angoli sonno in una proportione infra loro. Commo<lb></lb> |
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663 sienno .2. triangoli .abc. e .def. e sia l’ angolo .a. iguali al’ angolo .d. e l’ angolo .b. iguali a-<lb></lb> |
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664 l’ angolo .f. e l’ angolo .c. iguali al’ angolo .e., dico che una medesima proportione sia il<lb></lb> |
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665 lato .ab. al lato. df. con quella del lato .bc. alo lato .fe. con quello delo lato .ac. al lato .de.<lb></lb> |
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666 D’ ogni .2. triangoli de’ quali ciascun lato al suo relativo á una medesima proportione, gli an-<lb></lb> |
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667 goli che sonno contenuti da ditti lati sonno simili infra loro. Questa è conversa ala pas-<lb></lb> |
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668 sata: cioé sia la proportione del .ab. al .df. comme .de. al .ca. e commo .fe. al .bc. Dico<lb></lb> |
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669 l’ angolo .d. esser simile al’ angolo .a. e l’ angolo .f. al’ angolo .b. e l’ angolo .e. al’ angolo .c. .6. |
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670 </p> |
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671 <p class="main"> |
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672 Sienno .2. triangoli de’ quali uno angolo del’ uno sia iguali al’ angolo del’ altro e gli .2. lati<lb></lb> |
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673 che contengono l’ uno angolo del’ uno triangolo abbino una medesima proportione<lb></lb> |
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674 agli altri .2. lati che contengono l’ altro angolo del’ altro triangolo. Dico i ditti .2. triango-<lb></lb> |
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675 li essere equiangoli infra loro. Commo sienno .2. triangoli .abc. e .def. e sia l’ angolo .a. si-<lb></lb> |
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676 mile al’ angolo .d. e sia una medesima proportione quella del .ab. al .de. commo quella del .ac. al .df., di-<lb></lb> |
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677 co li .2. triangoli esser equiangoli: cioé che l’ angolo .b. è iguale al’ angolo .e. e l’ angolo .c. al’ angolo .f. .7. |
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678 </p> |
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679 <p class="main"> |
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680 Se saranno .2. triangoli de’ quali uno angolo del’ uno a uno angolo del’ altro sia eguale<lb></lb> |
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681 e gli .2. lati degli altri angoli sienno proportionali: e uno degli angoli del’ uno sia magio-<lb></lb> |
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682 re over menore del retto: e così l’ angolo del’ altro sia magiore over menore del ret-<lb></lb> |
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683 to. Dico che e detti .2. triangoli sonno equiangoli. Commo sia .2. triangoli .abc. e .def. e<lb></lb> |
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684 sia l’ angolo .a. iguale al’ angolo .d. e la proportione del .ac. al .df. sia commo .bc. al .ef. e niuno degli an-<lb></lb> |
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685 goli .b. e .e. sia menore del retto over e sienno menori: cioé non voglio sienno retti. Dico l’ uno trian-<lb></lb> |
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686 golo essere equiangolo al’ altro. .8. |
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687 </p> |
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688 <p class="main"> |
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689 Se dal’ angolo retto del triangolo ortogonio una perpendiculare si muove in sula basa,<lb></lb> |
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690 dividerá il triangolo in .2. triangoli simili al triangolo grande, comme sia il triangolo<lb></lb> |
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691 .abc. e sia l’ angolo .a. retto: dal quale si meni la perpendiculare .ad., dico che ’l triangolo .abd.<lb></lb> |
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692 <lb></lb> |
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693 <pb></pb></p> |
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694 <p class="folio"> |
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695 folio 5r |
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696 </p> |
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697 <p class="main"> |
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698 <lb></lb> |
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699 </p> |
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700 <p class="runhead"> |
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701 Distinctio prima. Capitulum quartum. 5 |
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702 </p> |
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703 <p class="main"> |
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704 <lb></lb> |
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705 è simile al triangolo .abc. e il triangolo .adc. e simile al triangolo .abc. E che il triangolo. adc. è<lb></lb> |
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706 simile al triangolo .adb. E per questo se manifesta lo lato .ad. essere in proportione con ciascuna dele<lb></lb> |
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707 parti .bd. e .dc. cioé sia in medio loco proportion ale ditte. .9. |
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708 </p> |
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709 <p class="main"> |
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710 Sienno proposte .2. linee infra le quali bisogni trovare un’ altra linea che sia la terza in con-<lb></lb> |
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711 tinua proportione a quelle. E commo si facia il mostraró. Sienno .2. linee proposte .ab. e .c.<lb></lb> |
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712 infra le quali voglio una linea nella proportion continua trovare. Agiongneró l’ una di<lb></lb> |
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713 quelle con l’ altra. E sia quella che è composta d’ amendoi .ad. imperoché io porró .bd. igua-<lb></lb> |
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714 li al .c. e sopra tutta descriveró uno semicirculo .aed. e produrró .be. perpendicular sopra la linea<lb></lb> |
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715 .ad. e dico che la linea .be. è quella che noi cierchiamo. <lb></lb> |
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716 .10. |
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717 </p> |
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718 <p class="main"> |
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719 Sienno date .2. linee alle quali voglio trovare una linea che sia in continua proportione con<lb></lb> |
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720 quelle .2. che comme si facia il dimostraró. Sienno .2. linee proposte .ab. e .c. alle quali vo-<lb></lb> |
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721 glio sugiongnere una linea in continua proportione: congiongneró la linea .ab. angula-<lb></lb> |
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722 re colla linea .c. e sia .ad. cioé .ad. sia iguali alla linea .c. e produrró la linea .ab. infino<lb></lb> |
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723 al .e. e sia fatto .be. iguali al .ad. E meneró la linea .bd. e dal ponto .e. meno la linea .ef. equedistan-<lb></lb> |
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724 te alla linea .bd. E meneró la linea .ad. infino al ponto .f. dico che .fe. sia quella linea. .11. |
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725 </p> |
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726 <p class="main"> |
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727 Sia assegnata una linea di quanto voi. Dela quale sia de bisogno torre una parte. Comme<lb></lb> |
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728 diciamo sia assegnata la linea .ab. e da quella voglio torre una parte. Comme a dire il<lb></lb> |
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729 terzo. Io congiongneró a quella una linea angularmente commo viene di quanto vorró che<lb></lb> |
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730 sia .ac. La quale taglio in .3. parti iguali: che sienno .ad.de.ec. e le linee .cb. e .df. produco<lb></lb> |
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731 equedistanti. Dico che .af. è il terzo del .ab. commo volavamo. .12. |
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732 </p> |
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733 <p class="main"> |
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734 E sienno proposte .2. linee dele quali una sia divisa in parti. L’ altra voglio dividere secon-<lb></lb> |
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735 do quelle parti. Che comme si facia mostraró. Sienno le ditte linee .ab. e .ac. le quali con-<lb></lb> |
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736 giongneró nel ponto .a. angularmente: e sia la linea .ab. divisa in .3. iguali portioni a-<lb></lb> |
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737 segnati in quelle li ponti, cioé .e. e .d., voglio secondo quelle parti dividere la linea .ac. Quando<lb></lb> |
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738 l’ aró congionta angularmente, meneró la linea .bc. e a quella meneró le quedistanti .df. e .eg. le quali li-<lb></lb> |
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739 nee equedistanti dico che dividono la linea .ac. in parti proportionali alle parti dela linea .ab., cioé<lb></lb> |
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740 la linea .ab. sia divisa commo volavamo ne’ ponti .fg. .13. |
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741 </p> |
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742 <p class="main"> |
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743 E se fienno .2. superficie iguali e d’ equedistanti lati dele quali uno angolo del’ una sia simile<lb></lb> |
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744 a uno angolo del’ altra, e lati che contengono quelli angoli fienno nella proportione mu-<lb></lb> |
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745 tua over mutukefia. E ancora quando e lati continenti gli angoli iguali fienno nella pro-<lb></lb> |
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746 portione mutukefia le .2. superficie fienno iguali. Sienno .2. superficie .abcd. e .cgef. e-<lb></lb> |
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747 quedistanti e iguali. E sia l’angolo .c. del’ una iguale al’ angolo .c. del’ altra. Dico che tal parte è il lato .bc.<lb></lb> |
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748 a il lato .cg. commo .ce. al .dc. E ancora, quando .bc. è tal parte del .cg. commo .ce. al .cd., allora quel-<lb></lb> |
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749 le .2. superficie sonno iguali e equedistanti che era da mostrare. .14. |
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750 </p> |
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751 <p class="main"> |
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752 E se fienno .2. triangoli iguali. De’ quali uno angolo del’ uno sia iguali a uno angulo del’ al-<lb></lb> |
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753 tro. Dico che ‘lati che contengano quello angolo iguale sonno in proportione mutua o-<lb></lb> |
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754 ver mutukefia. E se i lati di .2. triangoli che contengano l’ angolo simile sonno in<lb></lb> |
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755 proportione mutua allora e detti .2. triangoli sonno simili. Sienno .2. triangoli .abc. e<lb></lb> |
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756 .cde. iguali e sia l’ angolo .c. del’ uno iguale e simili al’ angolo .c. del’ altro. Dico la proportione del .ac.<lb></lb> |
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757 al .ce. essere commo .dc. al .cb. E così quando la proportione del .ac. al .ce. è commo .dc. al .cb., allora que’ <lb></lb> |
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758 .2. triangoli sarebbono simili commo ditto è. .15. |
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759 </p> |
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760 <p class="main"> |
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761 Se fienno .4. linee proportionali quello che è fatto dalla prima e ultima linea: cioé la su-<lb></lb> |
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762 perficie rettangula dela prima e ultima linea è iguale alla superficie rettangula ch’ é fatta dal-<lb></lb> |
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763 l’ altre .2. linee. E sse quello ch’ é fatto dalla prima e ultima linea è iguale a quello ch’ é fatto<lb></lb> |
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764 dell’ altre .2., alhora quelle linee fienno proportionali. Sienno .4. linee proportionali .a.b.c.d.<lb></lb> |
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765 E sia la proportione del .a. al .b. commo il .c. al .d., dico che la superficie rettangula fatta dal .b. e .c. É<lb></lb> |
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766 commo quella fata dal .a. e dal .d. E, se la superficie fatta dal .a. al .d. é commo quella fatta dal .b. al .c., alo-<lb></lb> |
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767 ra quelle .4. linee sonno proportionali: cioé tal parte è .a. al .b. commo .c. al .d. et cetera. .16. |
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768 </p> |
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769 <p class="main"> |
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770 Se siranno .3. linee proportionali, quello ch’ é fatto dalla prima e dala terza è iguale al quadrato dela se-<lb></lb> |
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771 conda. E, se quello ch’ é fatto dala seconda linea in sé è iguale a quello ch’ é fatto dela prima nella .3a., allora <lb></lb> |
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772 quel-<lb></lb> |
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773 le linee sonno proportionali. Commo sienno .3. linee proportionali .a.b.c. Dico che la superficie rettangula fatta<lb></lb> |
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774 dal .a. e .c. è iguale al quadrato fatto del .b. e ancora, se ’l quadrato fatto dala linea .b. è iguale alla superficie<lb></lb> |
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775 rettangula fatta dal .a. in .c. Allora quelle .3. linee .abc. sonno proportionali ch’ era bisogno mostrare. .17. |
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776 </p> |
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777 <p class="main"> |
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778 Se fienno .2. triangoli simili, la proportione del’ uno al’ altro è commo la proportione del’ uno lato al’ al-<lb></lb> |
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779 tro in sé multiplicata. E per questo se manifesta che quando sonno .3. linee proportionali la super-<lb></lb> |
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780 ficie fatta dalla prima è alla superficie fatta dalla seconda commo la proportione della prima linea<lb></lb> |
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781 <lb></lb> |
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782 <pb></pb></p> |
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783 <p class="folio"> |
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784 folio 5v |
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785 </p> |
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786 <p class="main"> |
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787 <lb></lb> |
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788 </p> |
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789 <p class="runhead"> |
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790 Distinctio prima. Capitulum quintum. |
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791 </p> |
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792 <p class="main"> |
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793 <lb></lb> |
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794 alla terza: intendi le superficie sienno simili nelle linee e de figure. Comme sienno .2. triangoli simi-<lb></lb> |
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795 li .abc. e .def. e gli angoli intendi sienno iguali. E l’ uno lato, cioé .ab., sia al .de. comme .2/5. Dico che<lb></lb> |
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796 il triangolo .abc. sia al triangolo .def. e .4/25., cioé la multiplicatione di .2/5. in sé. E sse fussino .3. linee pro-<lb></lb> |
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797 portionali .a.b.c. dico che lla superfice fatta dal .a., cioé o quadrato o triangolo che sia per cia-<lb></lb> |
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798 scuna facia il lato .a., è alla superficie fatta del .b., cioé che habia per facia il .b., comme é .a. al .c. et cetera. <lb></lb> |
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799 .18. |
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800 </p> |
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801 <p class="main"> |
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802 Ogni .2. superficie simili di molti angoli sonno divisibili in triangoli simili e in tan-<lb></lb> |
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803 ti l’ una quanto l’ altra. E la proportione del’ uno al’ altro è comme ciascuno lato al’ al-<lb></lb> |
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804 tro lato del’ altro: cioé al suo relativo in sé multiplicata. Comme sienno .2. pentago-<lb></lb> |
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805 ni simili .acb. e .fgh. Dico che sonno divisibili e pentagoni ditti in triangoli si-<lb></lb> |
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806 mili: e in tanti triangoli l’ uno quanti l’ altro comme sienno divisi in tre triangoli ciascuno. Di-<lb></lb> |
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807 co la proportione del’ uno triangolo al’ altro suo simile: è comme la proportione del lato del’ u-<lb></lb> |
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808 no al lato del’ altro suo relativo in sé multiplicata. .19. |
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809 </p> |
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810 <p class="main"> |
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811 Sia data una linea sopra la quale voglio scrivere una superficie simile ala superficie da-<lb></lb> |
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812 ta: comme sia data la linea .ab. sopra la quale voglio fare una superficie simile alla super-<lb></lb> |
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813 ficie pentagona .cdefg. Divideró el pentagono ditto in triangoli menate le li-<lb></lb> |
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814 nee .df. e .dg. e sopra il ponto .a. faró l’ angolo iguale al’ angolo .c. menata la linea<lb></lb> |
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815 .ah. e sopra il ponto .b. faró l’ altro angolo che sia .abh. iguale a langolo .cdg. e menato la li-<lb></lb> |
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816 nea .bh. infino s’ agionga con .ah. nel ponto .h. e haró fatto per la .32a. del primo l’ angolo<lb></lb> |
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817 .ahb. iguale al’ angolo. cgd. E per la .4a. di questo e lati de’ .2. triangoli sienno proportionali.<lb></lb> |
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818 Dapoi faró l’ angolo .hbk. menato la linea .bk. iguale al’ angolo .gdf. E l’ angolo .kbl. me-<lb></lb> |
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819 nata la linea .bl. iguale al’ angolo .fde. E così haremo fatto il pentagono che era da fare che<lb></lb> |
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820 è il proposito. .20. |
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821 </p> |
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822 <p class="main"> |
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823 Se .2. o piú superficie sonno simili a una superficie, certamente infra loro saranno<lb></lb> |
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824 simili. Comme sienno l’ uno e l’ altro di .2. pentagoni: cioé .abc.def. simili al pentago-<lb></lb> |
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825 no .ghk., dico che ’l pentagono .abc. è simile al pentagono .def. che è il propo-<lb></lb> |
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826 sito. .21. |
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827 </p> |
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828 <p class="main"> |
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829 Se fienno molte linee proportionali le superficie fatte in sule .2. prime linee sonno<lb></lb> |
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830 proportionali, comme le superficie simili fatte in sule due seconde. E così comme le su-<lb></lb> |
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831 perficie simili fatte in sule terze. Comme sienno .4. linee proportionali .a.b.c.d. E facia-<lb></lb> |
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832 si in sule .2. prime: cioé in sul .a. et .b. due superficie simili: comme .2. pentagoni. E in su-<lb></lb> |
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833 le seconde si facia .2. superficie simili comme .2. triangoli. Dico che tal proportione è de’ pentagoni<lb></lb> |
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834 l’ uno al’ altro: comme de’ triangoli l’ uno al’ altro e ancora, quando sonno date .4. linee e sopra le<lb></lb> |
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835 .2. prime si faciano .2. superficie simili e sopra l’ altre .2. si faciano .2. superficie simili, comme ó dit-<lb></lb> |
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836 to in sule prime .2. pentagoni e in sule seconde .2. triangoli, e tal parte sia il pentagono al pen-<lb></lb> |
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837 tagono comme il triangolo al triangolo, allora quelle .4. linee sonno proportionali che è il<lb></lb> |
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838 proposito. .22. |
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839 </p> |
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840 <p class="main"> |
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841 Se sará nello spacio d’ un paralello un altro paralello simile a quello e contenga parte<lb></lb> |
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842 del grande. Dico che quello paralello é intorno al diametro del gran paralello, commo sia<lb></lb> |
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843 nel paralello .bd. il paralello .fg. simili. E contenga parte del gran paralello comme .1/5.<lb></lb> |
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844 Dico il paralello .fg. stare intorno al diametro del paralello .bd. el quale diame-<lb></lb> |
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845 tro è .aec. .23. |
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846 </p> |
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847 <p class="main"> |
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848 Tutte le superficie d’ equedistannti lati che stanno intorno al diametro del paralello a<lb></lb> |
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849 tutto il paralello sonno simili superficie. Comme sia il paralello .bd., intorno al quale dia-<lb></lb> |
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850 metro sia la superficie .gh. e .fk. d’ equedistanti lati. Dico che ciascuna è simile al gran<lb></lb> |
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851 paralello: cioé tal proportione è del .cg. al .bc. comme del .ch. al .cd. .24. |
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852 </p> |
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853 <p class="main"> |
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854 D’ ogni .2. superficie d’ equedistanti lati dele quali uno angolo del’ una al’ uno angolo del’ <lb></lb> |
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855 altra è iguale, la proportionne del’ una superficie al’ altra superficie è quella ch’ é fatta delle<lb></lb> |
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856 .2. proportioni de’ suoi lati continenti l’ angolo iguale: comme sienno .2. superficie d’ equedistan-<lb></lb> |
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857 ti lati .ac. e .ed. E sia l’ angolo .b. del’ una iguale al’ angolo .b. del’ altra, dico che la propor-<lb></lb> |
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858 tione del’ una al’ altra: è fatta dela proportione del .ab. al .bd. e di quella che á .cb. al .be. E acio-<lb></lb> |
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859 ché bene intenda sia .ab.6. e .cb. sia .2. e sia .be.10. e .bd. sia .8. che la proportione del .ab. al .bd. è .3/4.<lb></lb> |
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860 E la proportione del .cb. al .be. è .1/5. e multiplicato .1/5. via .3/4. fanno .3/20. E tal proportione ala superfi-<lb></lb> |
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861 cie .ac. alla superficie .ed. cioé comme .3. è a .20.così è la superficie .ac. alla superficie .ed. che è il<lb></lb> |
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862 proposito. .25. |
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863 </p> |
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864 <p class="main"> |
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865 Sienno date .2. superficie dele quali sia de bisogno fare una superficie iguale alla proposta e simile<lb></lb> |
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866 alla data: commo sienno proposte .2. superficie una .a. e l’ altra .b., voglio fare una superficie iguali al .b.<lb></lb> |
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867 <lb></lb> |
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868 <pb></pb></p> |
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869 <p class="folio"> |
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870 folio 6r |
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871 </p> |
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872 <p class="main"> |
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873 <lb></lb> |
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874 </p> |
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875 <p class="runhead"> |
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876 Distinctio prima. Capitulum quartum. 6 |
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877 </p> |
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878 <p class="main"> |
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879 <lb></lb> |
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880 Sopra alla mittá d’ una linea data faciendo uno paralello: sia maggiore d’ un pa-<lb></lb> |
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881 ralello el quale sia aplicato alla data linea al quale manchi uno paralello simile e stan-<lb></lb> |
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882 te intorno al diamitro. Comme sia data la linea .ab. e sopra la mitá si facia il para-<lb></lb> |
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883 lello .cd. Del quale il diametro sia .be. e alla linea .ab. si facia il paralello .af. del quale<lb></lb> |
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884 uno lato seghi .ec. nel ponto .g. e manchi a finire tutta la linea .ab. la superficie .bf. che sia simi-<lb></lb> |
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885 le alla superficie .cd. e consista sopra il diametro. Dico alora che ’l paralello .cd. è magiore del pa-<lb></lb> |
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886 ralello .af. .27. |
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887 </p> |
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888 <p class="main"> |
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889 Sia proposta una superficie trilatera e voglio sopra una linea data designare uno<lb></lb> |
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890 paralello iguali al detto triangolo al quale manchi a finire la linea uno paralello si-<lb></lb> |
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891 mile a uno dato paralello. E sia in modo che ’l triangolo sia minore d’ un paralello si-<lb></lb> |
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892 mile al dato collocato nella mitá dela linea data altramente se lavorerebbe allo in-<lb></lb> |
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893 possibile. Sia dato il triangolo .c. e il paralello .d. e la linea .ab.; divideró la linea .ab. in .2. parti<lb></lb> |
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894 iguali sopra il ponto .e. e sopra la mittá .eb. faró el paralello .ef. simile al paralello .d. e finiró so-<lb></lb> |
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895 pra tutta la linea .ab. il paralello .bg. e perché .c. nonn é magiore del paralello .ef., ma è iguale o<lb></lb> |
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896 minore, commo è posto, e, se sará iguale, sará il paralello .eg. iguale. Et se il .c. è minore del .ef. in<lb></lb> |
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897 alcuna superficie la quale si conponga simile al .d., che sia .h. e sará .h. simile al .ef. e, alla misura della<lb></lb> |
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898 superficie .h. che sia rettangula, risegheró .fk. e .ek., menate le linee .lm. e .no. equedistanti alle linee<lb></lb> |
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899 de’ lati dela superficie .ef. Le quali si segano nel ponto .p. comme la superficie .kp. e sia iguale alla<lb></lb> |
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900 superficie .h. e sia per la .23a. di questo il ponto .p. sopra il diametro .kpb. E meneró .on. infino al<lb></lb> |
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901 .ag., dico che ’l paralello .pa. è quello vogliamo. Non vole dire altro questa conclusione se none che<lb></lb> |
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902 si voglia fare uno paralello sopra una linea data: el quale paralello habia per uno lato una linea<lb></lb> |
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903 data e sia piú che l’ area d’ un triangolo dato uno paralello simile al dato paralello. Onde il<lb></lb> |
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904 paralello .oa. è piú che ’l triangolo dato: il paralello .pb. simile al paralello .d. Ma diciamo<lb></lb> |
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905 con numeri, sia la linea data .ab. 24. E il triangolo dato .c. sia l’ area sua .84. et il paralello .d.<lb></lb> |
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906 sia l’ area sua .24., cioé per l’ uno lato .6. per l’ altro .4. Dico che voglio fare sopra la linea .ab. uno pa-<lb></lb> |
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907 ralello al quale a finire tutta la linea .ab. manchi uno paralello simile al paralello .d. e quello pa-<lb></lb> |
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908 ralello fatto sia l’ area sua iguali al triangolo .c., dove faró il paralello .kb. che sia .eb.12. e .bf.<lb></lb> |
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909 .8. imperoché gli á a essere simile al paralello ditto. Dove tutto l’ area del paralello .f. e sia .96.<lb></lb> |
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910 che manca al triangolo .12. del quale comporrai una superficie iguale a .12. e simile al paralel-<lb></lb> |
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911 lo .d. La quale supeficie é .h.: sia per l’ uno lato .R. de .8. e per l’ altro .R. de .18. La quale super-<lb></lb> |
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912 ficie porrai nel ditto paralello in questo modo. Che farai .km. iguale al lato che è .R. di .18.<lb></lb> |
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913 e .kn. sia iguale al lato che è .R. di .8. E, menate le linee .lm. e .nop. equedistanti a’ lati del para-<lb></lb> |
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914 lello .fe., harai il paralello .kp. iguale ala superficie .h. e la superficie .kp. è simile ala superfi-<lb></lb> |
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915 cie .pb., perché sonno circa il diametro .kpb. Adunque é fatto il paralello .ap. che l’ area è .84.<lb></lb> |
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916 comme è il triangolo dato. E manca a finire il paralello .ao. uno paralello .lo. simile al para-<lb></lb> |
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917 lello .d. dato. Imperoché li lati sonno in proportione comme .4. al .6., che sonno i lati del pa-<lb></lb> |
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918 ralello .d., imperoché .po. è .12. meno .R.18. e .bo. è .8. men. R.8. questo era bisogno mostrare .28. |
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919 </p> |
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920 <p class="main"> |
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921 Sopra alla data linea voglio desegnare uno paralello iguale a uno triangolo da-<lb></lb> |
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922 to el quale triangolo agionga al finimento della ditta linea uno paralello simile al<lb></lb> |
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923 paralello dato. Questa è differentiata dalla passata imperoché qui avanza el<lb></lb> |
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924 triangolo e in quella mancava. Adunque sia la linea data .ab. E il dato triangolo .c. e ‘l dato<lb></lb> |
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925 paralello .d.; voglio sopra la linea .ab. fare uno paralello iguale al triangolo .c. che agionga sopra tut-<lb></lb> |
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926 ta la linea .ab. el paralello simile al .d. Io divideró la linea .ab. in .2. parti iguali nel ponto .e. e<lb></lb> |
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927 sopra quella mittá faró .ef. paralello simile al paralello .d., secondo la .19a. di questo e secondo la<lb></lb> |
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928 .25a., faró .kl., del quale il diametro .gh. e sia simile al .d. E sia iguale ale .2. superficie .ef. e .c. dove,<lb></lb> |
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929 per la .20a. di questo, .kl. sia simile al .ef. Posta adunque la superficie .ef. di sopra alla superfi-<lb></lb> |
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930 cie .kl., in modo si communichino insieme nell’ angolo .g., e sia, per la .22a. di questo, la superficie<lb></lb> |
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931 .ef. stante intorno al diametro della superficie .kl. Onde il ponto .b. sia in sul diametro .gh.<lb></lb> |
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932 e finito il paralello .ah. Il quale paralello è quello volavamo: cioé iguale al triangolo dato<lb></lb> |
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933 e avanza alla linea data uno paralello .mn. simile a uno paralello dato che è il proposito. .29. |
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934 </p> |
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935 <p class="main"> |
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936 Sia proposta quale voi linea la quale si voglia dividere secondo la proportione ha-<lb></lb> |
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937 vente il mezo e .2. extremi, commo sia la data linea .ab. la quale voglio dividere secondo la<lb></lb> |
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938 proportione havente mezo e .2. extremi. De quella scriveró uno quadrato che sia .bc. e al<lb></lb> |
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939 lato .ac., secondo la passata, faró il paralello .cd. iguale al quadrato .bc. in modo che gli<lb></lb> |
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940 agiongi al finimento dela linea .ac. il quale sia simile allo .bc. e il lato del paralello sia equedistan-<lb></lb> |
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941 te e seghi la linea .ab. nel ponto .f. Dico la linea .ab. essere divisa in .af. e .bf. comme era proposto.<lb></lb> |
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942 <lb></lb> |
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943 <pb></pb></p> |
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944 <p class="folio"> |
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945 folio 6v |
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946 </p> |
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947 <p class="main"> |
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948 <lb></lb> |
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949 </p> |
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950 <p class="runhead"> |
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951 Distinctio prima. Capitulum quintum. |
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952 </p> |
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953 <p class="main"> |
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954 <lb></lb> |
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955 Se faranno .2. triangoli fatti sopra uno angolo de’ quali e .2. lati che tengono quello ango-<lb></lb> |
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956 lo ae .2. altri lati loro sienno equedistanti e sienno quelli .4. lati proportionali, quelli .2. triango-<lb></lb> |
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957 li fienno constituti sopra una medesima linea. Comme sienno .2. triangoli .abc. e .dce. fatti<lb></lb> |
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958 sopra l’ angolo .acd. e sia .ac. equedistante al .de. e .cd. equedistante al .ba. E sia la proportione di<lb></lb> |
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959 .ac. al .de. comme .ab. al .dc. dico che lle .2. base loro: cioé .bc. e .ce. sonno una medesima linea retta.<lb></lb> |
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960 .31. |
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961 </p> |
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962 <p class="main"> |
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963 In ogni triangolo rettangolo la superficie laterata che è oposta al’ angolo retto è igua-<lb></lb> |
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964 le a .2. superficie degli altri due lati insiemi prese. Quello che è detto nella penultima del<lb></lb> |
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965 primo de’ quadrati, qui si mostra d’ ogni superficie. E peró questa è piú utile, quanto le superficie<lb></lb> |
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966 d’ ogni specie sonno piú che ’quadrati. Dico adunque che sia uno triangolo .abc.<lb></lb> |
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967 del quale l’ angolo .a. sia retto. Dico che lla superficie fatta dal lato .bc. comme uno triangolo è iguali a<lb></lb> |
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968 .2. superficie constitute sopra i lati .ab. e .ac., cioé ancora a .2. triangoli fatti da’ ditti lati. Comme sia<lb></lb> |
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969 uno triangolo per ciascuna facia iguali al lato .bc. e faciasi .2. altri triangoli uno che sia per ciascuna<lb></lb> |
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970 facia quanto lo lato .ab. e un altro che sia per ciascuna facia quanto lo lato .ac. Dico il triangolo fatto<lb></lb> |
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971 dal .bc. esser iguale a’ .2. triangoli. E similmente quando il triangolo fatto dal .bc. è igual agli altri .2.<lb></lb> |
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972 triangoli alora l’ angolo .a. é retto che è conversa alla passata parte. .32. |
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973 </p> |
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974 <p class="main"> |
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975 Se ne’ cerchi iguali, over sopra il centro over sopra la loro circunferentia, si fará uno<lb></lb> |
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976 angolo, sará la loro proportione comme quella degli archi che contengono quelli angoli. Comme sien-<lb></lb> |
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977 no e cerchi .abc., del quale il centro sia .d., e .efg., del quale il centro .h., e sienno iguali. Sopra<lb></lb> |
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978 li quali centri sienno fatti .2. angoli: cioé .bdc. e .fhg. E sopra la loro circunferentia altri<lb></lb> |
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979 .2. che sienno .bac. e .gef. Dico che la proportione degli angoli, così quelli che sonno sopra la circunfe-<lb></lb> |
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980 rentia comme quelli che sonno sopra li centri, sonno comme gli archi .bc. al’ arco .fg. e questo è da notare.<lb></lb> |
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981 Havendo veduto con brevitá il primo, secondo e sexto d’ Euclide, mi pare de dare opera<lb></lb> |
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982 al capitolo .5o. Adunque diremo.<lb></lb> |
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983 <lb></lb> |
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984 Qualiter more Tusco seu Florenteno metiantur agri et possessiones. Capitulum quintum prime distinctionis.<lb></lb> |
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985 Benché lo strumento con che s’ usa de misurare l’ area delle superficie sia diverso<lb></lb> |
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986 secondo la costumanza de’ paesi, non mi pare de necessitá demostrare altro che quel-<lb></lb> |
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987 lo col quale el contado de Toscana si va misurando e maxime quel de Firenza. El quale è<lb></lb> |
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988 una certa longhezza nominato bracio over piede. Onde, quando diciamo questo terreno<lb></lb> |
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989 è .100.bracia. quadre, diciamo che in quel terreno entrarebbe una figura quadrata per ogni verso uno<lb></lb> |
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990 bracio .100. volte. E peró L. P. diffinendo quello che era a trovare l’ area d’ una superficie dici: tro-<lb></lb> |
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991 vare l’ area d’ una superficie é una superficie quadrata nota sapere quante volte entra nella superficie<lb></lb> |
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992 che vuoi misurare. E per lo contado de Firenza si vende el terreno a staiora, che uno staioro è<lb></lb> |
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993 .1728. bracia quadre da terra. Dico bracia da terra, perche è differentiato alcuna cosa da quello del<lb></lb> |
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994 panno. E, benché alcuni dichino che .1600.bracia quadre da panno sieno uno staioro nonn’ é<lb></lb> |
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995 peró vero: che io lo volsi a questi dí provare in questo modo: che quello che ‘l bracio del panno avanza al<lb></lb> |
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996 bracio da terra, e .1/18. del bracio da panno é .1/17. di quello da terra, dove tanto è a multiplicare .17. bra-<lb></lb> |
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997 cia da panno per sé quanto .18. da terra per sé. Adunque .289.bracia. quadre da panno sonno quanto<lb></lb> |
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998 .324.bracia. quadre da terra. Onde .1728.bracia. quadre da terra sonno, alla detta ragione, .1541. o<lb></lb> |
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999 circa, cioé una piccola parte di bracio. Lo staioro, adunque, è .1728.bracia. da terra quadre e divide-<lb></lb> |
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1000 si in .12. parti che sonno iguali e ciascuna se dice panoro. E uno panoro se divide in .12. parti igua-<lb></lb> |
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1001 li e ciascuna se dice pugnoro. E ancora il pugnoro se divide in .12. parti e ciascuna se dice bracio<lb></lb> |
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1002 da terra quadro. Onde uno panoro è .144.bracia. da terra quadre. E uno pugnoro è .12. di quel-<lb></lb> |
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1003 li bracia. E ancora lo staioro è .144. pugnora. |
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1004 </p> |
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1005 <p class="main"> |
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1006 Havendo detto la divisione e il modo che s’ usa de misurare, é da dire comme si multiplica<lb></lb> |
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1007 gran quantitá di bracia infra loro. Dico che havendo a multiplicare una summa di<lb></lb> |
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1008 gran numero, comme a dire .3940.bracia. via .3940.bracia, dove poi tenere il modo dele<lb></lb> |
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1009 caselle over per bericuocolo e harai .15523600.bracia quadre. Deli quali farai staiora in<lb></lb> |
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1010 questo modo: che, prima, ne farai pugnora, che sonno .1293633. pugnora .4.bracia. E dapoi, de .1293633.<lb></lb> |
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1011 pugnora, farai panora partendo in .12. e harai .107802. panora .9. pugnora .4.bracia. E<lb></lb> |
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1012 dapoi ne farai staiora dividendo per .12. e harai che sonno .8983. staiora .6. panora .9. pu-<lb></lb> |
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1013 gnora .4.bracia. E per tal modo si pó ancora fare. |
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1014 </p> |
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1015 <p class="main"> |
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1016 E ancora potresti, per l’ altro modo, fare la ditta multiplicatione. Ma, prima, daremo questa<lb></lb> |
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1017 ordinatione nel multiplicare panora e gli altri nomi infra loro, cioé:<lb></lb> |
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1018 Multiplicando bracia per bracia fanno bracia quadre. |
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1019 </p> |
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1020 <p class="main"> |
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1021 Multiplicando bracia per pugnora fanno pugnora. |
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1022 </p> |
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1023 <p class="main"> |
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1024 Multiplicando bracia per panora fanno panora; multiplicando per staiora fanno staiora<lb></lb> |
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1025 Multilpicando pugnora per pugnora fanno panora. Comme dicendo .6. pugnora via .8. pu-<lb></lb> |
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1026 <lb></lb> |
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1027 <pb></pb></p> |
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1028 <p class="folio"> |
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1029 folio 7r |
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1030 </p> |
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1031 <p class="main"> |
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1032 <lb></lb> |
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1033 Distinctio prima. Capitulum sextum. 7<lb></lb> |
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1034 <lb></lb> |
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1035 gnora fanno .48. panora che sonno .4. staiora. |
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1036 </p> |
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1037 <p class="main"> |
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1038 Multiplicando pugnora per panora fanno staiora; comme dicendo .6. pugnora via .8. pa-<lb></lb> |
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1039 nora fanno .48. staiora. |
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1040 </p> |
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1041 <p class="main"> |
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1042 Multiplicando pugnora via staiora fanno, per ogni unitá .12. staiora. Comme a multiplica-<lb></lb> |
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1043 re .6. pugnora via .8. staiora, fanno .48. volte .12. staiora, che sonno .576. staiora.<lb></lb> |
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1044 Multiplicando panora per panora fanno, per ogni unitá, .12. staiora. Comme multiplicando<lb></lb> |
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1045 .6. panora per .8. panora, fanno .48. volte .12. staiora, che sonno .576. staiora.<lb></lb> |
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1046 Multiplicando panora via staiora fanno, per ogni unitá, .144. staiora; che havendo a mul-<lb></lb> |
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1047 tiplicare .6. staiora via .8. panora, faranno .48. volte .144. staiora che sonno .6912. staiora.<lb></lb> |
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1048 Multiplicando staiora per istaiora fanno, per ogni unitá, .1728. staiora comme multiplicando .6.<lb></lb> |
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1049 staiora via .8. staiora fanno .48. volte .1728. staiora che sonno .82944. staiora.<lb></lb> |
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1050 E acioché meglio s’ abia lo intendimento, diremo: io voglio multiplicare .3940.<lb></lb> |
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1051 bracia via .3940.bracia. da terra. Intendi, dove farai prima, de .3940.bracia, sta-<lb></lb> |
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1052 iora e panora, pugnora e bracia, che sonno .2. staiora e .3. panora e .4. pugnora e<lb></lb> |
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1053 .4.bracia. Adunque a multiplicare .3940.bracia. via .3940.bracia. è comme a multipli-<lb></lb> |
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1054 care .2. staiora .3. panora .4. pugnora .4.bracia. Dove, per questo fare, segnerai queste quantitá a mo-<lb></lb> |
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1055 do di caselle over crocetta: cioé .2. staiora .3. panora .4. pugnora e .4.bracia, segnando le spe-<lb></lb> |
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1056 cie iguali: cioé le staiora sotto le staiora e le panora sotto le panora e le pugnora sotto le pu-<lb></lb> |
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1057 gnora e le bracia, sotto le bracia, comme io ó fatto nella dispositione da lato. E incomencia al mi-<lb></lb> |
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1058 nore grado, sempre ascendendo per ordine. E diremo: .4.bracia. via .4.bracia. fanno .16.bracia,<lb></lb> |
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1059 che sonno uno pugnoro e .4.bracia. E queste segna. E dapoi multiplicarai le bracia contro<lb></lb> |
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1060 ale pugnora, dicendo: .4.bracia. via .4. pugnora fanno .16. pugnora. E un’ altra volta, per la cro-<lb></lb> |
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1061 cetta, .4.bracia. via .4. pugnora fanno .16. pugnora che, agionte alle .16. pugnora de prima, fan-<lb></lb> |
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1062 no .32. pugnora, che sonno .2. panora e .8. pugnora. Le quali segna comme ordinando uno ca-<lb></lb> |
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1063 stellucio. Dipoi multiplicarai li bracia contro alle panora, multiplicando sempre per canto, dicen-<lb></lb> |
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1064 do: .4.bracia. via .3. panora fanno .12. panora. E, un’ altra volta, per la crocetta, .4.bracia. via .3. pa-<lb></lb> |
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1065 nora fanno .12. panora che, agionte alle .12. panora, fanno .24. panora. E a queste agiongni la<lb></lb> |
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1066 multiplicatione di .4. pugnora via .4. pugnora che fanno .40. panora, che sonno .3. staiora e<lb></lb> |
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1067 .4. panora. E segnale. Dipoi multiplicarai li bracia contro alle staiora, dicendo: .4.bracia. via<lb></lb> |
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1068 .2. staiora fanno .8. staiora. E con, un’ altra volta, per la crocetta, .4.bracia. via .2. staiora, fanno .16.<lb></lb> |
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1069 staiora. E a questo agiongni la multiplicatione di .4. pugnora in .3. panora e, ancora, .4. pugno-<lb></lb> |
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1070 ra in .3. panora che è .24. staiora: che, con .16. staiora, fanno .40. staiora, le quali segna. Dipoi mul-<lb></lb> |
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1071 tiplicarai .4. pugnora via .2. staiora e, un’ altra volta, per la crocetta, .4. pugnora via .2. staiora<lb></lb> |
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1072 e a questo agiongni la multiplicatione di .3. panora in .3. panora e haremo .25. volte .12. staiora,<lb></lb> |
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1073 cioé staiora .300., le quali segna. E dapoi multiplicarai .3. panora via .2. staiora: e .2. staiora per .3.<lb></lb> |
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1074 pugnora e haremo .12. volte .144. staiora: cioé .1728. staiora: e segna. E dapoi multiplica .2.<lb></lb> |
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1075 staiora via .2. staiora, fanno .4. volte .1728. staiora: cioé .6912. staiora, le quali segnerai. E da poi<lb></lb> |
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1076 agiongni le ditte summe, fanno .8983. Staiora .6. Panora .9. Pugnora e .4. Bracia quadre, comme<lb></lb> |
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1077 di sopra trovammo. Onde, qual modo voi, poi usare. |
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1078 </p> |
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1079 <p class="main"> |
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1080 E questo basti quanto a quello che, circa lo strumento del misurare, è da dire; dove seguen-<lb></lb> |
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1081 do, è da dare ordine alla pratica.<lb></lb> |
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1082 <lb></lb> |
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1083 <lb></lb> |
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1084 De dimensione figurarum quadratarum. capitulum sextum. |
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1085 </p> |
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1086 <p class="main"> |
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1087 Giá dicemmo che ’l quadrato d’ una linea era una superficie d’ equedistanti lati e de angoli<lb></lb> |
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1088 retti. La quale superficie si ha per lato quanto è la ditta linea. E, ancora, è ditto che, a volere<lb></lb> |
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1089 trovare l’ area d’ una superficie e volere sapere una superficie quadrata nota quante volte<lb></lb> |
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1090 entra nella superficie che voi misurare. E, peró, a volere trovare l’ area d’ una superficie qua-<lb></lb> |
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1091 drata nota quante volte entra in quella, e perché noi habiamo ditto che secondo l’ uso fiorentino quella<lb></lb> |
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1092 misura se dice bracio, adunque diciamo che noi vogliamo trovare l’ area del quadrato .abcd. Pri-<lb></lb> |
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1093 ma ti conviene havere uno de’ suoi lati, che pongo sia per facia .10.bracia., dove multiplicarai il<lb></lb> |
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1094 .10. in sé, fanno .100. E .100.bracia. é quadra la ditta superficie quadrata: cioé in ditta superficie entra-<lb></lb> |
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1095 rá .100. volte una superficie quadrata che sia per ogni verso uno bracio. Over multiplica uno de’ soi<lb></lb> |
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1096 lati per l’ altro a quello contiguo: ma, perché i lati sonno iguali una cosa e in se medesimo et cetera.<lb></lb> |
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1097 E, a ció ché el si manifesti, quella superficie detta contenere una superficie d’ un bracio .100.<lb></lb> |
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1098 volte, faremo la figura. Sia adunque il quadrato .abcd. e sia comme ó detto per ciascun<lb></lb> |
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1099 lato .10.bracia. Dove il lato .ab. divideró in .10. parti iguali. E, similmente, divideró .cd.<lb></lb> |
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1100 e menise dai ponti delle divisioni del .ab. a’ ponti delle divisioni del .cd. le linee eque-<lb></lb> |
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1101 <lb></lb> |
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1102 <pb></pb></p> |
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1103 <p class="folio"> |
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1104 folio 7v |
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1105 </p> |
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1106 <p class="main"> |
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1107 <lb></lb> |
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1108 </p> |
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1109 <p class="runhead"> |
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1110 Distinctio prima. Capitulum septimum. |
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1111 </p> |
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1112 <p class="main"> |
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1113 <lb></lb> |
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1114 distanti che fieno iguali ala linea .ac. Dipoi si divida il lato .ac. e .bd. ciascun in .10. parti iguali: e<lb></lb> |
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1115 da’ dettig ponti delle divisioni si meni le linee che fieno equedistanti alla linea .ac. e così haremo di-<lb></lb> |
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1116 viso el grande quadrato in picoli quadrati in de’ quali ciascun sia per facia uno bracio. Imperoché .ae., che è il<lb></lb> |
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1117 primo ponto, è uno bracio. Adunque il quadrato .aegi. è per ogni verso uno bracio e ciascun degli al-<lb></lb> |
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1118 tri picoli quadrati è iguale a quello, li quali picoli quadrati sonno .100. Imperoché fra la linea .ab. e la<lb></lb> |
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1119 linea .gh. n’ é .10. E, per simil modo, numerando, ne troveremo .100., comme evidentemente appare.<lb></lb> |
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1120 E dicendo e gli é uno quadrato che è per ogni verso 10.bracia. Vo’ sapere: una figura<lb></lb> |
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1121 quadrata che sia per ogni verso .2.bracia. quante volte v’ entrará. É de bisogno trovare<lb></lb> |
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1122 l’ area del’ una e l’ altra figura. E harai per la grande .100.bracia. quadre e per l’ altra .4. bra-<lb></lb> |
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1123 cia quadre. Onde partirai l’ area del’ una superficie, cioé dela magiore, per l’ area del’ al-<lb></lb> |
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1124 tra: viene .25. e .25. volte entrará la minore superficie nella magiore. Over uno de’ lati del gran<lb></lb> |
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1125 quadrato misura con uno de’ lati del minore: cioé .10.bracia. misura con .2.bracia., harai .5. volte,<lb></lb> |
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1126 el qual .5., in sé multiplicato, fa .25. e .25. volte entrará il picolo quadrato nel grande, comme dicemmo.<lb></lb> |
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1127 E questo basti quanto a questo capitulo.<lb></lb> |
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1128 <lb></lb> |
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1129 </p> |
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1130 <p class="head"> |
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1131 De dimensione omnium triangulorum capitulum septimum. |
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1132 </p> |
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1133 <p class="main"> |
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1134 <lb></lb> |
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1135 Le figure triangulari: cioé le superficie che si dicano triangoli. Sonno alcune dette trian-<lb></lb> |
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1136 goli ortogonii. Cioé quel triangolo che á l’ angolo retto. Alcuni si dicano oxigonii,<lb></lb> |
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1137 cioé quello triangolo che á tutti gli angoli acuti. Alcune ampligoni: cioé quelle che hano<lb></lb> |
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1138 uno angolo ampio. E questi nomi pigliano le dette figure dagli angoli. E possano<lb></lb> |
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1139 ancora ricevere nomi da’ lati. Imperoché alcuni si dicano isopleuri:, cioé equilateri, che sonno quel-<lb></lb> |
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1140 li che hano e lati infra loro iguali. Alcuni ysocheli, cioé equicurii sonno quelli che hano so-<lb></lb> |
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1141 lamente .2. lati iguali infra loro. Alcuni scaleni, cioé diversilateri che sonno quelli che hano cia-<lb></lb> |
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1142 scuno lato ineguale. |
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1143 </p> |
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1144 <p class="main"> |
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1145 L’ area di ciascun triangolo s’ á certamente della multiplicatione dela mitá del ca-<lb></lb> |
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1146 tetto in tutta la basa over della multiplicatione di tutto il catetto nella mitá dela<lb></lb> |
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1147 basa. Che tutto, con dimostrationi, lo faró certo. |
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1148 </p> |
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1149 <p class="main"> |
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1150 Catetto, over perpendiculare del triangolo, è una linea retta che si muove dal’ ango-<lb></lb> |
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1151 lo del triangolo e cade in sulla facia oposta a quello angolo: e fa in sulla detta fa-<lb></lb> |
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1152 cia .2. angoli retti. E, per questo, da ogni angolo solamente una perpendiculare, over ca-<lb></lb> |
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1153 tetto, si mena. Vuogliamo dire che solo uno catetto si pó menare. Imperoché,<lb></lb> |
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1154 se .2. o piú se ne menasse e ciascuno fusse catetto, comme nel triangolo .abc. ch’ é il catetto .ad., di-<lb></lb> |
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1155 co che dal ponto .a. non si menará se none la linea .ad. che sia catetto. E, se se ne menasse un’ al-<lb></lb> |
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1156 tra, comme .ae., seguitarebbe lo triangolo .aed. essere piú che .2. angoli retti, che è impossibile per<lb></lb> |
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1157 la .32a. del primo de Euclide. |
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1158 </p> |
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1159 <p class="main"> |
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1160 E acioché si vegga comme e catetti debbano o infra ’l triangolo o fora cadere lo mo-<lb></lb> |
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1161 straremo. Se da uno angolo maggiore che gli altri angoli d’ uno triangolo si me-<lb></lb> |
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1162 na il catetto, sempre dentro al triangolo cadrá. E acioché chiaramente si consenta.<lb></lb> |
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1163 Sia il triangolo .abg., del quale l’ angolo .a. sia magiore over iguali al’ angolo .b. over<lb></lb> |
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1164 al’ angolo .g. Dico che, se dal ponto .a. ala linea .bg. si meni una perpendiculare, che lla cadrá dentro<lb></lb> |
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1165 al triangolo e non di fora. Ma, se possibile é, caggia di fuora per l’ aversario dal lato del .b. in sul pon-<lb></lb> |
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1166 to .z. E menise .gb. diritto infino al .e. e passerá per lo ponto .z. e haremo il triangolo .azb. ortogo-<lb></lb> |
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1167 nio, che hará l’ angolo .z. retto. E l’ angolo .b. del triangolo .abz. per la .32a. del primo, è iguale a ciascun<lb></lb> |
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1168 degli angoli .a. e .g. del triangolo .abg. insiemi gionti e noi dicemmo l’ angolo .a. essere magiore o<lb></lb> |
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1169 iguali che alcuno degli angoli .b. over .g. E peró l’ angolo .b. del triangolo .abz. è magiore<lb></lb> |
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1170 che ’l retto. Onde, nel triangolo .azb. v’ é uno angolo retto e uno magiore che ’l retto, che è im-<lb></lb> |
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1171 possibile per la .32a. del primo. E peró il catetto .az. non sará fuori del triangolo .agb. dala par-<lb></lb> |
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1172 te del .b. E, se andasse fuori dal lato del .g., ne perverrebbe il medesimo inconveniente. E peró, de<lb></lb> |
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1173 necessitá, cade dentro. |
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1174 </p> |
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1175 <p class="main"> |
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1176 E per questo si manifesta che ’l catetto che si muove da ciascun angolo del trian-<lb></lb> |
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1177 golo oxigonio e il catetto che si muove dal’ angolo retto del triangolo ortogo-<lb></lb> |
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1178 nio e il catetto che si muove dal’ angolo obtuso del triangolo ampligonio, sem-<lb></lb> |
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1179 pre cadrá dentro al triangolo. |
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1180 </p> |
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1181 <p class="main"> |
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1182 Li .2. lati del triangolo ortogonio, cioé quegli che contengono l’ angolo retto, sonno<lb></lb> |
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1183 catetti over perpendiculari del detto triangolo. Comme sia il triangolo ortogonio<lb></lb> |
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1184 .bgd. havente l’ angolo .g. retto. Dico la retta .bg. essere perpendiculare sopra .gd. e lla<lb></lb> |
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1185 retta .gd. essere perpendiculare sopra la linea .gb. E, se non fussino (comme ó detto)<lb></lb> |
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1186 <lb></lb> |
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1187 <pb></pb></p> |
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1188 <p class="folio"> |
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1189 folio 8r |
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1190 </p> |
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1191 <p class="main"> |
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1192 <lb></lb> |
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1193 </p> |
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1194 <p class="runhead"> |
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1195 Distinctio prima. Capitulum septimum. 8 |
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1196 </p> |
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1197 <p class="main"> |
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1198 <lb></lb> |
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1199 perpendiculari, menise la linea che sia perpendiculare dal puncto .b. e sia .ae. e caggia di fuora in<lb></lb> |
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1200 sul puncto .a. e sia .ag. continuati con .gd., sará adunque nel triangolo .bag.2. angoli retti, imperoché<lb></lb> |
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1201 l’ angolo .g. del triangolo .abg. è retto, imperoché così si pose. Onde e .3. angoli del triangolo<lb></lb> |
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1202 .abg. sonno magiori di .2. retti, che è impossibile per la .32a. del primo. Ancora, quando il catetto ca-<lb></lb> |
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1203 desse dentro, se n’ andrebbe nel predetto errore, imperoché nel triangolo .bge. farebbe .2. ango-<lb></lb> |
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1204 li retti. E peró ne’ fuora ne’ dentro puó cadere. Onde, per necessitá, cadrá in sulla detta linea e fará<lb></lb> |
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1205 una con quella, anzi quella medesima. E, similmente, se dal’ angolo .d. si muove la perpendiculare so-<lb></lb> |
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1206 pra la faccia .bg., sará per quel medesimo la detta perpendiculare .dg. E così abbi a mente in simili.<lb></lb> |
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1207 Dal minore angolo del triangolo oxigonio una perpendiculare menata cade<lb></lb> |
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1208 dentro al triangolo. Exempli causa: sia il triangolo oxigonio .abg., del quale l’ an-<lb></lb> |
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1209 golo .a. sia minore. Dico che, se dal’ angolo .a. si mena la perpendiculare insino al la-<lb></lb> |
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1210 to .bg., cadrá dentro al triangolo. E, se possibil fusse (per l’ aversario), caggia di fo-<lb></lb> |
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1211 ri sopra il puncto .d. E, perché .ad. é catetto, sará l’ angolo .adb. retto e l’ angolo .abd. è maggiore<lb></lb> |
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1212 che ’l retto, imperoché, per la .32a. del primo, e gli é iguali hai .2. angoli .a. e .g. del triangolo .agb. E, per-<lb></lb> |
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1213 che ciascuno degli angoli del triangolo .agb. è minore che ’l retto (ex ypotesi), e peró e .2. ango-<lb></lb> |
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1214 li .a. e .g. sono magiori che ’l retto. E peró nel triangolo .adb. v’ é uno angolo retto e uno magio-<lb></lb> |
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1215 re che ’l retto, ch’ é impossibile. Conciosiacosaché ogni triangolo sia soluto in .2. angoli retti.<lb></lb> |
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1216 E peró cadrá dentro.<lb></lb> |
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1217 <lb></lb> |
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1218 E, similmente, se dal’ angolo ampio del triangolo ampligonio si mena la perpendi-<lb></lb> |
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1219 culare, quella cadrá dentro. Comme sia il triangolo ampligonio .abg., del quale l’ an-<lb></lb> |
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1220 golo .a. sia obtuso. Dico che la perpendiculare menata dal puncto .a. ala linea .bg.<lb></lb> |
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1221 cadrá dentro. E, se possibile sia (per l’ aversario), caggia di fuora in sul puncto .e. de-<lb></lb> |
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1222 la linea .gbe. Dove l’ angolo .abe. é magiore di .2. angoli .a. e .g. separati del triangolo .abg. per<lb></lb> |
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1223 la .16a. del primo. E peró nel triangolo .abe. v’ é uno angolo obtuso e uno retto, che è impossibile per<lb></lb> |
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1224 la .17a. del primo. Adunque cadrá dentro al triangolo<lb></lb> |
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1225 <lb></lb> |
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1226 E, menando la perpendiculare da ciascuno degli altri angoli acuti al lato oposto<lb></lb> |
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1227 al detto angolo, quella perpendiculare cadrá di fuori. Comme sia il triangolo ampli-<lb></lb> |
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1228 gonio .bgd. havente l’ angolo .d. obtuso. Dico che, se dal’ angolo .b. si mena la perpen-<lb></lb> |
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1229 diculare sopra lo lato .bg., che la detta perpendiculare cadrá di fuori del detto trian-<lb></lb> |
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1230 golo. E, se fosse possibile cadesse dentro (per laversario), caggia in sul puncto .a. Onde segui-<lb></lb> |
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1231 tarebbe nel triangolo .bad. essere .2. angoli, uno retto e uno magiore che ’l retto, imperoché l’ an-<lb></lb> |
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1232 golo .a. è retto e l’ angolo .d. è obtuso (ex ypotesi). E peró è inconveniente e cosa impossibile e que-<lb></lb> |
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1233 sto se doveva mostrare. |
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1234 </p> |
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1235 <p class="main"> |
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1236 Se in .2. linee contenenti uno angolo un’ altra linea retta si meni, la quale seghi ambe-<lb></lb> |
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1237 dui le dette linee, dala mitá dela linea che è in mezo dele sectioni si segni uno pun-<lb></lb> |
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1238 cto e da quel puncto una linea si meniinfino al’ angolo che è contenuto da quelle li-<lb></lb> |
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1239 nee, la qual linea sia iguale ala mitá dela linea, la quale è infra le .2. Allora l’ angolo<lb></lb> |
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1240 dato è retto. Comme sia la linea .ab. e .gb. che contenghino l’ angolo .b. e dal punto .e. (posto ne-<lb></lb> |
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1241 la linea .gb) si meni un’ altra linea infino al puncto .d. posto nella linea .ab. E sia la linea .hed.<lb></lb> |
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1242 E nel puncto .z. che sia la mitá dela linea .ed. si meni la linea .zb. la qual se lla è iguale ala linea<lb></lb> |
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1243 .ze., alora l’ angolo .b. è retto. E, quando la linea .ze. fosse magiore che la linea .zb., alora l’ ango-<lb></lb> |
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1244 lo .b. è obtuso. E, quando la linea .zb. è minore che la linea .ze., alora l’ angolo .b. è acuto. Comme per<lb></lb> |
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1245 la .30a. del terzo d’ Euclide appare. |
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1246 </p> |
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1247 <p class="main"> |
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1248 Questo detto, è da sapere che, a volere l’ area d’ una figura di lati equedistanti e canti<lb></lb> |
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1249 retti, che la diciamo quadrilatero over paralello d’ angoli retti, è da multiplicare l’ uno<lb></lb> |
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1250 lato per l’ altro a quello contiguo. Comme a dire: e gli é un paralello d’ angoli retti longo .10.<lb></lb> |
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1251 bracia e largo .8.bracia. Dico che, a volere l’ area, multiplicarai .10. via .8., fanno<lb></lb> |
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1252 .80. e .80.bracia quadre è la detta figura. E questo, comme facemmo nel quadrato, si puó chiari-<lb></lb> |
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1253 re, ma non bisogna. Ora ritorniamo a’ triangoli.<lb></lb> |
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1254 . |
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1255 </p> |
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1256 <p class="main"> |
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1257 Comme habiamo detto e sono di .3. maniere triangoli: cioé ortogonii, oxigoni e<lb></lb> |
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1258 ampligonii. Dove, per ordine dovendo dire, diremo prima il modo di trovare l’ area de’ <lb></lb> |
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1259 triangoli ortogonii e poi degli oxigonii e poi degli ampligonii, comme vedrai.<lb></lb> |
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1260 Li triangoli ortogonii sonno alcuna volta di .2. lati iguali e alcuna volta di .3. lati. non<lb></lb> |
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1261 iguali. E, comme è detto, l’ area di ciascuno triangolo s’ à dela multiplicatione dela mitá del<lb></lb> |
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1262 catetto in tutta la basa o dela mitá dela basa in tutto il catetto, la qual cosa faremo chiara.<lb></lb> |
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1263 Sia il triangolo ortogonio di .2. lati iguali .abg. e sia .ab. e .bg. ciascun .10.bracia, dove<lb></lb> |
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1264 <lb></lb> |
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1265 <pb></pb></p> |
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1266 <p class="folio"> |
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1267 folio 8v |
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1268 </p> |
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1269 <p class="main"> |
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1270 <lb></lb> |
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1271 </p> |
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1272 <p class="runhead"> |
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1273 Distinctio prima. Capitulum septimum. |
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1274 </p> |
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1275 <p class="main"> |
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1276 <lb></lb> |
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1277 .bg. sia la radice de. 200. braccia, dove il catetto che cade in sula basa .bg. e .ab. Aduncha, a mul-<lb></lb> |
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1278 tiplicare la mitá del .ab. in .bg., s’ ará l’ area del detto triangolo, che sia .50. braccia quadre, che, in<lb></lb> |
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1279 questo modo, lo proveremo. Dal ponto .a. si meni la linea .ad. equedistante e iguale ala linea<lb></lb> |
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1280 .bg. e sia l’ angolo .bad. retto e la linea .ad. sia .10. braccia. E, dipoi, dal puncto .d. si meni la linea<lb></lb> |
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1281 .dg. iguale e equedistante ala linea .ab. e sará .10. braccia e haremo constituto uno quadrato<lb></lb> |
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1282 .abgd. del quale la linea .ag. è il diametro. Aduncha il triangolo .abg. è la mitá. El quale qua-<lb></lb> |
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1283 drato viene dela multiplicatione del .ab. in .bg. Aduncha, per volere quadrare il triangolo, multipli-<lb></lb> |
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1284 carai la mitá del .ab. in tutta .bg. over la mitá del .bg. in tutta .ab. E questo era da mostrare.<lb></lb> |
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1285 Ancora diremo: e gli é un triangolo ortogonio diversilatero .bcd. del quale il lato<lb></lb> |
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1286 .bc. fará .8. braccia è il lato .cd. é .6. braccia. Dove il lato .bd. sia .10. braccia e sia l’ ango-<lb></lb> |
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1287 lo .c. retto. Dico che l’ area del ditto triangolo s’ á della multiplicatione dela mi-<lb></lb> |
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1288 tá dela linea .bc. che è catetto in tutta .cd. over dela mitá dela basa .cd. in tutta<lb></lb> |
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1289 .cb. che sieno .24. braccia quadre. Che in questo modo lo mostraró. Faciasi la linea .ba. iguale<lb></lb> |
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1290 e equedistante ala linea .cd. E faciasi .ad. iguale e equedistante ala linea .bc. E haremo constitu-<lb></lb> |
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1291 to il paralello d’ angoli retti .abcd., del quale il diametro è .bd., il quale lo divide per lo mezzo<lb></lb> |
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1292 per la .34a. del primo. Aduncha il triangolo .bcd. è la mitá. E l’ area del detto paralello viene<lb></lb> |
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1293 della multiplicatione del .bc. in .cd. Aduncha l’ area del triangolo viene dela multiplicatio-<lb></lb> |
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1294 ne dela mitá del .bc. in .cd. over dela mitá del .cd. in .bc. che così conveniva mostrarse et cetera.<lb></lb> |
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1295 Ancora altrimente dividasi .bc. in .2. parti iguali sopra il poncto .e., dal quel si me-<lb></lb> |
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1296 ni la linea .ef. iguale e equedistante ala linea .cd. e conpise .df. E, perché la linea .cd. è eque-<lb></lb> |
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1297 distante e iguale ala linea .fe., sará la linea .df. iguale e equedistante ala linea .ec., commo<lb></lb> |
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1298 per la .34a. del primo è manifesto. Onde la linea .df. è .4. bracci. Aduncha il trian-<lb></lb> |
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1299 golo .bhe. è iguale al triangulo .hdf., imperoché l’ angolo .e. è retto e .be. è iguale al .fd. e .fh. è igua-<lb></lb> |
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1300 le al .he. Aduncha tutto el triangolo .bcd. è iguale al quadrilatero .ecdf. che á gli angoli retti. El qua-<lb></lb> |
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1301 le quadrilatero è fatto dela multiplicatione dela linea .ec. nella linea .cd., cioé di .4. in .6. E peró<lb></lb> |
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1302 el triangolo .bcd. è fatto della multiplicatione di .4. in .6. Simelmente si mostrarebbe se dal<lb></lb> |
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1303 ponto .i., che è nel mezzo del .cd., si menasse la linea .ia. equedistante e iguale alla linea .be. e nel<lb></lb> |
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1304 medesimo modo hai operare et cetera. |
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1305 </p> |
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1306 <p class="main"> |
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1307 Ancora, se uno lato fosse non saputo e, per gli altri saputi, lo volessi trovare. Com-<lb></lb> |
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1308 me diciamo che ’l lato .bd. non sia dato noto. Multiplicarai .bc. in sé e .cd. in sé; agion-<lb></lb> |
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1309 gne quei .2. quadrati, che fano .100., la cui radici è il lato .bd., che è .10. E chiamasi<lb></lb> |
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1310 quello lato ypotemissa. E, se la ypotemissa .bd. è .10. e la basa .dc. è .6., sará da multi-<lb></lb> |
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1311 plicare .10. in sé e .6. in sé e trare l’ uno quadrato del’ altro, che rimane .64., per lo quadrato del la-<lb></lb> |
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1312 to .bc. Ancora la ypotemissa sia .10. e il lato .bc. è .8. e vogliamo il lato .cd. Multiplicarai .10. in sé<lb></lb> |
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1313 e .8. in sé e harai .100. e .64. e trarai .64. di .100., rimane .36., la cui radici è .6., per lo lato .dc.<lb></lb> |
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1314 L’ area di tutti e triangoli oxigonii (comme è ditto) s’ á dela multiplicatione del ca-<lb></lb> |
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1315 tetto nela mitá dela basa over dela mitá del catetto in tutta la basa. E acioché<lb></lb> |
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1316 questo appaia lo dimostraremo. E gli é da sapere che gli triangoli oxigonii sonno<lb></lb> |
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1317 di .3. lati iguali over di .2. lati equali over di .3. lati non iguali. sia prima il trian-<lb></lb> |
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1318 golo .abc. oxigonio: e per ciascun lato sia .10.bracia. Dico che l’ area sua s’ á dela multiplicatione<lb></lb> |
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1319 del catetto .ad. nela mitá dela basa .bc. over dela basa .bc. nela mitá del catetto .ad. che cosí<lb></lb> |
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1320 il proveró. La linea .ad. che è catetto fa sopra la linea .bc.2. angoli retti. E peró ciascuno de’ .2.<lb></lb> |
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1321 triangoli .abd. e .acd. è ortogonio e sonno iquali infra loro. Imperoché lo lato .ab. dell’ uno è<lb></lb> |
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1322 iguale al lato .ac. del’ altro e lo lato .ad. è commune. Dove la basa .bd. è iguale ala basa .de. con-<lb></lb> |
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1323 ciosiacosaché l’ angolo .d. di ciascuno triangolo è retto. E noi habiamo giá mostro che l’ area<lb></lb> |
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1324 de’ triangoli ortogonii s’ á dela multiplicatione dela mitá della linea che tiene l’ angolo ret-<lb></lb> |
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1325 to in tutta l’ altra linea che tiene quello angolo over è converso. Aduncha l’ area del triango-<lb></lb> |
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1326 lo .adc. s’ á dela multiplicatione dela mitá del catetto .ad. in tutta la basa .bd. E, simelmente, l’ a-<lb></lb> |
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1327 rea del triangolo .adc. s’ a dela multiplicatione del catetto .ad. nela mitá dela basa .de. over de-<lb></lb> |
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1328 la mitá del catetto .ad. nela basa .dc. E peró l’ area di tutto il triangolo .abc. s’ á dela multiplica-<lb></lb> |
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1329 tione dela mitá del catetto .ad. in tutto la basa .bc. ch’ era bisogno mostrare. La longheza adon-<lb></lb> |
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1330 cha del catetto .ad. (comme si mostra) è .R. di .75., quasi poco meno di .8 2/3., che, multiplicato per la<lb></lb> |
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1331 mitá del .bc. over la mitá di .8 2/3., multiplicato per .bc., haremo poco meno di .43 1/3. e<lb></lb> |
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1332 poco meno di .43. braccia .1/3. sia quadro il detto triangolo. Over, multiplicando la<lb></lb> |
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1333 .R.75. per .5., fanno R.1875., per l’ area del deto triangolo, che la .R. è poco meno di .43 1/3.<lb></lb> |
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1334 Ancora sia un triangolo oxigonio avente .2. facie iguali e l’ altra non iguale che sia<lb></lb> |
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1335 <lb></lb> |
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1336 <pb></pb></p> |
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1337 <p class="folio"> |
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1338 folio 9r |
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1339 </p> |
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1340 <p class="main"> |
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1341 <lb></lb> |
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1342 </p> |
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1343 <p class="runhead"> |
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1344 Distinctio prima. Capitulum septimum. 9 |
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1345 </p> |
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1346 <p class="main"> |
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1347 <lb></lb> |
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1348 .def. e il lato .de., il lato .df., ciascun sia .10. e il lato .ef. sia .12.; voglio trovare il catetto cadente sopra<lb></lb> |
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1349 la basa .ef., che sia .dg. Dico adunque che l’ area del triangolo. def. s’ á della multiplicatione del catetto<lb></lb> |
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1350 .dg. nela mitá dela basa .ef. Imperoché, quando si multiplica il catetto per la mitá dela basa, el constituirá<lb></lb> |
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1351 uno quadrilatero rettiangulo fatto dal chateto .dg. e dala mitá dela basa: cioé dal .gf., che chiara-<lb></lb> |
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1352 mente appare. Menise la linea .dh. iguale e equedistante ala linea .gf., che è iguale ala linea .ge. e com-<lb></lb> |
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1353 pise la retta .fh., che sará iguale al catetto .dg. per la .34a. del primo. Dove il quadrilatero .dgfh.<lb></lb> |
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1354 sará iguale al triangolo .def. Imperoché ’l triangolo .dfh. è iguale al triangolo .deg. Imperoché lo<lb></lb> |
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1355 lato .df. è iguale alo lato .de. (ex ypotesi), perché ciascuno è posto .10. braccia. E lo lato .dg.<lb></lb> |
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1356 iguale alo lato .hf. E lo lato .eg. è iquale alo lato .gf. E il quadrilatero .dgfh. è fatto dela multiplicatio-<lb></lb> |
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1357 ne del catetto nella mitá dela basa che si doveva mostrare. Lo catetto .dg. (comme sia mostro)<lb></lb> |
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1358 è .8., dove l’ area del detto triangolo è .48.bracia quadre. |
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1359 </p> |
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1360 <p class="main"> |
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1361 Ancora sia uno triangolo oxigonio diversilatero .abc., del quale il lato .ab. sia .13. e<lb></lb> |
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1362 .bc.14. e .ac.15.braccia, del quale il catetto sia in sulla facia dele .14.bracia, cioé in la facia<lb></lb> |
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1363 .bc. el quale catetto sia .ad. Dico che l’ area del detto triangolo s’ á di multiplicare la mitá del<lb></lb> |
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1364 detto catetto .ad. per tutta la basa .bc. overo di multiplicare la mitá dela basa .bc. per<lb></lb> |
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1365 lo catetto .ad., che chiaro il dimostaró. Faremo nel detto triangolo uno quadrilatero rettango-<lb></lb> |
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1366 lo avente, in lunghezza, la quantitá dela basa .bc. e, per larghezza, la mitá del chatetto. Dove si di-<lb></lb> |
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1367 mostará con veritá el detto quadrilatero essere iguale al detto triangolo. Dividase il catetto .ad. in<lb></lb> |
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1368 due parti iguali sopra il punto .e. e, per lo detto punto, passerá la linea .fg., dove aremo .fb. e. gc.<lb></lb> |
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1369 iguali e equedistanti ala linea .ed. E la linea .fg. sia iguali e equedistante ala linea overo ala basa .bc.<lb></lb> |
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1370 Dove il triangolo .abc. è diviso in .2. triangoli: l’ uno .abd. e l’ altro .adc. Dove si proverá il triango-<lb></lb> |
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1371 lo .abd. essere iguale al quadrilatero .fedb. in questo modo. Tragasi da ciascuna parte la figura di .4. la-<lb></lb> |
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1372 ti, cioé .hedb., rimarrá il triangolo .bfh. dall’ una parte e dall’ altra il triangolo .aeh., li quali dimo-<lb></lb> |
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1373 staró che sonno iguali in questo modo. Lo lato .he. è iguale alo lato .fh. per la .39a. del primo. Imperoché’ l<lb></lb> |
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1374 lato .ad. del triangolo .adb. è diviso per la linea .ef. in .2. parti iguali e quella linea .ef. è equestistante al-<lb></lb> |
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1375 la linea .bd. per la .2a. del .6o. e peró la linea .ab. è segata per lo mezzo nel punto .f. dalla linea .fe. E peró<lb></lb> |
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1376 .ha. è iguali al .hb. E ’l lato .fb. è iguale al lato .ae., dove lo lato .fh. è iguale al lato .he. Imperoché<lb></lb> |
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1377 l’ angolo .e. è iguale al’ angolo .f., perché ciascuno è retto. Dove tanto rimane a trare il quadrato<lb></lb> |
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1378 del lato .bf. del quadrato del lato .bh. quanto rimane a trare el quadrato dello lato .ae. del quadra-<lb></lb> |
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1379 to del lato .ah. E il quadrato del lato .he. overo .fh. è quello che rimane per la .46a. del po.<lb></lb> |
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1380 Adunque è provato e lati del triangolo .ahe. essere iguali alli lati del triangolo .bfh. E peró tut-<lb></lb> |
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1381 to el triangolo è iguale a tutto il triangolo. E peró el quadrilatero .fedb. è iguale al triangolo .abd. E, per simil <lb></lb> |
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1382 modo, el triangolo .adc. è iguale al quadrilatero .edcg., ch’ era bisogno mostrare.<lb></lb> |
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1383 Adunque l’ area del triangolo detto s’ á di multiplicare la mitá del .ad., che comme mostaró è<lb></lb> |
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1384 .6., via tutta .bc., che è .14., che fanno .84. per l’ area detta. Imperoché ’l quadrilatero .fbcg. è igua-<lb></lb> |
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1385 le al detto triangolo .abc. e questo chiaro apare per la detta figura.<lb></lb> |
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1386 Ancora, per quelle cose che si dissano nel triangolo ortogonio, si prova lo trian-<lb></lb> |
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1387 golo .adb. essere iguale a quello ch’ é fatto della mitá del .ad. in .bd. e, similmente,<lb></lb> |
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1388 l’ area del triangolo .adc. venire dela mitá del .ad. in .dc. E perché li .2. triangoli or-<lb></lb> |
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1389 togonij, cioé .abd. e .adc., insiemi gionti, sono iguali al triangolo .abc., segue il gran<lb></lb> |
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1390 triangolo .abc. essere iguale al quadrilatero rettangolo fatto del .bc. e dela mitá del .ad., che é<lb></lb> |
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1391 quello che habiamo detto. |
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1392 </p> |
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1393 <p class="main"> |
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1394 Se l’ area vuoi del triangolo ampligonio el quale è di .2. facie iguali overo di .3.<lb></lb> |
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1395 facie non iguali. Comme sia il triangolo .abc. e l’ angolo .a. sia obtuso e il lato .ab. e .ac.<lb></lb> |
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1396 sienno iguali infra lloro. Dico che, se il catetto si menerá dal’ angolo .a. in sulla facia<lb></lb> |
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1397 .bc. e sia .ad., che l’ area del detto triangolo si truova dela multiplicatione dela mi-<lb></lb> |
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1398 tá del .ad. in tutta .bc. overo dela multiplicatione dela mitá del .bc. in tutta .ad. E questo chia-<lb></lb> |
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1399 ro apare per le cose dette. Imperoché ’l triangolo .abc. è diviso in .2. triangoli ortogonij, cioé<lb></lb> |
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1400 .adb. e .adc. e l’ area del triangolo .adb. s’ á dela multiplicatione del .bd. nela mitá del .ad. E, simile, il trian-<lb></lb> |
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1401 golo .adc., se ne truova l’ area nel multiplicare la mitá del .ad. in .dc. E peró, a multiplicare tutta .bc. nela mitá<lb></lb> |
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1402 del .ad. haremo l’ area del gran triangolo .abc. che si conveniva. E, con numeri, sia .ab.12. e il<lb></lb> |
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1403 .bc. sia .26. e o .ad.5. che, multiplicato .bc. nela mitá del .ad., cioé .26. in .2 1/2., overo .ad. nela mitá<lb></lb> |
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1404 del .bc., haremo .65. per l’ area di detto triangolo.<lb></lb> |
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1405 <lb></lb> |
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1406 E similmente, quando il triangolo ampligonio fusse diversilatero. Commo sia<lb></lb> |
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1407 el triangolo ampligono .abg., del quale l’ angolo .b. sia obtuso: e sia .ab.13. bracie e .bg.<lb></lb> |
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1408 sia .4. e il .ga. sia .15. Dico che, dal’ angolo .b. mosso il catetto .bd. che, per quello che s’ é<lb></lb> |
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1409 <lb></lb> |
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1410 <pb></pb></p> |
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1411 <p class="folio"> |
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1412 folio 9v |
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1413 </p> |
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1414 <p class="main"> |
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1415 <lb></lb> |
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1416 </p> |
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1417 <p class="runhead"> |
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1418 Distinctio prima. Capitulum octavum. |
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1419 </p> |
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1420 <p class="main"> |
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1421 <lb></lb> |
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1422 detto, l’ area del detto triangolo haversi dela multiplicatione del .ga. in la mitá del .bd. overo<lb></lb> |
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1423 dela multiplicatione del .bd. nela mitá del .ag., che si manifesta per quello che habiamo detto.<lb></lb> |
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1424 E peró non è da replicare. E con numeri sia .bd.3 1/5. che, multiplicato .3 1/5. via .7 1/2., fanno .24. per l’ area <lb></lb> |
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1425 del<lb></lb> |
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1426 detto triangolo. Overo, multiplicato la mitá de .3 1/5., cioé .1 3/5. via .15., fanno ancora el .24., comme se disse.<lb></lb> |
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1427 Ma, se dal’ angolo .a. overo dal’ angolo .g. vorrai menare il catetto sopra quel trian-<lb></lb> |
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1428 golo, non cadrá dentro ma cadrá fuor del triangolo comme dimostrammo. Onde<lb></lb> |
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1429 menise dal’ angolo .a. il catetto .ad. E continuise .gb. con .d. e haremo .gbd. Dico che<lb></lb> |
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1430 l’ area del detto triangolo .abg. s’ á del multiplicare .ad. nela mitá del .gb. overo del<lb></lb> |
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1431 .gb. nela mitá del .ad., che si conviene dimostrare. Lo triangolo .adg. è ortogonio e l’ angolo .d.<lb></lb> |
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1432 è retto. Adunca l’ area di quello s’ á de multiplicare .ad. nela mitá del .gd. E il triangolo .agd.<lb></lb> |
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1433 è diviso in .2. triangoli .adb. e .abg. Dove il triangolo .agd. avanza al triangolo .abg. l’ area del<lb></lb> |
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1434 triangolo .adb. E l’ area del triangolo .adb. s’ á del multiplicare .ad. in mitá del .bd. Onde, se<lb></lb> |
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1435 trarremo la mitá del .bd. dela mitá del .dg., rimarrá la mitá del .gb. Adunca l’ area del triango-<lb></lb> |
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1436 lo .abg. s’ á del multiplicare .ad. nela mitá del .gb., che se doveva mostrare. E per numeri sia .cd.12.<lb></lb> |
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1437 che, multiplicato per la mitá del .gb., fanno .24. Overo, mulplicato .gb. per la mitá di .12., hare-<lb></lb> |
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1438 mo ancora .24. per la detta area, che è il proposito. |
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1439 </p> |
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1440 <p class="main"> |
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1441 E accioché perfetta dottrina in questo libro sia a misurare e triangoli, è da mostrare<lb></lb> |
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1442 comme ciascuno triangolo, senza investigatione del catetto, si possa misurare. Li<lb></lb> |
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1443 lati de ciascun triangolo agiongni insiemi: e di quel togli la mitá, del qual trae<lb></lb> |
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1444 per ordine e lati del triangolo e multiplica l’ avanzo del’ un lato per l’ altro avanzo<lb></lb> |
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1445 del’ altro lato e la somma per l’ avanzo del’ altro lato multiplica. E tutto multiplicarai per la mitá<lb></lb> |
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1446 de’ .3. lati. E dela somma truova la radici, la qual radici sia l’ area del detto triangolo. Verbi<lb></lb> |
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1447 gratia, agionti insiemi li lati del triangolo .abg., del quel il lato .ab. è .15. braccia e .bg. 14. e .ag. è .13.<lb></lb> |
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1448 che, in uno agionti, fanno .42., del qual la mitá è .21., dal qual il maggior lato è distante .6. braccia<lb></lb> |
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1449 e l’ altro .7. e l’ altro .8. braccia. Dove multiplicarai .6. per .7. e tutto per .8., fanno .336. E questo<lb></lb> |
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1450 multiplica per .21., cioé per la mitá de’ lati, fanno .7056. La cui radici è .84. per l’ area del detto tri-<lb></lb> |
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1451 angolo, comme giá di sopra mostrammo. E questo basti quanto al presente capitolo: e seguen-<lb></lb> |
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1452 do del’ ultimo capitolo di questa distintione diremo. |
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1453 </p> |
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1454 <p class="main"> |
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1455 Qualiter inveniantur catheti. siue perpendiculares. cuiuscunque trianguli. capitulun. octavum.<lb></lb> |
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1456 Havendo mostro comme l’ area di ciascuno triangolo si truova per la multiplicatione<lb></lb> |
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1457 dela mitá del catetto per tutta la basa: over la mitá dela basa per tutto il catetto, è ra-<lb></lb> |
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1458 gionevole cosa de dimostrare comme tale catetto si truova. Detto adunca habia-<lb></lb> |
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1459 mo che tutti i triangoli o e sonno di .3. lati iguali overo di .2. lati iguali. e uno non<lb></lb> |
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1460 iguale overo di .3. lati non iguali. El catetto che si muove da ciascuno angolo del triangolo<lb></lb> |
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1461 di .3. lati iguali sempre cade in sul ponto del mezzo lato oposto a quello angolo. Comme sia il<lb></lb> |
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1462 triangolo .abc. che ciascuno lato sia .10.bracia. Dico che ’l catetto che si muove dal’ angolo .a. ca-<lb></lb> |
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1463 de nel ponto del mezzo lato .bc. el quale è il ponto .d. e che questo sia vero. Ponga l’ aversario<lb></lb> |
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1464 che caggia in sul ponto .f. e non sia .f. nel mezzo dela linea, ma sia meno .bf. che .cf. Dove, a tro-<lb></lb> |
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1465 vare quanto sia .fa. (secondo la .46a. del primo d’ Euclide), tu trarai el quadrato del .bf. del quadra-<lb></lb> |
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1466 to del .ab. overo el quadrato del .fc., del quadrato .ac. E tanto debia essere l’ uno rimanente quan-<lb></lb> |
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1467 to l’ altro. E questo non puó essere, imperoché meno rimane a trare il quadrato .fc. del qua-<lb></lb> |
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1468 drato .ac. che non rimane a trare el quadrato .fb. del quadrato .ba. E peró, di necessitá, il pon-<lb></lb> |
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1469 to dove cade il catetto è in sul mezzo lato .bc., cioé in sul ponto .d. Adunca .bd. è .5.bracia. Dove,<lb></lb> |
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1470 a trovare quanto è .ad., trarai el quadrato .bd. overo .dc. del quadrato .ba. overo del quadra-<lb></lb> |
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1471 to .ac. e quello rimane è il quadrato del .ad. Adunca trarai .25. di .100., riman .75. e dirai che’ l<lb></lb> |
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1472 quadrato del .ad. sia .75. E peró il catetto .ad. è radici di .75. che è circa .8 2/3. E questo volemo di-<lb></lb> |
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1473 mostrare. |
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1474 </p> |
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1475 <p class="main"> |
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1476 Se ’l triangolo è di .2. lati iguali e l’ altro non iguale. Comme sia il triangolo .dfe., del<lb></lb> |
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1477 quale il lato .de. e .df. sia ciascuno .10.bracia. e il lato .ef. sia .12. E noi volessimo me-<lb></lb> |
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1478 nare il catetto dal ponto .d. in sula faccia .ef., dico il ponto dove cade detto catet-<lb></lb> |
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1479 to essere aponto nel mezzo della linea .ef. che, per lo modo dela passata, si pruova, el qual pon-<lb></lb> |
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1480 to è il ponto .g. E, se el ponto .g. è nel mezzo dela linea .ef. e tu voglia la perpendiculare .dg.,<lb></lb> |
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1481 trai el quadrato dela linea .eg. del quadrato dela linea .ed. overo trai el quadrato dela linea<lb></lb> |
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1482 .gf. del quadrato dela linea .fd. E quello che rimane è il quadrato dela perpendiculare .dg.,<lb></lb> |
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1483 cioé trarai .36. di .100., che rimane .64., la cui radici è la linea .dg., che è .8. et cetera.<lb></lb> |
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1484 <lb></lb> |
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1485 <pb></pb></p> |
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1486 <p class="folio"> |
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1487 folio 10r |
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1488 </p> |
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1489 <p class="main"> |
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1490 <lb></lb> |
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1491 </p> |
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1492 <p class="runhead"> |
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1493 Distinctio prima. Capitulum octavum. 10 |
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1494 </p> |
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1495 <p class="main"> |
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1496 <lb></lb> |
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1497 E se ’l triangolo è di diversilatero con gli angoli. Comme sia il triangolo .abc., del<lb></lb> |
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1498 quale il lato .ab. sia .13.bracia. E lo lato .bc. sia .14. el lato .ca. sia .15. E vogliasi sape-<lb></lb> |
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1499 re quanto è la perpendiculare che si muova dal ponto .a. e caggia in sula linea .bc.<lb></lb> |
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1500 Questo tale catetto non si puó havere se non s’ á prima il ponto dove tale chatet-<lb></lb> |
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1501 to caggia. Cioé quel ponto quanto el caggia presso al’ angolo .b. e quanto caggia presso al’ an-<lb></lb> |
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1502 golo .c. Imperoch’ el nel ponto di mezzo non cade: provando per la .46a. del primo di Euclide.<lb></lb> |
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1503 Ma dove caggia, cioé dove sia tale ponto, per .3. modi lo dimostraró. El primo modo è<lb></lb> |
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1504 che agionga la potentia d’ uno lato con la potentia dela basa. Cioé con la potentia di quello<lb></lb> |
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1505 lato dove tale ponto debba essere. E di quella summa si traga la potentia del’ altro lato. E<lb></lb> |
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1506 quel che rimane parti per lo doppio dela basa overo la mitá di quel che rimane parti per la<lb></lb> |
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1507 basa e quel che ne viene sia la distantia che è dal’ angolo fatto dal lato che la sua potentia,<lb></lb> |
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1508 cioé il suo quadrato, s’ agionse al quadrato dela basa, fino al detto cadimento, cioé al ponto<lb></lb> |
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1509 dove tale catetto cade. Comme nello exemplo del dato triangolo agiongnerai la potentia<lb></lb> |
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1510 del lato .ab. con la potentia del lato .bc., cioé .169. con .196., fanno .365. Dela qual summa tra’ la<lb></lb> |
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1511 potentia del lato .ac. cioé .225., rimane .140. Dove dividerai .140. per lo doppio dela basa, cioé<lb></lb> |
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1512 per .28. overo la mitá de .140. dividerai per .14. che ne viene .5., che sono la distantia che<lb></lb> |
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1513 è dal’ angolo .b. al cadimento del detto catetto, cioé infino al ponto .d. E, dal ponto .d. infino<lb></lb> |
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1514 al’ altro angolo, sia l’ avanzo infino in .14. braccia, che sia .9.bracia. Overo per l’ altro modo, cioé agion-<lb></lb> |
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1515 gnendo la potentia del lato .ac. con la potentia dela basa .bc., cioé .225. con .196., fanno .421., di<lb></lb> |
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1516 quali tra’ la potentia del lato .ab., cioé .169., rimane .252. e questo parti per lo doppio dela basa,<lb></lb> |
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1517 vienne .9. E .9.bracia. è dal’ angolo .c. infino al ponto .d. El qual ponto .d. lo diciamo cadimento<lb></lb> |
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1518 di detto catetto. Lo secondo modo è che agionga le bracia degli lati dal lato dala basa do-<lb></lb> |
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1519 ve tal catetto cade: che in questa agiongnerai lo lato .ac. con lo lato .ba., fanno .28. Li quali<lb></lb> |
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1520 sempre per .2. dividi, vienne .14., li quali multiplica per la diferentia che è da uno di detti lati a-<lb></lb> |
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1521 la ditta mitá overo avenimento. La quale differentia in questa è .uno., fanno .14., li quali divi-<lb></lb> |
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1522 di per la mitá dela basa, cioé per .7., vienne .2., li quali agiongni ala mitá dela basa: fanno .9. El<lb></lb> |
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1523 qual .9. è la distantia dal’ angolo del maggiore lato infino al cadimento di detto catetto, cioé<lb></lb> |
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1524 dal’ angolo .c. al ponto .d. Overo el .2. che ne viene tra’ dela mitá dela basa e rimane .5. e tan-<lb></lb> |
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1525 to é dal’ angolo dela minore ypotemissa overo del minor lato infino al detto ponto: cioé da-<lb></lb> |
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1526 l’ angolo .b. infino al ponto .d. L’ altro modo è che traga la potentia dela minore ypotemissa<lb></lb> |
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1527 dela potentia dela maggiore: cioé in questa .169. di .225., rimane .56. e l’ avanzo, che è .56., parti<lb></lb> |
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1528 per la basa, che ne viene .4., li quali agiongni ala basa, sonno .18. De’ quali la mitá è .9. per la<lb></lb> |
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1529 distantia del .c. al .d. Overo il detto .4. trarai dela basa, rimangono .10., de’ quali la mitá è .5.<lb></lb> |
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1530 per la distantia del .b. al .d., comme abiamo detto. Onde, adunca, se vorrai la perpendiculare<lb></lb> |
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1531 .ad., trarai la potentia del minore cadimento, cioé .25. dela potentia dela ypotemissa .ab., rima-<lb></lb> |
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1532 ne .144. per la potentia dela detta perpendiculare. Adunca la perpendiculare é la radici di<lb></lb> |
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1533 .144. che è .12. Overo la potentia del maggior cadimento trae dela potentia dela maggiore<lb></lb> |
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1534 ypotemissa, cioé .81. de .225., rimane .144. per la pontetia delo catetto .ad., che è .12. il detto catetto.<lb></lb> |
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1535 Mostrasi nel secondo de Euclide per .12am.13am. onde prociede il modo dela pri-<lb></lb> |
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1536 ma inventione del cadimento del catetto nel triangolo predetto. E noi mostra-<lb></lb> |
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1537 remo onde prociede il modo dato, nel secondo e terzo modo, con figure geo-<lb></lb> |
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1538 metriche.. Scrivasi ancora el triangolo .abc. e menisi in quello il catetto .ad. E,<lb></lb> |
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1539 per gli ponti .b. e .c., a retti angoli, si meni la retta .be. e .cf. E sia la retta .eb. iguali ala retta .ca., cioé<lb></lb> |
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1540 sia .15.bracia. E la retta .fc. sia iguali ala retta .ab., cioé sia .13.bracia. E faciase .fe.fd.ed. E divi-<lb></lb> |
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1541 dasi la retta .ef. in .2. parti iguali sopra il ponto .g. E, dal ponto .g., si meni la retta .gh. equedi-<lb></lb> |
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1542 stante ala linea .cf. e .be. E dal ponto .f. si meni la retta .fik. equedistante ala linea .bc. Ancora,<lb></lb> |
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1543 dal ponto .g., si meni la retta .gl. equedistante e iguali ala retta .ki. E, perché e triangoli .adc.<lb></lb> |
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1544 e .adb. sonno ortogonij, imperoché gli hano uno angolo retto per ciascuno: cioé l’ angolo<lb></lb> |
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1545 .d., la potentia delo lato .ac. è iguale a .2. potentie, cioé ala potentia dela linea .ad. e dela linea.<lb></lb> |
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1546 .dc. E la potentia dela linea .ab. è iguale a .2. potentie, cioé ala potentia dela linea .ad. e dela li-<lb></lb> |
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1547 nea .db. Dove, se comunamente si trae la potentia dela linea .ad., potrá la potentia de maggior<lb></lb> |
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1548 cadimento .dc. piú che la potentia del minore .bd. quanto puol la potentia dela linea .ac. piú<lb></lb> |
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1549 che la potentia .ab. Adunque la potentia dela linea .ab. e dela linea .dc. è quanto la potentia<lb></lb> |
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1550 <lb></lb> |
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1551 <pb></pb></p> |
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1552 <p class="folio"> |
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1553 folio 10v |
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1554 </p> |
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1555 <p class="main"> |
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1556 <lb></lb> |
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1557 </p> |
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1558 <p class="runhead"> |
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1559 Distinctio prima. Capitulum Octavum. |
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1560 </p> |
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1561 <p class="main"> |
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1562 <lb></lb> |
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1563 dela linea .ac. e dela linea .bd. Ma la retta .fc. è iguale ala retta .ab. E la retta .eb. è iguale a-<lb></lb> |
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1564 la retta .ac. Onde la potentia dele linee .fc. e .cd. sonno iguali ale potentie dele linee .ab. e .dc.<lb></lb> |
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1565 E la potentia dela linea .fd. è iguali ale .2. potentie, cioé del .fc. e .cd., perché l’ angolo .c. è retto.<lb></lb> |
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1566 Adunque la potentia dela linea .fd. è iguale ale .2. potentie .ab. e .dc. E, per simil modo, la po-<lb></lb> |
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1567 tentia dela linea .ed. è iguali a .2. potentie .ac. e .bd. Per la qual cosa .ed. e .df. sonno infra lo-<lb></lb> |
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1568 ro iguali. Adunque il triangolo .fde. á .2. facie iguali. E, perché la basa .ef. é divisa in .2. parti<lb></lb> |
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1569 iguali sopra il ponto .g., la linea .dg. è certamente catetto sopra la linea .fe. Onde l’ angolo<lb></lb> |
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1570 .egd. è retto e l’ angolo .fgd. è ancora retto. Ancora, perché la retta .gh. è equedistante ala<lb></lb> |
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1571 retta .fc. e cade sopra la retta .cb., gli angoli adunque .fch. e .ghc. sonno iguali a .2. retti. Ma<lb></lb> |
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1572 l’ angolo .fch. è retto. Onde l’ angolo .ghc. è ancora retto. E l’ angolo .ghd. è retto, perché è<lb></lb> |
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1573 di fuori; catetto è adunque la linea .gh. sopra la retta .bc. E il quadrilatero .kbcf. è paralel-<lb></lb> |
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1574 lo di retti angoli. Dove i lati oposti sonno iguali. Iguali è adunque la retta .fk. alla retta .bc.<lb></lb> |
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1575 E la retta .kb. ala retta .cf. Adunque .bk. è .13. e .ke. è .2. Imperoché la retta .kb. è iguali a-<lb></lb> |
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1576 la retta .ih. E il paralello di retti angoli .kbhi. E il paralello .ihcf. ancora di retti angoli.<lb></lb> |
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1577 Onde la retta .ih. è iguali ala retta .ba. che è ancora .13. Ancora, perché nelle equedistanti<lb></lb> |
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1578 .eb. e .gh. la retta .ef. passa, sará l’ angolo .fgi. di fuora iguale al’ angolo .gel. dentro. E l’ ango-<lb></lb> |
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1579 lo .fig. è iguale al’ angolo .gle., imperoché sonno in infra due linee equedistanti e ciascuno è<lb></lb> |
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1580 retto. E l’ angolo .gfi. è iguale al’ angolo .egl. E la retta .fg. è iguali alla retta .ge. Onde gli al-<lb></lb> |
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1581 tri lati sonno igual agli altri lati. E peró il lato .gi. è iguali al lato .el. E il lato .fi. è igual al<lb></lb> |
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1582 lato .gl. Ma la retta .gl. è iguale ala retta .ik., perché il paralello rettangolo è di lati equedi-<lb></lb> |
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1583 stanti. Cioé il paralello .lkig. Adunque la retta .fi. è iguali ala retta .ik. E tutta .ch. è igua-<lb></lb> |
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1584 le ala linea .hb. Adunque è divisa la basa in .2. iguali parti sopra il ponto .h. Onde .ch. è .7. |
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1585 </p> |
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1586 <p class="main"> |
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1587 Ancora, perché il paralello .lkig. è rettangolo, iguali è la retta .lk. ala retta .ig. Ma la retta<lb></lb> |
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1588 .ig. è mostro che è iguali ala retta .le. Onde è iguali al .kl. E tutta .ke. è .2. Adunque ciasca-<lb></lb> |
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1589 na dele linee .kl.le.ig. è uno bracio. Dove tutta .hg. è .14.bracia. Ancora, perché retto è l’ an-<lb></lb> |
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1590 golo .dgf., e .2. angoli che sonno .dgh. e .hgf. sonno iguali a uno retto. Ancora, perché il tri-<lb></lb> |
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1591 angolo .gif. è rettangolo, che l’ angolo .i. è retto, gli altri .2. angoli, che sonno .igf. e .ifg., son-<lb></lb> |
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1592 no iguali a uno retto. Adunque gli angoli .dgh. e .hgf. sonno iguali agli angoli .hgf. e .gfi.<lb></lb> |
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1593 Onde, se comunamente si trae di ciascuno l’ angolo .hgf., rimarrá l’ angolo .dgh. iguali al’ an-<lb></lb> |
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1594 golo .gfi. E l’ angolo .gif. al’ angolo .ghd. E l’ altro angolo .igf. al’ altro .hdg. Simile é a-<lb></lb> |
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1595 dunque il triangolo .fig. al triangolo .ghd., onde è comme .fi. al .gi., cioé comme .7. a .1. co-<lb></lb> |
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1596 sí .gh., cioé .14., al .hd. Onde la multiplicatione del .gi. in .gh., divisa per .fi. fa .hd. E questo è<lb></lb> |
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1597 quando facemo che agiongnemo .ab. collo lato .ac., cioé .fc. con .be. e havemo .28., de’ quali la<lb></lb> |
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1598 mitá è .14., cioé la retta .gh. La qual multiplicamo per .ig., che è quello nel quale la retta .gh.<lb></lb> |
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1599 avanza la retta .fc., cioé .el. over .gi. E .gi. è quello che la retta .ac., cioé .eb., avanza la linea .gh.<lb></lb> |
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1600 Dela qual multiplicatione havemo .14., la qual dividemo per la mitá dela basa, cioé per .ch.,<lb></lb> |
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1601 cioé per .fi. che è .7. e haremo .2. per la quantitá .hd., la quale agiongneremo ala mitá dela ba-<lb></lb> |
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1602 sa, cioé al .ch. e haremo .9. per lo .cd., che è il maggiore cadimento, over traremo .hd. del .hb.,<lb></lb> |
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1603 cioé .2. di .7., rimarano .5. per lo minor cadimento .db., ch’ era di bisognio se dimostrasse et cetera.<lb></lb> |
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1604 Ancora mostraremo la figura necessaria al terzo modo, cioé la dimostratione del<lb></lb> |
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1605 trovare el cadimento per lo terzo modo. Faciasi ancora el detto triangolo e meni-<lb></lb> |
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1606 se adunque sopra la basa .bc. el catetto .ad. e, perché maggiore è il lato .ac. che .ab.,<lb></lb> |
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1607 maggiore è il cadimento .dc. che .db. Onde del .dc. si tolga .dg. che sia iguali a-<lb></lb> |
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1608 la retta .db. e sará la retta .bg. divisa in .2. parti iguali sopra il ponto .d. ala quale, per lo dritto,<lb></lb> |
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1609 s’ agiongne la retta .gc., onde la multiplicatione del .gc. in .bc., col quadrato dela linea .bd., è<lb></lb> |
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1610 iguali al quadrato dela linea .cd.ex. 6a. secundi. Adunque el quadrato dela linea .dc. avan-<lb></lb> |
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1611 za el quadrato dela linea .db., cioé el quadrato del maggior cadimento, nela quantitá dela<lb></lb> |
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1612 multiplicatione dela linea .gc. nela linea .bc. Ma disopra è mostro, nel’ altra figura, che l’ a-<lb></lb> |
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1613 vanzo del quadrato del cadimento .dc. al quadrato del cadimento .db. è comme la soprabun-<lb></lb> |
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1614 dantia del quadrato delo lato .ac. al quadrato del lato .ab. e comme la multiplicatione del<lb></lb> |
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1615 .gc. nel .bc. Ma il quadrato del .ac. è .225., che avanza il quadrato del .ab., cioé .169., in .56.; on-<lb></lb> |
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1616 de la multlplicatione del .gc. in .bc. è .56. Dove, se ’l .bc. è .14., sará .gc.4., che tratto il detto .4.<lb></lb> |
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1617 dela basa, cioé di .14., rimangono .10. per lo .bg. De’ quali la mitá è .5. che è il minore cadimen-<lb></lb> |
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1618 to, ch’ era bisogno dimostrare.<lb></lb> |
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1619 <lb></lb> |
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1620 <lb></lb> |
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1621 <pb></pb></p> |
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1622 <p class="folio"> |
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1623 folio 11r |
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1624 </p> |
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1625 <p class="main"> |
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1626 <lb></lb> |
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1627 </p> |
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1628 <p class="runhead"> |
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1629 Distinctio prima. Capitulum octavum. 11 |
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1630 </p> |
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1631 <p class="main"> |
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1632 <lb></lb> |
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1633 Havendo mostro in che modo el catetto di ciascuno triangolo si truova, mi pare<lb></lb> |
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1634 di necessitá dimostrare la cagione e il perché, nel trovare l’ area de’ triangoli per lo<lb></lb> |
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1635 secondo modo dato, li lati s’ agiongono insiemi e dela somma se ne pigli la mitá<lb></lb> |
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1636 e aoperare comme nel capitolo passato mostrammo. E a questa indurremo una<lb></lb> |
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1637 figura triangulare .abg. Ove dividerai l’ angolo .b. e l’ angolo .g. in .2. parti iguali dale rette<lb></lb> |
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1638 .bt. e .tg. E dal ponto .t. si meni li catetti .te.th.tz. E compisi .at. E, perché l’ angolo .thg. e .tzg. è<lb></lb> |
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1639 retto, iguale é l’ angolo .thg. al’ angolo .tzg. E l’ angolo .tgh. è iquali al’ angolo .tgz. Perché po-<lb></lb> |
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1640 nemmo l’ angolo .g. esser diviso in .2. parti iguali dala linea .gt. Onde seguita l’ angolo .gtz. esse-<lb></lb> |
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1641 re iguali al’ angolo .gth. Adunque il triangolo .ztg. è iguali al triangolo .htg. E, perché il la-<lb></lb> |
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1642 to .gt. è comune, gli altri lati del’ uno fienno iguali agli altri lati dell’ altro, cioé il lato .th. al<lb></lb> |
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1643 lato .tz. E il lato .hg. al lato .gz. Similmente si mostra la retta .hb. ala retta .be. essere iguali e<lb></lb> |
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1644 il triangolo .thb. essere iguali al triangolo .teb. E, perché l’ una e l’ altra dele rette .te. e .tz. sonno<lb></lb> |
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1645 iguali ala retta .th. fienno infra loro iguali. Onde iguale è la retta .te. ala retta .tz. E, per comu-<lb></lb> |
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1646 ne è la retta .ta. E peró .te. e .ta. sonno iguali al .tz. e .ta. E l’ angolo .aet. al’ angolo .azt. è iguale<lb></lb> |
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1647 e il lato .at. è comun. Onde equilatero e equiangolo é il triangolo .aet. al triangolo. azt. E<lb></lb> |
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1648 peró il lato .az. è iguali al lato .ae. E, perché e gli é iguali la retta .az. ala retta .ae., se s’ agiongni<lb></lb> |
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1649 a ogni parte la retta .eb., sará la retta .ab. iguali a .2. rette, cioé .az. e .eb., cioé al .az. e .bh. Ancora,<lb></lb> |
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1650 perché la retta .zg. è iguali ala retta .gh., saranno le .2. rette .ag.hb. iguali a .2. rette .ab. e .gh. Im-<lb></lb> |
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1651 peroché .ab. é quanto .az. e .bh. e il .hz. è quanto .gh. E peró, agiongnendo al .ab. il .gh., ha-<lb></lb> |
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1652 remo .ab. e .gh. iguali al .ag. e .hb., comme dicemmo. Adunque .ag. e .hb. sonno la mitá de’ detti<lb></lb> |
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1653 lati de’ triangoli posto. Onde .eb. è quello che la mitá de’ detti lati avanza el lato .ag. E, simil-<lb></lb> |
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1654 mente, .ae. è quello che la mitá de’ detti lati avanza al lato .bg. E il .gz. è quello che la mitá de’ <lb></lb> |
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1655 detti lati avanza el lato .ab. Onde la retta .ab. e .hg. sonno la mitá de’ lati del triangolo .abg. e<lb></lb> |
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1656 sonno le .3. differentie. E ancora .ag. e .hb. sonno la mitá de’ .3. lati di detto triangolo. Meni-<lb></lb> |
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1657 se adunque la retta .ab. e .ag. per lo diritto: ne’ ponti .l. e .m. E sia .bl. iguali ala retta .hg. E il<lb></lb> |
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1658 .gm. sia iguale ala retta .hb. Sirá adunque l’ una e l’ altra retta .al. e .am. quanto che la mitá de’ <lb></lb> |
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1659 lati del triangolo. E poi si produca .at. nel ponto .k. e faciasi la retta .lk. e .km. E sia l’ angolo<lb></lb> |
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1660 .alk. retto. E retto sia ancora .amk. e, perché le .2. rette .al. e .ak. sonno iguali ale .2. rette. ak. e<lb></lb> |
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1661 .am. e l’ angolo .lak. è iguali al’ angolo .mak. Onde il lato .lk. è iguali al lato .mk. e gli altri la-<lb></lb> |
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1662 ti e angoli sonno infra loro iguali. Seghisi adunque la linea .gb. in .2. parti: una iguali ala li-<lb></lb> |
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1663 nea .bl. e sia .bn. e compisi .nk.kg.kb. e, perché .gh. è l’ avanzo dela mitá de’ lati del triangolo<lb></lb> |
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1664 .abg., alo lato .ab., iguali é .al.bn., cioé al .bl. Onde .ng. è iguali al .gm., cioé al .hb. Onde e trian-<lb></lb> |
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1665 goli .gmk. e .blk. sonno ortogonij e la potentia dela linea .kg. è iguali a .2. potentie di .2. li-<lb></lb> |
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1666 nee .gm. e .mk. e la potentia dela linea .bk. è iguale a .2. potentie di .2. linee .kl. e .bl., cioé del .kl.<lb></lb> |
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1667 e .bn. Ma la potentia dela linea .lk. è iguali ala potentia .km. Onde quanto la potentia dela<lb></lb> |
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1668 linea .kg. soprabunda la potentia dela linea .kb. tanto la potentia .ng. avanza la potentia .nb.<lb></lb> |
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1669 Onde la linea .kn. è catetto sopra la linea .bg. che chiaro appare. Imperó, quando si negasse,<lb></lb> |
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1670 dirá l’ aversario sia il catetto .ko. E, perché la potentia del .kg. avanza la potentia del .kb., mag-<lb></lb> |
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1671 giore adunque .kg. del .kb. E, se ’l .ko. è catteto, avanzerá la potentia del .gk. la potentia del .bk.<lb></lb> |
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1672 quanto la potentia del .go. avanza la potentia del .bo. E noi habiamo mostro che la poten-<lb></lb> |
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1673 tia del .gk. avanza la potentia del .bk. quello che la potentia del .gn. avanza la potentia del<lb></lb> |
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1674 .nb. Adunque .ob. e .nb. fienno iguali e cosí .gn. e .go., che è impossibile. E peró .kn. è catetto e<lb></lb> |
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1675 non altro. E ancora .kn. è iguali al .kl. imperoché il .kb. è comune infra .2. triangoli ortogonii<lb></lb> |
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1676 .klb. e .knb. e .bn. e .bl. sonno iguali. E peró seguita .kn. e .kl. essere iguali. E, perché gli an-<lb></lb> |
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1677 goli .knb. e .klb. sonno retti, rimarranno gli angoli .nbl. e .lkn. iguali a .2. angoli retti. Ma<lb></lb> |
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1678 gli angoli .ebn. e .nbl., similmente, sonno iguali a .2. angoli retti, per la .13a. del primo, impero-<lb></lb> |
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1679 ché la linea .nb. cade sopra la linea .el. Onde l’ angolo .ebn. è iguali al’ angolo .lkn. E l’ angolo<lb></lb> |
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1680 .lkb. è la mitá del’ angolo .lkn., perché la linea .kb. divide e .2. triangoli iguali. Adonca è igua-<lb></lb> |
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1681 le l’ angolo .ebt. (che è la mitá del’ angolo .ebh.) al’ angolo .lkb. e l’ angolo .e. è retto, ch’ é iguale<lb></lb> |
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1682 al’ angolo .l. retto. Adunque l’ angolo .etb. è iguale al’ angolo .lbk. e ‘l triangolo adunque .kbl.<lb></lb> |
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1683 sia simile al triangolo .bte. La proportione adunque del .kl. al .lb. è commo la proportione<lb></lb> |
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1684 del .be. al .et. Multiplicato adunque .kl. in .et., fa quanto .lb. in .be. Ma la proportione del<lb></lb> |
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1685 tetragono .et. a quello che fa .et. in .kl. è commo la proportione del .et. al .lk. e la proportione<lb></lb> |
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1686 del .et. al .lk. è comme .ae. alo .al., per la seconda del sexto. Imperoché .te. e .lk. sonno equedi-<lb></lb> |
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1687 stanti. La proportione adunque del .ae. alo .al. è commo la proportione del tetragono .et.<lb></lb> |
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1688 <lb></lb> |
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1689 <pb></pb></p> |
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1690 <p class="folio"> |
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1691 folio 11v |
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1692 </p> |
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1693 <p class="main"> |
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1694 <lb></lb> |
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1695 </p> |
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1696 <p class="runhead"> |
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1697 Distinctio prima. Capitulum octavum. |
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1698 </p> |
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1699 <p class="main"> |
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1700 <lb></lb> |
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1701 a quello che fa .et. in .kl. E quello che fa .et. in .kl. è iguale al ditto del .eb. in .bl. La proportio-<lb></lb> |
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1702 ne adunque del .ae.ab.al. é comme la proportione del tetragono .et. a quello ch’ é fatto del .eb.<lb></lb> |
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1703 in .bl. E, multiplicato adunque el tetragono .et. in .al., è comme multiplicato .ae. nel produt-<lb></lb> |
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1704 to del .be. in .bl. E la multiplicatione del tetragono .et. nel tetragono .al. è comme la multipli-<lb></lb> |
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1705 catione del .ae. nel produtto del .eb. in .bl. e quello che fanno in .al. Ma la multiplicatione del<lb></lb> |
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1706 tetragono .et. nel tetragono .al. è comme el tetragono dela superficie del triangolo .abg. com-<lb></lb> |
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1707 me dimostraremo. Onde la multiplicatione del .ae., che è la soprabundantia dela mitá de’ la-<lb></lb> |
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1708 ti del triangolo .abg. al lato .bg. nello .eb., che è la soprabundantia dela mitá de’ lati del detto<lb></lb> |
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1709 triangolo alo lato .ag. e quello che fanno, multiplicato in .bl., che è la soprabundantia dela mi-<lb></lb> |
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1710 tá deli lati .agb. del triangolo al lato .ba. E li produtti, multiplicati in .al., cioé nela mitá de’ la-<lb></lb> |
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1711 ti del triangolo .abg., fará el tetragono del’ area del triangolo .abg. Ora ci resta a mostrare in<lb></lb> |
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1712 che modo, a multiplicare el quadrato del .et. nel quadrato del .al., fa el tetragono, cioé el quadrato del’ area<lb></lb> |
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1713 del triangolo .abg. Perché il triangolo .abg. è risoluto in .3. triangoli dal ponto .t., che sonno .atb. e<lb></lb> |
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1714 .atg. e i catetti di ciascuno triangolo provammo erano infra loro iguali è sonno .te.tz.th.<lb></lb> |
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1715 Adunque, multiplicato .et. nela mitá dela basa .ab., fará l’ area del triangolo .atb. Similmen-<lb></lb> |
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1716 te, multiplicato .th., cioé .te., nela mitá del .gb., fará l’ area del triangolo .btg. E ancora, multipli-<lb></lb> |
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1717 cato .tz., cioé .te., nela mitá delo .ag., fará l’ area del triangolo .atg. Onde, multiplicato .te. in .al., cioé<lb></lb> |
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1718 multiplicato .te. nela mitá de’ lati del triangolo .abg., fará l’ area del triangolo .abg. Onde,<lb></lb> |
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1719 multiplicato el quadrato del .et. nel quadrato .al., fará il quadrato del’ area del detto triango-<lb></lb> |
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1720 lo. E questo era bisogno mostrare. E, medianti questi .3. modi demostrati, ciascuna sor-<lb></lb> |
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1721 ta de triangoli sempre si poterá quadrare. E per numero determinatamente dire loro super-<lb></lb> |
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1722 ficie avenga che a le volte non discretamente ma per via de radici. Ma uno altro degno<lb></lb> |
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1723 modo ci dá esso Euclide nell’ ultima del .2o., a saperli quadrare con precisione e senza nume-<lb></lb> |
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1724 ro quando el dici: “Dato trigono equum quadratum ei describere”. Ove al triangolo, per la<lb></lb> |
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1725 .42a. del primo, ci fa fare un pararello rettangolo equale, mediante la inventione de una li-<lb></lb> |
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1726 nea media proportionale trovata col semicirculo, sí comme per la .9a. del suo .6o. ci dechiara,<lb></lb> |
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1727 commo tu, per te, ivi legendo, potrai intendere, che è una ligiadra conclusione. In modo che,<lb></lb> |
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1728 se fossero proposti mille triangoli, subito ci dá verso farne un sol quadrato, trovando a cia-<lb></lb> |
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1729 scuno il suo lato tetragonico e quelli componere asiemi mediante la penultima del primo,<lb></lb> |
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1730 quam tibi dimito et cetera. |
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1731 </p> |
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1732 <p class="main"> |
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1733 E, se d’ alcuno triangolo e .2. lati fussino solamente noti e volesse per quelli have-<lb></lb> |
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1734 re la misura del’ altro lato e ancora l’ area del detto triangolo. Comme sia uno tri-<lb></lb> |
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1735 angolo .abg., del qual e lati .ab. e .bg. sienno noti e voi noto il lato .ag. Prima è<lb></lb> |
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1736 da considerare se l’ angolo contento dale .2. linee o vogliammo dire da’ noti lati<lb></lb> |
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1737 .ab. e .bg. sia retto: overo minore che ’l retto overo maggiore che ’l retto. Sia primo retto.<lb></lb> |
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1738 Ove la retta .ab. è catetto sopra la retta .bg. e, multiplicato adunque .ab. in .bg., ne pervie-<lb></lb> |
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1739 ne .2. cotanti del’ area del triangolo .abg. E, se agiongniamo insiemi e quadrati dele linee<lb></lb> |
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1740 .ab. e .bg. note, ne perverrá el quadrato dela linea .ag. Del quale quadrato la radicie è<lb></lb> |
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1741 la linea .ag. |
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1742 </p> |
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1743 <p class="main"> |
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1744 Ma, se l’ angolo .abg. è minore che ’l retto, alora piglia nela linea .ab. un ponto che<lb></lb> |
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1745 sia .d., dal qual sopra la linea bg. si meni uno catteto che sia .de. E, se lla proportio-<lb></lb> |
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1746 ne del .be. al .bg. è iguali ala proportione del .bd. al .ba., l’ angolo .g. è retto, per-<lb></lb> |
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1747 ché la linea .de. è equedistante ala linea .ag., per la .2a. del .6o. Onde se ’l quadrato del lato .bg.<lb></lb> |
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1748 si togli del quadrato del lato .ab., rimarrá el quadrato delo lato .ag. Overo, perché .de. è eque-<lb></lb> |
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1749 distante al .ag., sará cosí .bd. al .ba., come .de. al .ag. Onde, se multiplicheremo il lato .ba. in<lb></lb> |
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1750 .ed. e divideremo la somma per .db., verranne .ag. Verbi gratia: sia .ab.20. e .bg. sia .12. E sia<lb></lb> |
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1751 l’ angolo .agb. retto e la linea .bd. sia .5. e .de. sia .4., sará .eb.3. Sará adunque cosí .bd. al .ba.,<lb></lb> |
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1752 cioé .5. a .20., cosí .3. al .12., cioé .be. al .bg. Onde la retta .de. è equedistante alla retta .ag. E pe-<lb></lb> |
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1753 ró l’ angolo .agb. è retto, imperoché retto è l’ angolo .deb. Onde, se ’l quadrato delo lato .bg.<lb></lb> |
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1754 si trae del quadrato del lato .ba., rimarrá il quadrato delo lato .ag., cioé trahendo del .400.<lb></lb> |
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1755 el .144., rimarram .256., la cui radici è .16., per lo lato .ag. Overo, se multiplicaremo .ba. in .ed.,<lb></lb> |
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1756 cioé .20. per .4., e partiremo per .db., cioé per .5., verranne .16. per lo detto lato del quale la mi-<lb></lb> |
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1757 tá, cioé .8., multiplicato per .12., fanno .96. per l’ area del triangolo .agb.<lb></lb> |
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1758 E, se la proportione del .be. al .bg. sará in minore proportione che ’l .bd. al .ba.,<lb></lb> |
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1759 comme si dimostra nel’ altro triangolo .agb., alora l’ angolo .agb. è minore che ’l retto.<lb></lb> |
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1760 <lb></lb> |
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1761 <pb></pb></p> |
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1762 <p class="folio"> |
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1763 folio 12r |
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1764 </p> |
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1765 <p class="main"> |
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1766 <lb></lb> |
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1767 </p> |
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1768 <p class="runhead"> |
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1769 Distinctio prima. Capitulum octavum. 12 |
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1770 </p> |
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1771 <p class="main"> |
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1772 <lb></lb> |
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1773 Onde oxigonio sará il triangolo .abg. E dal ponto .a. caggia il catetto .af. in sula basa .bg.<lb></lb> |
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1774 e acioché habiamo .fb., cioé il cadimento di detto catetto dala parte del .b., dirai cosí essere<lb></lb> |
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1775 il .bd. al .ba., cosí .be. al .bf. E, perché .af. è catetto sopra .bg. e diciamo con numeri. sia .ab.<lb></lb> |
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1776 .20. e .bg. sia .17. e .fb. sia .12. e .bd. sia .5. e .de.4. e .eb.3., sará per quello che habiamo detto, .af.<lb></lb> |
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1777 .16. El cui quadrato è .256. Agionto al quadrato .fg., cioé con .25., fanno .281., la cui radici é il la-<lb></lb> |
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1778 to .ag. Dico che .fg. è .5., perché .bf. fu fatto .12. E, se multiplicarai la mitá del .af., cioé .8., per la<lb></lb> |
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1779 basa, cioé per .17., fanno .136. per l’ area del detto triangolo.<lb></lb> |
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1780 E, se la proportione del .be. al .bg. sará in maggiore proportione che ’l .bd. al .ba.,<lb></lb> |
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1781 sará l’ angolo .abg. maggiore che ’l retto. Onde il catetto dal ponto .a. cade fuori<lb></lb> |
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1782 del triangolo .abg. E sia il catetto .ah. e menise .bg. infino al .h., sará .be. alo .bh.<lb></lb> |
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1783 comme .bd. al .ba. E con numeri, sia .ab.20. E .bg. sia .7. E .bd.5. E .de.4. E .eb.<lb></lb> |
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1784 .3. Dove .bh. sará .12., che lo troverai se la multiplicazione del .be. in .ba. dividerai per .db., imperoché<lb></lb> |
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1785 gli é .be. al .bh. comme .bd. al .ba. E sia ancora .ed. al .ah. comme .bd. al .ba. Onde, divisa multiplicazi-<lb></lb> |
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1786 one del .de. in .ba. per .db., ne perverrá il catetto .16., cioé .ah. E, se del .bh., cioé di .12., si togli .bg.,<lb></lb> |
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1787 cioé .7., rimarrá .5. per lo cadimento, cioé .gh. Del quale el quadrato, se s’ agiongni al quadrato .ah., cioé a<lb></lb> |
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1788 .256., fanno .281. per lo quadrato dela linea .ag. Onde il lato .ag. è radici di .281. e, multiplica-<lb></lb> |
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1789 to la mitá del .ah. in .gb., ne perviene .56. per l’ area del triangolo .abg.<lb></lb> |
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1790 E, se solamente uno lato del triangolo si pone noto: e vorrai per quello gli altri<lb></lb> |
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1791 lati del triangolo e l’ area havere. Comme del triangolo .dez. Del quale il lato .dz.<lb></lb> |
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1792 é noto. Torró nela retta .dz. una parte che sia .az. E, dal ponto .a. sopra la retta<lb></lb> |
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1793 .ez., meneró la linea .ab. equedistante ala linea .de. E misureró .ab. e .bz. acio-<lb></lb> |
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1794 ché sia manifesto al lato del triangolo .abz. E, perché la retta .ab. è equedistante ala retta<lb></lb> |
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1795 .de. proportionalmente é comme .za. al .zd. cosí la retta .zb. ala retta .ze. e ancora .ab. al .de.<lb></lb> |
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1796 Ma la proportione del .za. al .zd. è nota. Onde e lati .de.ze. saranno noti. Verbi gratia.<lb></lb> |
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1797 Sia il lato .dz.18. e .az. sia .6. e .ab. sia .7. e .bz. sia .5. Dove .dz. é al .az.3. cotanti. Onde .de. é<lb></lb> |
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1798 .3. cotanti delo .ab. e lo .ez. del .zb. é similmente .3. cotanti. Onde il lato .de. è .21. ez. é .15. e quan-<lb></lb> |
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1799 do e lati del triangolo sonno noti fienno note l’ area di detti triangoli. Adunque l’ area del<lb></lb> |
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1800 triangolo .dez. sará nota. comme pigliando l’ area del triangolo .azb., che è la radici di .216. |
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1801 </p> |
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1802 <p class="main"> |
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1803 E, perché simile è il triangolo .dez. al triangolo .abz., sará l’ area del triangolo .dez. al’ area<lb></lb> |
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1804 del triangolo .abz. comme il quadrato delo lato .zd. al quadrato delo lato .za. Comme nela<lb></lb> |
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1805 .17a. del sexto de Euclide è manifesto. E, perché la proportione del .dz. al .az. è .3. cotanti, se<lb></lb> |
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1806 multiplichi la radici di .216. per .9., harai l’ area del triangolo dato, cioé la radici di .17496., che<lb></lb> |
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1807 é apresso a .132 1/4. E cosí usa in simili. |
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1808 </p> |
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1809 <p class="main"> |
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1810 Ancora daremo modo di misurare e triangoli secondo uno uso vulgare, el quale<lb></lb> |
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1811 debbe usare quel che misura el terreno, in questo modo. Stia il misuratore in sul-<lb></lb> |
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1812 l’ angolo maggiore del detto triangolo. E con l’ ochio guardi dove il catetto che<lb></lb> |
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1813 si muove da quello angolo debba cadere. E, non sapendo discernere coll’ ochio,<lb></lb> |
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1814 habia uno filo de spago o di corda. E l’ uno capo fichi in sul angolo detto e l’ altro filo disten-<lb></lb> |
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1815 da infino in sula facia dove tale catetto debba cadere e meni quel filo intorno infino a tan-<lb></lb> |
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1816 to che tochi quella facia colla men corda che pó. E quel sia il cadimento del catetto. Ove-<lb></lb> |
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1817 ro prostenda quel spago o corda in modo che tochi in .2. luoghi quella facia e quello spatio,<lb></lb> |
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1818 che è infra ’ due ponti del tocamento, divida in .2. parti iguali e nel ponto delle due parti, s ‘á<lb></lb> |
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1819 il cadimento del catetto. E alora, colla misura nota, misuri quel catetto. E, misurato anco-<lb></lb> |
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1820 ra la facia dove tale catetto cade, multiplichi lo catetto per la mitá di detta facia overo la mi-<lb></lb> |
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1821 tá di detto catetto per tuta la facia. E harai l’ area di detto triangolo. E, acioché questo chia-<lb></lb> |
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1822 ro apaia, sia il triangolo .abc. avente il lato .bc. maggiore che gli altri, cioé maggiore che l’ <lb></lb> |
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1823 .ab. e’ l .bc. Onde dal’ angolo .a. é da menare il catetto sopra il lato .bc. e, se ’l lato .ab. è iguale al<lb></lb> |
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1824 lato .ac., cadrá il catetto nel mezzo del lato .bc. E, se sará minore, cadrá inverso il .b. E, se sará<lb></lb> |
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1825 maggiore, cadrá inverso il .c. Onde stará il misuratore sopra il ponto .a. e considererá ove dal<lb></lb> |
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1826 detto ponto .a. el catetto debba cadere sopra linea .bc. La qual cosa fatto ficchi lo filo in sul<lb></lb> |
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1827 ponto .a. e istendilo sopra la linea .bc. E terala con mano sopra il ponto .d., cioé un poco fuo-<lb></lb> |
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1828 ri del triangolo. E menala inverso il .c., infino a tanto che ’l ponto .d. tocchi la linea .bc. E sia<lb></lb> |
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1829 il loro contatto il ponto .e., dove lo filo .ae. sia quel medesimo che ’l filo .ad. Dipoi mena il fi-<lb></lb> |
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1830 lo in verso il .b. infino a tanto che ’l ponto .d. tocchi la linea .bc. che sia il ponto .f., cioé che ’l filo<lb></lb> |
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1831 .af. sia quel medesimo che ’l filo .ad. E dividasi la linea .ef. in .2. parti sopra il ponto .h. E dal<lb></lb> |
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1832 <lb></lb> |
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1833 <pb></pb></p> |
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1834 <p class="folio"> |
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1835 folio 12v |
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1836 </p> |
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1837 <p class="main"> |
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1838 <lb></lb> |
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1839 </p> |
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1840 <p class="runhead"> |
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1841 Distinctio secunda. Capitulum primum. |
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1842 </p> |
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1843 <p class="main"> |
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1844 <lb></lb> |
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1845 ponto .a. al ponto .h. menerai la linea .ah., che dico essere catetto sopra la linea .bc., impero-<lb></lb> |
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1846 ché è iguali la linea .af. ala linea .ae. E iguale è .fh. al .he. E peró misurerai il catetto .ah. e an-<lb></lb> |
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1847 cora il lato .bc. e multiplica la mitá del catetto per la basa .bc. overo la mitá della basa per tut-<lb></lb> |
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1848 to il catetto e harai l’ area del detto triangolo .abc. |
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1849 </p> |
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1850 <p class="main"> |
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1851 E, se ’l dito triangolo fosse tanto grande in modo che ’l filo che gli havesse fosse mi-<lb></lb> |
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1852 nore del catetto overo fosse terra vignata o alberata overo piena di biada, in<lb></lb> |
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1853 modo che ’l catetto, per lo modo detto, non si potesse havere, alora farai una mi-<lb></lb> |
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1854 sura picola, la quale sia .1/20. di braccio o circa. E quanti braccia sonno nella linea<lb></lb> |
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1855 .ac. tanti ne togli dela misura piccola che fienno .ak. E quanti bracia sonno nela linea .ab.<lb></lb> |
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1856 tanti ne togli della misura picola che sia .al. E faciasi la linea .lk. E truovisi il catetto col fi-<lb></lb> |
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1857 lo per lo detto modo nel triangolo .alk. in sul lato .lk. Ove per lo dritto menerai .ai. infi-<lb></lb> |
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1858 no al ponto .m. in sul lato .bc. e sia .am. catetto del triangolo .abc.<lb></lb> |
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1859 Quando .2. lati d’ uno triangolo sonno noti e a quelli si traga l’ altro lato equedi-<lb></lb> |
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1860 stante al terzo lato e le septioni d’ un lato fienno note, alora le septioni del’ altro la-<lb></lb> |
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1861 to fienno note e la linea menata sia nota per la .2a. del .6o. de Euclide.<lb></lb> |
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1862 Ancora volendo, secondo uno uso vulgare, misurare l’ area d’ uno scudo equilate-<lb></lb> |
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1863 ro, multiplicarai uno de’ suoi lati in se medesimo: e di quella somma piglia .13/30.<lb></lb> |
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1864 Comme sia il triangolo .abc., che ciascuno suo lato sia .10.bracia. e voglisi l’ area.<lb></lb> |
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1865 Multiplicarai .10. in sé, fanno .100. Del qual piglia .13/30., che sonno .43.bracia.1/3. E<lb></lb> |
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1866 .43.bracia. è .1/3. non peró aponto, ma piccola cosa manco, imperoché ’l quadrato di .43 1/3. è<lb></lb> |
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1867 .1877 7/9. e il triangolo é la su’ area aponto la radici di .1875., che è molto frivola cosa meno<lb></lb> |
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1868 di .43 1/3. E peró questo modo poi usare quando in sul terreno fossi overo quando secondo l’ appressa-<lb></lb> |
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1869 mento ái a ffare: conciosiacosaché nulla o poco sia distante ala veritá. E peró fo da dimostra<lb></lb> |
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1870 al’ algrimensore et cetera. |
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1871 </p> |
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1872 <p class="main"> |
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1873 E questo basti quanto al detto capitolo. E ancora quanto ala prima distintione e dela<lb></lb> |
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1874 seconda vederemo. Seconda distinctio. |
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1875 </p> |
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1876 <p class="main"> |
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1877 De modo inveniendi quantitatem unius linee ab uno puncto protracte intra vel ex-<lb></lb> |
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1878 tra quencuque triangulum. Capitulum primum .2a.di. |
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1879 </p> |
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1880 <p class="main"> |
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1881 Infino a qui è mostro a trovare l’ area de’ triangoli assai amplamente e ancora,<lb></lb> |
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1882 quando una linea si mena equedistante ad alcun lato del triangolo, quanto è la<lb></lb> |
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1883 sua longhezza è copiosamente mostro. Ove, in questa, intendo demostrare, quan-<lb></lb> |
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1884 do una linea è menata e terminata neli noti termini di doi lati d’ uno triangolo<lb></lb> |
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1885 (la quale non sia equedistante al terzo lato), la sua quantitá. E, di questa parte, non si fará al-<lb></lb> |
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1886 cuna divisione, ma solo un capitolo e peró starai atento. |
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1887 </p> |
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1888 <p class="main"> |
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1889 Sia un triangolo .abg. e la linea menata sia .ez. E sia .eb. gli .2/3. dela linea .ba. E il<lb></lb> |
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1890 ponto .z. sia in mezzo dela linea .bg. E sia .ab.13. e .ag. sia .14. E .bg. sia .15. Ove .be.<lb></lb> |
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1891 sia .8 2/3. e .ea. sia .4 1/3. e adimandasi la quantitá del .ez. Menise il catetto .bd. El qua-<lb></lb> |
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1892 le giá è mostro essere .12. e .ad., cioé il minore cadimento, sia .5. E il maggiore sia .9.,<lb></lb> |
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1893 cioé .dg. E dal ponto .c. si meni .ci. ala basa .bg., el quale sia equedistante ala basa .ag. E dal<lb></lb> |
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1894 ponto .z. si meni el catetto .zk., che sia equedistante al catetto .bd. E, perché la linea .zk. è eque-<lb></lb> |
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1895 distante ala linea .bd. sia cosí .gz. al .gb., cosí .gk. al .gd. Ma .gz. del .gb. è la mitá. Onde .gk.<lb></lb> |
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1896 del .gd. è la mitá. E, similmente, .zk. è la mitá del .bd. Adunque .kg. è .4 1/2. e .kd. ancora .4 1/2.<lb></lb> |
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1897 e il .zk. è .6. Ancora si meni .ef. equedistante al catetto .bd., sia cosí .ae. al .ab. comme .af. al .ad.<lb></lb> |
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1898 e lo .ef. al .bd. E noi sappiamo che .ae. è il .1/3. del .ab. E peró .af. è .1 2/3. E .fd. è .3 1/3. E .fe. è .4., cioé<lb></lb> |
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1899 la terza parte del .bd. E, perché le linee .ef. e .tk. sonno equedistanti al catetto .bd., fienno in-<lb></lb> |
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1900 fra loro equedistanti per la .30a. del primo. E faciasi .et. e .fk. equedistanti, sará .ef. iguale al<lb></lb> |
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1901 .tk. E la linea .et. sia iguale ala linea .fk. Ma .fk. è iguale a .2. linee che sonno .kd. e .fd., cioé<lb></lb> |
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1902 .4 1/2. e .3 1/3., dove tutta .kf. è .7 5/6. E peró .te. è .7 5/6. Similmente .tk. è .4., imperoché la è iguale al .ef.<lb></lb> |
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1903 Rimane .tz.2. E perché in .2. linee equedistanti .ei. e .ag. v’ é la retta .zk. l’ angolo di fuori .zte.<lb></lb> |
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1904 è iguale al’ angolo oposto dentro .zkf., per la .29a. del primo. Ma l’ angolo .zks. è retto, im-<lb></lb> |
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1905 peroché gli é retto ancora l’ angolo .zte. ortogonio adonca é lo triangolo .zte. e gli e .2. lati son-<lb></lb> |
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1906 no noti, cioé .zt. e .te. che contengono l’ angolo retto. Onde il terzo lato, cioé .ez., sia noto quan-<lb></lb> |
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1907 do multiplicarari .zt. in sé e .te. in sé che agionti insiemi haremo .65 13/36, la cui radici è la linea .ez.,<lb></lb> |
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1908 che è poco meno de .8 1/12. e questo era da mostrare.<lb></lb> |
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1909 <lb></lb> |
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1910 <pb></pb></p> |
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1911 <p class="folio"> |
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1912 folio 13r |
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1913 </p> |
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1914 <p class="main"> |
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1915 <lb></lb> |
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1916 </p> |
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1917 <p class="runhead"> |
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1918 Distinctio secunda. Capitulum primum. 13 |
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1919 </p> |
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1920 <p class="main"> |
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1921 <lb></lb> |
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1922 Ancora possiamo altramente ala notitia dela linea .zt. e .te. venire, imperoché’ l<lb></lb> |
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1923 triangolo .zti. è simile al triangolo .bdg., perché la retta .ti. è equedistante ala li-<lb></lb> |
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1924 nea .dg. e il .zt. equedistante al .db. E il lato .zi. alo lato .bg. simile. É adunque<lb></lb> |
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1925 cosí .zi. al .bg., cosí .it. al .gd. E il .ct. al .bd. E il .gi. è la terza parte del .gb.,<lb></lb> |
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1926 perché .ti. è la terza parte del .bd. E il .gz. è la mitá del .gb. Onde .iz. è la sexta parte del<lb></lb> |
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1927 .gb., cioé .2 1/2. Onde .zt. è la sexta parte del .db., cioé .2. E .ti. è la sexta parte del .gd., cioé .1 1/2. che,<lb></lb> |
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1928 tratto del .ie., che è .9 1/3., rimane .te.7 5/6, comme dicemmo et cetera.<lb></lb> |
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1929 A ncora altramente menise .ie. equedistante ala linea .bd. E sia il triangolo .icg.<lb></lb> |
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1930 simile al triangolo .bdg. e .zti. É adunque comme .gi. al .gb., cosí .gc. al .gd. <lb></lb> |
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1931 E .ic. al .bd., per la .2a. del .6o. E il .gi. è la terza parte del .gb. E il .ge. è la terza par-<lb></lb> |
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1932 te del .gd. E .ic. è la terza parte del .bd. Adunque .ig. è .5. E .ge. è .3. E .ic. è .4. E,<lb></lb> |
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1933 perché il triangolo .zti. è simile al triangolo .tcg., é comme .zi. al .zg., cosí .ti. al .gk. E cosí<lb></lb> |
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1934 comme .zi. è al .ig., cosí .it. è al .cg. Ma .zi. è la mitá del .ig. E peró .it. è la mitá del .gc. E il .zt.<lb></lb> |
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1935 è la mitá del .tk., cioé del .ic., ch’ era de bisogno mostrare et cetera.<lb></lb> |
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1936 Ma, se le linee .ze. e .ga. fuori del triangolo nel ponto .h. meneremo e vorremo sa-<lb></lb> |
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1937 pere la quantitá dele linee .ah. e .eh., multiplicaremo la linea .te. in .ef. e dividere-<lb></lb> |
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1938 mo la somma in .zt. e haremo la linea .fh. E questo faremo perché il triangolo .zte.<lb></lb> |
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1939 è simile al .efh. E l’ angolo .zte. è iguale al’ angolo .efh., imperoché amendui son-<lb></lb> |
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1940 no retti; l’ altro .tze. al’ altro .feh. è iguale. Onde è comme .zt. al .te., cosí .ef. al .fh. Onde la mul-<lb></lb> |
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1941 tiplicatione del .te. in .ef., divisa per .zt., fa la linea .fh. overo, per la permutata proportione, é comme il<lb></lb> |
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1942 .zt. al .ef., cosí .te. è al .fh. E il .zt. è la mitá del .ef. Onde il .te. sará la mitá del .fh., cioé che ’l .fh.<lb></lb> |
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1943 è doppio del .te. Ma .te. è .7 5/6., dove .fh. è .15 2/3., de’ quali tratto .fa., rimane .ah.14. et cetera.<lb></lb> |
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1944 Ancora e gli é cosí .zt. al .ef., cosí .ze. al .eh. Onde .eh. é doppio del .ez., imperoché .zt.<lb></lb> |
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1945 è la mitá del .ef. Ma, perché .ez. nonn’ é numero ratiocinato, tirremo la proportio-<lb></lb> |
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1946 ne neli quadrati loro, cioé è cosí il quadrato dela linea .ze. al quadrato dela linea<lb></lb> |
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1947 .ef., comme .4. a .16. E cosí il quadrato dela linea .ze. è al quadrato dela linea .eh.<lb></lb> |
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1948 Onde il quadrato dela linea .eh. è .4. cotanti del quadrato .ez. Overo agiongnisi in uno li qua-<lb></lb> |
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1949 drati dele linee .ef. e .fh. e haremo li quadrati dela linea .eh. |
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1950 </p> |
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1951 <p class="main"> |
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1952 Ancora altramente menise la linea .hz. fuori del triangolo infino a tanto che si con-<lb></lb> |
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1953 giunga colla linea .bl. nel ponto .l. E sia .bl. equedistante ala linea .hg. E, perché ne-<lb></lb> |
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1954 le equedistanti .bl. e .hg. é la retta .lh., sará l’ angolo .blh. iguale al’ angolo .ghl., che<lb></lb> |
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1955 sono coalterni. E l’ angolo .lbg. al’ angolo .hgb. Ove gli angoli che sonno al .z. sonno infra lo-<lb></lb> |
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1956 ro iguali, che sonno contraposti, per la .15a.del .po. Adonca el triangolo .lbz. è simile al triangolo<lb></lb> |
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1957 .hgz. E pero è cosí .gz. al .zb., cosí .hg. al .bl. E il .gz. è iguale al .zb. E .hg. è iguale al .bl. Ancora, per<lb></lb> |
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1958 ché e sonno simili li triangoli .elb. e il triangolo .eha., sará comme .ae. al .eb., cosí .ah. al .bl. E .ae. è il<lb></lb> |
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1959 terzo del .ba. Onde .ae. è il mezzo del .cb. E cosí .ha. sará il mezzo del .bl., cioé del .hg. E cosí .ag. sia<lb></lb> |
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1960 la mitá del .hg. E la linea .ag. è .14. E peró .ah. sia similmente .14. E ciascuna dele rette .hg. E<lb></lb> |
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1961 .bl. sia .28. Onde, se ’l congiunto dela linea .ze. e .eh. vogliamo havere, perché e gli é cosí .zt. al .ef.,<lb></lb> |
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1962 cosí .ze. al .eh. ma il .zt. è al .ef. comme la mitá del .zt. ala mitá del .ef., cioé comme .1o. a .2. Faciase<lb></lb> |
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1963 adonca una linea .mo. E sia divisa in .mn. e .no. E sia .mn. 1o. e .no.2. E, perché e gli é cosí .mn. al<lb></lb> |
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1964 .no., cosí .zt. al .ef. E, per la proportionalitá congiunta, sia cosí .mn. al .mo., cosí .ez. al .zh. Onde sia com-<lb></lb> |
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1965 me il quadrato dela linea .mn. al quadrato dela linea .mo., cioé cosí .1o. a .9., cosí el quadrato dela<lb></lb> |
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1966 linea .ze. al quadrato dela linea .zb. Onde il quadrato dela linea .zh. sia .9. tanti del quadrato dela li-<lb></lb> |
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1967 nea .ze. Onde, se multiplicaremo per .9. lo quadrato dela linea .ze., cioé .65 13/36., haremo .588 1/4., per lo qua-<lb></lb> |
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1968 drato dela linea .zh. Ancora altramente, perché el triangolo .zth. è ortogonio, avente l’ angolo .k.<lb></lb> |
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1969 retto. Agiongnise li quadrati dele linee .zt. e .th. e haremo il quadrato dela linea .zh. E la linea .hd. è<lb></lb> |
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1970 trovata essere .4 1/2. e .da. è .5. e .ha. è .14. Ove tutta .ak. è .23 1/2., el cui quadrato è .552 1/4. E il quadrato de-<lb></lb> |
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1971 la linea .zk. è .36. E cosí haremo, per lo quadrato dela linea .zh., 588 1/4. comme dicemmo.<lb></lb> |
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1972 Ancora sia uno triangulo .bgd. del quale il lato .bg. sia .13. E il lato .gd. sia .14. |
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1973 </p> |
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1974 <p class="main"> |
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1975 E il lato .bd. sia .15. E menise .dg. infino al ponto .a. E sia .ga.10. E sopra il .gb.<lb></lb> |
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1976 si pigli il ponto .z. E sia .zg.5. e rimarrá .zb.8. Adimandasi, se si mena .az. infino ala<lb></lb> |
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1977 linea .bd., cioé al ponto .e., e questa sia .ed. e .eb. menise per lo ponto .b., la linea .bi. equidistante<lb></lb> |
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1978 ala linea .ad. e faciasi .ai. Saranno e .2. triangoli .bzi. e .azg. infra loro simili. Onde e gli é cosí .ag. al<lb></lb> |
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1979 .gz. comme .ib. al .bz. Ma .ag. è doppio al .gz. Onde .bi. sará .16., cioé doppio al .bz. Ancora, perché el<lb></lb> |
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1980 triangolo .aed. è simile al triangolo .eib., sia cosí .ad. al .bi. comme .de. al .cb. el lato .ad. a .bi. é comme.<lb></lb> |
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1981 <lb></lb> |
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1982 <pb></pb></p> |
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1983 <p class="folio"> |
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1984 folio 13v |
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1985 </p> |
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1986 <p class="main"> |
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1987 <lb></lb> |
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1988 </p> |
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1989 <p class="runhead"> |
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1990 Distinctio secunda. Capitulum primum. |
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1991 </p> |
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1992 <p class="main"> |
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1993 <lb></lb> |
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1994 La octava parte del .bi. e l’ ottavo del .ad. è .3. e .2. è la ottava parte del .bi. Adunque cosí è .3.<lb></lb> |
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1995 a .2. comme .de. al .eb. E, per la congionta propotionalitá, sará .3. a .5. cosí .de. al .db. Onde<lb></lb> |
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1996 .de. è li .3/5. del .db., cioé .9. e .eb. è .6., ch’ era de bisogno mostrare et cetera.<lb></lb> |
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1997 Sieno ancora note le parti .bz. e .gz. e .be. e .ed. E sieno in detto modo e il detto<lb></lb> |
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1998 triangolo e non si sappia .ag. e .bi. E voglinse investigare. Prima, perché e gli é co-<lb></lb> |
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1999 sí .de. al .eb., cioé .3. a .2., cosí .ad. al .bi. Adonca .ad. è una volta e .1/2. il .bi. Ancora, perché<lb></lb> |
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2000 e gli é cosí .gz. al .zb., cioé comme .5. è .a.8., cosí .ag. al .bi. Adonque .ag. è li .5/8. del .bi.<lb></lb> |
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2001 E tutta .ad. è trovata essere li .12/8. del .bi., imperoché tutta .ad. è una volta e .1/2. el .bi. Donde,<lb></lb> |
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2002 tratto .5/8. di .12/8., rimangono .gd.7/8. del .bi. Adunque è comme .7. a .8., cosí .gd. al .bi. Onde, multi-<lb></lb> |
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2003 plicato .8. per .14., cioé per .gd., e diviso per .7., overo il .1/7. di .14., multiplicato per .8., vienne .16.<lb></lb> |
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2004 per la linea .bi. De’ quali, preso li .5/8., haremo per la linea .ag.10., commo era di bisogno dedure.<lb></lb> |
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2005 Ancora sia il medesimo triangolo .abg. e sia .ab.13. e il .bg.14. e .ga.15. E piglise<lb></lb> |
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2006 il .d., ponto che non sia nel diritto dela linea .bg. E per lo ponto .d. si meni la linea<lb></lb> |
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2007 .de. equedistante ala basa .bg. E sia .de. e .eb. nota, cioé sia .de.3. e .dz. sia .4. e .eb.<lb></lb> |
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2008 sia .5. e .ea. sia .8. E piglise nel .ab. il ponto .z. e sia .ze. 1o. e .dz. sia .4. E menise .dz.<lb></lb> |
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2009 infino al ponto .i. Adimandase quanto è .ai. e .ig. Compise .at. e .td. E, perché il triangolo<lb></lb> |
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2010 .dze. è simile al triangolo .zat. sia cosí .ez. al .za., che sonno note, cioé cosí .1o. a .7., cosí .de. e al<lb></lb> |
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2011 .at. E il .de. si pose .3., adunque .at. sia .21. Ancora, perché simili sonno li triangoli .hzb. e .zat.<lb></lb> |
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2012 sia cosí .az. al .zb., cosí .at. al .bh., cioé comme .7. a .6., cosí .21. a .18. Adunque .bh. sia .18. E tutta<lb></lb> |
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2013 .hg. sia .32. E, perché simili sonno e triangoli .hig. e .iat. é cosí .hg. nota e ’l .ta., cioé .32. a .21.,<lb></lb> |
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2014 cosí .gi. al .ia. Adunque cosí .gi. al .ia. E cosí .32. a .21. Dove, comme .32. a .53., cosí .gi. al .ga. Do-<lb></lb> |
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2015 ve .ga. è .15., adunque .gi. fienno li .32/53. di .15., che sonno .9 3/53. E tanto è .gi. E .ia. sia l’ avanzo infi-<lb></lb> |
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2016 no in .15., che sia .5 50/53. ch’ era de bisogno mostrare. |
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2017 </p> |
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2018 <p class="main"> |
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2019 Ancora sia il triangolo .gab. e sia .ga.13.ab.14. e .gb.15. Del quale il catetto<lb></lb> |
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2020 .gd. E piglise in quello il ponto .e., noto cioé che .ed. sia .4. e .eg. sará .8. E per lo<lb></lb> |
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2021 ponto .e. si meni la linea .aez. Dico adunque che la proportione .bz. al .cg. sará no-<lb></lb> |
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2022 ta. Menise adunque la linea .gi. equedistante alla linea .ab. E menise .az. nel pon-<lb></lb> |
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2023 to .i. Fienno li triangoli .aed. e .eig. infra lloro simili. E sia cosí .de. al .eg., cosí .ad. al .gi. E<lb></lb> |
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2024 peró .ag. sará noto, imperoché .ad. sonno note. E sará .gi.10., imperoché .ad. è .5. Onde è<lb></lb> |
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2025 cosí .de. al .eg., cioé .4. a .8., comme .da., cioé .5., al .gi. E peró .gi. è .10. E, perché simili sonno e tri-<lb></lb> |
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2026 angoli .abz. e .igz., sia cosí .ab. cioé .14., al .gi., cioé .a.10., comme .bc. al .zg. E, per la propor-<lb></lb> |
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2027 tionalitá congionta: sia .14. a .24., cioé .7. a .12., comme .bz. al .bg. E peró .bc. sia li .7/12. del .bg.,<lb></lb> |
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2028 cioé .8 3/4. e il .zg. sia l’ avanzo infino in .15., che sia .6 1/4., ch’ era de bisogno mostrare.<lb></lb> |
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2029 E, se il ponto per lo quale passa la linea non sia in sul catetto, ma sia inn’ altra linea.<lb></lb> |
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2030 Comme nel triangolo .dez., nel quale è dato il ponto .a. nela linea .dg. che non<lb></lb> |
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2031 n’ é catetto. Per lo quale ponto passa la linea .eab. e sia nota la proportione del<lb></lb> |
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2032 .ga. al .ad. e sia .ga.5. e sia .ad.8. Imperoché tutta .dg. pongo sia .13. e .ge. sia .10. |
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2033 </p> |
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2034 <p class="main"> |
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2035 Dico che la proportione del .zb. al .bd. sia nota. Compise .di. e faciasi .ebi. E, perché li trian-<lb></lb> |
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2036 goli .ida. e .ega. sonno simili, sia cosí .ag. al .ad., cosí .eg. al .di. Onde .di. sia .16., imperoché<lb></lb> |
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2037 .ag. al .ad. è comme .5. a .8. E peró .eg., che è .10., sia al .di., comme .5. a .8. Adunque .ad. sia .16.<lb></lb> |
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2038 e sia cosí .ez. al .di., cioé .14. a .16., cosí .zb. al .bd. E sia ancora cosí .14. a .30. comme .zb. al .cd.<lb></lb> |
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2039 E il .zd. è .15., adunque .zb. sia li .14/30., cioé .7/15. del .zd. E peró sia .7. e il .bd. sia .8. ch’ era de biso-<lb></lb> |
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2040 gno mostrare. |
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2041 </p> |
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2042 <p class="main"> |
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2043 E, se la proportione del segamento dela linea che passa per lo ponto dato sia in<lb></lb> |
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2044 una linea data equedistante al catetto, sará nota la quale linea sia terminata, da<lb></lb> |
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2045 una parte, in sula basa e, dal’ altra parte, sia terminata in sula linea equedistante<lb></lb> |
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2046 ala basa. Dico la proportione dele septioni del lato del triangolo essere note.<lb></lb> |
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2047 Comme sia un’ altra volta il triangolo .dez. E sia dato il ponto .a. nela linea .tg., che sia eque-<lb></lb> |
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2048 distante al catetto. La qual linea sia in sula basa in sul ponto .g. E in sula linea equedistante a-<lb></lb> |
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2049 la basa, che sia la linea .di., in sul ponto .t. E sia .ag.8. e .at. sia .4. e sia .eg.10. E, perché queste<lb></lb> |
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2050 cose sonno manifeste, dico che la proportione del .zb. al .bd. sia manifesta in questo modo.<lb></lb> |
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2051 Perché li triangoli .eag. e .iat. sonno simili: sia cosí .ga. al .at. cosí .eg. al .ti. Adunque .ti. sia<lb></lb> |
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2052 .5. E menise il catetto nel triangolo .dez. che sia .dk. E sia .dt. iguali al .kg. Adunque .dt. sia<lb></lb> |
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2053 .5. Adunque tutta .di. sia .10. E, perché gli é cosí .ez. al .di. cosí .zb. al .bd. cioé comme .14. a .10.,<lb></lb> |
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2054 cosí .zb. al .bd. E, per la congionta proportionalitá, sia cosí .14. a .24., cosí .zb. al .zd. Adunque<lb></lb> |
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2055 .zb. sia .8 3/4. e .bd. sia l’ avanzo in .15., che sia .6 1/4. E questo era de bisogno mostrare.<lb></lb> |
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2056 <lb></lb> |
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2057 <pb></pb></p> |
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2058 <p class="folio"> |
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2059 folio 14r |
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2060 </p> |
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2061 <p class="main"> |
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2062 <lb></lb> |
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2063 </p> |
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2064 <p class="runhead"> |
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2065 Distinctio secunda. Capitulum primum. 14 |
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2066 </p> |
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2067 <p class="main"> |
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2068 <lb></lb> |
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2069 E, se ’l ponto dato nel triangolo sará infra ’l catetto e angolo, dal quale la linea<lb></lb> |
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2070 s’ extende e va infino alo lato oposto, siranno, similmente, le septioni del’ altro la-<lb></lb> |
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2071 to note. sia adunque nel triangolo .abg., del quale el catetto è .ad. E sia .ab.<lb></lb> |
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2072 13., bg.14., ga.15. e .ad. sia .12. E il dato ponto sia .e. E sia fata la linea .hei. eque-<lb></lb> |
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2073 distante e iguale al .ad. E sia .eh.3. e .bh. sia .4. sia. hd. 1o. E menise la linea .bez. Dico che .gz.<lb></lb> |
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2074 e .za. fienno note. Menise per lo ponto .a. la linea .it. equedistante ala linea .bg. e faciasi la li-<lb></lb> |
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2075 nea .bt. E, perché noi habiamo posto che .he. sia .3., sia .ei.9., imperoché .hi. é .12., cioé lo egua-<lb></lb> |
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2076 le al .ad. E, perché el triangolo .beh. è simile al triangolo .eit., é cosí .he. al .ei., cosí .bh. al .it.<lb></lb> |
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2077 Adunque è noto .it., imperoché .he. è .3. e .ei. è .9. E peró .he. è il terzo del .ei. e cosí .bh., che è<lb></lb> |
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2078 .4., sia il .1/3. del .it., adunque .it. sia. .12. Del quale, tratto .ia., che è iguali al .hd., che è .1o., rimane .at.<lb></lb> |
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2079 E, perché e gli é cosí .bg. al .at., cosí .gz. al .za., adunque comme .14. a .11., cosí .gz. al .za. E<lb></lb> |
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2080 cosí .14. a .25., comme .gz. al .ga. Adonque pigliaremo e .14/25. del .15., che .è .ga., che sonno .8 2/5. Adon-<lb></lb> |
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2081 que .gz. è .8 2/5. e .za. è l’ avanzo infino in .15., che è .6 .3/5. E cosí fa le simili.<lb></lb> |
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2082 Similmente, se per lo ponto dato in uno lato del triangolo una linea si mena infi-<lb></lb> |
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2083 no al’ altro lato del triangolo, si troverá la proportione del segamento. Comme<lb></lb> |
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2084 sia il triangolo .bgd. e sia .bg.13, .gd.14., db.15. E il catetto .bi. sia .12. E sia da-<lb></lb> |
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2085 to un ponto infra ’l triangolo, che sia .a. E sia .az.8. Sonno in sulla basa .gd. che<lb></lb> |
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2086 sta .e. E sia .ed.10. E il ponto .a. sia noto nella linea .zk. equedistante e iguale al catetto .bi. E<lb></lb> |
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2087 sia .az.8. e .ak. sia .4. E per li ponti .ea. si meni la linea .eat. Dico la proportione del .td. al .tb.<lb></lb> |
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2088 essere nota. Faciase la linea .bh. E la linea .eath. e sia .bh. equedistante ala basa .gd., sia .bk.<lb></lb> |
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2089 iguale al .iz. E il .zk. è fatto iguale al catetto che sia .12. e .az. è posto .8. Adonque .ak. è .4. E,<lb></lb> |
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2090 perché e gli é cosí .za. al .ak., cosí .ez. al .hk., adonque .hg. è nota, imperoché .ez. è .8. E peró<lb></lb> |
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2091 è cosí .az., cioé .8., al .ka., cioé a .4., comme .ez., che è .8. Adonque .hk. sia .4., al quale, agionta<lb></lb> |
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2092 .kb. iguale al .zi., che è .7., imperoché tutta .ed. è .10. comme ponemmo. E .ez. è .8. Adonque<lb></lb> |
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2093 .zd. è .2., che è tratto del .zi., che è .9., rimane .zi.7. Adonque .bh. sia .11. E, perché simili sonno<lb></lb> |
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2094 e triangoli .etd. e .tbh., sia cosí .ed. al .bh., cioé.10. a .11. cosí .td. al .bt., adonque cosí .10. a .21.,<lb></lb> |
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2095 cosí .td. al .db. E peró .td. è li .11/21. di .15., che sonno .7 6/7. E peró .td. sia .7 6/7. e .tb. sará l’ avanzo in-<lb></lb> |
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2096 fino in .15., che sia .7 1/7., ch’ era besogno mostrare. |
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2097 </p> |
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2098 <p class="main"> |
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2099 Ancora sia el triangolo .abg., del quale .ab. sia .13. e .bg. sia .14. e .ag.15., del quale<lb></lb> |
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2100 il catetto sia .ad., che è .12. E piglise il ponto .e. fuori del triangolo, dal quale si me-<lb></lb> |
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2101 ni la retta .ez. equedistante al catetto .ad. e sia .ez.2. e il .zb. sia .1o., dove .zd. sia .4. |
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2102 </p> |
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2103 <p class="main"> |
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2104 E piglise ancora infra il triangolo il ponto .i. e sia .it.3. e sia equedistante al catet-<lb></lb> |
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2105 to .ad. E sia .tz.9. E per gli ponti .ei. se meni la linea .eik. Dico che la proportione del .gk. al<lb></lb> |
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2106 .ka. é nota. Menise la linea .al. equedistante ala basa .bg. E menise .ek. infino al ponto .l. E<lb></lb> |
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2107 faciasi .ti. infino al .h. e infino al .m. E sia .tm. iguali .ez. E compise .em. E, perché la retta .ez.<lb></lb> |
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2108 e .it. sonno equedistanti al catetto .ad., sará la retta .tm. equedistante ala retta .ze. E, perché<lb></lb> |
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2109 sonno iguali, sia la retta .em. iguale e equedistante ala retta .zt. Adunque .em. è .9. e equedi-<lb></lb> |
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2110 stante ala retta .al. e .th. è .12., imperoché la è iguale al catetto .ad. Adonque tutta .mh. è<lb></lb> |
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2111 .14. E certamente .mi.5., adonque .ih. è .9. Adonque sia .mi. al .ih., cosí .em. al .hl., cioé comme<lb></lb> |
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2112 .5. é a .9., cosí .9. è al .hl. Adonque .hl. è .16. e tutta .al. è .21 1/5. Ancora e gli é equedistante .ez. al<lb></lb> |
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2113 .ti: simili sonno e triangoli .enz. e .int. Onde e gli é cosí il .ti. al .ez. cosí .tn. al .nz. Onde. tn. è<lb></lb> |
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2114 del .tz. gli .13/5., cioé .5 2/5. Overo, altrimente, perché i triangoli .nit. e .lih. sonno simili infra loro,<lb></lb> |
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2115 é cosí .ti. al .ih., cioé comme .3. è a .9., cosí .nt. é .al.hl., adonque .nt. è il terzo di .16 1/5., che è .5 1/5., al<lb></lb> |
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2116 quale, agionto .tg., che è .4., fanno per lo .ng.9 2/5. e sia cosí .ng. al .al., cosí .gk. al .ka. E noi sappia-<lb></lb> |
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2117 mo .al. esser .21 1/5., che sonno .106/5. E .ng. sonno .47/5. Adonque cosí .47. è al congionto di .47. e di<lb></lb> |
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2118 .106., cioé .a.153., cosí .gk. è al .ga., cioé a .15. Onde è .47. al terzo di .153., cioé a .51., cosí .gk. è al ter-<lb></lb> |
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2119 zo di .15., cioé a .5. Onde, se multiplicaremo .47. per .5. e divideremo il produtto per .51., hare-<lb></lb> |
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2120 mo .4 31/51. per la linea .gk. L’ avanzo che è infino in .15., cioé .10 20/51., sará la linea .ak. Ancora altra-<lb></lb> |
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2121 mente menise la linea .em. e .ag. infino al ponto .o., sará cosí .eo. al .al., cosí .ok. al .ka. E la li-<lb></lb> |
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2122 nea .eo. è .14 1/2. E la linea .al. è .21 1/5. Adonque comme .eo. alo .al., cioé comme el .14 1/2. è al .21 1/5.,<lb></lb> |
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2123 cioé comme .145. è a .212. cosí .ok. al .ka. e cosí comme .145. è a .357., cosí .ok. è al .oa.; onde<lb></lb> |
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2124 multiplicarai .145. via .17 1/2. e parti in .357., viene .7 11/102. Haremo che .ok. è .7 11/102., del quale tra’<lb></lb> |
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2125 .iog. che è .2 1/2., haremo .4 31/51. per la linea .gk., comme volavamo.<lb></lb> |
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2126 <lb></lb> |
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2127 <pb></pb></p> |
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2128 <p class="folio"> |
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2129 folio 14v |
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2130 </p> |
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2131 <p class="main"> |
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2132 <lb></lb> |
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2133 </p> |
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2134 <p class="runhead"> |
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2135 Distinctio secunda. Capitulum secundum. |
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2136 </p> |
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2137 <p class="main"> |
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2138 <lb></lb> |
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2139 E la noticia dela linea .go. e .oe. s’ á per le cose dette. Perché el triangolo .adg. è simi-<lb></lb> |
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2140 le al triangolo .afo. Ancora, se la noticia del ponto .c., per lo quale la linea .el. sega<lb></lb> |
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2141 el catetto, vogliamo havere, perché el triangolo .ncd. è simile al triangolo .lca.,<lb></lb> |
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2142 sia cosí .nd. alo .al. cioé come .2/5. sonno a sé e .21 1/5., cioé .2/5. sonno a .21 3/5., cosí. dc. é al<lb></lb> |
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2143 .da. Onde il .de. e gli é .2/9. d’ uno intero. E il .ta. sia .11 7/9. Overo, se meneremo il catetto .nx., sia<lb></lb> |
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2144 cosí .ex. al .xn., cosí .nd. al .dc. E gli é certamentente .ef. iguali al .dz. E il .dn. iguali al .xf. E la li-<lb></lb> |
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2145 nea .ef. è .4. e il .dn. è .2/5. E lo .ex. è .3 3/5. e .xn. è comme diciamo .2. E .nd. è .2/5. Adunque, multipli-<lb></lb> |
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2146 cando .2. per .2/5. e partendo per .3 3/5., ne vienne .2/9. per la linea .dc., comme dicemmo.<lb></lb> |
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2147 Ancora sia il medesimo triangolo. E sia dato il ponto fuori .e. Dal qual, menan-<lb></lb> |
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2148 do una linea equedistante al catetto .ad., caggia in sul ponto .g. e sia .2. E dipoi si<lb></lb> |
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2149 dia un ponto dentro e sia .1. Dal quale, menando una perpendiculare al .bg., sia<lb></lb> |
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2150 .it. e sia .3. e .tg. sia .4. E meno una linea .eif. Adimando nota .bf. e .fa. Menerai<lb></lb> |
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2151 .eif. infino al .c. E .ti. infino alo .l. E sia .tl. iguali e equedistanti al .da. Conciosiacosaché .cl.<lb></lb> |
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2152 sia equedistante al .bg. Sia adunque .eg. al .ti., cosí .gz. al .zt., cioé comme .eg. e .ti. al .ti., cioé com-<lb></lb> |
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2153 me .5. a .3., cosí .gt. al .zt. E il .gt. è .4. Adunque .zt. sia .2 2/5. Ancora e gli é cosí .ti. al .il., cosí .zt.<lb></lb> |
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2154 al .lc., cioé commo .3. é .a.g., cosí .zt., che è .2 2/5., é al .lc., adunque .lc. è .7 1/5., del quale tratto .la., che è<lb></lb> |
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2155 iguali al .dt., che è .5., rimane .ac. 2. Ancora e gli é cosí .dz. al .ac., cioé cosí .7 2/5. al .2 1/5., comme<lb></lb> |
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2156 .dh. al .ha. e cosí è .dz. e .ac., cioé .9 3/5., al .ca., cioé .2 1/5., cosí .da., ch’ é .12., al .ha. Adunque .ha. sia .2 3/4.<lb></lb> |
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2157 e .dh. sia .9 1/4. Ancora, perché e gli é cosí .bz. al .ca., cosí .bf. al .fa., cioé perché gli é cosí .12 2/5. a .2 1/5.<lb></lb> |
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2158 cosí .bf. al .fa. e cosí sia .bz. al .ca., cioé .14 3/5. al .2 1/5., cosí .ba. al .fa. Adunque .fa. sia .1 70/73 e .bf.<lb></lb> |
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2159 sia .11 3/73. E cosí fa sempre. |
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2160 </p> |
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2161 <p class="main"> |
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2162 De vi et potentia ypotumisse protracte extrinsice in triangulis orthogoniis. Capitulum secundum.<lb></lb> |
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2163 Se nel triangolo ortogonio di noti lati si mena fuora del triangolo el lato oposto<lb></lb> |
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2164 al’ angolo retto per longitudine nota e dal termine di quella linea al’ angolo ret-<lb></lb> |
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2165 to si menerá una linea, sirá quella linea retta ancora nota. Exempli gratia. Sia<lb></lb> |
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2166 il triangolo ortogonio .abc. di noti lati. E sia .ab.4. e .bc.3. e .ac. sia .5. E sia l’ an-<lb></lb> |
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2167 golo .abc. retto. E menise il lato .ac. infino al ponto .d. fuor del triangolo. E sia .ad. 20. E dal<lb></lb> |
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2168 ponto .d. al ponto .b. si meni una linea retta .bd., la qual linea dico che la sia manifesta.<lb></lb> |
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2169 Che cosí il manifesteró. Meneró la retta .ab., secondo la longhezza, quanto vorró per lo ponto .e.<lb></lb> |
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2170 E per lo ponto .d. meneró la retta .de. equedistante ala linea .bc. E sia l’ angolo .aed. retto.<lb></lb> |
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2171 Imperoché gli é iguali al’ angolo .abc. per la .29a. del primo. Imperoché, quando in .2. linee<lb></lb> |
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2172 rette equedistanti la linea retta cade, sará l’ angolo dentro iguali al’ angolo di fuora a quello<lb></lb> |
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2173 oposto. E peró nelle equedistanti .bc. e .ed. la retta cade .ae. E peró l’ angolo .aed. è iguali a-<lb></lb> |
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2174 l’ angolo .abc. E, per quella medesima, el angolo .ade. è iguali al’ angolo .acb. E l’ angolo .a.<lb></lb> |
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2175 è comune. Adunque e triangoli .abc. e .aed. sonno equiangoli infra loro e simili. E gli tri-<lb></lb> |
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2176 angoli simili hano e lati che sonno intorno a’ simili angoli proportionali, per la diffinitione<lb></lb> |
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2177 dele superficie simili, posta nel sexto de Euclide. E peró è cosí .ac. al .cb., cosí .ad. al .de. E, per<lb></lb> |
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2178 la permutata proportione, sia cosí .ad., che è nota, al .ac., nota, cosí .ed. al .bc., nota. Onde la ret-<lb></lb> |
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2179 ta .ed. sia nota. Cioé, multiplicando .ad. in .cb. e partendo in. ac., e haremo .ed.12. Ancora<lb></lb> |
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2180 sia cosí .ad. al .ac., cosí .ac. al .ba. E peró la retta .ae. sia nota che, multiplicando .ad. in .ba. e<lb></lb> |
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2181 partendo in .ac., e haremo .ae. essere .16., dela quale, tratto .ab. nota, harai .be. essere .12. Del<lb></lb> |
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2182 quale il quadrato, agionto al quadrato dela linea .de., haremo il quadrato dela linea .bd., com-<lb></lb> |
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2183 me volavamo, cioé .288. Adunque .bd. è la radici di .288., comme volavamo.<lb></lb> |
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2184 Dici. L.P. che di questa figura ne risulta la solutione d’ una quistione propostagli<lb></lb> |
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2185 da uno veronese: che propose uno arbore esser ritto sopra una ripa d’ uno fiume.<lb></lb> |
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2186 E ffo la longhezza dell’ abore .40. La quale lunghezza pongo la linea .bg. E lo<lb></lb> |
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2187 spatio ch’ era dappié del’ albore infino al fiume pose essere .5. Lo quale spatio<lb></lb> |
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2188 sia<lb></lb> |
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2189 la linea .bc. E fo nell’ albore preso uno ponto comme il ponto .a. E fo .ba.10. E nel ponto .a.<lb></lb> |
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2190 fu tagliato l’ albero e cade la parte. ag., che è .30.bracia., sopra lo ponto .c. E fo la linea .ad. Adi-<lb></lb> |
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2191 mandase la quantitá dela linea .bd., cioé quanto é dal ponto dela sommitá del’ albero, cioé de-<lb></lb> |
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2192 la vetta , infino al ponto del pedale di quello. Onde, quando volse tal quistione asolvere, in-<lb></lb> |
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2193 tese la figura passata e agionse li quadrati dele linee .ba. e .bc., cioé .100. e .25. E hebbe .125. per<lb></lb> |
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2194 lo quadrato dela linea .ac. E, perché e gli é cosí .ad. al .ac., cosí .ed. al .bc., sia comme il qua-<lb></lb> |
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2195 drato dela linea .ad. al quadrato dela linea .ac., cioé cosí .900. a .125., cosí el quadrato dela<lb></lb> |
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2196 linea .ed. a .25. Ma la proportione di .900. a .125., neli numeri minori, è comme .36. a .5.<lb></lb> |
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2197 <lb></lb> |
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2198 <lb></lb> |
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2199 <pb></pb></p> |
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2200 <p class="folio"> |
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2201 folio 15r |
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2202 </p> |
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2203 <p class="main"> |
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2204 <lb></lb> |
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2205 </p> |
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2206 <p class="runhead"> |
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2207 Distinctio secunda. Capitulum secundum. 15 |
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2208 </p> |
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2209 <p class="main"> |
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2210 <lb></lb> |
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2211 Adunca comme .36. a .5., cosí el quadrato dela linea .ed. è a .25. E, per la permutata proportio-<lb></lb> |
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2212 ne, sia cosí el quadrato dela linea .ed. a .36., comme .25. a .5. Ma .5. è la quinta parte di .25., adun-<lb></lb> |
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2213 ca .36. fu il quinto del quadrato dela linea .ed. Onde multiplichise .36. per .5. e haremo .180.<lb></lb> |
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2214 per lo quadrato dela linea .ed. El quale quadrato si cavi delo quadrato dela linea .ad., cioé<lb></lb> |
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2215 di .900. Rimangano .720. per lo quadrato dela linea .ae. Overo, altramente, perché e gli é<lb></lb> |
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2216 cosí .ad. al .ac., cosí .ae. al .ab., fo adunca cosí .36. a .5., cosí el quadrato dela linea .ea. al quadrato<lb></lb> |
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2217 dela linea .ba., cioé a .100. Onde, per la permutata proportionalitá, è cosí .5. a .100., cosí .36. al<lb></lb> |
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2218 quadrato dela linea .ae. Onde multiplicarai .36. per .20. (Perché .5. sonno .1/20. di .100.) e, simil-<lb></lb> |
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2219 mente, haremo .720. per lo quadrato dela linea .ae. Adunca .ae. è radici di .720. Dela quale<lb></lb> |
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2220 cava la linea .ab., che è .10., riman la radici di .720. meno .10. per la linea .eb. La qual multipli-<lb></lb> |
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2221 ca in sé e fa .820. meno radici di .288000. per lo quadrato dela linea .eb. Al qual agiongni<lb></lb> |
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2222 el quadrato dela linea .ed., cioé .180., e fa .1000. meno radicie di .288000. per lo quadrato dela<lb></lb> |
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2223 linea .bd. Onde .bd. è radici di .1000. meno radici di .288000., cioé, preso la radici di .288000.<lb></lb> |
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2224 e tratta di .1000. e, di quel che restará, preso la radici. Ma acioché se reduchino a numero ra-<lb></lb> |
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2225 tiocinato, preso la radicie di .288000., che è preso .a.536 2/5. meno .1/96., che, tratto di .1000., rimane<lb></lb> |
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2226 .463 11/12. Del qual ancora preso la radicie sia .24 1/2. piú alcuna cosa per la quantitá dela linea .bd.<lb></lb> |
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2227 E volendo mostrare comme el quadrato dela linea .eb. è .820. meno radici di .288000. La li-<lb></lb> |
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2228 nea .eb. è residuo. Imperoché la è la differentia che è infra due linee solamente comunican-<lb></lb> |
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2229 ti in potentia, cioé infra .ae. e .ab. Dele quali .ae. è radici di numero ratiocinato, cioé di .720. |
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2230 </p> |
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2231 <p class="main"> |
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2232 E .ab. è .10., che è numero. E, perché la retta .ae. è divisa in .2. parti nel ponto .b., li quadrati de-<lb></lb> |
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2233 le linee .ae. e .ab. sonno iguali a .2. cotanti dela multiplicatione del .ab. in .ae. col quadrato de-<lb></lb> |
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2234 la linea .eb., per la septima del .2o. de Euclide. Onde, se deli quadrati .ae. e .ab., cioé di .820., si to-<lb></lb> |
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2235 gli el doppio dela multiplicatione del .ab. in .ae., el quale doppio è iguali a .20. radici de .720.,<lb></lb> |
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2236 che ancora è iguali a una radici che perviene dela multiplicatione del quadrato de .20., cioé<lb></lb> |
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2237 de .400. in .720., el qual numero è .288000., rimarranno .820. meno la radici di .288000., com-<lb></lb> |
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2238 me habiamo detto.<lb></lb> |
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2239 <lb></lb> |
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2240 </p> |
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2241 <p class="head"> |
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2242 Distinctionis tertie divisio. |
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2243 </p> |
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2244 <p class="main"> |
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2245 <lb></lb> |
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2246 Conciosiacosaché l’ area di campi di .4. facie che hano tutti gli angoli retti,<lb></lb> |
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2247 sia fatta comme è detto, cioé, quando hano amendui e lati dove si contengano<lb></lb> |
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2248 iguali, cioé sonno le facie tutte iguali, alora se multiplica una dele facie per sé e<lb></lb> |
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2249 quella multiplicatione è la sopra detta area. Overo, quando e lati che contengono<lb></lb> |
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2250 la detta superficie non sonno iguali, alora se multiplica l’ uno lato per l’ altro e l’ area è la<lb></lb> |
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2251 detta multiplicatione. E ’l perché è da notare, che Euclide, parlando dele figure de .4. lati, nel<lb></lb> |
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2252 principio del primo, non prese di quelle se non quatro spetie, cioé quadrato, tetragono lon-<lb></lb> |
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2253 go, elmuaym, che ‘ vulgari chiamo rombo e, simile, elmuaym, che lo chiamano capo taglia-<lb></lb> |
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2254 to. Tutte l’ altre sorti e spetie de figure de .4. lati le chiamó sotto questo solo nome: elmuarisse.<lb></lb> |
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2255 Hoc est irregulari: e quasi imperfette respetto al quadrato al qual tutte, mediante el triangolo, sonno reduci-<lb></lb> |
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2256 bili. La quantitá imperfectum ad suum perfectum. E peró la presente distinctione, per lor mesure, in .6. capitoli voio <lb></lb> |
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2257 divi-<lb></lb> |
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2258 dere. Nel primo mostraremo la solutione d’ alcuna questione proposta sopra le figure di .4. facie che<lb></lb> |
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2259 hano gli angoli retti, la quale, per le .6. regole de algebra, si truova, le quale regole principalmente fo-<lb></lb> |
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2260 ron trovate e fabricate per rispetto della quantitá continua, cioé geometria. Perché in lei<lb></lb> |
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2261 sonno, ale volte, quantitá sorde che, per forza numerale, discretamente non si possan dare, ma<lb></lb> |
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2262 conviense respondere e operare per radici e per linee e quadrati e cubi, commo praticando, te<lb></lb> |
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2263 avirrá neli sequenti casi e molti altri. E sonno tutte fondate nel secondo de Euclide. Me-<lb></lb> |
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2264 diante la forza e virtú del quinto, cioé dele proportioni e proportionalitá che maxime in lo-<lb></lb> |
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2265 ro se recercano, commo de tutto sopra in l’ arithmetica intendesti a suo luogo et cetera. Nel se-<lb></lb> |
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2266 condo mostraremo il modo a trovare l’ area dele figure dette rombi. Nel terzo il modo a<lb></lb> |
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2267 trovare l’ area de’ romboidi. Nel quarto il modo a trovare l’ area dele figure dette caput<lb></lb> |
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2268 abscisum, cioé capo tagliato. Nel quinto il modo a trovare l’ area dele figure dette diver-<lb></lb> |
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2269 silarete. Nel sexto e ultimo il modo a trovare l’ area dele figure di piú di .4. lati.<lb></lb> |
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2270 Modus solvendi varios casus figurarum quadrilaterarum rettangularum per viam al-<lb></lb> |
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2271 gebre. Capitulum primum. |
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2272 </p> |
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2273 <p class="main"> |
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2274 Benché nela parte de arithmetica dicessimo dela regola d’ algebra assai copisa-<lb></lb> |
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2275 mente, nientedimeno é necessario alcuna cosa qui dirne. Sonno certamente<lb></lb> |
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2276 li numeri cosí rotti comme interi over radici over quadrati over numeri sempli-<lb></lb> |
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2277 ci. Quando li numeri si multiplicano in sé, alora quei numeri si dicano radici.<lb></lb> |
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2278 <lb></lb> |
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2279 <pb></pb></p> |
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2280 <p class="folio"> |
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2281 folio 15v |
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2282 </p> |
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2283 <p class="main"> |
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2284 <lb></lb> |
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2285 </p> |
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2286 <p class="runhead"> |
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2287 Distinctio tertia. Capitulum primum. |
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2288 </p> |
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2289 <p class="main"> |
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2290 <lb></lb> |
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2291 E quegli produtti si dicano quadrati over censi. E, quando li numeri non hano rispetto ale<lb></lb> |
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2292 radici overo ai quadrati, alora si dicano numeri semplici. Adunca, secondo questa diffinitio-<lb></lb> |
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2293 ne, ogni numero è alcuna volta radici over quadrato over numero semplici. E, di queste<lb></lb> |
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2294 .3. essentie, .3. regole semplici e altretante composte si truovano. Le .3. regole semplici sonno quan-<lb></lb> |
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2295 do nelle questioni arithmetici over geometrici si truova le radici essere iguali a’ quadrati over le<lb></lb> |
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2296 radici al numero over i quadrati al numero. Le composite sonno quando e si trova le .R. aguagliarse a’ quadrati e<lb></lb> |
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2297 al numero over e quadrati aguagliarse ale radici e al numero over el numero aguagliarse ale .R. e a’ censi. El<lb></lb> |
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2298 quadrato che s’ aguaglia ale radici è comme a dire el quadrato s’ aguaglia ale .4. radici sue.<lb></lb> |
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2299 Ove la radici del quadrato è .4. e il quadrato è .16., cioé il lato dela superficie quadrata è<lb></lb> |
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2300 .4. e l’ area è .16. Imperoché quante unitá sonno in ciascuno de’ lati di quel, tante radici in<lb></lb> |
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2301 quell’ area, cioé nel’ area di quello quadrato, sonno. Comme sia il quadrato .abcd., che ha per<lb></lb> |
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2302 ciascuno lato .4. Dove l’ area sua è iguale a .4. radici. De’ quali una è il quadrilatero .ae., l’ altra<lb></lb> |
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2303 .zt. l’ altra .ik. e l’ altra è .kd. Onde il quadrilatero .lkcd. è .16. E, similmente, se il lato del quadra-<lb></lb> |
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2304 to è .5., sará quadrato .25. E, quando diciamo .4. quadrati sonno iguali a .24. radici, alo-<lb></lb> |
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2305 ra uno quadrato è iguali a .6. radici e una di quelle radici è .6. e il quadrato è .36. Et, quan-<lb></lb> |
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2306 do diciamo la mitá d’ uno quadrato overo di censo è iguali a .4. radici, alora uno censo è igua-<lb></lb> |
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2307 li a .8. radici. E sia la radici .8. E il censo sia .64. Ancora el .1/5. del quadrato è iguali a .3. radi-<lb></lb> |
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2308 ci. Onde tutto il quadrato è iquali a .15. radici. Adunca la radici è .15. e il quadrato è .225. |
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2309 </p> |
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2310 <p class="main"> |
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2311 Similmente, quando è piú d’ un quadrato overo meno d’ un quadrato, ad un quadrato si<lb></lb> |
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2312 riduca. E cosí harai quando el quadrato, cioé censo, è iguali ale radici. E, quando se dici<lb></lb> |
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2313 un quadrato è iquali al numero. Comme a dire un quadrato è iguali a .36. Alora l’ area è .36.<lb></lb> |
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2314 e il lato suo è .6. E, quando .5.censi. sonno iguali a .125. per numero, alora il quadrato è .25. e<lb></lb> |
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2315 la sua radici è .5. Over dicendo la quarta parte d’ un censo è iquali ale .16. dramme (Dico dram-<lb></lb> |
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2316 me, perché cosí, secondo l’ arithmetica, ale volte, se dicano li numeri.) Ove adunca el quadra-<lb></lb> |
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2317 to sia iguali a .64. e la sua radici sia .8. Similmente ogni censo acresciuto over diminuto a un<lb></lb> |
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2318 censo lo ritorna. E, per quel medesimo modo, si fa, quando el numero s’ aguaglia agli censi.<lb></lb> |
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2319 Le radici che s’ aguagliano al numero è comme a dire una radici è iguali a .16. per nume-<lb></lb> |
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2320 ro. Alora la radici è .16. e il quadrato è .256. E, se .6. radici fienno iguali a .30., alora una radi-<lb></lb> |
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2321 ci è iguali a .5. Ancora, dicendo la mitá d’ una radici è iguali a .9., adunca una radici è .18. e<lb></lb> |
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2322 il censo è .324. Le radici che sonno iguali a’ quadrati e a’ numeri: sonno comme a dire .36.<lb></lb> |
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2323 radici sonno iguali a .3.censi. e .105. dramme, cioé .12. radici sonno iguali a un censo e a .35.<lb></lb> |
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2324 dramme. E dicendo la mitá d’ un censo e .12. dramme sonno iguali a .5. radici. É questo com-<lb></lb> |
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2325 me a dire .10. radici sonno iguali a un censo e .24. dramme. E quadrati che sonno iguali<lb></lb> |
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2326 ale radici e al numero: sonno comme a dire .3. quadrati sonno iguali a .12. radici e .36. dram-<lb></lb> |
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2327 me. E questo è comme a dire uno censo essere iguali a .4. radici e .12. per numero. E, se dirai<lb></lb> |
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2328 la quarta parte d’ un censo è iguali a .2. radici e .12. dramme: che è quanto a dire un quadrato<lb></lb> |
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2329 è iguali a .8. radici e .48. dramme. El numero che è iguali a’ quadrati e ale radici. E comme<lb></lb> |
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2330 a dire .78. dramme è iguali a .2. quadrati e .10.R., cioé un quadrato e .5.R. è iguali a .39. dramme. E,<lb></lb> |
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2331 se dirai .32. è iguali ala mitá d’ un quadrato e .a.6.R., sonno, cioé, un quadrato e .12.R. iguali a .64.<lb></lb> |
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2332 dramme. E cosí sempre dobiamo le quistione redure a un censo. E, secondo quel che propor-<lb></lb> |
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2333 tionalmente viene, quella reduttione de piú quadrati: overo le parti d’ un quadrato a un<lb></lb> |
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2334 quadrato: cosí è da redure in quella medesima proportione le radici e le dramme. E, accio-<lb></lb> |
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2335 ché troviamo quello è il quadrato e quello é la radici, lo mostraremo nel dire sequente.<lb></lb> |
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2336 Se nel quadrato .abgd., che per ciascun lato è .10., e il diametro .ag. over .bd. vuoi<lb></lb> |
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2337 havere, radoppirai l’ area del detto quadrato, la quale area è .100. che, duplica-<lb></lb> |
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2338 ta, sonno .200. De’ quali piglia la radici e harai la longhezza d’ uno de’ detti diame-<lb></lb> |
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2339 tri. Verbi gratia. Perche gli é retto l’ angolo .abg., ortogonio è il triangolo .abg.<lb></lb> |
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2340 Dove la multiplicatione delo lato del .ag. in sé è iguali agli .2. quadrati dele linee .ab. e .bg., per la penultima<lb></lb> |
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2341 del .primo. Ma il quadrato del lato .bg. è l’ area del quadrato .abgd. e il quadrato del lato .bg. Onde e<lb></lb> |
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2342 .2. quadrati dele linee .ab. e .bg. sonno doppi al quadrato del lato .bg. e il quadrato del lato diamitrale<lb></lb> |
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2343 .ag. è iguali a’ .2. quadrati de’ .2. lati .ab. e .bg. Adunca el quadrato del diametro .ag. è doppio al quadrato<lb></lb> |
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2344 del lato .bg. e il quadrato delo lato .bg. è l’ area del tetragono .abgd. e peró il quadrato delo lato .ag. è<lb></lb> |
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2345 doppio ala misura del tetragono .abgd., ch’ era bisogno mostrare. E questo è quello in che il cen-<lb></lb> |
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2346 so è iguale al numero, cioé l’ area è .100. Onde il quadrato del diametro sia iguale al doppio del’ area,<lb></lb> |
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2347 cioé .200. Dico anchora il diametro .ag. essere iguali al diametro .bd., perché la retta .bg. è iguale<lb></lb> |
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2348 <lb></lb> |
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2349 <pb></pb></p> |
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2350 <p class="folio"> |
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2351 folio 16r |
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2352 </p> |
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2353 <p class="main"> |
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2354 <lb></lb> |
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2355 </p> |
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2356 <p class="runhead"> |
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2357 Distinctio tertia. Capitulum primum. 16 |
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2358 </p> |
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2359 <p class="main"> |
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2360 <lb></lb> |
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2361 ala retta .ad. E, se comunamente se piglia la retta .ab. fienno .2. rette .ab. e .bg. iquali a .2.<lb></lb> |
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2362 rette .ba. e .ad. e l’ angolo .abg. è iguale al’ angolo .bad. Onde il diametro .ag. è iguale al dia-<lb></lb> |
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2363 metro .bo. E ancora dico che ’l diametro .ag. e .bd. si segano insiemi per iqual parti nel ponto<lb></lb> |
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2364 .e. Perché infra loro sonno iguali le rette .ad. e .bg. e ne’ loro termini sonno compilate .ab. e .dg., che<lb></lb> |
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2365 infra loro sonno iguali, per la .34a. del primo, saranno certamente infra loro equedistanti le rette .ad.<lb></lb> |
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2366 e .bg. E, perché in quelle le rette .bd. e .ag. le tagliano, sia iguale l’ angolo .adb. al’ angolo .dbg. E l’ an-<lb></lb> |
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2367 golo .dag. al’ angolo .agb. e l’ altro angolo .aed. al’ altro .beg. è iguali, per la .15a. del primo. Per la qual cosa<lb></lb> |
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2368 il triangolo .aed. è iguali al triangolo .beg. E la retta .be. ala retta .ed. è iguali. Similmente e la ret-<lb></lb> |
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2369 ta .ge. ala retta .ea. è iguale. Adunca per igual parti infra loro si segano e diamietri .ag. e .bd.,<lb></lb> |
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2370 ch’ era bisogno mostrare. |
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2371 </p> |
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2372 <p class="main"> |
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2373 E, se i diametri del tetragono dato fienno la radici di .200. e non sappia quanto sia<lb></lb> |
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2374 l’ area e il suo lato, togli il .1/2. di .200., che è .100., che sia l’ area del detto quadrato e la<lb></lb> |
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2375 radici di .100., cioé .10., haremo per lo suo lato, cioé .2.censi. sonno iguali a .200. On-<lb></lb> |
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2376 de un quadrato, cioé un censo, è .100., cioé la misura del detto quadrato è .100. e<lb></lb> |
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2377 il suo lato è .10. E, se ’l quadrato del diametro con l’ area del tetragono fienno .300., adunca .3.<lb></lb> |
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2378 quadrati son iguali a .300. Onde la terza parte di quelli, cioé .100., sia l’ area e l’ altro rimanen-<lb></lb> |
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2379 te, cioé .200., sia il quadrato del diametro del detto tetragono. E la radici del’ area sia il lato del<lb></lb> |
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2380 detto quadrato. |
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2381 </p> |
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2382 <p class="main"> |
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2383 E, se gli .4. lati d’ uno quadrato con l’ area del detto quadrato fussino .140. e vuoi<lb></lb> |
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2384 sapere quanto sia lo lato del detto quadrato, comme per figura si deba fare, il mo-<lb></lb> |
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2385 straremo. Sia il tetragono .ezit. A quello s’ agionga la superficie .ae. d’ angoli ret-<lb></lb> |
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2386 ti. E sia .ai. per lo diritto del .it. e .il.be. sia per lo diritto del .ez. E sia ciascuna dele<lb></lb> |
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2387 rette .be. e .ai.4. per numero. Ove la superficie .ae. è iguale agli .4. lati. E tutta la superficie .az.<lb></lb> |
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2388 sia .4. lati e il quadrato delo lato .it., cioé tutta la superficie .az., sia .4. lati e il quadrato del lato<lb></lb> |
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2389 del tetragono .ezit. Adunca la superficie .za. è .140. comme proponenmo. E questo é quello che<lb></lb> |
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2390 dicemmo di sopra: cioé quando el censo e .4. radici sono iguali a .140. E il censo è il tetragono<lb></lb> |
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2391 .iz. e le .4. radici è la superficie .ae. Dividasi adunca la retta .ai. in due parti iguali sopra il ponto<lb></lb> |
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2392 .g. E, perché la linea .it. è agionta ala linea .ai., sará la superficie rettangula del .it. in .at., col quadra-<lb></lb> |
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2393 to dela linea .gi., iguale al tetragono .gt., per la .6a. del .2o. E la superficie del .it. in .at. è comme la superfi-<lb></lb> |
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2394 cie del .at. in .tz., imperoché .it. è iguale al .tz. Adunca la superficie .ztab., col quadrato de-<lb></lb> |
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2395 la linea .gi., è iguale al quadrato dela linea .gt. Ma il .zt. in .at. è la superficie .za., che è .140., a’ <lb></lb> |
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2396 quali agionto il quadrato dela linea .gi., cioé .4., fanno .144. per lo quadrato dela linea .gt.<lb></lb> |
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2397 Onde .gt. è .12., cioé la radici di .144. Onde, se del .gt. se ne trae .gi., rimarrá .it.10., che è il la-<lb></lb> |
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2398 to del tetragono, imperoché l’ area sua (che è .100.), agionta a’ .4. suoi lati, fanno .140. com-<lb></lb> |
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2399 me bisogna. E cosí è da ffare in tutte le quistioni dove li numeri sonno iguali al quadrato e<lb></lb> |
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2400 radici. Cioé sopra il detto numero s’ agionga il quadrato dela mitá dele radici e dela summa<lb></lb> |
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2401 si truovi la radici e di quella si tolga la mitá dele radici: rimarrá la radici delo adimandato cen-<lb></lb> |
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2402 so, che, in sé multiplicata, sará il detto censo. Comme ancora dicendo .133. dramme sonno igua-<lb></lb> |
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2403 li a un censo e .12. radici. Onde, se ’l quadrato dela mitá dela radici s’ agiongne a .133., faranno<lb></lb> |
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2404 .169., de’ quali la radici è .13., de’ quali, trattone .6., cioé la mitá dele radici, rimaranno .7. per la ra-<lb></lb> |
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2405 dici del censo e il censo sia .49. |
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2406 </p> |
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2407 <p class="main"> |
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2408 Ancora e gli é un tetragono dela cui area, se se ne toglie .4. suoi lati, rimaranno .77. |
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2409 </p> |
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2410 <p class="main"> |
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2411 Adimando quale è quel censo. Ove piglise uno tetragono .bd. E piglise il ponto<lb></lb> |
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2412 .a. nella linea .gd. E sia .ga.4. E per lo ponto .a. si meni la linea .az. equedistante<lb></lb> |
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2413 a ciascuna dele rette .gb. e .de. E, perché .ga. é .4., la superficie .ab. sia .4. lati, cioé .4.<lb></lb> |
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2414 radici, del tetragono .bd. Onde, se del tetragono detto si trae el quadrilatero .ba. (che é .4.<lb></lb> |
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2415 suoi lati), rimarrá la superficie .zd.77. E, perché le .2. superficie .ab. e .ae., cioé .ba. e .zd., son-<lb></lb> |
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2416 no iguali al tetragono .bd., adunca il censo è iguale ale radicie e al numero: cioé il quadrato<lb></lb> |
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2417 .bd. è iguali a .4. sue radici e .77. dramme. Onde, a trovare quanto è il censo, dividasi .ga. in<lb></lb> |
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2418 .2. parti iguali sopra il ponto .i. E a lei sará agionta per lo dritto la retta .ad. Sará la multiplicatione del .ad.<lb></lb> |
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2419 in .gd., col quadrato dela linea .ai., iguale al quadrato .di., per la .6a. del .2o. Ma la multiplicatione del .ad. in .dg. è<lb></lb> |
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2420 comme la multiplicatione del .da. in .de., imperoché .de. è iguale ala linea .dg. Ma il produtto del .da. in .sé. è<lb></lb> |
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2421 la superficie .zd., che è .77. Adunca .da. in .de. fanno .77., a’ quali, agionto el quadrato del .ai., che è .4.,<lb></lb> |
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2422 <lb></lb> |
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2423 <pb></pb></p> |
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2424 <p class="folio"> |
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2425 folio 16v |
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2426 </p> |
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2427 <p class="main"> |
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2428 <lb></lb> |
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2429 </p> |
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2430 <p class="runhead"> |
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2431 Distinctio tertia. Capitulum primum. |
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2432 </p> |
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2433 <p class="main"> |
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2434 <lb></lb> |
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2435 fanno .81. per lo quadrato della linea .di. Del quale la radici, cioé .9., è la linea .di., ali quali, agionto la<lb></lb> |
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2436 linea .ig., che è .2., fanno .11. per lo lato del censo .dg. Onde l’ area del quadrato .bd. è undeci vol-<lb></lb> |
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2437 te .11., cioé .121., e ‘.4. suoi lati sonno .44. che, tratti di .121., rimangano .77. per la superficie .zd., comme<lb></lb> |
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2438 era de bisogno. E cosí è da ffare in tutte le quistioni nele quali el censo s’ aguaglia ale radici e al<lb></lb> |
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2439 numero, cioé sopra il numero s’ agionga el quadrato dela mitá dele radici e dela summa si pigli la radici e<lb></lb> |
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2440 quella s’ agionga ala mitá dele radici: e cosí haremo la radici del quadrato che, in sé multipli-<lb></lb> |
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2441 cata, fará l’ adimandato quadrato, comme se si propone che ’l censo sia iguali a .10. radici e .39. dramme.<lb></lb> |
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2442 Agiongase il quadrato di .5., cioé .25., sopra .39., fanno .64., sopra le radici dele quali agiogni .5.,<lb></lb> |
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2443 cioé sopra .8. agiongni .5., fanno .13. per la radici del censo e il censo sia .169., dove .10. radici son-<lb></lb> |
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2444 no .130., che sonno fatti del .10. in .13. che, con .39., fanno .169. per lo quadrato del censo.<lb></lb> |
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2445 Anchora e gli é un tetragono la cui area, se si tra’ degli .4. suoi lati, rimane .3. Adi-<lb></lb> |
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2446 mandasi quali sonno e lati e il tetragono. E, per questo trovare, tolghise il tetragono<lb></lb> |
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2447 .ge. del quale ciascuno suo lato sia meno di .4. E ala linea .de. s’ agionga la linea .da.<lb></lb> |
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2448 E sia tutta .ae.4. E dividase .ae. in .2. parti iguali: sopra il punto .b. E menise la ret-<lb></lb> |
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2449 ta .az. equedistante e iguale ala retta .dg. E menise .fg. infino .al.z. E, perché .ae. è .4. e .ef. è il la-<lb></lb> |
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2450 to del tetragono .ge., sará adunque .fe. in .ea., cioé la superficie .ze. iguale a .4. radici: cioé a’ la-<lb></lb> |
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2451 ti del tetragono .ge. De’ quali, se si togli el tetragono .ge., rimarrá la superficie .ga.3. Ma il te-<lb></lb> |
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2452 tragono .ge. e la superficie .ga. sono iguali ala superficie .ze. Adunque .4. radici sonno iguali al<lb></lb> |
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2453 censo e .3. É di bisogno, adunque, troviamo el censo e la sua radici. Perché la linea .ae., che è .4.,<lb></lb> |
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2454 è divisa in .2. parti iguali sopra il punto .b. e in .2. parti non iguali sopra il punto .d., sia la mul-<lb></lb> |
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2455 tiplicatione del .ed. in .da. col quadrato dela linea. bd. iguale al quadrato dela linea .be., che<lb></lb> |
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2456 è fatto dala mitá dela linea .ae., per la .5a. del .2o. Ma multiplicatione del .ed. in .da. fa la<lb></lb> |
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2457 superficie .ga., che è .3., imperoché .dg. è iguale al .de. Adunque la superficie fatta dal .gd. in<lb></lb> |
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2458 .da., col quadrato dela linea .bd., è iguale al quadrato dela linea .be., che è .4. Adunque il qua-<lb></lb> |
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2459 drato .bd. è .1o. Dela quale, se si togli la radici, sia .1o. E tragase dela linea .be., rimarrá la linea<lb></lb> |
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2460 .de .1o., che è il lato del tetragono .ge. del quale l’ area è .1o., cioé el censo è .1. Ancora, se lla mitá de-<lb></lb> |
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2461 la linea .ae. sará infra ’l .de. in sul punto .b., comme si manifesta in questa altra figura. Agiongni<lb></lb> |
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2462 sopra .eb., cioé sopra .2., la linea .bd. e sia tutta .de.3., che è la radici del quadrato .ge. adimandato.<lb></lb> |
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2463 E il quadrato sia .9. E la superficie .ag. sará similmente .3., comme adimandammo. E cosí é da ffare<lb></lb> |
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2464 in ogni quistioni nele quali la radici è iguale al quadrato e al numero: cioé del quadrato dela mi-<lb></lb> |
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2465 tá dele radici si tolga el numero e delo rimanente si pigli la radici e tolghise dela detta mittá. O-<lb></lb> |
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2466 vero s’ agionga sopra quella e harai la radice delo adimandato quadrato. Comme a dire .12. ra-<lb></lb> |
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2467 dici sono iguali a un censo e .27. per numero. Dove il quadrato dela mitá dele radici è .36., del qua-<lb></lb> |
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2468 le tra’ .27., rimane .9. De’ quali la radici, che è .3., agiongnise sopra .6., cioé sopra la mitá dele radici,<lb></lb> |
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2469 haremo .9. per la radici adimandata. E il censo sia .81. Overo si traga il .3. dela mitá dele radi-<lb></lb> |
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2470 ci, cioé di .6., rimarrá .3. per la radici. E il censo è .9. E cosí sempre, quando le radici sono iguali al<lb></lb> |
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2471 censo e al numero, sonno absolute le quistioni in .2. modi. Nientedimeno in alcune quistioni quan-<lb></lb> |
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2472 do chade .1a. asolutione e quando l’ altra. |
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2473 </p> |
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2474 <p class="main"> |
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2475 Ancora e gli é un tetragono che ’l quadrato del diametro, con l’ area sua e con .4. suoi<lb></lb> |
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2476 lati, sonno .279. Adimandase el lato del detto tetragono. Perché il quadrato del<lb></lb> |
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2477 diametro è doppio al’ area del detto tetragono. Adunque el quadrato del diame-<lb></lb> |
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2478 tro, con l’ area del detto diametro, è .3. cotanti al’ area del tetragono. E peró .3. quadra-<lb></lb> |
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2479 ti e .4. radici sonno iguali a .279. Onde, acioché riduchi questo a uno censo, togli el terzo di<lb></lb> |
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2480 queste quantitá e troverai .1.censo. e .1a. radici e .1/3. iguali a .93., dove piglia la mitá dele radici,<lb></lb> |
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2481 che sonno .2/3., multilplica in sé, fanno .4/9. Le quali agiogni a .93., fanno .93 4/9. dela cui radici togli .2/3., cioé<lb></lb> |
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2482 la mitá dele radici, rimangono .9. per lo lato del tetragono, dove l’ area è .81. E il quadrato del<lb></lb> |
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2483 diamitro è .162. |
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2484 </p> |
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2485 <p class="main"> |
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2486 Ancora .4. lati del tetragono sonno iguali ae .2/9. di tutto el tetragono. Adimanda-<lb></lb> |
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2487 se il lato del tetragono. Tolghise il tetragono .abgd. E piglise in quello a diritto<lb></lb> |
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2488 i punti .ez. E sia ciascuna dele rette .be. e .az.4. E compise la retta .ez. Fienno adun-<lb></lb> |
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2489 que e pararelli .ae. e .zg. sopra le equedistanti .ad. e .bg. Onde, comme el paralello .ae. è al paralello<lb></lb> |
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2490 .zg., cosí la basa .be. é ala basa .ge., per la prima. del .6o. Ma il paralello .ae. è .4. radici del tetragono<lb></lb> |
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2491 .ag. Adunque la superficie .ae. è .2/9. del tetragono .ag. Onde il paralello .zg. è .2/9. del .ag. Adun-<lb></lb> |
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2492 que la superficie .ae. è ala superficie .zg. comme .2. a .7. Per la qual cosa, e gli é cosí .2. a .7., cosí .be.,<lb></lb> |
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2493 cioé .4., è al .eg. Per la qual cosa, multiplica .4. per .7. e dividi per .2., vienne .14. per la linea .eg.<lb></lb> |
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2494 Over altramente, ‘l perché gli é doppio .4. a .2., cosí è doppio .eg. del .7. Adunque .eg. è .14.<lb></lb> |
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2495 <lb></lb> |
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2496 <pb></pb></p> |
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2497 <p class="folio"> |
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2498 folio 17r |
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2499 </p> |
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2500 <p class="main"> |
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2501 <lb></lb> |
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2502 </p> |
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2503 <p class="runhead"> |
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2504 Distinctio tertia. Capitulum primum. 17 |
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2505 </p> |
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2506 <p class="main"> |
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2507 <lb></lb> |
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2508 a’ quali, agionto .4., fanno .18., cioé .eb. Haremo per .bg.18., cioé per lo lato del tetragono .ag.,<lb></lb> |
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2509 ch’ era de bisogno mostrare. |
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2510 </p> |
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2511 <p class="main"> |
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2512 Over altramente, secondo la computatione de algebra. Perché la superficie .ae. è<lb></lb> |
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2513 .4. radici over .2/9. del tetragono .ag., adunca .4. radici sonno iguali .2/9. del censo.<lb></lb> |
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2514 Onde, acioché se riduchi a un censo, multiplica .9. per .4. e. dividi per .2. Over la<lb></lb> |
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2515 mittá di .9. multiplica per .4., perché quante volte .2/9. sonno in .9. noni, cioé nel cen-<lb></lb> |
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2516 so, tante volte .4. é nela radici del tetragono .ag. Onde il quadrato .ag. è iguale a .18. radi-<lb></lb> |
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2517 ci, commo dicemmo, e contiene per area .324. Ancora .4. lati e .3/8. del’ area d’ uno censo sonno igua-<lb></lb> |
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2518 li a .77 1/2. Dove reduci .3/8. di censo e harai che uno censo e .10. radici .2/3. è iguali a<lb></lb> |
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2519 .206 2/3. E questo troveremo multiplicando .4. radici e .77 1/2. per .8. e dividendo l’ una e l’ altra mul-<lb></lb> |
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2520 tiplicatione per .3. Dimezza adonca le radici, sonno .5 1/3. De’ quali il quadrato togli, che è<lb></lb> |
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2521 .28.1/9. El quale numero agiongni con .206 2/3. Fanno .235 1/9., dela cui la radici (che è .15 1/3.) se ne traga<lb></lb> |
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2522 la mittá dele radici, che è .5 1/3. Rimane .10. per lo lato del censo e l’ area è .100. E, se .4. suoi<lb></lb> |
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2523 lati sonno iguali al’ area del censo, alora .4. sue radici sonno iguali al censo. Dove ciascuno<lb></lb> |
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2524 suo lato è .4. E il censo è .16. E, se .4. lati d’ un censo sonno iguali al doppio del’ area sua, al-<lb></lb> |
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2525 lora .4. lati, cioé .4. radici, sonno eguali a .2. censi. Onde .1o. censo è iguale a .2. radici. Adonca<lb></lb> |
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2526 il lato del tetragono è .2. El tetragono è .4. Ancora, se del’ area del tetragono si tolga .3. suoi<lb></lb> |
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2527 lati, rimarrá .40. Adunca .3. radici e .40. sonno iguali a uno censo. La mittá adonca dele ra-<lb></lb> |
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2528 dici in sé multiplica e fienno .2 1/4. El quale agiongni a .40., fanno .42 1/4. Sopra la cui radici,<lb></lb> |
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2529 cioé sopra .6 1/2., agiongni la mittá dele radici, fienno .8., che sonno il lato del tetragono. An-<lb></lb> |
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2530 cora, diviso l’ area del quadrato per lo diametro suo, ne perviene .10. Adimandase quanto è<lb></lb> |
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2531 il diametro e il suo lato. Perché dividendo l’ area per lo diametro ne perviene .10. |
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2532 </p> |
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2533 <p class="main"> |
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2534 Adonca, a<lb></lb> |
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2535 multiplicare .10. per lo diametro, fará la quadratura. Onde, a multiplicare el diametro nel<lb></lb> |
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2536 doppio di .10., ne perviene doppio del’ area. Ma il doppio del’ area è iguale al quadrato del<lb></lb> |
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2537 diametro. Adonca, a multiplicare il diametro per .20., ne perviene il quadrato del diametro.<lb></lb> |
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2538 E, a multiplicare el diametro in sé, ancora ne perviene el preditto quadrato. Unde il diame-<lb></lb> |
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2539 tro è .12. E il suo quadrato è .400. E l’ area è la mitta, cioé .200. El suo lato ala sua radice, cioé<lb></lb> |
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2540 radici de .200., che è poco meno de .14 1/7. et cetera. |
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2541 </p> |
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2542 <p class="main"> |
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2543 Ancora, tratto .4. lati del’ area, remangano .4. Adimando quanto è il suo lato. Pi-<lb></lb> |
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2544 glise il tetragono .abgd. E piglise li ponti .ez. E sia ciascuna dele rette .az. e .de.<lb></lb> |
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2545 .4. E compisse .ze. Sará adonca la superficie .dz. iguale a’ .4. lati del tetragono .db., ri-<lb></lb> |
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2546 marrá la superficie .eb.4. Adonca .4. radici e .4. sonno iguali al tetragono .db. Di-<lb></lb> |
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2547 vidase adonca .az. in .2. parti iguali sopra ’l ponto .i. Sará adonca la multiplicatione del .cb.<lb></lb> |
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2548 in .ab., col quadrato dela linea .iz., comme il tetragono dela linea .ib., per sexta secondi. Ma<lb></lb> |
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2549 il .zb. in .ab. è como .zb. in .ez. Ma il .zb. in .ze. fa la superficie .eb., cioé .4. Adonca del .zb.<lb></lb> |
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2550 in .ab. ne perviene .4. a’ quali, agionto el tetragono dela linea .iz., fanno .8. per lo quadrato de-<lb></lb> |
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2551 la linea .ib. Adonca la retta .ib. è radici di .8. ala quale, agionto .2., cioé la retta .ia., haremo .2.<lb></lb> |
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2552 e radici di .8., per tutta la linea .ab., che è il lato del tetragono .db., che era bisogno vedere.<lb></lb> |
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2553 E, se il diametro d’ un quadrato avanza a ciascun lato .6., adimandasse quanto è<lb></lb> |
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2554 il lato del ditto quadrato. Multiplica adonca .6. in sé, fanno .36. Lo quale radop-<lb></lb> |
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2555 pia, fanno .72., sopra la qual radici agiongni .6. E haremo la radici di .72. e .6. E di-<lb></lb> |
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2556 remo ciascun lato essere .6. e radici di .72., che mostraremo la cagione. Sia fatta la<lb></lb> |
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2557 retta .ab. E sia iguale al dato diametro del ditto quadrato e il suo lato sia .bg. Dove .ga. è .6.,<lb></lb> |
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2558 cioé quello che ’l diametro agiongne: over avanza al lato. E faciase sopra la retta .ab. el tetra-<lb></lb> |
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2559 gono .ad. E menise in quello el diametro .eb. E, per lo ponto .g., si meni la retta .gz. equedi-<lb></lb> |
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2560 stante ala retta .ae. e al .bd. E, per lo ponto .i. si meni la retta .tk. equedistante alla retta .de.ba.<lb></lb> |
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2561 Dipoi se pigli nela retta .bg. il ponto .l. E sia .gl. iguale al .ga. e compise la figura medesima<lb></lb> |
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2562 nel tetragono .gk. E, perché il quadrato .ad. è quadrato: quadrati sonno quelli che son-<lb></lb> |
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2563 no intorno al diametro suo, cioé .tz. e .gk., per lo corelario dela .4a. del secondo. E il lato del te-<lb></lb> |
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2564 tragono .tz. è .ti., che è iguale ala retta .ag. Adonca .ti. è .6. E il tetragono .tz. è .36. Ancora<lb></lb> |
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2565 tetragoni sonno le superficie .om. e .lp. e sonno intorno al diametro del tetragono .gk. E il<lb></lb> |
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2566 tetragono .om. è iguale al tetragono .zt. Perché la retta .on. è iguale ala retta .gl. E il .gl. è<lb></lb> |
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2567 iguale ala retta .ga. E la retta .ga. ala retta .ti. facemmo iguali. E, perché .bg. è il lato del tetra-<lb></lb> |
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2568 gono del quale il .ba. è diametro. Onde il tetragono fatto dal .ba. è doppio al tetragono .gk.<lb></lb> |
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2569 Onde il supplemento .q. e il supplemento .s., col quadrato .r., é iguale al quadrato: cioé al tetra-<lb></lb> |
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2570 gono .gk. Unde del supplemento .q. e .s. e del quadrato .r. se traga el quadrato .r., rimarrá lo sup-<lb></lb> |
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2571 <lb></lb> |
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2572 <pb></pb></p> |
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2573 <p class="folio"> |
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2574 folio 17v |
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2575 </p> |
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2576 <p class="main"> |
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2577 <lb></lb> |
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2578 </p> |
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2579 <p class="runhead"> |
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2580 Distinctio tertia. Capitulum primum. |
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2581 </p> |
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2582 <p class="main"> |
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2583 <lb></lb> |
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2584 plemento .s. e lo supplemento .q. E tragase lo quadrato .tz., cioé .om., del quadrato .gk. Ri-<lb></lb> |
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2585 marrá lo gnomone .c.h.f. iguali a’ doi supplementi .q. e .s., cioé le superficie .ai. e .id. sonno iguali<lb></lb> |
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2586 ale superficie .gn. e .nk. e .lp. e la superficie .ai. è iguale ala superficie .ai.li. Imperoché .gl.<lb></lb> |
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2587 è iguale al .ag. e .lm. è iguale al .ta. E ancora la superficie .ok. è iguale ala superficie .id. Per-<lb></lb> |
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2588 ché .op. è iguale al .zd. e .iz. al .io. Adonca le .2 .superficie, cioé .li. e .ok., sonno quanto el gno-<lb></lb> |
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2589 mone .c.f.h. Onde, se da ciascuno si tolga la superficie .gn. e .ng., rimará il tetragono .lp. igua-<lb></lb> |
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2590 le a .2. tetragoni, che l’ uno sia quanto .om., cioé rimarrá il quadrato .lp. iguale a .2. cotanti del<lb></lb> |
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2591 quadrato .om. Ma il tetragono .om. è .36. Onde il tetragono .lp. è .72. Del quale il lato è .bl.,<lb></lb> |
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2592 che è radici di .72. All quale, agionto .lg., che è .6., haremo per tutta .bg.6. e radici di .72., commo<lb></lb> |
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2593 era bisogno mostrare. Al quale, agionto .ga., sará tutta .ab., cioé il diametro del dato quadra-<lb></lb> |
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2594 to .12. e radici di .72. Over altramente, perché la retta .ag. è .6. e .gi. è il lato del tetragono, del<lb></lb> |
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2595 quale il diametro è iguale ala linea .ab., sará, per questo, la superficie .ai. iguale a .6. radici del<lb></lb> |
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2596 tetragono .gk. E il quadrilatero .id. è iguale al quadrilatero .ai. E peró el quadrilatero .id.<lb></lb> |
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2597 è .6. radici del quadrato .gk. E il quadrato .tz. è .36. Onde tutto lo gnomone .q.r.s. è igua-<lb></lb> |
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2598 le al tetragono .gk., commo è mostro. Donde, se porremo la retta .bg. una cosa, sará .gk. uno<lb></lb> |
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2599 censo, che è iguale a .12. sue radice e .36. dramme. E operarai in questo secondo che s’ é ditto: cioé<lb></lb> |
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2600 quando il censo è iguale ale radici e al numero et cetera. |
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2601 </p> |
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2602 <p class="main"> |
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2603 Ancora io ó multiplicato il diametro per uno lato e feci .100. Adimandasi quan-<lb></lb> |
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2604 to è il diametro e il lato del ditto tetragono. Quando e si multiplica il lato per lo<lb></lb> |
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2605 diamettro e fanno .100. Onde, multiplicando el quadrato del diametro per lo qua-<lb></lb> |
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2606 drato del lato, cioé per l’ area del tetragono, ne perviene il quadrato de .100., cioé<lb></lb> |
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2607 .10000., commo per la .64a. conclusione dele quantitá semplici dicemmo sopra in la parte de a-<lb></lb> |
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2608 rithmetica. Onde, multiplicando el quadrato del diametro per lo doppio del’ area, cioé per<lb></lb> |
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2609 lo quadrato del diametro, ne proviene il doppio di .10000., cioé .20000. Onde, a multiplicare l’ a-<lb></lb> |
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2610 rea del ditto tetragono in sé, fanno la radici di .5000., cioé il .1/4. di .2000., imperoché l’ area del<lb></lb> |
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2611 ditto tetragono è la mitá del quadrato del diametro. Adonca il lato è la radici di radici di<lb></lb> |
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2612 .5000. e il diametro è la radici de radici di .20000. |
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2613 </p> |
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2614 <p class="main"> |
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2615 Ancora io ó multiplicato l’ area d’ uno tetragono per lo suo diametro e feci .500. |
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2616 </p> |
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2617 <p class="main"> |
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2618 Adimando quanto è il suo lato. Perché a multiplicare l’ area per lo suo diame-<lb></lb> |
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2619 tro fanno .500., adonca, a multiplicare il diametro per lo doppio del’ area, fará il<lb></lb> |
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2620 doppio di .500., cioé .1000. E il doppio del’ area è il quadrato del diametro. Onde,<lb></lb> |
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2621 a multiplicare il diametro per lo suo quadrato fa .1000. E, quando el si multiplica alcun qua-<lb></lb> |
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2622 drato per lo suo lato, fanno il cubo di quel lato. Adonca il ditto diametro era la radici cu-<lb></lb> |
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2623 bica di .1000., la quale radici è .l0. Adonca il diametro è .l0., il suo quadrato è .100. e l’ area sua<lb></lb> |
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2624 è la mitá, cioé .50. e il lato suo sia la radici di .50. |
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2625 </p> |
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2626 <p class="main"> |
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2627 Sopra de’ quadrati habiamo asai ditto, dove de’ quadrilateri detti parte altera lon-<lb></lb> |
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2628 giore, per al presente, ne daremo alcun caso. E quadrilatero parte altera lon-<lb></lb> |
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2629 giore è quella superficie che á e lati opposti equedistanti e gli angoli retti, ma i l-<lb></lb> |
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2630 lati non sonno iguali infra loro, se non li opposti, per la .34a. del primo. E que-<lb></lb> |
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2631 sta tal chiama Euclide tetragono longo nel principio del primo. E l’ area sua si truova del<lb></lb> |
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2632 multiplicare lo lato dela longhezza per quel dela larghezza. Commo sia uno quadrilatero par-<lb></lb> |
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2633 te altera longiore .abcd. che, per ciascun lato de’ minori, è .6. e sonno e lati .ab. e .cd. E, per cia-<lb></lb> |
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2634 scun de’ maggiori, .8., che sonno .ad. e .bc. Dico che l’ area sua s’ á del multiplicare del .ab. in .bc.<lb></lb> |
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2635 Dove l’ area sua è .48. E, volendo il suo diametro, cioé .ac., agiongni insiemi e quadrati .ab.<lb></lb> |
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2636 e .bc., cioé .36. e .64., fanno .100. De’ quali la radici, che è .10., é il diametro .ac.<lb></lb> |
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2637 Sia adonca uno quadrilatero commo il ditto e il diametro sia .10. E il lato .cb. sia<lb></lb> |
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2638 .8. e voglio sapere quanto è lo lato .ab. Del quadrato dela linea .ac. tra’ el quadra-<lb></lb> |
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2639 to dela linea .bc., e rimarrá el quadrato dela linea .ab. E cosí volendo de con-<lb></lb> |
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2640 verso. Commo nel triangolo ortogonio dicemmo di sopra. E, per simile modo, el<lb></lb> |
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2641 diametro .bd. sia ancora .10., per la equalitá de’ triangoli .abc. e .bad. E ancora dico che ’dit-<lb></lb> |
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2642 ti triangoli .abc. è .bad., e ancora dico che ’ditti triangoli si segano infra loro per igual par-<lb></lb> |
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2643 ti sopra il ponto .e. Sonno certamente equedistanti le linee .ad. e .bc. Onde l’ angolo .ade.<lb></lb> |
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2644 è iguale al’ angolo .cbe. E ancora l’ angolo .dae. è iguale al’ angolo .bce. L’ altro .aed. al’ al-<lb></lb> |
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2645 tro .bec. E hano e lati .ab. e .bc. infra loro iguali. Ove gli altri lati del’ uno triangolo agli al-<lb></lb> |
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2646 tri lati del’ altro triangolo fienno iguali, cioé il lato .be. al lato .ed. E il lato .ce. al lato .ea. E, per-<lb></lb> |
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2647 ché il diametro .ac. è iguale al diametro .bd., e sonno segatosi per igual parti nel ponto .e. E<lb></lb> |
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2648 <lb></lb> |
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2649 <pb></pb></p> |
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2650 <p class="folio"> |
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2651 folio 18r |
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2652 </p> |
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2653 <p class="main"> |
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2654 <lb></lb> |
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2655 </p> |
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2656 <p class="runhead"> |
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2657 Distinctio tertia. Capitulum primum. 18 |
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2658 </p> |
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2659 <p class="main"> |
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2660 <lb></lb> |
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2661 peró ciascuna parte sia .5., ch’ era bisogno mostrare. |
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2662 </p> |
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2663 <p class="main"> |
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2664 E, dicendo che l’ area sia .48. e, agionto un de’ lati magiori conn’ un de’ minori, fanno<lb></lb> |
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2665 .14. E adimandase quanto sia il lato longo: over il breve. La mitá de’ .14., in sé mul-<lb></lb> |
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2666 tiplica, fanno .49. Del quale togli l’ area, cioé .48., rimane .1o. Del quale la radici agion-<lb></lb> |
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2667 gni a .7., fanno .8. per lo lato magiore, che infino in .14., v’ é .6. per lo lato minore.<lb></lb> |
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2668 Verbi gratia. Sia el quadrilatero .bgde. parte altera longiore. E sia .bg. il lato breve e il<lb></lb> |
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2669 .gd. sia il magiore e menise la retta .bg. infino al ponto .a. E sia la retta .ga. iguale ala retta<lb></lb> |
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2670 .gd. E dividase la retta .ab. in .2. parti iguali sopra il ponto .c. Sia la linea .ba.14., dove .ac.<lb></lb> |
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2671 over .bc. sienno .7. E, perché la retta .ba. è divisa in .2. parti iguali e in .2. parti non iguali sopra e<lb></lb> |
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2672 ponti .g. e .c., sia .bg. in .ga., col quadrato dela linea .gc., iguale al quadrato dela linea .ca., per<lb></lb> |
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2673 quinta secundi Euclidis. Ma il .bg. in .ga. è commo .bg. in .gd. e del .bg. in .gd. ne perviene l’ a-<lb></lb> |
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2674 rea, cioé .48. Adonca .bg. in .ga. è .48. A’ quali, agiongnendo el quadrato dela linea .gc., fa-<lb></lb> |
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2675 rá .49. Adonca il quadrato del .gc. è .1o. Del quale la radici è uno, che tanto è la linea .gc. Adon-<lb></lb> |
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2676 ca .bc. è .8., che è igual al lato .gd. per lo magiore lato. |
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2677 </p> |
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2678 <p class="main"> |
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2679 A ncora, se l’ area è .48., il lato magiore agionga sopra il minore .2., adimandase quan-<lb></lb> |
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2680 to é il lato magiore è quanto il minore. Togli la mitá di .2., che è uno, e in sé multi-<lb></lb> |
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2681 plica, fanno .1o. Agiongni a .48., fanno .49., la cui radici é .7., che, al detto .1o., ch’ é mi-<lb></lb> |
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2682 tá de .2., agionto, fanno .8. per lo lato magiore e .6. per lo minore, che ancora il pos-<lb></lb> |
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2683 siamo comprendere nela figura qui scritta. Sia il magiore lato commo dicemmo .gd., che è igua-<lb></lb> |
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2684 le ala retta .bc. e .ga. E tolgase dala retta .ga. la retta .gc., che sia .2., dove la retta .ac. sia iguale<lb></lb> |
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2685 ala retta .bg. Dividase adonca la retta .gc. in .2. parti iguali sopra il ponto .f. Sia la retta .af.<lb></lb> |
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2686 iguale ala retta .bc. Adonca la retta .ab. è divisa, in .2. parti iguali, sopra .f., in .2. parti non igua-<lb></lb> |
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2687 li, sopra il ponto .g. Dove, per la .5a. del .2o. de Euclide, la multiplicatione del .ag. in .gb., col qua-<lb></lb> |
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2688 drato dela linea .gf., è iguale al quadrato dela linea .af. Ma .bg. in .ga. è .48. E il quadrato<lb></lb> |
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2689 del .gf. è uno, adonca .bg. in .ga., col quadrato dela linea .gf., è .49. De’ quali la radici è .7. per<lb></lb> |
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2690 la linea .af. Al qual, agionto .fg., ch’ é uno, sia tutta .ag.8., che è il lato magiore. Over, se del .fb.,<lb></lb> |
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2691 che è .7., si trae .fg., che è uno, rimane .bg.6. per lo lato minore. Over altramente poni el la-<lb></lb> |
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2692 to brieve una cosa, sará el lato magiore una cosa e .2. E, perché a multiplicare el magiore la-<lb></lb> |
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2693 to per lo minore fanno .48., adonca, a multiplicare una cosa è .2., fanno similmente .48. E, a<lb></lb> |
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2694 multiplicare una cosa in una cosa e .2. fanno uno censo e .2. cose. E questo è iguale a .48. Do-<lb></lb> |
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2695 ve opra secondo la regola del’ algebra: harai la cosa valer .6., dove il lato minore è .6. e il ma-<lb></lb> |
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2696 giore è .8., commo volavamo. |
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2697 </p> |
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2698 <p class="main"> |
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2699 A ncora e gli é un quadrilatero parte altera longiore del quale l’ area è .48. e il diame-<lb></lb> |
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2700 tro è .10., adimandase quanto è il lato suo. Sopra il quadrato del diametro agion-<lb></lb> |
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2701 gni el doppio del’ area, cioé sopra .100. agiongni .96., fanno .196. Del quale la radi-<lb></lb> |
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2702 ci è .14., per l’ agiontione d’ amendoi e llati. Verbi gratia. Sia il quadrilatero par-<lb></lb> |
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2703 te altera longiore .abgd. Del quale il diametro .ag. è .10. E menise la retta .ab. infino al ponto<lb></lb> |
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2704 .e. E sia .be. iguali al .bg. E, perché la retta .ae. è divisa in .2. parti sopra il ponto .b., fienno e .2.<lb></lb> |
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2705 quadrati dele parti .ab. e .be., col doppio del .ab. in .bc., iguali al quadrato di tutta .ae., per la<lb></lb> |
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2706 quarta del .2o. de Euclide. Ma la parte .bc. è iguale al lato .bg. Adonca e quadrati dele li-<lb></lb> |
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2707 nee .ab. e .bg. è iguale al quadrato di tuta .ae. Ma e quadrati de’ lati .ab. e .bg. fanno quan-<lb></lb> |
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2708 to el quadrato del diametro. E il doppio del .ab. in .bg. fa il doppio del’ area, cioé .96. che, con<lb></lb> |
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2709 .100. agionti, fanno .196. per lo quadrato dela linea .ae. Adonca .ae. è la radici di .196., che è<lb></lb> |
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2710 .14., commo dicemmo. Dapoi, a trovare ciascun lato per sé, fa commo dicemmo di sopra: cioé dove<lb></lb> |
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2711 dicemmo l’ area è .48. e il lato magiore col minore fanno .14. |
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2712 </p> |
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2713 <p class="main"> |
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2714 Ancora e gli é un quadrilatero parte altera longiore del quale il diametro è .10. e,<lb></lb> |
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2715 ragionti, uno de’ lati minori con un de’ lati magiori, fanno .14. Adimando quanto<lb></lb> |
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2716 è l’ area e quanto è ciascuno lato. Multiplica .14. in sé, fanno .196., del quale togli el<lb></lb> |
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2717 quadrato del diametro, cioé .100., rimangano .96., de’ quali la mitá è .48. per l’ area<lb></lb> |
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2718 del ditto quadrilatero. E diremo ora e gli é un quadrilatero del quale l’ area è .48. E il diame-<lb></lb> |
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2719 tro è .10., che, commo di sopra dicemmo, farai. E harai il lato minore .6. e il magiore .8. Che ancora<lb></lb> |
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2720 nella sopraditta figura con quest’ ordine. El quadrato del diametro .ag. è iguale a’ .2. quadra-<lb></lb> |
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2721 ti de’ .2. lati .ab. e .bg., cioé a’ quadrati delle parti .ab. e .bc. Ma gli quadrati dele parti .ab. e<lb></lb> |
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2722 .be., col doppio del .ab. in .be., è iguale al quadrato dela linea .ae. Adonca il quadrato del<lb></lb> |
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2723 diametro .ag., col doppio del .ab. in .bc., è iguale al quadrato .ae. Onde, se si togli el quadra-<lb></lb> |
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2724 to del diametro, cioé .100., del quadrato dela linea .ac., cioé di .196., rimarranno .96. per lo doppio<lb></lb> |
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2725 <lb></lb> |
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2726 <pb></pb></p> |
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2727 <p class="folio"> |
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2728 folio 18v |
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2729 </p> |
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2730 <p class="main"> |
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2731 <lb></lb> |
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2732 </p> |
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2733 <p class="runhead"> |
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2734 Distinctio tertia. Capitulum primum. |
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2735 </p> |
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2736 <p class="main"> |
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2737 <lb></lb> |
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2738 del .ab. in .be., cioé per lo doppio del .ab. in .bg., adonca .ab. in .bg. fanno .48., cioé la mitá di<lb></lb> |
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2739 .96. Ma .ab. in .bg. fa l’ area del ditto quadrilatero. Adonca l’ area è .48., commo dicemmo, dove, vo-<lb></lb> |
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2740 lendo e lati, farai per lo modo dato e harai il magiore .8. e il minore .6. |
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2741 </p> |
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2742 <p class="main"> |
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2743 Ancora e gli é un quadrilatero parte altera longiore. Del quale il diametro agion-<lb></lb> |
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2744 to al lato minore fanno .16. E l’ altro lato è .8. Adimandase quanto è il lato mino-<lb></lb> |
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2745 re e quanto il diametro. Multiplica .16. in sé, fanno .256., del quale togli el quadra-<lb></lb> |
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2746 to del dato lato, cioé .64., rimangano .192. El quale dividi per lo doppio di .16., cioé<lb></lb> |
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2747 per .32., vienne .6. che è il lato che è congionto col diametro. Adonca il diametro è .10., cioé trat-<lb></lb> |
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2748 to .6. del .16. Verbi grattia. Sia il quadrilatero preditto .bgdc. E sia il diametro .bd. col la-<lb></lb> |
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2749 to .bg.16. E il lato .dg. sia .8. Poni adonca il lato .bg. una cosa, rimarrá il quadrato .dg.16. me-<lb></lb> |
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2750 no una cosa, che lo multiplica in sé e haremo .256. e uno censo meno .32.cose. per lo quadra-<lb></lb> |
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2751 to del diametro .bd. Ma il quadrato del .bd. è iguale a’ .2. quadrati .bg.e .gd., che sonno e la-<lb></lb> |
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2752 ti del ditto quadrilatero. Ove ristora le parti dando a ogni parte .32. radici. Haremo che<lb></lb> |
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2753 gli .2. quadrati de’ .2. lati .dg. e .bg., con .32. radici, sonno quanto .256. e uno censo. Dove multiplica<lb></lb> |
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2754 il lato .dg. in sé, che è .8., fanno .64. E multiplica il lato .bg. in sé, fa uno censo e harai uno censo<lb></lb> |
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2755 e .64. e .32.cose. iguali a uno censo è .256. Ove ristora le parti levando a ogni parte uno cen-<lb></lb> |
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2756 so e .64. Haremo .192. iguali a .32. cose. Ove la cosa vale .6. E tanto è il lato .bg. che, tratto di<lb></lb> |
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2757 .16., rimangano .10. per lo diametro .db. |
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2758 </p> |
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2759 <p class="main"> |
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2760 Ancora e gli é un quadrilatero parte altera longiore che ’l suo diametro è .4. piú<lb></lb> |
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2761 che llato minore. E il lato magiore é .8. Adimando quanto è il lato minore e quan-<lb></lb> |
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2762 to il magiore. Multiplica .8. in sé, fanno .64. A’ quali agiogni lo quadrato del .4.,<lb></lb> |
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2763 fanno .80. E radoppia el detto .4., sonno .8., per li quali dividi .80., vienne .10. per la<lb></lb> |
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2764 quantitá del diametro. Del quale trai .4., rimangano .6. per lo lato minore. E questo modo è<lb></lb> |
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2765 trovato, quando reducesse questa questione a una dele .6. regole del’ algebra. In questo mo-<lb></lb> |
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2766 do prima è manifesto che gli .2. quadrati, cioé quello del lato minore e quello del lato ma-<lb></lb> |
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2767 giore, insiemi agionti, sonno iguali al quadrato del diametro. Onde poni il diametro una co-<lb></lb> |
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2768 sa che in sé multiplica, sará el quadrato del diametro. E multiplica il lato minore in sé, che è<lb></lb> |
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2769 una cosa men .4., fanno uno censo e .16. men .8. cose che, agiontovi il quadrato del lato minore,<lb></lb> |
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2770 haremo uno censo e .80. men .8.cose. E questo sonno iguali a uno censo. Ove raguaglia le par-<lb></lb> |
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2771 ti: haremo che .8.cose. sonno iguali a .80., dove la cosa vale .10. E tanto è il diametro e il lato<lb></lb> |
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2772 minore sia .6. |
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2773 </p> |
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2774 <p class="main"> |
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2775 Ancora e gli é un quadrilatero parte altera longiore che, multiplicato il lato magiore<lb></lb> |
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2776 per l’ area sua, fanno .384. E il lato minore è .6. Adimandase quanto è il lato magio-<lb></lb> |
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2777 re. E gli é certa cosa che l’ area è la multiplicatione del minore lato nel magiore.<lb></lb> |
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2778 Onde, multiplicando el magiore lato per l’ area, é quanto multiplicare el mino-<lb></lb> |
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2779 re lato per lo quadrato del magiore, cioé la multiplicatione del minore lato per lo magiore<lb></lb> |
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2780 e tutto per lo magiore è quanto la multiplicatione del magiore lato nel magiore e tutto nel<lb></lb> |
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2781 minore. Onde, dividendo .384. in .6., ne viene .64. per la multiplicatione del lato magiore in<lb></lb> |
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2782 sé. Adonca il lato magiore fu .8. |
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2783 </p> |
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2784 <p class="main"> |
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2785 Ancora e gli é uno quadrilatero parte altera longiore. Del quale l’ area è .48. E, di-<lb></lb> |
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2786 viso lo lato magiore per lo minore, ne viene .1 1/3. Adimandase quanto è il lato mino-<lb></lb> |
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2787 re e quanto el magiore. Poni il lato brieve .3. per .1/3., che è denominato da quello e<lb></lb> |
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2788 multiplica per .1 1/3., fanno .4., che lo poni per lo lato magiore. Onde in che propor-<lb></lb> |
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2789 tione è .3. a .4., in medesima proportione è lo lato minore al lato magiore. Onde multiplica-<lb></lb> |
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2790 rai .3. per .4., fanno .12., per gli quali dividi .48., vienne .4., la cui radici è .2. e multiplicalo per la<lb></lb> |
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2791 medesimo .3. e .4. posta, fanno .6. e .8. per ciascuno lato, cioé .6. per lo minore e .8. per lo magio-<lb></lb> |
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2792 re. Over multiplica .3. per .48. e dividi per .4. e haremo .36. De’ quali la radici è il lato minore.<lb></lb> |
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2793 E, per lo magiore, multiplica .4. per .48. e dividi per .3., vienne .64. per lo quadrato delo lato<lb></lb> |
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2794 magiore, adonca il lato magiore è .8. E, se a partire il lato minore per lo lato magiore ne vien<lb></lb> |
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2795 .3/4., porrai per lo lato magiore .4., che denomina .3/4., e per lo minore .3. e dapoi segui commo è ditto.<lb></lb> |
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2796 Ancora e gli é un quadrilatero parte altera longiore del quale il lato minore col<lb></lb> |
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2797 diametro è .16. E il lato magiore è piú .2. che ’l lato minore. Adimando quanto è<lb></lb> |
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2798 ciascuno lato e quanto è il diametro. Dirai, se si ragiogni il lato magiore col dia-<lb></lb> |
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2799 metro, fanno .18. Onde multiplica .16. in sé e .18. in sé e haremo .256. e .324. E agion-<lb></lb> |
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2800 gni le ditte multiplicationi insiemi, fanno .580., del quale togli il quadrato de .2., rimangano .576. |
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2801 </p> |
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2802 <p class="main"> |
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2803 De’ quali la radici è la summa de doi lati e del diametro. Del quale togli il diametro e<lb></lb> |
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2804 <lb></lb> |
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2805 <pb></pb></p> |
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2806 <p class="folio"> |
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2807 folio 19r |
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2808 </p> |
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2809 <p class="main"> |
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2810 <lb></lb> |
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2811 </p> |
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2812 <p class="runhead"> |
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2813 Distinctio tertia. Capitulum primum. 19 |
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2814 </p> |
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2815 <p class="main"> |
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2816 <lb></lb> |
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2817 il lato brieve, cioé .16., rimangano .8. per lo lato magiore. E, se voi il diametro, trai .8. di .18., ri-<lb></lb> |
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2818 mangano. 10. per lo diametro. E il minore lato sia .6. Onde, se al computamento del’ alge-<lb></lb> |
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2819 bra la voi redurre, poni il lato minore una cosa, rimarrá il diametro .16. meno una cosa. Mul-<lb></lb> |
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2820 tiplica adonca una cosa per una cosa, fanno uno censo. El quale agiongni al quadrato del la-<lb></lb> |
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2821 to magiore, cioé al quadrato d’ una cosa e .2., che è uno censo e .4.cose. e .4., per lo quadrato de-<lb></lb> |
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2822 lo lato magiore, che, con .1. censo agionto, fanno .2.censi. e .4.cose. e .4. E questo è iguale al qua-<lb></lb> |
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2823 drato del diametro, cioé a .16. meno una cosa, che è .256. e .1. censo men .32. cose. Restora le par-<lb></lb> |
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2824 ti dando a ogni parte .32.cose. e levando da ogni parte .4. per numero e .1. censo. E haremo<lb></lb> |
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2825 che uno censo e .36.cose. sonno iguali a .252. Dimezza le cose, sonno .18., in sé multiplica, fanno<lb></lb> |
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2826 .324., poni sopra .252., fanno .576. la cui radici è .24., dela quale tra’ .18., rimangano .6. per lo la-<lb></lb> |
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2827 to minore. E il lato magiore sia .8. e il diametro sia .10. |
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2828 </p> |
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2829 <p class="main"> |
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2830 Ancora .2. lati che l’ area fanno .62. e il lato magiore agionga al minore .2. Adiman-<lb></lb> |
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2831 do quanto è ciascuno lato del quadrilatero parte altera longiore. El modo a que-<lb></lb> |
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2832 sto trovare è questo questo. Che traga .2. di .62., rimangano .60. e il .2., il quale è la mitá de’ <lb></lb> |
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2833 .4. lati, cioé del numero de’ .4. lati, in sé multiplica, fanno .4. Agiongni a .60., fanno<lb></lb> |
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2834 .64., la cui radici è .8., sia il lato magiore. O, se voi il lato minore, tra’ .2. di .8., rimangano .6. per lo<lb></lb> |
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2835 lato mimore. E prociede questo modo ditto di questa regola. Poni el lato minore una cosa<lb></lb> |
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2836 e il lato magiore sia una cosa e .2. E, a multiplicare il lato breve per lo lato magiore, fanno .1o.<lb></lb> |
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2837 censo .2. cose. E questo è l’ area. Alla qual summa, agiongnendo e lati, cioé .2.cose. e .2., fanno .1.<lb></lb> |
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2838 censo e .4.cose. e .2. iguali a .62. Ove tra’ .2. di ciascuna parte, harai .1. censo e .4.cose. iguali a .60. |
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2839 </p> |
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2840 <p class="main"> |
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2841 Dove dimezza le cose e in sé multiplica e agiongni a .60. Haremo .64. La cui radici è .8. che, tra-<lb></lb> |
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2842 tone .2., rimangon .6. per lo lato minore et cetera. |
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2843 </p> |
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2844 <p class="main"> |
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2845 Ancora e gli é uno quadrilatero parte altera longiore del quale, multiplicato il dia-<lb></lb> |
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2846 metro per lo lato magiore, fanno .80. E il lato breve è .6. Adimandase quanto è cia-<lb></lb> |
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2847 scun lato, cioé quanto è il lato magiore. Multiplica .80. in sé, fanno .6400. e .6. in sé,<lb></lb> |
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2848 fanno .36.; togli la mitá de .36., che è .18., del quale il quadrato è .324. Ragiongni<lb></lb> |
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2849 a .6400., fanno .6724., de’ quali la radici è .82. che, con la mitá del .36., fanno .100. De’ quali la ra-<lb></lb> |
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2850 dici, cioé .10., è il diametro. Per gli quali .10. dividi .80., vienne .8., che sonno il lato magiore. O-<lb></lb> |
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2851 vero delo .82. tra’ .18., rimarrá .64., de’ quali la radici è il lato magiore, che è .8. E, se ala computa-<lb></lb> |
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2852 tione del’ algebra la voi ridure, perché dela multiplicatione del diametro nel magiore lato<lb></lb> |
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2853 pervienne .80., adonca, a multiplicare e quadrati loro l’ uno contro l’ altro fanno el quadrato<lb></lb> |
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2854 del .80., cioé .6400. Ma il quadrato del diametro è iguale a .2. quadrati: cioé al quadrato del<lb></lb> |
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2855 lato minore e al quadrato del lato magiore. E il quadrato del lato minore è .36., adonca, a mul-<lb></lb> |
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2856 tiplicare il quadrato del magiore lato in sé e in .36., fanno .6400. Onde poni el quadrato del<lb></lb> |
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2857 magiore lato una cosa che, in sé multiplicato e in .36., fanno .36.cose. e uno censo, che sonno igua-<lb></lb> |
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2858 li a .6400., dove opra secondo la regola: cioé dimezza le cose, sonno .18., multiplica in sé, fanno<lb></lb> |
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2859 .324., pon sopra .6400., fanno .6724. Del quale la radici è .82., trane .18., rimangano .64., del quale<lb></lb> |
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2860 la radici è il lato magiore, cioé .8. imperoché si pose il quadrato essere una cosa, cioé il quadra-<lb></lb> |
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2861 to del magiore lato. |
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2862 </p> |
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2863 <p class="main"> |
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2864 Per la qual regola faresti, dicendo multiplicato el minore lato del quadrilatero<lb></lb> |
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2865 parte altera longiore per lo diametro fanno .60. e il lato magiore è .8. Che haresti<lb></lb> |
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2866 che uno censo e .64. radici sonno iguali a .3600., che multiplicaresti la mitá di .32.<lb></lb> |
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2867 in sé e agiongni a .3600. e haresti .4624., la cui radici è .68., tratone .32., rimanga-<lb></lb> |
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2868 no .36., la cui radici è .6. per lo lato minore et cetera. |
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2869 </p> |
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2870 <p class="main"> |
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2871 Ancora e gli é uno quadrilatero parte altera longiore che ’l lato minore, con la sua<lb></lb> |
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2872 area, è .54. E il lato magiore è .2. piú che ’l minore. Adimandase quanto è ciascun<lb></lb> |
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2873 lato. Perché a multiplicare el magiore lato per lo minore ne perviene l’ area sua,<lb></lb> |
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2874 se porremo il lato brieve una cosa, fará il lato magiore una cosa e .2. che, multi-<lb></lb> |
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2875 plicato per una cosa, cioé per lo lato minore, fanno .1. censo e .2.cose. per l’ area, a’ quali, agionto<lb></lb> |
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2876 una cosa, cioé il lato minore, haremo .1. censo .3.cose. iguali a .54. Dove multiplica .1 1/2., cioé la mi-<lb></lb> |
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2877 tá dele cose, in sé, fanno .2 1/4., agiongni a .54., fanno .56 1/4. la cui radici è .7 1/2. che, tratone .1 1/2., riman-<lb></lb> |
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2878 gano .6. per lo lato minore e il magiore è .8. |
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2879 </p> |
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2880 <p class="main"> |
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2881 E dicendo e gli é un quadrilatero parte longiore del quale il lato magiore con<lb></lb> |
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2882 l’ area è .56. E il lato minore è .2. meno che ’l magiore. Adimando quanto é ciascun<lb></lb> |
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2883 lato. Poni il lato magiore una cosa, sia il minore una cosa meno .2. Multiplica il<lb></lb> |
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2884 magiore per lo minore, fanno uno censo men .2. cose. Agiongni il lato magiore, cioé<lb></lb> |
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2885 <lb></lb> |
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2886 <pb></pb></p> |
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2887 <p class="folio"> |
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2888 folio 19v |
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2889 </p> |
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2890 <p class="main"> |
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2891 <lb></lb> |
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2892 </p> |
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2893 <p class="runhead"> |
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2894 Distinctio tertia. Capitulum primum. |
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2895 </p> |
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2896 <p class="main"> |
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2897 <lb></lb> |
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2898 una cosa, fanno .1. censo meno una cosa: per lo lato iguale a .56. Dimezza una cosa e .1/2., multipli-<lb></lb> |
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2899 ca in sé, fa .1/4., poni sopra .56., fanno .56 1/4. la cui radici è .7 1/2., poni su .1/2, fanno .8. per lo lato magio-<lb></lb> |
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2900 re e il lato minore sia .6. |
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2901 </p> |
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2902 <p class="main"> |
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2903 E dicendo e gli é uno quadrilatero parte altera longiore del quale ’.4. suoi lati, con l’ a-<lb></lb> |
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2904 rea sonno .76. E il lato magiore avanza al lato minore .2. Adimando quanto è cia-<lb></lb> |
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2905 scun lato. Poni il lato minore una cosa, sia il lato magiore una cosa e .2. che multi-<lb></lb> |
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2906 plica una cosa via una cosa e .2., fanno un censo e .2. cose. Agiongnivi li .4. lati,<lb></lb> |
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2907 cioé .4.cose. e .4., fanno un censo, .6.cose. e .4., che sonno iguali a .76. Dove da ogni parte leva .4.,<lb></lb> |
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2908 rimarrá un censo e .6.cose. iguali a .72. Dimenzza le cose, sonno .3., multiplica in sé, fanno .9., poni<lb></lb> |
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2909 sopra .72., fanno .81. del quale la radici è .9. che, trattone .3., rimangano .6. per lo lato minore e<lb></lb> |
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2910 il magiore è .8. |
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2911 </p> |
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2912 <p class="main"> |
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2913 Ancora e gli é un quadrilatero parte altera longiore del quale, tratto del’ area sua<lb></lb> |
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2914 el minore lato, rimane .42. E avanza el magiore lato al minore in .2. Adimando<lb></lb> |
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2915 quanto è il minore lato e quanto il magiore. Poni il lato minore una cosa, si-<lb></lb> |
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2916 rá il magiore una cosa e .2. Dove multiplica una cosa via una cosa e .2., cioé il mi-<lb></lb> |
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2917 nore lato per lo magiore, fanno un censo e .2.cose. e tanto è la sua area. Dela quale trai il mi-<lb></lb> |
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2918 nore lato, cioé una cosa. Rimangano un censo e una cosa. E questo è iguale a .42. Dimez-<lb></lb> |
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2919 za le cose e multiplica in sé e pon sopra .42. Harai .42 1/4. La cui radici è .6 1/2., trane .1/2., rimanga-<lb></lb> |
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2920 no .6. per lo lato minore e .8. sia il magiore. |
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2921 </p> |
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2922 <p class="main"> |
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2923 Ancora e gli é un quadrilatero parte altera longiore del quale, tratto del’ area il la-<lb></lb> |
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2924 to magiore, riman .40. E il lato magiore avanza al minore .2. Adimando quan-<lb></lb> |
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2925 to è ciascun lato. Dirai, se a trare il magiore riman .40., a trarne il minore rimar-<lb></lb> |
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2926 rá .42., dove opera commo nela passata e harai il lato minore .6. e il magiore .8. |
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2927 </p> |
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2928 <p class="main"> |
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2929 Ancora e gli é uno quadrilatero parte altera longiore del quale, tratto ’.4. suoi lati de-<lb></lb> |
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2930 la sua area, rimangano .20. Et il magiore lato avanza al minore .2. Adimando quan-<lb></lb> |
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2931 to è ciascun lato. Poni el lato minore una cosa, sia il magiore una cosa e .2. E<lb></lb> |
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2932 multiplica il lato magiore per lo minore, cioé una cosa per una cosa .2., fanno un<lb></lb> |
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2933 censo e .2. cose. E tanto è l’ area dela quale tra’ .4.cose. e .4., cioé e .4. suoi lati, rimangano un cen-<lb></lb> |
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2934 so men .2.cose. e .4. E questo è iguale a .20. Dove raguaglia le parti dando a ogni parte .2. cose<lb></lb> |
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2935 e .4. E haremo che un censo sonno iguali a .24. e .2. cose. Dove dimezza le cose e multiplica in<lb></lb> |
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2936 sé e poni sopra a .24. e dela summa piglia la radici e agiongnivi la mitá dele cose e haremo<lb></lb> |
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2937 .6. per lo lato minore e .8. per lo lato magiore. |
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2938 </p> |
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2939 <p class="main"> |
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2940 Ancora e gli é un quadrilatero parte altera longiore. Del quale, agionto un lato mi-<lb></lb> |
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2941 nore con un lato magiore e il diametro, fanno .24. E l’ area sua è .48. Adimanda-<lb></lb> |
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2942 se quanto é ciascuno lato. Multiplica .24. in sé fanno .576. Del quale tra’ el do-<lb></lb> |
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2943 pio del’ area sua, cioé .96., rimangano .480. Del quale piglia la mittá che è .240. e<lb></lb> |
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2944 partilo per la summa de’ .2. lati e del diametro, cioé per .24., vienne .10. per lo diametro che, tratto di<lb></lb> |
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2945 .24., rimangano .14. per li lati. Onde comprenderai che il minore lato è .6. e il magiore è .8., imperoché<lb></lb> |
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|
2946 dirai la summa è .14., cioé è de’ .2. lati e l’ area è .48. Dove, comme habiamo ditto, farai e harai<lb></lb> |
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2947 il proposito. E donde questo modo procieda voi cognoscere. Sia la retta .ab.24., cioé la summa de’ <lb></lb> |
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2948 .4. lati. e del diametro. E sia .ac. iguale al magiore lato del dato quadrilatero parte altera longio-<lb></lb> |
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2949 re e .cd. sia iguale al minore lato. Rimarrá adonca .db. iguali al diametro. Faciase adonca, so-<lb></lb> |
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2950 pra la retta .ab., il tetragono .ae. E menise el diametro .fb. e per gli ponti .c. e .d. si menino le rette<lb></lb> |
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2951 .cg. e .dh. equedistanti alle rette .af. e .be. e per gli ponti .i.k. si menino le rette .lim. e .npko. E, per-<lb></lb> |
|
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2952 ché e gli é tetragono la figura .ae., fienno ancora tetragoni le figur scritte intorno al diametro<lb></lb> |
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2953 .fb., per lo corelario dela .4a. del .2o. Adonca tetragono è .kdbo. e .knfh. Ancora, perché tetra-<lb></lb> |
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2954 gono è la figura quadrilatera .nh., fienno ancora tetragoni .pimk. e .ilfg. ed é il lato del tetra-<lb></lb> |
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2955 gono .do. la retta .db. Onde il tetragono .do. è iguale al quadrato del diametro. E il tetragono<lb></lb> |
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2956 .pm. il suo lato è la retta .pk., che è iguale ala retta .cd. e .cd. è iguale al minore lato. Adonca<lb></lb> |
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2957 il tetragono .pm. è il quadrato del minore lato. E il tetragono .lg. è iguale al quadrato del magio-<lb></lb> |
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2958 re lato. Perché la retta .li. è iguale ala retta .ac. e .ac. è fatta iguale al magiore lato del quadri-<lb></lb> |
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2959 latero parte altera longiore del quale parliamo. E, perché .pm. è tetragono, sará la retta .kp. iguale<lb></lb> |
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2960 ala retta .pi., adonca .pi. è iguale al minore lato e .il. è iguale al lato magiore. Adonca il supplemen-<lb></lb> |
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2961 to .ni. è iguale al’ area del dato quadrilatero. E il supplemento .ih. è iguale al supplemento .ni., com-<lb></lb> |
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2962 mo mostra Euclide nel primo per la .43a. Adonca li supplementi .ni. e .ih. sonno doppi al’ area del da-<lb></lb> |
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2963 to quadrilatero parte altera longiore. E quali supplementi sonno .96. che, tratti del’ area del ditto<lb></lb> |
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2964 <lb></lb> |
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2965 <pb></pb></p> |
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2966 <p class="folio"> |
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2967 folio 20r |
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2968 </p> |
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2969 <p class="main"> |
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2970 <lb></lb> |
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2971 </p> |
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2972 <p class="runhead"> |
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2973 Distinctio tertia. Capitulum tertium. 20 |
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2974 </p> |
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2975 <p class="main"> |
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2976 <lb></lb> |
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2977 tetragono, cioé de .576., rimangano .480., cioé il tetragono .lg. e .pm. e il gnomone .aoh. Ma<lb></lb> |
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2978 il tetragono .lg. col tetragono .pm. sonno iguali al tetragono .do. Imperoché gli quadra-<lb></lb> |
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2979 ti degli .2. lati del quadrilatero parte altera longiore sonno quanto il quadrato del diametro.<lb></lb> |
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2980 Adonca il tetragono .do. con lo gnomone .aoh. sonno .480. Ma la superficie .ao. è la mi-<lb></lb> |
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2981 tá del tetragono .do. e del gnomone .aoh., che chiaro appare, imperoché la mitá del gnomo-<lb></lb> |
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2982 ne .aoh. e lo quadrilatero .nk.ba. è la mitá del tetragono .do. e il triangolo .kbo. E, agionto il<lb></lb> |
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2983 triangolo .kbo. al quadrilatero .nkba., fanno la superficie .ao. La quale è fatta del .ab. in .an., cioé<lb></lb> |
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2984 .db. che è iguale al diametro. Adonca partendo .240. per .24., che è .ab., vienne .10. per lo lato del te-<lb></lb> |
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2985 tragono .do. Adonca .do. è per lato .10. E peró il diametro è .10. e questo era bisogno mostrare.<lb></lb> |
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2986 Ancora e gli é uno quadrilatero parte altera longiore che, agionto e .2. lati, cioé<lb></lb> |
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2987 uno minore e uno magiore con lo diametro, fanno .24. e il lato magiore è piú che ’l<lb></lb> |
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2988 minore .2. Adimandasse quanto è ciascuno lato. Radopia il quadrato di .24.,<lb></lb> |
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2989 cioé .576., fanno .1152. sopra il quale agiongni il quadrato dela soprabundantia del<lb></lb> |
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2990 magiore lato al minore, cioé il quadrato di .2., fanno .1156. De’ quali la radici, cioé .34., tranne<lb></lb> |
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2991 .24., rimangono .10. che sonno il diametro. Dove infino in .24. v’ é .14. per gli .2. lati. Del quale<lb></lb> |
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2992 .14. tranne .2., rimangano .12., de’ quali la mitá è .6. per lo lato minore. E .8. per lo lato magio-<lb></lb> |
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2993 re. E, volendo mostrare la regola di tale dimostratione, sia uno tetragono .abcd. havente per<lb></lb> |
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2994 ciascun lato la quantitá degli lati e del diametro. E sia .be. iguale al magiore lato et .ef. sia igua-<lb></lb> |
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2995 le al minore. Rimarrá .fc. iguale al diametro e menise il diametro .ac. e per lo ponto .f. si meni<lb></lb> |
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2996 la linea .fgh. equedistanti a ciascuna dele linee .ba. e .cd. E per lo ponto .g. si meni la linea<lb></lb> |
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2997 .igk. equedistante ala retta .bc. e sia la retta .ig. equedistante e iguale e ala retta .bf. e tolghi-<lb></lb> |
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2998 se dala retta .gi. la retta .gl. che sia iguale ala retta .fe. Rimarrá adonca .li. iguale ala retta<lb></lb> |
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2999 .eb. Adonca .gl. è iguale al minore lato e .li. al magiore lato. E tolghise dal .li. la retta .im.<lb></lb> |
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3000 (cioé quello in che il magior lato avanza el minore) rimarrá .lm. igual al .lg., cioé al minor<lb></lb> |
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3001 lato. E menise .ad. nel ponto .n. E sia .dn. iguale al .fc., cioé al diametro. E faciasi sopra<lb></lb> |
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3002 .an. el tetragono .nopa. e, perché e sonno tetragoni .bd. e .pn. e sonno intorno a uno angolo,<lb></lb> |
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3003 cioé al .a. conn’ un diametro fienno. Adonca, menato il diametro .ac. nel ponto .o., sará la ret-<lb></lb> |
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3004 ta .ao. diametro. E menise la retta .dc. infino al .q. e la retta .bc. infino al .r. Onde .qr. sia<lb></lb> |
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3005 tetragono e contiene in sé il quadrato del diametro del dato quadrilatero parte altera<lb></lb> |
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3006 longiore detto. E tutto il tetragono .pn. è iguale al tetragono .bd. e al gnomone .bod.<lb></lb> |
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3007 E, questo inteso, lo mostraró commo lo gnomon .bod. è iguale al tetragono .bd. e al tetrago-<lb></lb> |
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3008 no fatto dela linea .im., che è quello che ’l magiore lato avanza al minore. Sonno certamente li<lb></lb> |
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3009 supplementi .pc. e .cn. iguali ale superficie .bk. e .fd. Ma le .2. superficie .bk. e .fd. sonno iguali al<lb></lb> |
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3010 gnomone .ich. e al tetragono .fk. a’ quali, agionto lo iguale del tetragono .qr., che è iguale al te-<lb></lb> |
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3011 tragono .fk., sirá il doppio del tetragono .fk., col gnomone .ich. iguale al gnomone .bod.<lb></lb> |
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3012 Restaci a mostrare el doppio del tetragono .fk. essere quanto il tetragono .ih. e al tetrago-<lb></lb> |
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3013 no fatto del .im. É certamente .gm. diviso in .2. parti al ponto .l., ala quale per lo diritto á agion-<lb></lb> |
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3014 to .im. dove, per la .6a. del .2o. de Euclide, sará cosí il tetragono fatto dal .gi. con quel tetragono ch’ é<lb></lb> |
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3015 fatto dal .im. iguale al doppio del tetragono fatto dal .lg. Ma il tetragono .gi. è il tetrago-<lb></lb> |
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3016 no .ih. Adonca il tetragono .ch., col tetragono fatto dal .im., è questo il doppio del tetragono fat-<lb></lb> |
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3017 to dal .lg. E il tetragono fatto dal .lg. è quanto il tetragono .fk. E peró il tetragono .ih. col te-<lb></lb> |
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3018 tragono fatto dal .im. è quanto il doppio del tetragono .fk. Adonca lo gnomone .bod. è quan-<lb></lb> |
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3019 to il tetragono .bd. e il tetragono fatto dal .im. Ora veniamo ala cagione. Multiplicammo<lb></lb> |
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3020 di sopra .24. per .24. e havemmo .676. per lo .bd. dove lo radoppiammo, cioé agiognemovi altretan-<lb></lb> |
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3021 to, feci .1152. E, dapoi, v’ agiongneremo .4., cioé il tetragono .im. per lo tetragono .pn., del qual il la-<lb></lb> |
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3022 to è .34., cioé la radici de .1156. del quale, tratto .pq., rimasono .10., cioé .qo. ala quale è iguale<lb></lb> |
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3023 la retta .cf. Adonca .cf. è .10. commo dicemmo et cetera. |
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3024 </p> |
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3025 <p class="main"> |
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3026 E possiamo ancora altramente venire ala noticia de’ detti lati, cioé che poniamo il<lb></lb> |
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3027 lato brieve una cosa, sia il magiore una cosa .2. che, tratti di .24., rimane il diame-<lb></lb> |
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3028 tro .22. meno .2. cose. Dapoi multiplica il minore lato in sé, fanno uno censo. E il ma-<lb></lb> |
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3029 giore lato in sé fanno uno censo e .4.cose. e .4. Agiongni a uno censo: fanno .2.<lb></lb> |
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3030 censi e .4.cose. e .4. E questo è iguale al quadrato de il diametro, cioé al quadrato de .22. men<lb></lb> |
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3031 .22.cose., che è .484. e .4.censi. meno .88.cose., ove .484. e .4.censi. men .88.cose. sonno iguali a .2.<lb></lb> |
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3032 censi .4.cose. e .4. Raguaglia le parti dando a ciascuna .88. cose e levando da ciascuna .4. e<lb></lb> |
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3033 .2. censi. Haremo che .2.censi. e .480. sonno iguali a .92. cose. Parti ne’ censi e harai che uno<lb></lb> |
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3034 censo e .240. é iguale a .46. cose. Dimezza le cose e in sé multiplica e tranne .240., rimarrá .289.<lb></lb> |
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3035 <lb></lb> |
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3036 <pb></pb></p> |
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3037 <p class="folio"> |
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3038 folio 20v |
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3039 </p> |
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3040 <p class="main"> |
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3041 <lb></lb> |
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3042 </p> |
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3043 <p class="runhead"> |
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3044 Distinctio tertia. Capitulum secundum. |
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3045 </p> |
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3046 <p class="main"> |
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3047 <lb></lb> |
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3048 la cui radici è .17. che, tratta di .23., rimane .6. per lo lato minore.<lb></lb> |
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3049 Ancora e gli é uno quadrilatero parte altera longiore che ’l suo diametro è .2. piú<lb></lb> |
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3050 che ’l magiore lato. E il magiore lato è .2. piú che ’l minor. Adimandasse quanto è<lb></lb> |
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3051 ciascuno lato. Sempre, quando le soprabundancie sonno iguali, multiplicarai quel-<lb></lb> |
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3052 la soprabundantia per .5. e harai il diametro. E, se per .4., harai il magiore. E, se per .3.<lb></lb> |
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3053 harai il minore. Verbi gratia: in questo exemplo multiplica .2. per .5., fanno .10. e questo è il<lb></lb> |
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3054 diametro che, multiplicato ancora .2. per .4., fanno .8. per lo lato magiore e, multiplicato .2. per .3.,<lb></lb> |
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3055 fanno .6. per lo lato minore. E questo aviene, perché il quadrilatero parte altera longiore del qua-<lb></lb> |
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3056 le il lato magiore è .4. e il lato minore è .5., il diametro sia .3. e di questi lati la soprabundantia è .1o.<lb></lb> |
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3057 E peró comme .1o. è a quali voi soprabundantia: cosí questi .3. lati fienno agli lati di quello quadri-<lb></lb> |
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3058 latero parte altera longiore del quale sia data la soprabundantia. Comme se la soprabundan-<lb></lb> |
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3059 tia sia .3., perché .3. sonno .3. cotanti del .1o., cosí .3. cotanti fienno e lati de’ lati preditti, cioé il diame-<lb></lb> |
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3060 tro sia .15. e il magiore lato .12. e il minore .9. E se ’l magiore lato fosse .20., parti per .4., vienne .5. per<lb></lb> |
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3061 la soprabondantia. E, se ’l diametro fosse .20., la soprabondantia sia .5. e, se ’l minore lato è .20., sará la soprabundantia <lb></lb> |
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3062 .6 <lb></lb> |
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3063 2/3. et cetera. |
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3064 </p> |
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3065 <p class="main"> |
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3066 E dicendo e gli é uno quadrilatero parte altera longiore del quale l’ abundantia del<lb></lb> |
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3067 diametro al lato magiore è .1o. e l’ abundantia del magiore lato al minore è .7. Alo-<lb></lb> |
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3068 ra opererai per l’ algebra. E porremo il lato minore una cosa: sará il lato magiore .1a.<lb></lb> |
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3069 cosa .7. E il diametro sia una cosa e .8. E multiplica una cosa in sé, fanno .1. censo e<lb></lb> |
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3070 una cosa e .7. in sé, fanno un censo .14.cose. e .49. che, insiemi agionti, fanno .2.censi.14.cose. e<lb></lb> |
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3071 .49., che sonno iguali al quadrato del diametro, cioé ala multiplicatione d’ una cosa e .8. in sé. La<lb></lb> |
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3072 qual multiplicatione è uno censo e .16.cose.64. Togli adonca da ogni parte .1o. censo e .14.cose. e .49. |
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3073 </p> |
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3074 <p class="main"> |
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3075 Haremo .1o. censo iguale a .2.cose. e .15., dove dimezza le cose e uno in sé multiplica, fanno, con .15.16. del<lb></lb> |
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3076 quale la radici è .4. che, con .1o., sonno .5. per lo lato minore. E il magiore .12. E il diametro .13. et cetera.<lb></lb> |
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3077 Ancora questa: e gli é el tetragono .abde. el cui diametro .eb. e li .2. supplementi<lb></lb> |
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3078 .ai. e .id., che ciascuno è .7. piú longo che largo, cioé .ag. piú che .gi. over .dk. piú che<lb></lb> |
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3079 .ik. e li quadrati deli lor lati, gionti asiemi, fanno .169. o voi dire che ’l diametro de<lb></lb> |
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3080 ciascuno è .13. Dimandase che siano per ciascuna facia ditti supplementi. Poni per<lb></lb> |
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3081 algebra che .gi. over .ik. sia una cosa, donca .ag. sia una cosa piú .7. e cosí .dk. Ora quadra ciascuno<lb></lb> |
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3082 de’ ditti lati, multiplica .1.co. in sé, fará .1.ce. e poi multiplica .1.co. piú .7., che è la loro longhezza, pur in sé, fará<lb></lb> |
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3083 .1.ce. piú .14.co. piú .49. per lo quadrato .hf. e a questo giongnici .1.ce. per lo quadrato .gk., faran .2.ce. piú .14.co.<lb></lb> |
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3084 piú .49. E questa summa sirá equale a .169. over la .R. de questa summa sia iguale a .13., cioé al diametro<lb></lb> |
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3085 de ciascuno de’ supplementi. Ora, per venire ala valuta dela cosa, cava .49. che è el quadrato dela<lb></lb> |
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3086 superficie habundantia in che el magior lato excede el minore, resta .120. iguale a .2.ce. piú .14.co., parti<lb></lb> |
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3087 la equatione nela quantitá deli censi, che son .2., ne ven .60. per uno extremo dela ditta equatione che sia la<lb></lb> |
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3088 superficie over area d’ uno de’ ditti supplementi. E, per l’ altro extremo, sia .1.ce. piú .7.co. e sonno equali<lb></lb> |
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3089 a .60. Seque el capitulo smezzando le cose, che son .7., ne ven .3 1/2. e multiplicando l’ una de ditte mitá in sé, fa-<lb></lb> |
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3090 rá .12 1/4. e, sopra questo ponendo el numero, cioé .60., fará .72 1/4. la cui .R., meno el dimezzamento dele co-<lb></lb> |
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3091 se, vene a valere la cosa, cioé .R.72 1/4. men .3 1/2., che vol dire .5., perché la .R. de .72 1/4. è .8 1/2. che, trattone<lb></lb> |
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3092 .3 /12., resta .5. per la valuta dela cosa e tanto dirai che fosse largo ciascun de’ ditti supplementi e la longhe-<lb></lb> |
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3093 zza dirai che fo .R.72 1/4. piú .3 1/2., cioé .12. aponto per lo .ag. over .dk. e fie facta unde per lo tema tu hai<lb></lb> |
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3094 che .ag. è piú .7. che non è .gi. over .ik., perché .ag. è .12. e lo .gi. è .5. Sí che s[e] tu ben guardi el tema è <lb></lb> |
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3095 satisfacto, conciosiaché ’l quadrato delo .ag., che è .144., gionto con lo quadrato delo .gi., che è .25.,<lb></lb> |
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3096 fanno in tutto .169., commo se vole, la cui .R. ene .13. per lo diametro deli supplementi. E lo dia-<lb></lb> |
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3097 metro .eb. de tutto el tetragono rettangolo sia .R.578., perché .ei. è .R.288. e .ib.R.50. che, gion-<lb></lb> |
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3098 te insiemi, fanno .R.578. per tutta la diagonale .eb. E tu l’ altre simili per te farai et cetera.<lb></lb> |
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3099 <lb></lb> |
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3100 De dimensione rumborum seu helmuaym Capitulum secundum<lb></lb> |
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3101 Gli altri quadrilateri in .4. capitoli se dividono: nel primo sonno rombi, li quali hano e .4. lati<lb></lb> |
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3102 infra loro iguali, ma gli angoli non sonno retti. Sia adonca el rombo .abcd., havente cia-<lb></lb> |
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3103 scun lato .13. El quale a noi bisogna misurare. E, questo volendo fare, é di bisogno a noi<lb></lb> |
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3104 havere uno de’ diametri. E sia adonca uno de’ diametri suoi. E sia il diametro brieve .bd.10. e<lb></lb> |
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3105 sia adonca el ditto rombo diviso in .2. triangoli iguali de’ quali ciascun è triangolo equicrurio. Im-<lb></lb> |
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3106 peroché gli .2. lati del’ uno sonno iguali agli .2. lati del’ altro: cioé e .2. lati .ab. e .ad. del triangolo<lb></lb> |
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3107 .abd. sonno iguali a’ .2. lati del triangolo .bcd., cioé agli lati .bc. e .dc. E il lato .bd. è commune. E<lb></lb> |
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3108 peró e .2. triangoli sonno infra loro iguali. Adonca, se l’ area di questo rombo voi, radopierai l’ area<lb></lb> |
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3109 del triangolo .abd. over .bcd. e aremo il proposito. E l’ area del triangolo .abd. è fatta dela multiplica-<lb></lb> |
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3110 tione del cateto .ae. nella mitá dela linea che è basa .bd., commo nel trattato de’ triangoli demostra-<lb></lb> |
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3111 <lb></lb> |
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3112 <pb></pb></p> |
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3113 <p class="folio"> |
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3114 folio 21r |
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3115 </p> |
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3116 <p class="main"> |
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3117 <lb></lb> |
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3118 </p> |
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3119 <p class="runhead"> |
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3120 Distinctio tertia. Capitulum secundum. 21 |
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3121 </p> |
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3122 <p class="main"> |
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3123 <lb></lb> |
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3124 to fo. Onde, multiplicando el catetto .ae. in tutto .bd., virrá el doppio del triangolo .abd., cioé l’ a-<lb></lb> |
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3125 rea del rombo. La quale linea perpendiculare .ae. volendo, multiplica .ab. in sé, che fanno .169., e<lb></lb> |
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3126 tranne la multiplicatione del .be. in sé, cioé .25., remangano .144. La cui radici è .12. per la linea<lb></lb> |
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3127 .ae. E ancora il catetto .ce. è ancora .12. e sonno in una medesima linea .ae. Onde tutta .ac.<lb></lb> |
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3128 è .24. Adonca l’ area del detto rombo è fatta dela mitá del diametro .ac. in tutto il diametro<lb></lb> |
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3129 .bd., la quale multiplicatione è .120. E, similmente, quando fosse dato lo diametro .ac.24. e<lb></lb> |
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3130 vorrai il diametro .bd., per lo detto modo lo troverai, cioé, perché equicurij sonno li triango-<lb></lb> |
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3131 li .bac. e .dac., e sonno infra loro iguali, dove se dela potentia del lato .ba. traremo la poten-<lb></lb> |
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3132 tia dela linea .ae., cioé .144. de .169., rimarranno .25. per la potentia del catetto .be. Adonca .be. è<lb></lb> |
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3133 .5. dove tutta è .10. E, perché se multiplicaremo el catetto .be., cioé la mitá del diametro .bd.<lb></lb> |
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3134 in tutto il diametro .ac., haremo l’ area del rombo, perché multiplicando .be. nella mitá dela ba-<lb></lb> |
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3135 sa .ac. fa l’ area del triangolo .abc. E, multiplicando .be. in .ac., cioé .5. in .24., fanno .120. commo di-<lb></lb> |
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3136 cemmo. Adonca l’ area de ciascun rombo è fatta dela multiplicatione d’ un diametro nella mi-<lb></lb> |
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3137 tá del’ altro. E questa è regola universale a tutti.. E dicendo e gli é uno rombo che ’l magiore<lb></lb> |
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3138 diametro è .24. e il minore .10. e voi trovare li lati del rombo. Li quadrati dela mitá de’ dia-<lb></lb> |
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3139 metri, cioé dele linee .ae. e .eb., agiongni insiemi, de’ quali la radici, cioé .13., harai per ciascuno<lb></lb> |
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3140 lato del rombo. |
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3141 </p> |
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3142 <p class="main"> |
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3143 E possiamo molti questioni sopra li rombi proporre le quali tutte si possono redure<lb></lb> |
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3144 agli quadrilateri parte altera longiore de’ qual il lato magiore è la mitá del ma-<lb></lb> |
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3145 giore diametro e il lato minore è la mitá del minore diametro. E, acioché chia-<lb></lb> |
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3146 ro appaia, sia un rombo .abcd. Dove menise la linea .af. equedistante ala linea .eb. E com-<lb></lb> |
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3147 pise .fb. Dico adonca che ’l quadriletero .ef. ala mitá del rombo .abcd. E sonno li lati soi iguali<lb></lb> |
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3148 ala mitá de’ diametri .ac. e .bd., imperoché .ae. è la mitá del. ac. e .bc. è la mitta del .bd. É adon-<lb></lb> |
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3149 ca il triangolo .abd. la mitá del rombo .abcd. Ma il triangolo .abd. è iguale al quadrilate-<lb></lb> |
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3150 ro .ef. É fatto adonca ciascuno del multiplicare .ae. in .eb. Adonca il quadrilatero .ef. è la mitá<lb></lb> |
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3151 del rombo .abcd. E il triangolo .abd. è iguale al quadrilatero .ef. e peró è la mitá del rombo com-<lb></lb> |
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3152 mo è detto. E commo le questioni de’ rombi si possino redure ali quadrilateri parte altera<lb></lb> |
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3153 longiore alcuna dele molte voglio proporró. |
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3154 </p> |
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3155 <p class="main"> |
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3156 Se dicesse io ó agionto li .2. diametri d’ uno rombo e fecero .34. e l’ area del detto rom-<lb></lb> |
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3157 bo è .120. Adimando quanto è ciascun diametro. Perché li diametri sonno .34., la<lb></lb> |
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3158 mitá di quelli, cioé .ae. e .be. sonno .17. e l’ area del quadrilatero .ef. è .60. Adonca<lb></lb> |
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3159 hai produtta questa questione a una dele questioni de’ quadrilateri parte altera longiore. A quella<lb></lb> |
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3160 nella quale se propone l’ area essere .60. e l’ agregatione de’ lati è .17. Dove delo quadrato dela mi-<lb></lb> |
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3161 tá de’ .17. tra’ .60., rimangano .12 1/4. La cui radici tra’ del .8 1/2. E haremo .5. per lo lato minore. E per<lb></lb> |
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3162 lo magiore .12. E, questo trovato, radoppia ciascun lato. E haremo per lo primo diametro, cioé<lb></lb> |
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3163 il minore, .10. E il magiore .24. |
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3164 </p> |
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3165 <p class="main"> |
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3166 E ancora dicendo e gli é un rombo che i soi diametri insiemi agionti sonno .34. e il magio-<lb></lb> |
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3167 re è piú che ’l minore .14. Adimando quanto é ciascuno e l’ area. Togli .14. di .34., rimanga-<lb></lb> |
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3168 no .20., de’ quali la mitá, cioé .10., è il diametro minore e l’ avanzo, cioé .24., è il diame-<lb></lb> |
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3169 tro magiore. Dove multiplica la mitá del’ uno diametro per tutto l’ altro e haremo .120. per l’ area.<lb></lb> |
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3170 E ancora e gli é un rombo che gli .2. diametri con l’ area del rombo detto fecero .154. e il<lb></lb> |
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3171 magiore diametro agiongne sopra il minore .14. Adimando quanto è ciascun diametro e quan-<lb></lb> |
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3172 to è l’ area. Perché li .2. diametri del rombo sonno iguali a’ .4. lati del quadrilatero parte<lb></lb> |
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3173 altera longiore che è .ef., poni el lato breve una cosa, sará el lato magiore .1.co. e .7. |
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3174 </p> |
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3175 <p class="main"> |
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3176 Multiplica .1.co. via .1.co. e .7., fanno .1.ce. e .7.co., che sonno l’ area del ditto quadrilatero. E, perché el quadrila-<lb></lb> |
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3177 tero .ef. è la .1/2. del rombo ditto, radoppiarai .1.ce.7.co., fanno .2.ce.14.co. E tanto è l’ area del rombo.<lb></lb> |
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3178 Ala quale agiongni li .2. diametri del rombo, cioé li .4. lati del quadrilatero, che fanno .2.ce.18.co. e .14.,<lb></lb> |
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3179 che sonno iguali a .154. Togli da ogni parte .154., rimangano .140. Dove harai .2. ce .18.co. igua-<lb></lb> |
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3180 li a .140., dove areca a un .ce. harai .1.ce. e .9.co. iguali a .70. Dove parti le cose in .2., sonno .4 1/2., multiplica<lb></lb> |
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3181 in sé, fanno .20 1/4. Agiongni a .70., fanno .90 1/4. La cui radici è .9 1/4. Dela quale tra’ .4 1/2., rimangano .5. per<lb></lb> |
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3182 lo lato .be., che è la mitá del diametro breve. Adonca il diametro breve è .10. e il magiore è<lb></lb> |
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3183 .24. E cosí fa sempre et cetera. |
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3184 </p> |
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3185 <p class="main"> |
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3186 Ancora io ó agionto el diametro breve e il lato del rombo e forono .23. e il diame-<lb></lb> |
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3187 tro magiore agiogne sopra el minore .14. Quanto è adonca el diametro e il lato<lb></lb> |
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3188 del rombo. Perché el diametro magiore agiongni .14. sopra il minore, adonca la<lb></lb> |
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3189 mitá del diametro magiore avanza .7. al diametro minore, cioé il lato .ae. alo la-<lb></lb> |
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3190 to sopra lo. lato .eb. Adonca .eb. con .ea. agiongni .7. sopra il diametro .bd. Ma il .bd. con lo lat-<lb></lb> |
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3191 <lb></lb> |
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3192 <pb></pb></p> |
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3193 <p class="folio"> |
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3194 folio 21v |
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3195 </p> |
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3196 <p class="main"> |
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3197 <lb></lb> |
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3198 </p> |
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3199 <p class="runhead"> |
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3200 Distinctio tertia. Capitulum secundum. |
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3201 </p> |
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3202 <p class="main"> |
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3203 <lb></lb> |
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3204 o .ab. sonno .22. adonca .be. e .ae. con .ab. sonno .30. Onde tal questione habiamo creata. Io<lb></lb> |
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3205 ó agionto doi lati d’ un quadrilatero col suo diametro e fanno .30. E il magiore lato agiongni al<lb></lb> |
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3206 minore lato .7. Fa como di sopra dicemmo e harai el diametro minore .10. e il magiore .24.<lb></lb> |
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3207 e il lato del rombo .13. |
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3208 </p> |
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3209 <p class="main"> |
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3210 Ancora e gli é uno rombo del quale multiplicato un diametro in sé e l’ altro in sé<lb></lb> |
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3211 e, agionte le ditte multiplicationi, fanno .676. e l’ area è .120. Adimando quanto è ciascu-<lb></lb> |
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3212 no diametro. Piglia il .1/4. di .676., che è .169. De’ quali la radici è .13., che è el diame-<lb></lb> |
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3213 tro del quadrilatero .ef., cioé il lato del rombo. E, perché dimostrammo di sopra el quadrato<lb></lb> |
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3214 del diametro del quadrilatero parte altera longiore, agiongni sopra al doppio del’ area del<lb></lb> |
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3215 detto quadrilatero el quadrato dela sopra habundantia de’ lati soi. Adonca, se traremo el dop-<lb></lb> |
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3216 pio del’ area del quadrilatero .ef. de .169., rimaranno .49., de’ quali la radici è quello che lo lato<lb></lb> |
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3217 magiore del quadrilatero avanza al lato minore che, radoppiato, fanno .14. per la habundan-<lb></lb> |
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3218 tia del magiore diametro del rombo al minore. Adonca l’ area del quadrilatero è .60. e il ma-<lb></lb> |
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3219 giore diametro avanza al minore .7. Adimando quanto è il lato, dove el modo dato ne’ qua-<lb></lb> |
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3220 drilateri terrai e harai sará .5. minore e il magiore .12. Adonca il magiore diametro è .24., il<lb></lb> |
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3221 minore .10. |
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3222 </p> |
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3223 <p class="main"> |
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3224 Ancora e gli é uno rombo del quale multiplicato il magiore diametro per lo mino-<lb></lb> |
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3225 re fanno .240. e il magiore diametro agiongni sopra il minore .14. Adimandase quan-<lb></lb> |
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3226 to è ciascun diametro. Dela multiplicatione dela mitá del’ uno adonca perverá<lb></lb> |
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3227 .120., cioé, a multiplicare la mitá d’ un diametro per tutto l’ altro, fará .120. El quale è l’ area del rombo. Adunca,<lb></lb> |
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3228 a multiplicare un diametro per l’ altro, fanno el doppio del’ area, cioé el doppio del’ area del rombo. Adon-<lb></lb> |
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3229 ca .240. sonno el doppio del’ area del rombo. Onde l’ area del rombo è .120., che è il doppio del’ area<lb></lb> |
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3230 del quadrilatero .ef. Adonca l’ area del quadrilatero .ef. è .60. Onde proporremo tal questione. E gli é un qua-<lb></lb> |
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3231 drilatero parte altera longiore del quale l’ area è .60. E il magiore lato avanza al minore .7. Adi-<lb></lb> |
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3232 mandase quanto è ciascuno lato. Dove oservando i modi passati e dati ne’ quadrilateri, ha-<lb></lb> |
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3233 rai el diametro del rombo, cioé il minore, essere .10. e ’l magiore essere .24. |
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3234 </p> |
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3235 <p class="main"> |
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3236 E ancora e gli é uno rombo del quale Io ó agionto e diametri e fecino .34. E, a mul-<lb></lb> |
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3237 tiplicare l’ uno diametro nell’ altro fecino .240. Adimandase quanto è ciascuno la-<lb></lb> |
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3238 to. E, acció che meglio sia, faciase la retta .ab., che sia .34., e sia divisa, in .2. parti<lb></lb> |
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3239 iguali, nel ponto .g. e, in .2. parti non iguali, nel ponto .d. E sia .ad. eguale al dia-<lb></lb> |
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3240 metro breve. Rimane adonca .bd. iguale al magiore. E dela multiplicatione del .ad. in .db.,<lb></lb> |
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|
3241 col quadrato dela linea .dg., esci il quadrato dela linea .gb., per la .5a. del 2o., cioé de .17., che è .289.,<lb></lb> |
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3242 de’ quali trarai la multiplicatione del .ad. in .db., rimane .49. per lo quadrato .dg. Adonca .dg. è .7. che,<lb></lb> |
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3243 agionto con .bg., sia tutta .bd., cioé il magiore diametro .24. Rimane .da.10., che è il minore diametro.<lb></lb> |
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3244 E ancora e gli é uno rombo che, diviso il magiore diametro per lo minore, ne perviene,<lb></lb> |
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3245 di quella divisione, .2 2/5. E l’ area del rombo è .120. Adimando quanto è ciascuno diame-<lb></lb> |
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3246 tro. Perché e gli é cosí il tutto al tutto cosí ciascuna parte ala medesima parte sará.<lb></lb> |
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3247 Adonca cosí el magiore diametro al minore, cosí la mitá del magiore ala mitá del<lb></lb> |
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3248 minore. E la mitá del magiore è il lato del quadrilatero parte altera longiore .ef., cioé la linea<lb></lb> |
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3249 .ae. E la mitá del minore diametro è lo lato minore del quadrilatero .ef., cioé il lato .eb., cioé il<lb></lb> |
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3250 minore lato. Imperoché tutti e numeri che hano una medesima proportione, se si dividono e ma-<lb></lb> |
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3251 giori per gli minori, sempre ne perverá una medesima divisione. Adonca, se divideremo el magio-<lb></lb> |
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3252 re lato del quadrilatero .ef. per lo minore, cioé .ef. per .eb., ne perverá similmente .2 12/5. Onde è tale questio-<lb></lb> |
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3253 ne: l’ area del quadrilatero è .60., cioé la mitá del’ area del rombo e, divisi il magior lato per lo mi-<lb></lb> |
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3254 nore e vennene .2 2/5. Multiplica adonca .1. per .2 2/5. e tutto per .60., fanno .144. De’ quali la radici è .12. e quali, divi-<lb></lb> |
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3255 si per numeri dela proportione, cioé per .1. e per .2 2/5., vienne .12. e .5. che sonno e lati del quadrilatero .ef., cioé son-<lb></lb> |
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3256 no la mitá degli diametri del rombo. Adonca il magior diametro è .24. e il minore .10. |
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3257 </p> |
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3258 <p class="main"> |
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3259 E donde questo venga voi cognoscere, sia unitá .a. e .b. sia .2 2/5. e multiplichise .a. per .b. e<lb></lb> |
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3260 pervengane .g. E sia el minore lato del quadrilatero parte altera longiore .d. e il ma-<lb></lb> |
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3261 giore .e. E l’ area sia .f. E, perché dela divisione del magior lato per lo minore ne pervie-<lb></lb> |
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3262 ne .2 2/5. é comme .a. al .2 2/5., cosí el minore lato è al magiore, cioé cosí .a. al .b., cosí .d. al .e.<lb></lb> |
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3263 Onde la multiplicatione del .a. in .e. è commo la multiplicatione del .b. in .d. Se adonca la multiplicatione del .a. in .e.<lb></lb> |
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3264 over del .b. in .d. fanno .h. e .h., se multiplichi in sé fanno .i. E multiplichisi .g. in .f. e faccino .k., dico il numero .k. essere<lb></lb> |
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3265 iguali al numero .i. Tu multiplicasti certamente .a. in .b. e feci .g. e .d. in .e. e feci .f. e .g. in .f. e feci .k.,<lb></lb> |
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3266 dico che ’l numero .k. è iguali al numero .i., imperoché gli é fatto dela multiplicatione del .a.<lb></lb> |
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3267 in .b. fatto in .d. e multiplicato tutto in .e. Ancora tu multiplicasti .a. in .e. e pervenene .h. e .b. in .d.<lb></lb> |
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3268 <lb></lb> |
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3269 <pb></pb></p> |
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3270 <p class="folio"> |
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3271 folio 22r |
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3272 </p> |
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3273 <p class="main"> |
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3274 <lb></lb> |
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3275 </p> |
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3276 <p class="runhead"> |
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3277 Distinctio tertia. Capitulum tertium. 19] |
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3278 </p> |
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3279 <p class="main"> |
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3280 <lb></lb> |
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3281 e pervennene ancora .h. e del .h. in .h. è fatto .i. Adonca .i. è fatto del multiplicare del .a. in .e. multiplicata<lb></lb> |
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3282 per .b. e multiplicato tutto in .d. Ma la multiplicatione del .a. in .b., multiplicata in .d. e tutto multiplicato in .e.<lb></lb> |
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3283 è iguale del .a. in .e. multiplicata in .b. e tutto in .d. Adonca .k. è iguali al .i. commo dicemmo. Multi-<lb></lb> |
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3284 plicammo adonca di sopra .a. in .b. e feci .2 2/5. El quale multiplicammo per l’ area, cioé .g. per .f. e havemmo il<lb></lb> |
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3285 numero .k., cioé il numero .i., che fa .144. Del quale togliemmo la radici che fo .h., perché a mul-<lb></lb> |
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3286 tiplicare .h. in sé ne perviene .i. Adonca .h. é .12. E, perché del .a. in .e. ne perviene .h., cioé .12. e .a. è .1. |
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3287 </p> |
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3288 <p class="main"> |
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3289 E peró dividemmo .12. per .1. e per .2 2/5. e vennene .12., che è .e., per lo lato magiore. Ancora del .b. in .d.<lb></lb> |
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3290 ne perviene .h., cioé .12. e .b. è .2 2/5. E peró dividemmo .12., cioé .h., per .b., cioé per .2 2/5. e havemmo .d., cioé il<lb></lb> |
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3291 lato minore che fo .5. E, se l’ area fosse .100., alora .k., cioé .i., sarebbe .240. Del quale la radici è .h.<lb></lb> |
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3292 Ma .240. non á radici. Onde, perché non possiamo dividere .h. per .a. e per .b., torremo i quadrati<lb></lb> |
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3293 loro e quelli divideremo per .i. Overo, altramente, torremo la proportione ne’ numeri sani che<lb></lb> |
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3294 hano .i. al .b., fienno adonca .5. e .12. sia adonca .a.5. e .b.12. e .60. sia il .g. Dove il .k. over .i. sia .6000.,<lb></lb> |
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3295 che lo divideremo per gli quadrati de’ numeri .a. e .b., cioé .25. per .144., vienne, per lo magiore<lb></lb> |
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3296 lato, la radici .240. e, il minore lato, la radici de .41 2/3. Overo, altramente, poni il lato menore .1a.<lb></lb> |
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3297 cosa. Sará il lato magiore .2.cose.2/5., multiplica adonca una cosa in .2 2/5.cose., che fienno .2.censi.2/5.,<lb></lb> |
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3298 che sonno iguali al’ area data. Sia adonca l’ area data .100. La quale dividerai per .2 2/5., ne vene<lb></lb> |
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3299 .41 2/3. De’ quali la radici è il lato minore. E, perché e gli é cosí .a. al .b., cosí .d. al .e., sará adonca com-<lb></lb> |
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3300 mo il quadrato .a. al quadrato del nunero .b. per lo quadrato dela quantitá del .d., cioé .144. per<lb></lb> |
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3301 .41 2/3. e la summa dividi per lo quadrato del numero .a., cioé per .25., vienne .240. per lo quadrato del<lb></lb> |
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3302 lato magiore. E questo basti quanto al dire de’ rombi e, seguendo, diremo dela .3a. parte,<lb></lb> |
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3303 dicendo de’ romboidi.<lb></lb> |
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3304 <lb></lb> |
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3305 </p> |
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3306 <p class="head"> |
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3307 Qualiter metiantur Romboides. Capitulum tertium. |
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3308 </p> |
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3309 <p class="main"> |
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3310 <lb></lb> |
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3311 Del .2o. genere de’ quadrilateri è detto assai e del terzo diremo che sonno detti romboi-<lb></lb> |
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3312 di. El romboide è una figura paralelograma non rectangola che ha solamente e lati e gli an-<lb></lb> |
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3313 goli opposti iguali per la .34a. del po. E, quando adonca gli voi misurare, menerai in<lb></lb> |
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3314 quelli il diametro, per lo quale la ditta figura sia divisa in .2. triangoli iguali. On-<lb></lb> |
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3315 de, se ’l catetto d’ uno per tutta la basa, cioé per lo diametro suo, multiplicaremo, e haremo l’ a-<lb></lb> |
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3316 rea di tutto il romboide. Ala quale demostratione. Sia il romboide .abcd. che per ciascun de’ <lb></lb> |
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3317 lati .ab. e .cd. á .30., li quali lati sonno oppositi e sonno equedistanti. Gli altri .2. lati .ac. e .bd. son-<lb></lb> |
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3318 no similmente iguali e equedistanti, aventi in ciascun lato .13. E il diametro .bc. sia .37. Dal quale dia-<lb></lb> |
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3319 metro el romboide .abcd. é diviso in .2. parti iguali che sonno .2. triangoli: cioé il triangolo .abc.<lb></lb> |
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3320 e .dbc. E ciascuno di queli è ampligonio. Perché la potentia del lato .bc. è piú che le .2. potentie<lb></lb> |
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3321 de’ lati .ba. e .ac. overo del .bd. e .dc. Onde, sopra la basa .bc., si meni il catetto dal ponto .a. nel<lb></lb> |
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3322 triangolo .abc. e sia .ae. E multiplicarai el catetto .ae. per la basa .bc. e harai l’ area di tutto il rom-<lb></lb> |
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3323 boide .abcd. Overo truova il catetto .df. nel triangolo .bcd., sopra la basa .bc., e multiplica-<lb></lb> |
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3324 rai il ditto catetto per la basa e harai l’ area di tutto il romboide .abcd. Verbi gratia. El romboi-<lb></lb> |
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3325 de .abcd. è doppio al triangolo .bcd., del quale triangolo l’ area s’ á dela multiplicatione del catet-<lb></lb> |
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3326 to .df. nella mitá dela basa .bc. Onde la multiplicatione del catetto .df. per tutta la basa .bc.<lb></lb> |
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3327 fa il doppio del’ area del triangolo .bcd. Adonca fa l’ area di tutto il romboide, che è doppio al<lb></lb> |
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3328 triangolo .bcd. Imperoché l’ uno e l’ altro catetto .ae. e .df. troverai essere .9 27/37. che, multiplicati<lb></lb> |
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3329 per lo diametro .bc., cioé per .37., fanno .360. per l’ area di tutto il romboide .abcd. Ancora, a multipli-<lb></lb> |
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3330 care el catetto .ch. per la basa .ab., fa ancora l’ area del detto romboide. Trovisi l’ uno e l’ altro ca-<lb></lb> |
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3331 tetto per la regola di sopra detta nel triangolo ampligonio. Cioé tratta la potentia de’ lati .ca. e<lb></lb> |
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3332 .ab. overo .bd. e .de., cioé .1069., che è l’ agiongnimento di .169. e .900., de .1369., cioé dela poten-<lb></lb> |
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3333 tia del diametro .bc., che è lato del triangolo .abc., rimangano .300. Del quale mezzo, se lo divi-<lb></lb> |
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3334 derai per la basa .cd., cioé per .30., haremo .5. per la quantitá dela linea .dg. overo .ab., del quale la poten-<lb></lb> |
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3335 tia, cioé .25., tratta dela potentia del .bd., cioé de .169., rimane .144. Dela quale la radici è .12. per lo<lb></lb> |
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3336 catetto .bg. El quale, multiplicato per la basa .cd., cioé .12. multiplicato per .30., fanno .360. per l’ area del<lb></lb> |
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3337 detto romboide commo di sopra dicemmo. |
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3338 </p> |
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3339 <p class="main"> |
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3340 Ancora si trova l’ area del romboide per gli .2. altri catetti che sonno .di. e .ak., che si<lb></lb> |
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3341 trovano per lo diametro .ad. e per gli lati del romboide. Perché, menato in quello il dia-<lb></lb> |
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3342 metro, è risoluto il detto romboide in .2. triangoli oxigonij, commo in questa altra figu-<lb></lb> |
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3343 ra si mostra. Deli quali l’ uno è il triangolo .acd. e l’ altro .abd., è il catetto .id. over .ak.12. E<lb></lb> |
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3344 nota perché dal .a. nel .i., dove il catetto si mena infra il romboide, infra ‘l .k. e .d., sopra la linea<lb></lb> |
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3345 .kd. cade. Imperoché ’l cadimento di quello potrai trovare per quello che infra ‘triangoli<lb></lb> |
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3346 oxigonij dicemmo. Overo con lo filo commo ne’ triangoli dicemmo et cetera.<lb></lb> |
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3347 <lb></lb> |
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3348 <pb></pb></p> |
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3349 <p class="folio"> |
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3350 folio 22v |
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3351 </p> |
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3352 <p class="main"> |
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3353 <lb></lb> |
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3354 </p> |
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3355 <p class="runhead"> |
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3356 Distinctio tertia. Capitulum quartum. |
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3357 </p> |
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3358 <p class="main"> |
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3359 <lb></lb> |
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3360 E alcuna volta el romboide che, per lo diametro menore, è risoluto in .2. triangoli<lb></lb> |
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3361 ortogonii, commo il romboide .bcde. De’ quali .bc. e .de. lati del ditto romboi-<lb></lb> |
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3362 de sonno .35., gli altri .2. lati .bd. e .ce. sonno .37. E il diametro .de. menore sia .12. |
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3363 </p> |
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3364 <p class="main"> |
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3365 Dico adonca il romboide .bcde. essere diviso in .2. triangoli ortogonii, per la po-<lb></lb> |
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3366 tentia dela linea .ec., che sonno iguali ale potentie dele linee .cb. e .be. Onde retto é l’ angolo<lb></lb> |
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3367 .cbe. Similmente è trovato essere retto l’ angolo .bed. e è iguale el triangolo .cbe. al triangolo<lb></lb> |
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3368 .bed. E, perché dela multiplicatione del catetto .be. nella basa .ed. haremo l’ area del rombo-<lb></lb> |
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3369 ide .bdec., adonca la ditta area è .420., per la detta area del romboide .bcde. E cosí de tut-<lb></lb> |
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3370 ti li romboidi é da ffare et cetera. E questo basti quanto al dire de’ romboidi e, seguendo, dire-<lb></lb> |
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3371 mo del’ altro genere.<lb></lb> |
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3372 <lb></lb> |
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3373 </p> |
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3374 <p class="head"> |
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3375 De modo metiendi figuras helmuariphas. Capitulum quartum. |
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3376 </p> |
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3377 <p class="main"> |
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3378 <lb></lb> |
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3379 Dividese questo quarto genere degli quadrilateri in molti diversi generi. E sonno<lb></lb> |
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3380 quelli che hano e .2. lati opposti equedistanti e non iguali. E gli primi si dicono<lb></lb> |
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3381 caput abscisum, de’ quali gli altri .2. lati sonno iguali infra loro. Commo nel qua-<lb></lb> |
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3382 drilatero .abcd. del qual il lato .ab.8. è equedistante alo lato .cd., che è .18., gli altri<lb></lb> |
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3383 .2. lati .ac. e .bd. sonno .13. In questa figura el lato .ab. se dici capo tagliato e il lato .bc. si dici<lb></lb> |
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3384 tagliamento di base. Dela qual figura l’ area s’ á del multiplicare el catetto nella mitá de’ lati<lb></lb> |
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3385 .ab. e .ed. E il catetto si mena dal capo ala basa. Onde, se dal ponto .a. overo dal ponto .b. el<lb></lb> |
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3386 catetto, sopra la basa .cd., voi dirizzare, el tagliamento del capo, cioé .8., del tagliamento dela<lb></lb> |
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3387 basa togli, cioé di .18., rimangono .10. Del quale la mitá, cioé .5., sia il cadimento .ce. over .df.<lb></lb> |
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3388 E, dal ponto .a., el catetto cade sopra .e. E, dal ponto .b., cade sopra .f. Onde, se la potentia del<lb></lb> |
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3389 .ce. dela potentia .ac. overo la potentia .df. dela potentia .bd., cioé .25. de .169. trarrai, riman-<lb></lb> |
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3390 gano .144. Del qual la radici, che è .12., é la perpendiculare .ae. overo .bf. Li quali .12. multi-<lb></lb> |
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3391 plicati in nella mittá de’ lati .ab. e .cd., cioé nella mitá di .26., cioé .13., fanno .156. per l’ area del<lb></lb> |
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3392 detto quadrilatero .abcd. Verbi gratia: menate le linee che sonno catetti .ae. e .bf., se ne fa il<lb></lb> |
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3393 quadrilatero parte altera longiore .aefb. Del quale l’ area se ne fa dela multiplicatione del<lb></lb> |
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3394 catetto .ae. in .ef. e la linea .ef. è iguale ala linea .ab. Adonca .ef. è .8. Per gli quali multiplica-<lb></lb> |
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3395 ti .12.bracia., fanno .96.bracia. per l’ area del quadrilatero .aefb. El qual quadrilatero cavato del quadrilate-<lb></lb> |
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3396 ro .abcd., rimangano .2. triangoli ortogonii iguali, che sonno .aec.bfd. E la multiplicatione del<lb></lb> |
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3397 catetto .ae. nella mitá del .ec. fanno l’ area del triangolo .aec. Onde la multiplicatione dela<lb></lb> |
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3398 linea .ae. in tutta .ec. fanno l’ area de’ .2. triangoli .aec. e .bfd. La qual multiplicatione è .90. che, con .96. agion-<lb></lb> |
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3399 ta, fanno .156., cioé agionto con .96. che è l’ area del quadrilatero .aefb., commo adonca dicen-<lb></lb> |
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3400 mo che gli era quadro el ditto quadrilatero .abcd.156., di sopra. E, se voi trovare il diame-<lb></lb> |
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3401 tro .da. e .cb., la potentia dela linea .de. overo .cf., cioé .169., con la potentia del catetto .ae.<lb></lb> |
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3402 overo .bf. agiongni, cioé con .144. Haremo .313. Del quale la .R. è lo diametro .da. ove-<lb></lb> |
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3403 ro .be., commo se dimandava et cetera. |
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3404 </p> |
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3405 <p class="main"> |
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3406 Ma, se il ponto dove s’ intersega tali diametri voi havere, agiongni el capo con la<lb></lb> |
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3407 basa, fienno .26., che in sé multiplica, fienno .676. E, dipoi, il capo .ab. in sé multipli-<lb></lb> |
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3408 ca, cioé .8., fanno .64. E la basa in sé, fanno .324. Multiplica adonca .64. per lo qua-<lb></lb> |
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3409 drato d’ uno de’ diametri, cioé per .313. E la summa dividi per .676. over il quar-<lb></lb> |
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3410 to di .64., cioé .16., per .313. multiplica e dividilo per lo quarto di .676., cioé per .169. e virranne<lb></lb> |
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3411 .29 107/169. Di quali la radici è la linea .ag. overo .bg. Similmente multiplica el quarto di .324.,<lb></lb> |
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3412 cioé .81., per .313. e dividi la summa per lo quarto di .676., cioé per .169. e aremo il quadrato de-<lb></lb> |
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3413 la linea .gc. overo .gd. E debbi sapere perché noi pigliamo e quadrati dele dette linee. Per-<lb></lb> |
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3414 che .313., cioé il quadrato d’ uno de’ diametri, non hano radici. Imperoché, se ’l diametro fos-<lb></lb> |
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3415 se rationale, lo multiplicaremo per .8. e per .18. e divideremo le summe per .26. E cosí hare-<lb></lb> |
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3416 mo li tagliamenti de’ diametri. Che tutto vogliamo geometricalmente demostrare. Perché<lb></lb> |
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3417 la retta .ab. è equedistante ala retta .dc. Simile è il triangolo .agb. al triangolo .dge. e l’ an-<lb></lb> |
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3418 golo .abg. al’ angolo .gcd. iguali e l’ angolo .bag. al’ angolo .gdc. Onde è cosí .ab. al .bg., cosí<lb></lb> |
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3419 .dc. al .cg. E, per la permutata proportione, è cosí .ab. al .cd., cosí .bg. al .gc. Ancora e similmen-<lb></lb> |
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3420 te un’ altra volta cosí .ab. al .cd., cosí .ag. al .gd. e certamente .ab. del .cd. è .4/9. Onde .bg. del .ga.<lb></lb> |
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3421 overo .ag. del .gd. sonno similmente e .4/9. E, perché per la disiuncta proportonalitá e sonno pro-<lb></lb> |
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3422 portionali, cioé quelle cose che sonno proportionali per la disiuncta proportione, saranno ancora, per la<lb></lb> |
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3423 congionta proportionalitá, in simile proportione. E peró sia adonca cosí .ab. a sé e al .cd., cioé commo<lb></lb> |
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3424 <lb></lb> |
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3425 <pb></pb></p> |
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3426 <p class="folio"> |
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3427 folio 23r |
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3428 </p> |
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3429 <p class="main"> |
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3430 <lb></lb> |
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3431 </p> |
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3432 <p class="runhead"> |
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3433 Distinctio tertia. Capitulum quintum. 23 |
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3434 </p> |
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3435 <p class="main"> |
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3436 <lb></lb> |
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3437 .8. a .26. over .4. al .13., che sonno e minori in detta proportione, cosí .bg. é a tutta .bc. e .ag. a tutta<lb></lb> |
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3438 .ad. Adonca .bg. è del .be. li .4/13. E .ag. del .ad. è similmente li .4/13. Onde si fará ratiocinato il<lb></lb> |
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3439 diametro .bc. Torremo li .4/13. e haremo la linea .bg. overo .ag. L’ avanzo adonca .gc. overo<lb></lb> |
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3440 .gd. sia .9/18. di tutto il diametro. Ma, perché li quadrati del diametro, cioé de radici di .313., è<lb></lb> |
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3441 .313. e posse anumerare, che in prima la radici de .313. non si poteva anumerare, pigliaremo li<lb></lb> |
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3442 quadrati de .4. e de .13., cioé .16. e .169. Perché e gli é cosí .ab. a sé e al .cd., cosí .bg. al .be., sia adon-<lb></lb> |
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3443 ca cosí el quadrato dele linee .ab. al quadrato del’ agiontione dele linee .ab. e .cd., cioé commo<lb></lb> |
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3444 .64. é a .676. Overamente commo il quadrato de .64. é al quadrato di .676., cioé .16. a .169., cosí<lb></lb> |
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3445 el quadrato dela linea .bg. é al quadrato dela linea diametrale .bc., cioé al .313. E in detta pro-<lb></lb> |
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3446 portione è al quadrato dela linea .ag. al quadrato dela linea .ad. Onde la retta .ag. è iguale<lb></lb> |
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3447 ala retta .bg. e hano quella medesima proportione. E, perché quando dele cose iguali le co-<lb></lb> |
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3448 se iguali si tolgano, quelle cose che rimangano sonno infra loro iguali. Eguale é adonca la<lb></lb> |
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3449 retta .gc. ala retta .gd. E hano proportione a tutto il diametro, cioé ala radici di .313., commo<lb></lb> |
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3450 .cd. ha a sé e al .ab. La qual proportione è commo .9. al .13. Onde la proportione del quadrato<lb></lb> |
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3451 dela linea .gc. overo .gd. è a .313. commo il quadrato de .9. è al quadrato de .13., cioé .81. a .169. |
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3452 </p> |
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3453 <p class="main"> |
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3454 Onde multiplicaremo, commo dicemmo di sopra, .81. per .313. e divideremolo per .169. e havemo il<lb></lb> |
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3455 quadrato dela linea .gc. over .gd., ch’ era de bisogno mostrare. |
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3456 </p> |
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3457 <p class="main"> |
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3458 E, se noi vorremo la linea .ca. e .bd. menare per l’ angolo infino a tanto si congion-<lb></lb> |
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3459 ghino al ponto .h. Commo in questa altra figura è manifesto. Nella quale la figu-<lb></lb> |
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3460 ra del quadrilatero è transmutata in uno triangolo .hcd. E vorrai sapere la quan-<lb></lb> |
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3461 titá dela linea .ah. overo .bh., la mitá del capo dela mitá dela basa tra’ , cioé<lb></lb> |
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3462 .4. di .9. E sopra l’ avanzo, cioé sopra .5., dividi la multiplicatione dela mitá dela tagliatura<lb></lb> |
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3463 del capo nella linea .ca., cioé de .4. in .13., vienne .10.2/5. per la quantitá dela linea .ah. over .bh.<lb></lb> |
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3464 E, se multiplicarai el detto .4. per lo catetto .ae., cioé per .12., e divideremo per .5., vienne .9 3/5. per<lb></lb> |
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3465 lo catetto del triangolo .hab., cioé per la linea .ih. la quale, menata infino al ponto .k., fará tut-<lb></lb> |
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3466 ta la linea .kh. catetto del triangolo .hcd. E, perché nel triangolo .hcd. è menata una linea<lb></lb> |
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3467 retta .ab. equedistante ala basa .cd., sia il triangolo .hai. simile al triangolo .hcd., cioé che hano<lb></lb> |
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3468 gli angoli eguali infra loro, cioé l’ angolo .hab. al’ angolo .hcd., cioé l’ angolo dentro al’ angolo de<lb></lb> |
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3469 fuori è iguale per la .29a. del primo. E l’ angolo .hba. al’ angolo .hdc. è similmente iguale. E l’ an-<lb></lb> |
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3470 golo .h. è commune, cioé l’ angolo .ahb. e l’ angolo .chd. é uno medesimo. Simili adonca certa-<lb></lb> |
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3471 mente sonno li triangoli. E gli triangoli che sonno simili hano e lati che sonno intorno agli<lb></lb> |
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3472 iguali angoli proportionali, comme nel sexto de Euclide s’ é dechiarato. Onde, commo il lato<lb></lb> |
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3473 .ha. è al .ab., cosí .hc. è al .cd. E cosí, commo .hb. è al .ba., cosí. hd. è al .dc. Onde, per la permu-<lb></lb> |
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3474 tata proportionalitá, cosí commo .ha. al .hc. cosí .hb. al .hd. e cosí .ab. è al .cd. Ancora commo<lb></lb> |
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3475 .ab., cioé la basa del triangolo .ahb. è ala basa .cd., cosí el lato .ha. è allo lato .hc. e lo lato .hb.<lb></lb> |
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3476 è allo lato .hd. E ancora in simile proportione è il catetto .ih. al catetto .hk. Adonca quella<lb></lb> |
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3477 parte che è .ab. del .cd., quella medesima sia .ha. del .hc., cioé che parte è .8. del .18., quella me-<lb></lb> |
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3478 desima parte è .10 2/3. di .23 2/5. E ancora quella medesima è .hb. del .bd. e il catetto .ih. del ca-<lb></lb> |
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3479 tetto .hk. Onde .8. di .18. sonno li .4/9. E ancora .hb. del .hd. sonno li .4/9. E ancora .hi. del .hk. son-<lb></lb> |
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3480 no li .4/9. E in questa figura ancora è il triangolo .cea. simile al triangolo .aih. E gli hano cer-<lb></lb> |
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3481 tamente gli angoli iguali: l’ angolo cioé .hia. al’ angolo .aec. Imperoché ciascuno di loro è ret-<lb></lb> |
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3482 to. E l’ angolo che è .c. al’ angolo .iah. è iguale. Imperoché gli é la linea .ab. equedistante ala<lb></lb> |
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3483 linea .cd. L’ altro angolo, cioé .ahi., al’ altro angolo .cae. iguale, perché e .3. angoli de cia-<lb></lb> |
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3484 scuno triangolo sonno a .2. angoli retti iguali per la .32a. del primo. Adonca è cosí .ce., cioé .5.,<lb></lb> |
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3485 al .ea., cioé a .12., cosí .ai., cioé .4., al .ih., cioé .9 3/5. El quale viene multiplicando .4. per .12. e divi-<lb></lb> |
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3486 dendo per .5. Adonca .hi. è .9 3/5. Ancora commo .ec. è al .ca., cioé .5. al .13., cosí .ia. è al .ab., cioé<lb></lb> |
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3487 .4. è a .10 2/5., el quale viene del multiplicare .4. per .13. e dividere per .5. E peró .ha. è .10 2/5. E, per<lb></lb> |
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3488 queste proportioni, si truovano le misure dele altezze e le longhezze e ancora le profunditá<lb></lb> |
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3489 di qual voi edificio, commo nel suo luogo chiaramente mostraremo et cetera. E questo basti<lb></lb> |
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3490 quanto al dire del misurare le figure quadrilatere, le quali si dicono caput abscisum, cioé ca-<lb></lb> |
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3491 po tagliato e, seguendo, diremo dela seconda specie dele figure di questa quarta differen-<lb></lb> |
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3492 tia<lb></lb> |
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3493 Alius modus metiendi figuras helmuariphas alterius speciei a predictis.<lb></lb> |
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3494 Capitulum quintum.<lb></lb> |
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3495 <lb></lb> |
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3496 <lb></lb> |
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3497 <pb></pb></p> |
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3498 <p class="folio"> |
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3499 folio 23v |
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3500 </p> |
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3501 <p class="main"> |
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3502 <lb></lb> |
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3503 </p> |
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3504 <p class="runhead"> |
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3505 Distinctio tertia. Capitulum quintum. |
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3506 </p> |
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3507 <p class="main"> |
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3508 <lb></lb> |
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3509 El secondo genere di questa differentia sonno figure le quali si dicono mezzo capo<lb></lb> |
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3510 tagliato de’ quali e .2. lati sonno equedistanti, ma non iguali. Gli altri .2. lati sonno<lb></lb> |
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3511 non iguali. De’ quali l’ uno si leva sopra la basa, secondo l’ angolo retto e, similmente, an-<lb></lb> |
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3512 golo retto col capo delo tagliamento. L’ altro lato se eleva dal’ altra parte dela<lb></lb> |
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3513 basa secondo l’ angolo acuto. Commo il quadrilatero .abcd. del quale il lato .ad., cioé il capo,<lb></lb> |
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3514 è quedistante ala basa .bc. Del quale la longhezza è .18. e la basa .bc. è .30. e il catetto .ab.16. e<lb></lb> |
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3515 dc.20. Dico adonca che, a volere l’ area de tutto il detto quadrilatero, ragionga il capo con la<lb></lb> |
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3516 basa, cioé .18. con .30., fanno .48. De’ quali piglia la mitá che è .24. E questo multiplica per la<lb></lb> |
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3517 linea .ab., cioé per .16. (imperoché la sta ritta ortogonalmente), fanno .384. per l’ area del detto qua-<lb></lb> |
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3518 drilatero semicaput abscisum, cioé mezzo capo tagliato .abcd. Verbi gratia: sopra la retta .bc.,<lb></lb> |
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3519 dal ponto .d., il catetto .de. si meni. Sará adonca il quadrilatero .abcd. in .2. parti diviso: cioé<lb></lb> |
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3520 dal quadrilatero .abed., parte altera longiore, e nel triangolo .dec., che è ortogonio. Et é .be. igua-<lb></lb> |
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3521 le al .ad. e .ad. è .18. Adonca .be. ancora è .18. e ancora il .de. catetto è iguale al catetto .ab. e<lb></lb> |
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3522 ciascuno di loro è .16. L’ area adonca del quadrilatero .abcd. è .288., la quale è fatta dela mul-<lb></lb> |
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3523 tiplicatione del .be. in .ab. over del .ad. in .ab., cioé del .16. in .18., che ben fanno .288. E l’ area<lb></lb> |
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3524 certamente del triangolo .dec. é fatta dela multiplicatione del catetto .de. nella mitá delo<lb></lb> |
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3525 .ec., cioé del .16. in .6., che fanno .96. li quali, agionti con .288., fanno .384., cioé agionti con l’ area<lb></lb> |
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3526 del quadrilatero .abed., fanno commo ó detto .384. E questa è l’ area de tutto il quadrilate-<lb></lb> |
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3527 ro .abcd. commo dicemmo. |
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3528 </p> |
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3529 <p class="main"> |
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3530 E se ’l diametro .ac. vorrai havere, perché ortogonio è il triangolo .abc., agiongni le<lb></lb> |
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3531 potentie dele linee .ab. e .bc. insiemi, cioé .256. e .900., fanno .1156. De’ quali la radi-<lb></lb> |
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3532 ci, che è .34., é il diametro .ac. E, a volere havere il diametro .db., agiongni la po-<lb></lb> |
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3533 tentia del catetto .de. con la potentia dela basa .eb., cioé .256. con .324., fienno .580. de’ <lb></lb> |
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3534 quali la radici é sorda e quella é la longhezza del diametro .bd. Onde diremo il diametro<lb></lb> |
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3535 .bd. essere la radici de .580. over il quadrato del diametro .bd. essere .580. |
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3536 </p> |
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3537 <p class="main"> |
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3538 E, volendo le intersegationi de’ diametri, faciamo commo di sopra: cioé agiongna-<lb></lb> |
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3539 mo el capo con la basa, cioé .18. con .30., fienno .48. Adonca commo .18. é a .48., cosí .af.<lb></lb> |
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3540 è a tutto il diametro .ac. e il .18. a .48. è certamente commo .3. a .8. Onde, commo .3. è a<lb></lb> |
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3541 .8., cosí .af. è al. ac. Adonca multiplicaremo .3. per .34. e divideremo per .8. e haremo .12 3/4.<lb></lb> |
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3542 per la linea .af. L’ avanzo che è infino in .34. é la linea .fc., che è .21 1/4. Adonca .af. è .12 3/4. e .fd. è<lb></lb> |
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3543 .21 1/4. Similmente, perché e gli é simile il triangolo .afd. al triangolo .bfc., sia cosí .af. al .ac.,<lb></lb> |
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3544 cioé .3. a .8., cosí .df. al .db. Adonca .df. del .db. é gli .3/8. Rimangono .fb.5/8. del .db. Ma, perché e gli é<lb></lb> |
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3545 sorda la linea .db., torremo la proportione loro infra ’quadrati loro. E, adonca, commo il quadrato<lb></lb> |
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3546 di .3. al quadrato del .8., cioé commo .9. al .64., cosí il quadrato del .df. al quadrato del .db., cioé al<lb></lb> |
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3547 .580. Adonca multiplicaremo .9. per .580. e divideremlo per .64. Over multiplicaremo .9. per lo quar-<lb></lb> |
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3548 to di .580., cioé per .145., e divideremo la suma per lo quarto di .64., cioé per .16. Perché sempre deb-<lb></lb> |
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3549 biamo schifare il modo del troppo grande multiplicamento e partimento: cioé togli i minori<lb></lb> |
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3550 numeri la medesima proportione e harai la medesima multiplicatione e divisione. É certamen-<lb></lb> |
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3551 te .580. a .64. commo il quarto di .580. al quarto de .64., cioé cosí lo ’ntero alo ’ntero, cosí la parte<lb></lb> |
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3552 ala parte: commo in Euclide appare. Adonca, dividendo la multiplicatione del .149. in .9. per .16.,<lb></lb> |
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3553 ne viene .83 13/16. per lo quadrato dela linea .df. Onde .df. è la radici de .83 13/16. Ancora, perché la<lb></lb> |
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3554 linea .fb. é gli .5/8. del .bd., multiplicaremo el quadrato di .5., cioé .25. per lo quadrato di radici di<lb></lb> |
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3555 .145., cioé per .145., e divideremo la summa per .16., vienne .226 9/16. per lo quadrato dela linea .fb.,<lb></lb> |
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3556 commo volavamo. |
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3557 </p> |
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3558 <p class="main"> |
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3559 E, se vorremo menare .ca. e .bd. in modo si tochino nel ponto .g., menando ciascun<lb></lb> |
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3560 lato diritto (commo in questa figura si manifesta)., e vorrai sapere la quanti-<lb></lb> |
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3561 tá del .ag., multiplicarai .ed. per .da., cioé .16. per .18., e la quantitá dividerai per .ce.<lb></lb> |
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3562 cioé per .12., e haremo .24. E gli é certamente simile il triangolo .deb. al triango-<lb></lb> |
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3563 lo .gad. Onde e gli é cosí .be. al .ed., cosí .da. al .ag., per la igual proportione. Adonca sará<lb></lb> |
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3564 cosí .eb. al .bd., cosí .ad. é al .dg. Onde la multiplicatione del .bd., cioé .20., per .da., cioé per .18.<lb></lb> |
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3565 e diviso per .eb., fanno .30. per la quantitá dela linea .dg. E questo chiaro appa-<lb></lb> |
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3566 re nella figura passata. |
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3567 </p> |
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3568 <p class="main"> |
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3569 E questo basti sopra il modo de misurare le seconde genere ditte figure che hano<lb></lb> |
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3570 mezzo il capo tagliato overo che si dicono mezzo capo tagliato e, seguendo, dire-<lb></lb> |
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3571 mo delo terzo genere.<lb></lb> |
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3572 <lb></lb> |
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3573 <pb></pb></p> |
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3574 <p class="folio"> |
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3575 folio 24r |
|
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3576 </p> |
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3577 <p class="main"> |
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3578 <lb></lb> |
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3579 </p> |
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3580 <p class="runhead"> |
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3581 Distinctio tertia. Capitulum quintum. 24 |
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3582 </p> |
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3583 <p class="main"> |
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3584 <lb></lb> |
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3585 El terco genere di questa differentia è una figura quadrilatera che è ditta diver-<lb></lb> |
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3586 so capo tagliato. Del quale il capo e la basa sonno equedistanti e non iguali e<lb></lb> |
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3587 gli altri due lati sonno elevati sopra la basa secondo uno angolo acuto e sonno non<lb></lb> |
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3588 iguali. Commo il quadrilatero .abcd. del quale il capo .ab. è .10. Et é equedistan-<lb></lb> |
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3589 te ala basa .cd., che è .24. E il lato .ac. è .13. E il lato .bd.15. Dela quale figura la sua area si pi-<lb></lb> |
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3590 glia del multiplicare il catetto, che è menato dal capo ala basa, nella mitá del capo e dela ba-<lb></lb> |
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3591 sa. E se ’l ditto catetto dal ponto .a. overo .b. sopra la basa .cd. vorrai menare, é de bisogno pri-<lb></lb> |
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3592 ma trovare e cadimenti de’ detti catetti. De’ quali il modo a volerli trovare é che traga il ca-<lb></lb> |
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3593 po dela basa, cioé .10. di .24., rimangano .14. Dapoi, tra’ la potentia del lato .ac., cioé .169., dela po-<lb></lb> |
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3594 tentia del lato .bd., cioé de .225., rimangano .56. el quale dividi per lo .14., che dicemmo rimane a trare il<lb></lb> |
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3595 capo dela basa, vengone .4. El quale agiongni con .14., fanno .18. Del quale la mitá, cioé .9.,<lb></lb> |
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3596 è il cadimento magiore dala parte del lato .bd. Da’ quali .9. infino in .14. sonno .5., che sonno<lb></lb> |
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3597 il ponto del cadimento brieve del .fc. dalo lato .ac., commo negli triangoli deli angoli acu-<lb></lb> |
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3598 ti dicemmo. Tratto adonca la potentia del menore cadimento dela potentia del .ac., cioé<lb></lb> |
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3599 tratto la potentia del cadimento .fc., che è .25., dela potentia del lato .ac., che è .169., rimanga-<lb></lb> |
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3600 no .144. De’ quali la radici è .12., che è il catetto .af. Overamente, tratto la potentia del cadi-<lb></lb> |
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3601 mento magiore .de., che è .81., dela potentia del lato .bd., che è .225., rimangano similmente .144.<lb></lb> |
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3602 per la potentia del .be. Onde .be. è .12. commo .af. E l’ agiontione adonca del capo e dela basa,<lb></lb> |
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3603 cioé del .10. e del .24. fanno .34. De’ quali la mitá è .17. che, per lo catetto .af. over per lo catet-<lb></lb> |
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3604 to .be. multiplicato, fanno .204. per l’ area del quadrilatero .acdb. El quadrato del diame-<lb></lb> |
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3605 tro .cb. harai, se insiemi agiongnerai el quadrato dele linee .eb. e .ec., cioé il quadrato del ca-<lb></lb> |
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3606 tetto .eb. col quadrato del catetto .ec., e la summa sia il quadrato del diametro .cb. Dove la<lb></lb> |
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3607 potentia del catetto .be. è .144. e la potentia del catetto .ec. è .225. che, con .144. agionti, fan-<lb></lb> |
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3608 no .369. De’ quali la radici è lo diametro .bc. E, se voi lo diametro .ad., agiongni la potentia<lb></lb> |
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3609 del .fd., che è .361., con la potentia del .af., che è .144., fanno .505. per la potentia del .ad. Adon-<lb></lb> |
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3610 ca .ad. è la radice de .505. |
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3611 </p> |
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3612 <p class="main"> |
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3613 E, se dove s’ intersegano el catetto .af. col diametro .cb. vorrai sapere, questo in .2.<lb></lb> |
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3614 modi puoi fare. El primo è questo. Perché certamente è equedistante la linea .ab.<lb></lb> |
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3615 e la linea .ef., è cosí .ab. a sé e al .cf., cosí .ag. e .af. Adonca .ag. é gli .2/3. del .af., cioé .8. |
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3616 </p> |
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3617 <p class="main"> |
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3618 Dove .gf. è .4. E sonno simili e triangoli .agb. e .cgf. e, similmente, .gb. é gli .2/3. del<lb></lb> |
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3619 .bc. Dove il quadrato suo è .4/9. del quadrato del detto diametro. Onde multiplica .4. per .369. e<lb></lb> |
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3620 dividi per .9., vienne .164. per lo quadrato dela linea .bg. E, perché .bg. e gli é .2/3. del .bc., riman-<lb></lb> |
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3621 gano .gc.1/3. del .bc. Dove il suo quadrato è .1/9., cioé .1/9. di .369., cioé .41. Similmente, perché si-<lb></lb> |
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3622 mili sonno i triangoli .ceb. e .cfg., é cosí .cf. al .ce. cosí .fg. al .eb., cioé la terza parte. Adonca<lb></lb> |
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3623 .fg. è .4. commo dissi. Ancora per quel medesimo el .cg. é .1/3. e del .cb. e cosí di tutti et cetera.<lb></lb> |
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3624 Possiamo ancora (per quelle cose che sonno dette in questa parte e ancora per<lb></lb> |
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3625 quello che se disse ne’ triangoli) trovare il diametro .da. e sapere in che luogo in-<lb></lb> |
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3626 tersega el catetto .be. E ancora dove e in che luogo s’ intersega col diametro .bc.<lb></lb> |
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3627 E, ancora, se dagli angoli .c.d. overo da alcuno altro ponto dato sopra la linea<lb></lb> |
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3628 retta .cd. over infra quello, cioé dentro over de fuora, si menerá la linea sopra le date parti de’ lati .db.<lb></lb> |
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3629 overo .ca. e mettase fuore dela figura parimente con la linea .ab. infino a tanto che insie-<lb></lb> |
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3630 mi si congiongnino. Potremo sapere il ponto dela congiontione di quelli e ancora la quan-<lb></lb> |
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3631 titá dele ditte linee le quali fienno menate. |
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3632 </p> |
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3633 <p class="main"> |
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3634 E, se vorrai menare le linee .db. e .ac. per lo deritto infino si congiungnino al pon-<lb></lb> |
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3635 to .h. el capo, cioé .10., dela basa, cioé .24., tra’ , remangano .14. E la multiplicatione<lb></lb> |
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3636 del capo, cioé del .10., cioé del .ab. nel .ac., dividi per lo ditto .14., cioé la multiplica-<lb></lb> |
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3637 tione del .10. in .13., vienne .9 2/7. per la linea .ah. Similmente, se dividerai la multi-<lb></lb> |
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3638 plicatione del .ab. in .bd., cioé del .10. in .15., per lo detto .14., haremo .10 5/7. per la linea .bh.<lb></lb> |
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3639 E questo basti quanto al terzo genere di questa differentia e, seguendo, diremo del mo-<lb></lb> |
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3640 do del misurare il quarto genere di questa differentia et cetera.<lb></lb> |
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3641 <lb></lb> |
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3642 </p> |
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3643 <p class="head"> |
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3644 Tertius modus mensurandi figuras helmuariphas alterius generis a predictis. |
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3645 </p> |
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3646 <p class="main"> |
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3647 <lb></lb> |
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3648 El quarto genere di questa differentia, el qual si chiama capo tagliato declinan-<lb></lb> |
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3649 te. Del quale il capo e la basa sonno non iguali e equedistanti e degli altri .2. lati<lb></lb> |
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3650 l’ uno è levato sopra la basa secondo l’ angolo acuto, l’ altro sopra la medesima basa<lb></lb> |
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3651 fa l’ angolo ampio. Commo nel quadrilatero .abcd. Del quale il capo .ad. è .12.<lb></lb> |
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3652 <lb></lb> |
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3653 <pb></pb></p> |
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3654 <p class="folio"> |
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3655 folio 24v |
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3656 </p> |
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3657 <p class="main"> |
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3658 <lb></lb> |
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3659 e la basa .bc.16. e il lato .ab.15. e .dc.13. Ancora di questa figura quadrilatera detta capo<lb></lb> |
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3660 tagliato declinante s’ á la sua quadratura dela multiplicatione del catetto nella mitá dela ba-<lb></lb> |
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3661 sa e del capo suo. Onde, se dal ponto .a. sopra la basa .bc., caderá el catetto infra il quadrilate-<lb></lb> |
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3662 ro .abcd. E dal ponto .d. el catetto caderá di fuora dela ditta figura. Onde, a ció che trovia-<lb></lb> |
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3663 mo la quantitá del catetto di fuora overo dentro, é de bisogno prima trovare il cadimemo lo-<lb></lb> |
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3664 ro. Del quale il modo a trovarlo è che si traga il capo dela basa, cioé .12. di .16., che rimangano<lb></lb> |
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3665 .4., de’ quali la potentia, cioé il quadrato loro, che è .16., s’ agionga col quadrato del lato .cd.,<lb></lb> |
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3666 cioé con .169., fanno .185., che si traga dela potentia del .ab., cioé di .225., rimangano .40. De’ qua-<lb></lb> |
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3667 li la mitá, cioé .20., sonno da essere divisi per lo detto .4., vienne .5. per lo cadimento di fuora .ce.,<lb></lb> |
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3668 cioé dico che dal ponto .c. al ponto .e.5., cioé terrai il modo che dimostrammo nel trovare e ca-<lb></lb> |
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3669 tetti ne’ triangoli ampligonij. Quando el catetto si muove da ciascun degli angoli acuti li qua-<lb></lb> |
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3670 li .5., agionti con la basa .cb., cioé con .16., fanno .21. per tutta la linea .eb. De’ quali, tratti .12., che<lb></lb> |
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3671 sonno la quantitá del capo, rimangano .9. per la quantitá del cadimento .fb., el qual cade den-<lb></lb> |
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3672 tro. Onde tratto la potentia del .bf., cioé .81., dela potentia delo lato .ab., cioé di .225., rimango-<lb></lb> |
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3673 no .144., de’ quali la radici è .12., che è il catetto .af. over il catetto .de., li quali .12., multiplicati per la mitá del<lb></lb> |
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3674 capo e dela basa, cioé per .14., fanno .168. per l’ area de tutto il quadrilatero detto capo tagliato<lb></lb> |
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3675 declinante .abcd. In simigliante figure, che si dicano capo tagliato declinante, li .2. catet-<lb></lb> |
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3676 ti che si menano dagli angoli del capo infino ala basa, alcuna volta, uno ne cade dentro e<lb></lb> |
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3677 uno di fuora, commo nella passata figura del capo tagliato declinante certamente mostran-<lb></lb> |
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3678 mo. Conciosiacosaché di tutte queste figure che dette sonno capo tagliato, sempre l’ area sua<lb></lb> |
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3679 s’ á del multiplicare el catetto nela mitá del capo e dela basa, commo hai veduto. Alcuna<lb></lb> |
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3680 volta uno de’ ditti catteti cade in sur l’ angolo dela basa, el quale angolo è opposto al capo:<lb></lb> |
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3681 commo nel quadrilatero detto capo tagliato declinante nel quale, se si mena el catetto dal pon-<lb></lb> |
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3682 to .a. ala basa, cade il detto catetto in sul ponto .e., commo nela figura preditta si manifesta.<lb></lb> |
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3683 L’ altro catetto che si muove dal ponto .b. cade fuora del detto quadrilatero .abcd. dove ca-<lb></lb> |
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3684 de in sul ponto .e. El quale ponto .e. è fuora del detto quadrilatero. Adonca in queste figure (det-<lb></lb> |
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3685 te caput abscisum declinans: cioé capo tagliato declinante) interviene uno de’ catetti cade-<lb></lb> |
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3686 re di fuora, e l’ altro in sur l’ angolo opposto al capo. |
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3687 </p> |
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3688 <p class="main"> |
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3689 Alcuna volta interviene che gli catetti menati dagli angoli del capo verso la bau-<lb></lb> |
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3690 sa caggiano amendoi di fuora, commo in questa figura chiaro appare. Impe-<lb></lb> |
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3691 roché l’ angolo .a. del capo menando lo catetto cade in sul ponto .f. fuor dela ba-<lb></lb> |
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3692 sa .dc. E ancora, menando dal’ angolo .b. del capo dela detta figura ditta caput<lb></lb> |
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3693 abscisum declinans, cioé capo tagliato inchinato in verso la basa, caderá in sul ponto .e., che<lb></lb> |
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3694 è fuora dela ditta figura, li quali catetti tutti si trovano secondo il modo dato di sopra.<lb></lb> |
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3695 Ora ritorniamo al quadrilatero primo, ditto capo tagliato declinante, dove el qua-<lb></lb> |
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3696 drato del .be., che è .441., e del .ed., che è .144., agiongni insiemi, fanno .585. per lo qua-<lb></lb> |
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3697 drato del diametro .ab. Adonca el diametro .bd. è la radici di .585. La quale neli<lb></lb> |
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3698 numeri discreti non si truova. E, volendo il diametro .ac., el quadrato dela linea<lb></lb> |
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3699 .cf., cioé la multiplicatione di .7. in sé, che fanno .49., agiongni al quadrato dela linea .af., cioé a<lb></lb> |
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3700 .144., fanno .193. e tanto è il quadrato dela linea .ac. Adonca il diametro .ac. é la radici di .193. |
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3701 </p> |
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3702 <p class="main"> |
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3703 E, volendo il diametro .ae., ragiongni la potentia del .fe., cioé .144., con la potentia del .af., cioé<lb></lb> |
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3704 con .144., haremo .288. E la radici de .288. è il diametro .ae. |
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3705 </p> |
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3706 <p class="main"> |
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3707 E, se voi menare ciascuna dele linee .ab. e .ed. per lo diritto infino si congionghino in-<lb></lb> |
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3708 siemi nel ponto .k., sará certamente .ka. al .kb. e .kd. al .kc. commo .ad. al .bc. per la<lb></lb> |
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3709 seconda del sexto. E noi habiamo ditto che .ad. è .12. e .bc. è .16., adonca, .ak. é gli .3/4.<lb></lb> |
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3710 del .kb. e il .dk. é gli .3/4. del .kc. Adonca .ab. è il .1/4. del .bk. E peró .ab. è il terzo di tut-<lb></lb> |
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3711 ta .bk. E, similmente, .dc. è il .1/3. di tutta .ck. Onde .ka. è .45. e .kd. è .39. |
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3712 </p> |
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3713 <p class="main"> |
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3714 E cosí ancora moltissime questioni si possono creare nele ditte figure, le quali tutte si<lb></lb> |
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3715 possono redurre al quadrilatero parte altera longiore. E peró, adonca, assai copiosa-<lb></lb> |
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3716 mente in quella parte ne dicemmo che, volendo qui replicare, sarebbono indarno.<lb></lb> |
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3717 Conciosiacosaché pure, quando non le sapesse appropiarle a quelle, e tu ricorrirai ala ma-<lb></lb> |
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3718 dre deli casi: cioe ala regola del algebra. La, cioé ala regola del’ algebra. La quale ti conseglio che, se non la sai la ’mpari <lb></lb> |
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3719 et <lb></lb> |
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3720 cetera. |
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3721 </p> |
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3722 <p class="main"> |
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3723 Se ‘quadrilateri non haranno i lati equedistanti, cioé che niuno de’ lati sia equedi-<lb></lb> |
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3724 stante al’ altro, commo nel quadrilatero .abcd., del quale il lato .ab. è .13. e .bc. è<lb></lb> |
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3725 .15. e .de.18. e .da.16. El quale quadrilatero si puó misurare, se s’ ará la longhezza<lb></lb> |
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3726 d’ un de’ diametri dal quale el ditto quadrilatero fosse diviso in .2. triangoli. Verbi gratia.<lb></lb> |
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3727 <lb></lb> |
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3728 <pb></pb></p> |
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3729 <p class="folio"> |
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3730 folio 25r |
|
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3731 </p> |
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3732 <p class="main"> |
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3733 <lb></lb> |
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3734 </p> |
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3735 <p class="runhead"> |
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3736 Distinctio tertia. Capitulum sextum. 25 |
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3737 </p> |
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3738 <p class="main"> |
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3739 <lb></lb> |
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3740 Sia il diametro .ac.20., dico che si puó arrecare a quadro il detto quadrilatero facilmente per<lb></lb> |
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3741 la cognitione del diametro .ac., imperoché ’l diametro .ac. divide el detto quadrilatero di-<lb></lb> |
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3742 versilatero in .2. triangoli, de’ quali l’ uno è il triangolo .abc. e l’ altro il triangolo .adc. E, perché<lb></lb> |
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3743 non hai congnitione del catetto che d’ alcuno angolo si muova, é de bisogno quadrare ciascu-<lb></lb> |
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3744 no de’ detti triangoli per lo modo del ragiongnere le facie, cioé, volendo quadrare il trian-<lb></lb> |
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3745 golo .abc., agiongnerai .13. e .15. e .20., fanno .48. De’ quali il 1/2. è .24. E saprai quanto é da ciascun<lb></lb> |
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3746 lato al .24. E harai quella differentia .11.9.4. Dove infra loro le multiplica, cioé .11. per .9. e tut-<lb></lb> |
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3747 to per .4., fanno .396. che, per .24. multiplicato, fanno .9504. E la radici de .9504. é quadro il<lb></lb> |
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3748 detto triangolo. E quadra, dipoi, l’ altro al medesimo modo. E harai che è quadro la radici de<lb></lb> |
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3749 .18711. E harai, per la detta figura, l’ area la radicie de .18711. piú radici de .9504. |
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|
3750 </p> |
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3751 <p class="main"> |
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|
3752 Ancora potresti menare la equedistante ala linea .bc., cioé la linea .ae. E misurare<lb></lb> |
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3753 quel quadrilatero .aebc. secondo la dottrina di quello che s’ é detto. E, di poi, misu-<lb></lb> |
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3754 rare el triangolo .aed. per lo modo detto. E quelli insieme agiongnerai e harai l’ a-<lb></lb> |
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3755 rea di tutto il detto quadrilatero diversilatero. |
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3756 </p> |
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3757 <p class="main"> |
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3758 Posseno essere de simile figure de quali i loro diamitri caggiono dentro tutti, alcune<lb></lb> |
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3759 de quali i loro diametri caggiono amendui de fuora, e di quelli che l’ uno cade di<lb></lb> |
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3760 fuora, e l’ altro dentro et cetera, commo fin qua hai veduto abastanza in tutte le sorte de’ <lb></lb> |
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3761 quadrilateri regulari e anche elmuariffi, cioé inregulari che cosí li chiama Eu-<lb></lb> |
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3762 clide for dele .4. spetie, cioé quadrato, tetragono longo, elmuaym e simile elmuaym in po.<lb></lb> |
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3763 Modus inveniendi aream figurarum multilaterarum. Capitulum sextum. |
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3764 </p> |
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3765 <p class="main"> |
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3766 El modo a misurare le figure di molti lati è che la divida in triangoli. E l’ area di<lb></lb> |
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3767 detti triangoli in una somma agiongni e cosí harai l’ area di ciascuna figura di<lb></lb> |
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3768 molti lati. Et è da notare che le figure avente .5. lati é soluta almeno in .3. triango-<lb></lb> |
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3769 li. E quelle che sonno di .6. lati sonno resoluti in .4. triangoli. E cosí ogni figura<lb></lb> |
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3770 di molti lati è absoluta in .2. triangoli meno che ’lati commo sopra la .32o. del primo fo ditto<lb></lb> |
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3771 in Euclide. La qual conclusione è una dele famose che vadi per le scole phylosophiche. E a-<lb></lb> |
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3772 la sua prova se ne recerca .31. passate. E per questo li oltramontani la sogliano chiamare Cli-<lb></lb> |
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3773 peus aristotelicus, peroché ut plurimum AR.. la induci a suoi exempli maxime in la posteriora ..<lb></lb> |
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3774 E molti altri luoghi dele sue opere, quando demostra la propria passione predicare del suo subietto.<lb></lb> |
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3775 Ut de triangulo .hre. tres equales duobus rectis per angulum extrinsecum equivalentem duobus intrinsecis<lb></lb> |
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3776 oppositis et cetera. E avenga che, per la reduttione di quelle in triangoli, le figure di molti lati si pos-<lb></lb> |
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3777 sino misurare. Nientedimeno, molte volte, piú sottilmeme, in alcune si puó procedere. Cioé<lb></lb> |
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3778 quando la figura sia pentagona: cioé di .5. lati iguali, che nne puoi fare .2. pezzi de’ quali l’ uno sia<lb></lb> |
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3779 triangolo e l’ altro quadrilatero, nel quale .2. lati sienno equedistati. Comme nel pentago-<lb></lb> |
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3780 no .abcde. Del quale, tagliato el triangolo .abe., rimane el quadrilatero .ebcd. capo asciso. Del<lb></lb> |
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3781 quale il lato .be. è equedistante al lato .cd. Onde, agiongnendo l’ area del triangolo .abe. con<lb></lb> |
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3782 l’ area del quadrilatero .bcde., harai l’ area del pentagono .abcde. Similmente del exagono é<lb></lb> |
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3783 possibile farne .2. quadrilateri, cioé dela figura di .6. lati, de’ quali uno á .2. lati equedistanti.<lb></lb> |
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3784 Overo ancora uno exagono si puó dividere in uno quadrilatero avente .2. lati equedistan-<lb></lb> |
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3785 ti e in .2. triangoli. E cosí in tutte l’ altre figure studierai di fare, le quali hano molti lati.<lb></lb> |
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3786 Vero è che, quando la figura sará di molti lati e equiangoli, la quale figura deside-<lb></lb> |
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3787 ri de misurare (altramente che quello che noi habiamo detto), potemo al’ area<lb></lb> |
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3788 di quella pervenire. Conciosiacosaché in quella caggia uno cerchio contingen-<lb></lb> |
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3789 te nel mezzo di ciascuno lato, al quale ponto, menato la perpendiculare dal centro<lb></lb> |
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3790 e la detta multiplicata contra ala mitá dele facie del detto pentagono, haremo per quella l’ a-<lb></lb> |
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3791 rea detta. |
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3792 </p> |
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3793 <p class="main"> |
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3794 Accioché chiaro appaia, sia uno pentagono equilatero .abcde. nel quale voglia-<lb></lb> |
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3795 mo descrivere uno cerchio contingente el lato di quel pentagono: che in questo<lb></lb> |
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3796 modo si fará. Divideró gli angoli .eab. e .abc. in .2. parti iguali dale .2. linee .af. e .fb.<lb></lb> |
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3797 E meneró le linee .fc.fd.fe. E segneró li ponti .g.h.i.k.l. nel mezzo de’ lati di quello. E compiró<lb></lb> |
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3798 le linee .fg.fh.fi.fk.fl. Le quali dimostraró che infra loro fienno iguali. Perché equiangolo<lb></lb> |
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3799 è il pentagono .abcde., sia l’ angolo .fab. iguale al’ angolo .fba. Conciosiacosaché ’l sia la mitá de-<lb></lb> |
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3800 l’ angolo del pentagono. Onde il triangolo .fab. è equicurio, cioé di .2. lati iguali e gli e angoli<lb></lb> |
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3801 sotto a quei .2. lati sonno iguali infra loro. E peró è iguale la retta .fa. ala retta .fb. e la retta<lb></lb> |
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3802 .fg. ala retta .fl., che è la retta .fg. catetto sopra la linea .ab., avenga caggia nel mezzo di quel-<lb></lb> |
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3803 la. E peró .la. è iguale ala retta .ag. Imperoché la è mitá della retta .ae. Ponghise adonca<lb></lb> |
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3804 <lb></lb> |
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3805 <pb></pb></p> |
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3806 <p class="folio"> |
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3807 folio 25v |
|
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3808 </p> |
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3809 <p class="main"> |
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3810 <lb></lb> |
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3811 </p> |
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3812 <p class="runhead"> |
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3813 Distinctio tertia. Capitulum sextum. |
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3814 </p> |
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3815 <p class="main"> |
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3816 <lb></lb> |
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3817 comunamente .fa., fienno .2. rette .ga. e .fa. iguali a .2. rette .fa. e .al. E l’ angolo .gaf. è iguale al’ an-<lb></lb> |
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3818 golo .fal. E la basa adunque .fl. è iguale ala basa .fg. E l’ angolo .afl. è iguale al’ angolo .afg. E<lb></lb> |
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3819 l’ angolo .alf. retto è iguale al’ angolo .agf. retto. Onde, perché la retta .fl. è catetto sopra la-<lb></lb> |
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3820 retta .ae. e perché .al. è iguale ala retta .el., onde, se si pone comunamente la retta .fl., fienno<lb></lb> |
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3821 .2. recte .fl. e .la. eguali a .2. rette .fe. e .le., e gli angoli fatti al .f. sonno iguali. Onde la retta .fe.<lb></lb> |
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3822 è iguale ala retta .fa. e il triangolo .afl. al triangolo .lfe. E tutto il triangolo .bfa. a tutto il<lb></lb> |
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3823 triangolo .afe. Similmente se mostrará ciascuna dele rette .fg. e .fl. Onde .f. è centro d’ un<lb></lb> |
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3824 cerchio che á di spazio la retta .fg. e .fh. e fasse il cerchio .ghikl. E sia il pentagono .abcde.<lb></lb> |
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3825 diviso in .5. triangoli iguali che sonno .fab.fbc.fed.fde.fea. e gli catetti cadenti a quelli<lb></lb> |
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3826 sonno infra loro iguali: che sonno .fg.fh.fi.fk.fl. E perché del multiplicare .fg. nella mitá<lb></lb> |
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3827 del .ab. ne perviene l’ area del triangolo .fah., onde, multiplicando la mitá del diametro del<lb></lb> |
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3828 cerchio cadente nel pentagono, cioé .fg., in .5. cotanti dela mitá del .ab., cioé nela mitá del la-<lb></lb> |
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3829 to del pentagono, ne perviene .5. cotanti dell’ area del triangolo .fab., cioé l’ area del pentago-<lb></lb> |
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|
3830 no .abcde., comme dicemmo di sopra. |
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|
3831 </p> |
|
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3832 <p class="main"> |
|
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3833 Similmente verrá in ogni figura equilatera e equiangula nela quale caggia uno<lb></lb> |
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3834 cerchio. E, per questo, è manifesto che la multiplicatione del mezzo il diametro<lb></lb> |
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3835 del cerchio in piú dela mitá dela linea circonferente fará piú che l’ area del detto<lb></lb> |
|
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|
3836 cerchio. |
|
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|
3837 </p> |
|
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3838 <p class="main"> |
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3839 Possiamo ancora uno pentagono equilatero e equiangolo altramente misurar-<lb></lb> |
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3840 lo. Conciosiacosaché caggia nel cerchio contingente ogni suo angolo in questo<lb></lb> |
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3841 modo facendo. Che si multiplichi la mitá e il .1/4. del detto diametro, cioé del dia-<lb></lb> |
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3842 metro di detto cerchio, per la mitá e .1/3. dela corda del’ angolo pentagonico. E quello fanno è<lb></lb> |
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|
3843 la detta area. |
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3844 </p> |
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3845 <p class="main"> |
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3846 E acioché chiaro appaia sia il pentagono .abgde. nel cerchio .abgde. del<lb></lb> |
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3847 quale il diametro sia .az. E il suo centro sia .c. E compise la retta .be., la quale è la cor-<lb></lb> |
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3848 da del’ angolo pentagonico, cioé del’ angolo .bae. E tolghise .ci., la mitá del mezzo<lb></lb> |
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3849 diametro, cioé il quarto del diametro .az. E faciase il ponto .k. in tal modo che sia<lb></lb> |
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3850 cosí .ai. al .ac., cosí .te.al.tk. É certamente .ac. del .ai. gli .2/3. E, similmente, .tk. é gli .2/3. del .te., che è<lb></lb> |
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3851 iguale del .tb. É iguale certamente .bt. del .te. Onde .tk. è il .1/3. di tutta .be. Onde .bk. è il .1/2. e<lb></lb> |
|
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|
3852 il .1/3. di tutta .be. Dico adunque che dela multiplicatione del .ai. in .bk. ne perviene l’ area del pen-<lb></lb> |
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|
3853 tagono .abgde., che cosí il proveró. Perché gli é cosí .ai. al .ac., cosí .te. al .tk., sará la multipli-<lb></lb> |
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|
3854 catione del .ca. in .te., cioé in .tb., iguali ala multiplicatione del .ia. in .tk. Ma dela multiplicatio-<lb></lb> |
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3855 ne del .ca. in .bt. ne perviene el doppio del’ area del triangolo .abc. Adunque, multiplicato<lb></lb> |
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|
3856 .ai. in .tk., ne perviene el doppio del triangolo .cba. E, perché .tk. è doppio del .ek., se multi-<lb></lb> |
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3857 plicaremo .ia. in .ek., ne perverrá lo iguale al’ area del triangolo .abc. che è la quinta parte di tut-<lb></lb> |
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3858 to il pentagono .abgde. Onde, multiplicando .ai. in .bk., cioé .5. cotanti del .ke., ne perverrá<lb></lb> |
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3859 ancora .5. cotanti del’ area del triangolo .abc., cioé lo eguale al’ area del pentagono .abgde.,<lb></lb> |
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|
3860 la qual cosa si convenia mostrare. |
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|
3861 </p> |
|
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|
3862 <p class="main"> |
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3863 Ed è da notare che, se ’l diametro del cerchio sia ratiocinato, alora il lato del pentago-<lb></lb> |
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3864 no cadente in quello sia la linea minor, cioé radici del quarto reciso. Lo quale re-<lb></lb> |
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3865 ciso è fatto del numero meno la radici. De’ quali due nomi el magiore puó sopra<lb></lb> |
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3866 el minore uno numero incomensurabile a quello in longitudine e la corda de-<lb></lb> |
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3867 l’ angolo pentagonico sia la linea maggiore, cioé la radice del quarto binomio che è fatto del<lb></lb> |
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|
3868 numero e radice. Del quale el magiore numero puó piú del minore uno numero incomen-<lb></lb> |
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3869 surabile a quello in longitudine. E sonno composti di .2. medesimi nomi la corda del’ angolo<lb></lb> |
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|
3870 pentagonico e lo lato del pentagono. Comme se ’l lato .ab. del pentagono .abgde. sia preso<lb></lb> |
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|
3871 la radici de .320. e tratta di .40. e di quel preso la radici. E la corda del’ angolo pentagonico<lb></lb> |
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3872 sia la radici di .320. posta sopra .40. e di quel preso la radici. E questo è quando il diametro<lb></lb> |
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3873 .az., cioé del cerchio dato, è .8., comme mostraremo nel suo luogo.<lb></lb> |
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|
3874 Se ’l campo fosse di .4. facie, dele quali le .2. oposte fossino equedistante e l’ altre .2.<lb></lb> |
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3875 facessino arco. Misura prima el quadrilatero che puoi e, dipoi, quello che rima-<lb></lb> |
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|
3876 ne dividi sotilmente in triangoli e harai quello vuoi. Comme sia la figura .abc<lb></lb> |
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|
3877 dez. De’ quali .az. e .cd. sonno de linee rette e .zed. e .abc. curve. Prima truova l’ a-<lb></lb> |
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|
3878 l’ area del quadrilatero .zdac. E poi divide l’ uno e l’ altro arco in triangoli comme vedi ha-<lb></lb> |
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|
3879 ver fatto a me nela figura qui posta. E l’ area de’ detti triangoli coll’ area del quadrilatero<lb></lb> |
|
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|
3880 <lb></lb> |
|
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|
3881 <pb></pb></p> |
|
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|
3882 <p class="folio"> |
|
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|
3883 folio 26r |
|
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|
3884 </p> |
|
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|
3885 <p class="main"> |
|
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|
3886 <lb></lb> |
|
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|
3887 </p> |
|
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|
3888 <p class="runhead"> |
|
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|
3889 Distinctio quarta. Capitulum primum. 26 |
|
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|
3890 </p> |
|
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|
3891 <p class="main"> |
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|
3892 <lb></lb> |
|
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|
3893 agiongnerai e harai l’ area dela predetta figura la quale si nomina, per alcuno, figura con ven-<lb></lb> |
|
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|
3894 tre, alcuno, figura con arco e comme vuoi nominala. E cosí de molte altre figure ái a ffare,<lb></lb> |
|
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|
3895 cioé dividendola sempre in triangoli e l’ area de’ triangoli insiemi agiongni e harai quello<lb></lb> |
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|
3896 vai cercando et cetera. E peró questo sia abastanza ala predetta distintione. E, seguendo, dire-<lb></lb> |
|
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|
3897 mo dela .4a. distintione seguente. E col nome di Dio starai atento.<lb></lb> |
|
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|
3898 <lb></lb> |
|
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|
3899 </p> |
|
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|
3900 <p class="head"> |
|
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|
3901 Distinctio quarta. |
|
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|
3902 </p> |
|
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|
3903 <p class="main"> |
|
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3904 <lb></lb> |
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3905 Conclusiones libri tertij Euclidis cum eius diffinitionibus. Capitulum primum.<lb></lb> |
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3906 E gli é il circulo una figura piana contenta d’ una linea sola, detta circonferentia<lb></lb> |
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3907 overo periferia, nel mezzo dela quale è uno ponto che se dici centro di cerchio.<lb></lb> |
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3908 Dal qual centro tutte le linee che si menano infino ala circonferentia sonno igua-<lb></lb> |
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3909 li. E la magiore linea che vi cape se dici diametro di cerchio. El qual diametro<lb></lb> |
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3910 passa pel centro e, da ogni parte, tocca la circonferentia e dividelo in doi parti iguali, com-<lb></lb> |
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3911 me, nela figura posta in principio del primo capitulo dela prima distintione de questo trat-<lb></lb> |
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3912 tato, apare. Dove intendo in questa distintione quello che s’ apertiene a’ cerchi dichiarare. E<lb></lb> |
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|
3913 peró la divideró in .3. capitoli. Nel primo ponendo quello che Euclide nel terzo libro<lb></lb> |
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|
3914 dice. E nel secondo ció che di pratica a quelli s’ apartiene. Nel terzo (accioché dele su-<lb></lb> |
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|
3915 perficie si dica apieno) diremo di quelle in luogo alto e basso collocate, cioé di quelle che sonno<lb></lb> |
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3916 in monte. E peró adonca starai atento e al .po. capitulo daremo opera. Prima conclusione.<lb></lb> |
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3917 Li cerchi, de’ quali i diametri sonno iguali, fienno ancora iguali. E, quando e non<lb></lb> |
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3918 fossino iguali, quel diametro che è magiore, il suo cerchio sará magiore. E, simile,<lb></lb> |
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|
3919 minore, comme nel’ exemplo di fuor appare. 2<lb></lb> |
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3920 El cerchio che è detto contingere, cioé tocare la linea, tocca il cerchio in modo che, menan-<lb></lb> |
|
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3921 dola da ogni parte, mai sega il cerchio. 3<lb></lb> |
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3922 E cerchi che si toccano, detti contingenti, sonno quelli che, tocandosi, non si segano. 4<lb></lb> |
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3923 Le linee rette, nel cerchio, equedistante si dicano equedistare dal centro, quando dal centro a<lb></lb> |
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3924 quelle, se si mena le perpendiculari, fienno iguali. 5<lb></lb> |
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|
3925 Piú distante, se dici, dal centro stare la linea dala quale, menata dal centro, la perpendicu-<lb></lb> |
|
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|
3926 lare è magiore. 6<lb></lb> |
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|
3927 La linea retta contenente la portione del cerchio si nomina corda. 7<lb></lb> |
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3928 E la portione dela circonferentia é nominata arco. 8<lb></lb> |
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3929 L’ angolo dela portione di cerchio è quello che è contenuto dala corda e dal’ arco. 9<lb></lb> |
|
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3930 L’ angolo fatto al’ arco è quello che è fatto da .2. linee che escono de’ ponti dela corda ter-<lb></lb> |
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|
3931 minali al’ arco e vanno rette infino al’ arco. 10<lb></lb> |
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3932 Settore di cerchio è una figura che sotto .2. linee menate dal centro al’ arco é con-<lb></lb> |
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|
3933 tenuta. 11<lb></lb> |
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3934 L’ angolo fatto dele dette .2. linee è detto contenersi sopra il centro. 12<lb></lb> |
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|
3935 Simili sonno dette le portioni del cerchio de’ quali l’ angolo che è fatto sopra al’ arco de-<lb></lb> |
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3936 l’ una è iguali al’ angolo fatto sopra al’ arco del’ altra. 13<lb></lb> |
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|
3937 Gli archi sonno simili quando hano simili angoli. 14<lb></lb> |
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3938 Diffinito quello è necessario a’ cerchi, é da dimostrare le conclusioni e dimostrationi del<lb></lb> |
|
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|
3939 terzo de Euclide in questo modo. Prima conclusio. |
|
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|
3940 </p> |
|
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|
3941 <p class="main"> |
|
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3942 Sia proposto un cerchio del qual si voglia trovare il centro. Onde è manifesto che .2.<lb></lb> |
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3943 linee rette, in uno cerchio terminate nela circonferentia, l’ una non segherá mai l’ altra ortogo-<lb></lb> |
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3944 nalmente per mezzo, se lla non passa per lo centro. Comme sia il cerchio .abce. Del quale vo-<lb></lb> |
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3945 glio trovare il centro. Meneró in quel cerchio la linea .ac. comme venga. La qual, da ogni<lb></lb> |
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3946 parte, tochi la circonferentia. La qual divideró per igual parte nel ponto .d. Dal qual mene-<lb></lb> |
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3947 ró la perpendiculare ala detta linea. La qual perpendiculare, da ogni parte, tochi la circonfe-<lb></lb> |
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3948 rentia la quale sia .bde. La quale ancora divideró per igual parti nel ponto .f. Onde dico che’ l<lb></lb> |
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3949 ponto .f. è centro del detto cerchio, ch’ era de bisogno mostrare. 2<lb></lb> |
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3950 Se si menerá una linea retta da .2. ponti segnanti in sula circonferentia dall’ uno al’ altro,<lb></lb> |
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3951 sempre quella linea retta segherá il cerchio. Comme sia il cerchio .abd. Del quale il centro .c.<lb></lb> |
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3952 E sienno .2. ponti in sula circunferentia, cioé .a. e .b. segnati. Dali quali voglio menare la linea<lb></lb> |
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3953 retta. Dico quella linea segherá quel cerchio, ch’ era bisogni mostrare. 3<lb></lb> |
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3954 Se gli é una linea collocata in uno cerchio, la quale non passi per lo centro. E un’ altra si<lb></lb> |
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3955 meni dal centro e seghi quella per .2. parti iguali. Quella linea menata dal centro è perpen-<lb></lb> |
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3956 diculare al’ altra. Comme sia nel cerchio .bgd., collocata una linea .bg. la quale non passa<lb></lb> |
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|
3957 <lb></lb> |
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|
3958 <pb></pb></p> |
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|
3959 <p class="folio"> |
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|
3960 folio 26v |
|
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|
3961 </p> |
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|
3962 <p class="main"> |
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|
3963 <lb></lb> |
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|
3964 </p> |
|
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|
3965 <p class="runhead"> |
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|
3966 Distinctio quarta. Capitulum primum. |
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|
3967 </p> |
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|
3968 <p class="main"> |
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|
3969 <lb></lb> |
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|
3970 per lo centro. Un’ altra ne meno dal centro.p. e passa per lo ponto .e., lo quale é in mezzo dela<lb></lb> |
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|
3971 linea .bg. Dico la linea .pe. essere perpendiculare ala linea .bg. E ancora, quando la linea .pe.<lb></lb> |
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3972 viene dal centro e sia perpendiculare ala linea .bg., dico alora la linea .bg. essere divisa per .2.<lb></lb> |
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|
3973 parti iguali, ch’ era bisogno intendere. 4 Quando in uno cerchio sonno .2. linee che in-<lb></lb> |
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|
3974 fra loro si seghino e alcuna di loro non passi pel centro, certamente non si segheranno per par-<lb></lb> |
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|
3975 ti iguali. Comme sia nel cerchio .abcd., del quale il centro sia .e., la linea .ac. e .db., le quali in-<lb></lb> |
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3976 fra loro si seghino nel ponto .f. e non passi alcuna di loro per lo centro. Dico che .bf. nonn’ é igua-<lb></lb> |
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|
3977 li al .fd. né il .ef. al .fa., comme chiaro apare. 5 E centri di .2. cerchi che infra loro se inter-<lb></lb> |
|
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|
3978 segano, nonn’ é uno medesimo, ma sonno diversi. Comme sienno .2. cerchi .acb. e .abd. se-<lb></lb> |
|
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3979 gantesi sopra .2. ponti .a. e .b. Dico che ’loro centri sonno diversi. Imperoché, se fossino uno,<lb></lb> |
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3980 sia quello il centro .e. E menise le linee ala circonferentia, cioé .ea.ef.ec. Dove, per la diffinitio-<lb></lb> |
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3981 ne de’ cerchi .ef. e .ec., fienno iguali la parte al tutto, che è impossibile. E peró hano diversi<lb></lb> |
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3982 centri, cioé .p. e .q., ch’ era bisogno mostrare. 6<lb></lb> |
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3983 E cerchi che sonno contingenti infra loro non hanno uno medesimo centro, ma<lb></lb> |
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3984 diversi. Comme sienno .2. cerchi .ab. e .ac. che si toccano nel ponto .a. Dico che i<lb></lb> |
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3985 loro sonno diversi centri. Imperoché, se fossino uno medesimo, che pongo fosse<lb></lb> |
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3986 .d., e menise .da. e .dc. e .db. iguali, che è impossibile, cioé la parte al tutto. Adonca<lb></lb> |
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3987 hano centri diversi, che è .d. e .p., ch’ era bisogno mostrare. 7<lb></lb> |
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3988 Se nel diametro d’ un cerchio si segna un ponto fuor del centro e da quello ala<lb></lb> |
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3989 circonferentia piú linee si menino, quella che passerá per lo centro sirá magiore<lb></lb> |
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3990 d’ alcuna del’ altre. E quella che compirá el diametro sirá di ciascuna del’ altre mi-<lb></lb> |
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3991 nore. E quelle che piú s’ acostano al centro sonno magiori del’ altre che meno s’ a-<lb></lb> |
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3992 costano. E, quanto piú sonno remote dal centro, tanto sonno minori. E le .2. linee equedistan-<lb></lb> |
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3993 ti collaterali al ponto, cioé che gli angoli dal ponto exaversi sonno iguali, quelle .2. linee fien-<lb></lb> |
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3994 no iguali. Comme sia nel cerchio .aef. segnato nel diametro .a. il ponto .k. fuori del centro.<lb></lb> |
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3995 Dal quale si meni .ka.kb.kc.kd.ke.kf.kg. Dico che .ka. (perché passa per lo centro) è di<lb></lb> |
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3996 ciascuna magiore. E, perché .kf. è il compimento del diametro, é di ciascuna minore. E la linea<lb></lb> |
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3997 .kb., ch’ é piú presso al centro dela linea .kc., è magiore di quella. E la linea .kd., perché è piú re-<lb></lb> |
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3998 mota dal centro che la linea .kc., è minore di quella. E, per quel medesimo, la linea .ke. è mino-<lb></lb> |
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3999 re del .kd. E, perché le .2. linee collaterali equedistantti .ek. e .kg. hano gli angoli iguali, son-<lb></lb> |
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4000 no ancora infra loro iguali. E questo chiaro appare nella figura che dal vulgo è chiama-<lb></lb> |
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4001 ta zampa d’ ocha. 8<lb></lb> |
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4002 Se fuori d’ un cerchio è segnato uno ponto e da quello ala circonferentia piú li-<lb></lb> |
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4003 nee si menino segando il cerchio, quella che passerá per lo centro sirá piú lon-<lb></lb> |
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4004 ga. E, quanto piú passeranno a presso al centro, tanto fienno magiori del’ altre<lb></lb> |
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4005 che meno s’ apresseranno al centro. E, di quelle linee de fuori che fienno mena-<lb></lb> |
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4006 te insino ala circonferentia, quella che va diritto al diametro sia minore d’ alcune del’ altre e,<lb></lb> |
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4007 quanto da quella si scosteranno, tanto fienno magiori. E le .2. linee brevissime che igualmen-<lb></lb> |
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4008 te da quella si scostono fienno iguali. Comme sia segnato il ponto .a. fuor del cerchio .kb.<lb></lb> |
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4009 Dal qual ponto si meni la linea .akb. che passi sopra il diametro .knb. E menise la linea.<lb></lb> |
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4010 .ahc. e .agd. e .afe. Dico che la linea .akb. (perché passa pel diametro) è magiore di ciascu-<lb></lb> |
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4011 na del’ altre. E la linea .ahc. (perché piú s’ appressa al diametro) è magiore che l’ altre che me-<lb></lb> |
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4012 no s’ accostano. E la linea .afe. (perché meno s’ accosta al diametro) è minore di ciascuna de-<lb></lb> |
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4013 l’ altre. E ancora dico che la linea .ak., perché si posa in sul ponto dela circonferentia dove si<lb></lb> |
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4014 posa il diametro .kb., è minore di ciascuna del’ altre linee, cioé .ah. e .ag. e .af. E che .ab. è mi-<lb></lb> |
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4015 nore del .ag., perche piú s’ appressa a quel ponto. E che .ag. è minore del .af., perché piú s’ ap-<lb></lb> |
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4016 presa al detto ponto. E, perché .kh. e .kl. sonno iguali, dico che .al. e .ah. sonno iguali. E que-<lb></lb> |
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4017 sto chiaro appare per la figura presente che dal vulgo è chiamata coda de pavone.. 9.<lb></lb> |
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4018 Se da un ponto dentro ad alcun cerchio si puó menare piú di .2. linee infino ala<lb></lb> |
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4019 circonferentia che sienno iguali, quel ponto certamente è centro di quel cerchio.<lb></lb> |
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4020 Comme sia dato un ponto nel cerchio .bcd., il quale sia il ponto .a. E da quel pon-<lb></lb> |
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4021 to .a. infino ala circonferentia si meni .3. linee iguali, cioé .ad.ab.ac. Dico il ponto<lb></lb> |
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4022 .a. essere centro di quel cerchio. E questo chiaro appare per la presente figura. 10<lb></lb> |
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4023 Se uno cerchio sega un altro cerchio solamente, lo segherá in .2. luoghi, cioé in .2.<lb></lb> |
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4024 ponti. Comme sia il cerchio .abd. che sega il cerchio .abc. Dico solamente lo segherá<lb></lb> |
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|
4025 in .2. luoghi, cioé in .2. ponti .a. e .b. E questo chiaro appare per la figura che é qui posta. 11<lb></lb> |
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|
4026 <lb></lb> |
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|
4027 <pb></pb></p> |
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|
4028 <p class="folio"> |
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|
4029 folio 27r |
|
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4030 </p> |
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4031 <p class="main"> |
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|
4032 <lb></lb> |
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4033 </p> |
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|
4034 <p class="runhead"> |
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|
4035 Distincito quarta. Capitulum primum. 27. |
|
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4036 </p> |
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4037 <p class="main"> |
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|
4038 <lb></lb> |
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4039 Se uno cerchio è contingente a uno cerchio e da’ centri di quelli .2. cerchij si meni<lb></lb> |
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4040 una linea retta, cioé l’ uno centro al’ altro, la detta linea passerá certamente per<lb></lb> |
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4041 lo ponto del contato, cioé dove e cerchij detti si toccano. Comme sieno .2. cerchij<lb></lb> |
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4042 contingenti .cd. e .ce. De’ quali il ponto del contatto è il ponto .c. Dico che, menan-<lb></lb> |
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4043 do una linea dal centro .a. al centro .b., quella linea passa per lo ponto del contatto, cioé so-<lb></lb> |
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4044 pra il ponto .c. E, quando e cerchij si toccassino dentro, alora mena la linea dal’ uno centro a-<lb></lb> |
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4045 l’ altro infinitamente, quella passerá sopra il ponto .c., comme chiaro appare per la presente figura. 12.<lb></lb> |
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4046 Se uno cerchio toca un cerchio dentro over di fuori, solamente lo toccherá in<lb></lb> |
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4047 uno luogo. Comme sia il cerchio .ab. contingente il cerchio .ad. Dico solamen-<lb></lb> |
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4048 te nel ponto .a. lo toccherá. E questo anchora è assai chiaro e peró non bisogna<lb></lb> |
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|
4049 altra dimostratione.13. |
|
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|
4050 </p> |
|
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|
4051 <p class="main"> |
|
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4052 Le linee rette che sonno in uno cerchio, quando e lle sonno iguali, e lle se discosto-<lb></lb> |
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4053 no igualmente dal centro. Comme sia nel cerchio .adbc. Del quale il centro sia<lb></lb> |
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4054 .e. E sienno .2. linee .ad. e .bc. iguali. Dico che le si scostono igualmente dal centro<lb></lb> |
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4055 del detto cerchio, cioé dal centro .e., che in questo modo si pruova. Menise la per-<lb></lb> |
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4056 perdinculare dal ponto .e. a ciascuna linea, fienno le dette perpendiculari .ef. e .eg. Le quali di-<lb></lb> |
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4057 vidino le ditte linee in due parti iguali per la .3a. di questa. Dove l’ angolo .agc. è iguale al’ an-<lb></lb> |
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4058 golo .efa. Imperoché ciascuno è retto. E .gc. è iguali al .fa. E .ea. è iguali al .ec., perché cia-<lb></lb> |
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4059 scuna è mezzo diametro. Onde seguita, per la .46. del primo, .eg. essere iguali al .ef., ch’ era bi-<lb></lb> |
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4060 sogno mostrare. 14. |
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4061 </p> |
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4062 <p class="main"> |
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4063 Se in un cerchio dato vi sonno piú linee rette, el diametro è magiore d’ alcuna de-<lb></lb> |
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4064 l’ altre. E, quanto al diametro s’ apressano, tanto sonno magiori. Comme sia nel cer-<lb></lb> |
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4065 chio .ads. date le linee .gf.bc.hk. Dico el diametro .ad. essere magiore de ciascu-<lb></lb> |
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4066 na del’ altre. E la linea .fg. (perché è piú presso al diametro) é magiore del’ altre<lb></lb> |
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4067 .2. linee. E cosí .bc. (perché è piú presso al diametro .ad. che non è .hk.) é magiore di quel-<lb></lb> |
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4068 la. E questo chiaro appare. 15. |
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4069 </p> |
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4070 <p class="main"> |
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4071 Se dal termine d’ alcuno diametro si mena una linea ortogonalmente fatta, di-<lb></lb> |
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4072 co che la cade di fuori del cerchio. E infra quella e il cerchio non puó capire altra<lb></lb> |
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4073 linea. E l’ angolo fatto da quella e dala circonferentia è minore di tutti gli ango-<lb></lb> |
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4074 li acuti. E l’ angolo dentro, fatto dal diametro e la circunferentia, è magiore di tutti<lb></lb> |
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4075 gli angoli acuti. Onde, per questo, si manifesta tutte le linee rette, cioé sempre quella linea ret-<lb></lb> |
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4076 ta, dal termine d’ alcuno diametro ortogonalmente menata, sará al cerchio contingente.<lb></lb> |
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4077 Comme sia il cerchio .abc. Del qual sia il diametro .ac. Voglio sopra il ponto .a., che è termi-<lb></lb> |
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4078 ne di detto diametro, menare una linea ortogonalmente fatta, che sia .ae. Dico che, infra la<lb></lb> |
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4079 circonferentia e la detta linea, altra linea retta non puó cadere. Imperoché, se la cade, overo<lb></lb> |
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4080 sará ortogonalmente sopra il ponto .a. o non. Se è ortogonalmente fatta, è impossibile che da<lb></lb> |
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4081 un punto medesimo .2. linee, da una parte menate, faccino .2. angoli retti. Imperoché l’ uno<lb></lb> |
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4082 conterrebbe l’ altro e sarebono gli angoli retti infra loro non iguali, che è contra la peti-<lb></lb> |
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4083 tione de Euclide. E, se non facesse quella linea angol retto sopra il ponto .a., la quale sia la linea<lb></lb> |
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4084 .af., menise dal centro .d. la perpendiculare sopra .af. e sia .dg., sia adunque .dg. minore del .ad.<lb></lb> |
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4085 E, per la .46. del primo, .ad. puó quanto .dg. e .ga. E questo è impossibile. Imperoché .da. e<lb></lb> |
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4086 .do. sonno iguali e .dg. è magiore del .do. E peró, infra la detta linea e la circonferentia, al-<lb></lb> |
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4087 tra linea non puó entrare. E, evidentemente, appare che l’ angolo fatto da quella e dela cir-<lb></lb> |
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4088 conferentia é minore di tutti gli angoli acuti di .2. linee rette. E, similmente, l’ angolo fatto da-<lb></lb> |
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4089 la circonferentia e dal diametro, cioé dentro di tutti gli angoli acuti di .2. linee rette, è magio-<lb></lb> |
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4090 re. E ancora è assai chiaro che ogni linea retta, dal termine del diametro menata ortogonal-<lb></lb> |
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4091 mente, è contingente al detto cerchio. E pero non bisogna piú dimostrationi. 16<lb></lb> |
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4092 Dal dato ponto al dato cerchio voglio menare una linea contingente. Comme<lb></lb> |
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4093 sia il dato ponto .d. e il dato cerchio sia .ab. Del quale il centro sia .c. Voglio, dal<lb></lb> |
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4094 ponto .d., menare una linea contingente al cerchio .ab. Produrró .dc., segante il<lb></lb> |
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4095 cerchio .ab. nella circonferentia nel ponto .a. E, sopra il centro .c., righeró il cerchio<lb></lb> |
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4096 secondo la quantitá del .dc. e dal ponto .e. meneró la perpendiculare .ea. sopra la linea .dc.<lb></lb> |
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4097 E, dipoi, meneró la linea .ce., segante il cerchio .ab. nel ponto .b., dal qual .b. meneró la linea .bd.<lb></lb> |
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4098 La qual dico ch’ é contingente al circulo .ab., che chiaro appare.<lb></lb> |
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|
4099 <lb></lb> |
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|
4100 <pb></pb></p> |
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4101 <p class="folio"> |
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|
4102 folio 27v |
|
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4103 </p> |
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4104 <p class="main"> |
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4105 <lb></lb> |
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4106 </p> |
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4107 <p class="runhead"> |
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4108 Distinctio quarta. Captulum primum. |
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4109 </p> |
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4110 <p class="main"> |
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4111 <lb></lb> |
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4112 Se gli é una linea retta contingente il cerchio e, dal contatto al centro, si meni una<lb></lb> |
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4113 linea retta, e gli é di necessitá che lla sia perpendiculare sopra quella contingente,<lb></lb> |
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4114 comme sia la linea .ab. contingente el cerchio .ce. nel ponto .c. E menise dal ponto .c. una<lb></lb> |
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4115 linea retta infino al centro .f. Dico che la linea cf. è perpendiculare sopra la linea<lb></lb> |
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4116 .ab. E questo, per le cose dette, chiaro appare. 18<lb></lb> |
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4117 Se una linea retta è contingente a uno cerchio e, dal toccamento al cerchio, una<lb></lb> |
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4118 linea ortogonalmente si meni, certamente e lla passerá per lo centro. Comme<lb></lb> |
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4119 sia .ab., contingente el cerchio .ce. nel ponto .c. Dico che, menando al ponto .c. la<lb></lb> |
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4120 perpendiculare, certamente la passerá pel centro .d. E questo chiaro, per le passat-<lb></lb> |
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4121 e appare e d’ altra dimostratione non è bisogno. 19<lb></lb> |
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4122 Se in uno cerchio si fará uno angolo sopra il centro e uno altro sopra la circonfe-<lb></lb> |
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4123 rentia e habino una medesima basa, l’ angolo del centro è doppio al’ angolo de-<lb></lb> |
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4124 la circonferentia. Comme sia il cercho .abcf., del quale il centro .d. E faccia-<lb></lb> |
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4125 se l’ angolo .adc. sopra il centro e l’ angolo .abc. sopra la circonferentia, stando com-<lb></lb> |
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|
4126 me vedi figurato nel primo circulo, havendo cadauno angolo una medesima basa, che è .ac.,<lb></lb> |
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4127 corda del’ arco .afc. Dico l’ angolo .adc. essere doppio al’ angolo .abc., che cosí è de bisogno pro-<lb></lb> |
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4128 varlo. Meneró la linea .bde. Dove, per la .32a. del primo, l’ angolo .ade. è iguali agli .2. angoli,<lb></lb> |
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4129 cioé .dab. e .dba. Li quali angoli sonno iguali per la .5a. del primo, perché .ad. e .bd. sonno<lb></lb> |
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4130 iguali. Per la qual cosa, l’ angolo .ade. è doppio al’ angolo .dba. Simigliantemente sará l’ an-<lb></lb> |
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4131 golo .cde. 2 cotanti del’ angolo .dbc. E gli .2. angoli .ade. e .cde. sonno quanto l’ angolo .adc.<lb></lb> |
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4132 e peró tutto .adc. è doppio a tutto .abc., ch’ era bisogno mostrare.<lb></lb> |
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4133 E, se l’ angolo .abc. sta comme nela seconda figuratione appare, dico che l’ angolo<lb></lb> |
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4134 .adc. è doppio al’ angolo .abc. In questo modo lo proveró. L’ angolo .cda. è igua-<lb></lb> |
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4135 li al’ angolo .dbc. e al’ angolo .dcb. per la .32a. del primo. E gli .2. angoli .dbc.<lb></lb> |
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4136 e .dcb. sonno iguali per la .5a. del primo, imperoché .db. e .dc. sonno iguali.<lb></lb> |
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4137 Adunque l’ angolo .adc. è doppio al’ angolo .abc., ch’ era bisogno mostrare.<lb></lb> |
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4138 E, se l’ angolo .abc. sta comme nela terza figuratione appare, cioé che la linea .ab.<lb></lb> |
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4139 seghi la linea .dc., producase la linea .bde. Sirá l’ angolo .ade. iguali agli angoli<lb></lb> |
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4140 .dab. e .dba. Li quali infra loro sonno iguali. E l’ angolo .cde. è, per la medesima,<lb></lb> |
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4141 iguali al’ angolo .dcb. e .dbc., che sonno infra loro iguali. Adunque l’ angolo .cde.<lb></lb> |
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4142 è doppio al’ angolo .dbc. E detto é che l’ angolo .ade. è doppio al’ angolo .abd. Adunque,<lb></lb> |
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4143 tratto l’ angolo .ade. del’ angolo .cde., rimarrá l’ angolo .adc. E, tratto l’ angolo .abd. del’ an-<lb></lb> |
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4144 golo .cbd., rimarrá l’ angolo .cba. Adunque l’ angolo .adc. è doppio al’ angolo .abc. ch’ era bi-<lb></lb> |
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4145 sognio mostrare. 20<lb></lb> |
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4146 Se in una portione di cerchio fienno molti angoli sopra la circonferentia fatti,<lb></lb> |
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4147 fienno infra loro iguali. Comme sia nela portione di cerchio .abcde., sienno mol-<lb></lb> |
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4148 ti angoli, cioé .acb. e .adb. e .aeb. Dico che infra loro sonno iguali. E questa, per<lb></lb> |
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4149 la passata, chiaro appare. E, volendo provare, facciase l’ angolo in sul centro, do-<lb></lb> |
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4150 ve l’ angolo del centro è doppio a ciascuno angolo. E peró non è de bisognio altra dimo-<lb></lb> |
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4151 stratione. |
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4152 </p> |
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4153 <p class="main"> |
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4154 Se in uno cerchio si scrive uno quadrilatero, e .2. angoli aversi sonno iguali a .2.<lb></lb> |
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4155 angoli retti. Sia il cerchio .abcd., nel quale sia collocato il quadrilatero .abcd.<lb></lb> |
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4156 Dico li .2. angoli aversi qual vuoi sonno iguali a .2. retti, che in questo modo lo<lb></lb> |
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4157 proveró. Menise nel detto quadrilatero il diametro .ac. e ancora il diametro .bd. Sirá, per<lb></lb> |
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4158 la passata, l’ angolo .cbd. iguali al’ angolo .cad. e l’ angolo .abd. iguali al’ angolo .acd. Onde<lb></lb> |
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4159 tutto .abc. è iguali a’ .2. angoli, che sonno .acd. e .cad. E, perché con l’ angolo .adc. sonno igua-<lb></lb> |
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4160 li a .2. retti per la .32a. del primo, e peró fienno li .2. angoli del quadrilatero, cioé .b. e .d., iguali<lb></lb> |
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4161 a .2. retti, che è il proposito. E, similmente, gli altri .2. angoli .a. e .c., per lo detto modo, sonno<lb></lb> |
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4162 iguali a .2. retti. Gran forza certamente se demostra per questa .21. conclusione, conciosia-<lb></lb> |
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4163 ché li angoli de un quadrilatero exadverso collocati (ut premittitur) si possino infra loro<lb></lb> |
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4164 infinitamente variare .vz. secundum omnes angulorum species .s. acutos obtusos e rectos,<lb></lb> |
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4165 nil minus semper idem quod dicitur eveniet. La qual evidentia in la pratica operativa molto vale. Ideo memorie manda <lb></lb> |
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4166 et <lb></lb> |
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4167 cetera. 22<lb></lb> |
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4168 Due portioni simili e inequali sopra una medesima parte cadere è impossibile.<lb></lb> |
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4169 Comme sia la corda .ab., sopra la quale si faccia una portione di cerchio .bca.<lb></lb> |
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4170 <lb></lb> |
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4171 <pb></pb></p> |
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4172 <p class="folio"> |
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4173 folio 28r |
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4174 </p> |
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4175 <p class="main"> |
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4176 <lb></lb> |
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4177 </p> |
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4178 <p class="runhead"> |
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4179 Distinctio quarta. Capitulum primum. 28 |
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4180 </p> |
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4181 <p class="main"> |
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4182 <lb></lb> |
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4183 Dico che gli é impossibile fare una portione di cerchio simile a quella e non iguale che ha-<lb></lb> |
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4184 bia la detta corda. E, se gli é possibile, facciase la portione .abd. E faciase l’ angolo .c. in sula cir-<lb></lb> |
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4185 conferentia .acb. e l’ angolo .adb. si facia in sula circonferentia .adb. E, perché le portioni son-<lb></lb> |
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4186 no simili, gli angoli fienno simili per la diffinitione. E questo per la .16a. del primo. L’ angolo .c.<lb></lb> |
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4187 magiore del’ angolo .d. potrebese dire, o se l’ angolo .c. fosse in sula faccia overo linea .ad. An-<lb></lb> |
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4188 cora, per la detta .16a., l’ angolo .c. sarebbe magiore del’ angolo .d. E, quando le portioni sonno<lb></lb> |
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4189 simili, hano simile proportione. Adunque non sarebono le portioni simili. Ancora potresti<lb></lb> |
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4190 dire per lo terzo modo: cioé che la linea .ab. dividesse overo segasse la linea .cb. nel ponto .f.<lb></lb> |
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4191 e tu rispondi non potere, per niun modo, le portioni essere simili. Imperoché io faró l’ ango-<lb></lb> |
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4192 lo .aeb. nella portione .abc. El quale angolo, per la .20. di questo, sia iguale al’ angolo .acb. E, per<lb></lb> |
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4193 la .16a. del primo, diremo l’ angolo .aeb. essere magiore del’ angolo .adb. E, per questo, l’ ango-<lb></lb> |
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4194 lo .acb. è magiore del’ angolo .adb. Adonque non sonno le portioni simili. 23<lb></lb> |
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4195 Se simili portioni di cerchij sonno sopra iguali linee e quelle portioni fienno igua-<lb></lb> |
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4196 li. Comme sienno .2. linee iguali .ab. e .cd. E sopra quelle si faccia .2. portioni di<lb></lb> |
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4197 cerchio a ciascuna la sua simile. Dico quelle portioni sonno iguali. E questo chia-<lb></lb> |
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|
4198 ro apare per lo detto dele passate. 24<lb></lb> |
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4199 Sia data una parte di cerchio. Voglio con quella compire il cerchio. Cioé, a ogni<lb></lb> |
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4200 arco, trovare il modo a compirlo. El quale è questo. Sia adunque .ab. alcuno ar-<lb></lb> |
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4201 co, del quale si voglia compire il cerchio. Meneró in quello .2. linee. comme vie-<lb></lb> |
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4202 ne, che sonno .ac. e .bd. intendi rette. Le quali divideró per igual parti, .ac. nel pon-<lb></lb> |
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4203 to .e. e .bd. nel ponto .f. E meneró .eg. perpendiculare ala linea .ac., faciendola magiore del<lb></lb> |
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4204 diametro del cerchio. E ancora meneró sopra la linea .db. la perpendiculare .hf., la quale seghe-<lb></lb> |
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4205 rá la perdendiculare .eg. nel ponto .k. El quale ponto .k., per la prima di questo, è centro, per-<lb></lb> |
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4206 ché ciascuna dele perpendiculari passano sopra il centro. E cosí hai trovato e finito il detto<lb></lb> |
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4207 tondo. Ma, alcuna volta, interviene che la parte del cerchio è magiore del mezzo cerchio<lb></lb> |
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4208 e le linee fatte date sonno equedistanti. Alora le perpendiculari fienno una linea insiemi.<lb></lb> |
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4209 Per la qual cosa è de bisogno pigliare il mezzo dela linea .ef. e sia il ponto .k. E cosí il ponto<lb></lb> |
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4210 .k. sia centro del detto cerchio. 25<lb></lb> |
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4211 Se in .2. cerchi iguali, over sopra la circonferentia overo sopra il centro,, si fan-<lb></lb> |
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4212 no angoli iguali, quelli archi fienno iguali. Comme sienno .2. cerchi .abc. e .efg.<lb></lb> |
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4213 E faciase in ciascun un angolo in sula circonferentia e sienno .c. e .f. overo in sul centro e<lb></lb> |
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4214 sienno .d. e .h. E sia l’ angolo .c. iguale al’ angolo .f. E l’ angolo .d. iguale al’ angolo .h.<lb></lb> |
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4215 Dico l’ arco .acb. essere iguale al’ arco .efg. E questo chiaro appare per lo detto dele passate. 26<lb></lb> |
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4216 Se infra .2. cerchi iguali si toglie archi iguali e faciase in quelli .2. angoli, overo<lb></lb> |
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4217 in sula circonferentia overo in sul centro, dico che li angoli del’ uno fienno iguali ali an-<lb></lb> |
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4218 goli del’ altro. Comme sienno .2. cerchi iguali .abc. e .gef. E piglise .2. archi igua-<lb></lb> |
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4219 li, cioé .abc. e .efg. E faciase .2. angoli in ciascuno: uno in sula circonferentia e si-<lb></lb> |
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4220 enno .f. e .b. Overo in sul centro e sienno .h. e .d. Dico l’ angolo .f. essere iguali al’ angolo .b. O-<lb></lb> |
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4221 vero l’ angolo .h. essere iguale al’ angolo .d. E questo ancora per le cose dette chiaro appare. 27<lb></lb> |
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4222 Se in cerchi iguali le linee iguali risegano e faccino archi, quelli archi fienno igua-<lb></lb> |
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4223 li. E, se le linee non iguali in cerchi iguali risegono e fanno archi, quelli archi fien-<lb></lb> |
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4224 no non iguali. Imperoché la magiore linea fará magiore arco e la minore mi-<lb></lb> |
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4225 nore arco. Comme sia .2. circoli: .abc., del quale sia il centro .d. e .efg., del quale il<lb></lb> |
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4226 centro .h. E sia la corda .ac. iguale ala corda .eg. Dico l’ arco .acb. essere iguale al’ arco .efg. E<lb></lb> |
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4227 similmente dico che la corda .ki., che è magiore della corda .pq. e ciascuna, nel’ iguale cerchio,<lb></lb> |
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4228 fa arco. Dico l’ arco .kio. essere magiore che l’ arco .pqr. E questo chiaro sanza altra dimo-<lb></lb> |
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4229 stratione appare. 28<lb></lb> |
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4230 Gli archi iguali de iquali cerchi è necessario habino iguali corde. Comme sien-<lb></lb> |
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4231 no .2. cerchi: .abc., del quale è il centro .d. e .efg., del quale è il centro .h. E sia l’ arco<lb></lb> |
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4232 .abc. iguale al’ arco .efg., dico che la corda .ac. è iguale ala corda .eg. Questa è con-<lb></lb> |
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4233 versa alla passata. E peró, per quella, chiaro questa se dimostra essere vera e peró<lb></lb> |
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4234 la demostratione lasceremo. E tutte quelle passioni che fin qua sonno state demostra-<lb></lb> |
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4235 te e provate essere vere de diversi cerchi, molto magiormente se convenceranno essere vere<lb></lb> |
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4236 de uno medesimo cerchio, arguendo sempre contra l’ aversario, commo nelle precedenti<lb></lb> |
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4237 s’ é fatto. Ideo et cetera. 29<lb></lb> |
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4238 <lb></lb> |
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4239 <lb></lb> |
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4240 <pb></pb></p> |
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4241 <p class="folio"> |
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4242 folio 28v |
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4243 </p> |
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4244 <p class="main"> |
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4245 <lb></lb> |
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4246 </p> |
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4247 <p class="runhead"> |
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4248 Distinctio quarta. Capitulum secundum. |
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4249 </p> |
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4250 <p class="main"> |
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4251 <lb></lb> |
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4252 Io voglio dividere uno arco in .2. parti iguali. Sia il dato arco .abc., del quale<lb></lb> |
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4253 la corda .ac. Voglio dividere il detto arco in .2. parti iguali, che in questo modo lo<lb></lb> |
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4254 faró. Divideró la corda .ac. in .2. parti iguali al ponto .d., ala quale si meni una per-<lb></lb> |
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4255 pendiculare, la quale caggia per lo ponto .d. e sia .db., la quale seghi la circonferen-<lb></lb> |
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4256 tia del dato arco nel ponto .b. La quale perpendiculare á diviso lo detto arco in .2. parti igua-<lb></lb> |
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4257 li, comme volavamo, che cosí te ’l proveró. Menise la linea .ab. e la linea .bc., le quali saranno,<lb></lb> |
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4258 per la quarta del primo, iguali, imperoché .bd. e .da., del triangolo .bda., é iguali al lato<lb></lb> |
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4259 .bd. e .dc., del triangolo .bdc. E l’ angolo .d. di ciascuno triangulo è retto. E peró lo lato .ab.<lb></lb> |
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4260 del’ uno sia iguale alo lato .bc. del’ altro. E, per la .27a. di questo, l’ arco .ab. è iguale al’ arco .bc.,<lb></lb> |
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4261 che è il proposito. 30<lb></lb> |
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4262 Se uno angolo de linee rette nel mezzo cerchio è fatto, el quale sia fatto nel ‘arco<lb></lb> |
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4263 cierto, quello angolo è retto. E, se la portione del cerchio dove el’ angolo è magio-<lb></lb> |
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4264 re del mezzo cerchio, alora quel angolo sia minore che ’l retto. E, se la portione<lb></lb> |
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4265 del cerchio dove el’ angolo è minore del mezzo cerchio, alora quello angolo è<lb></lb> |
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4266 magiore che ’l retto. E cosí è converso: quando l’ angolo fatto nel’ arco è retto, alora quella<lb></lb> |
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4267 portione di cerchio è mezzo cerchio. E, se l’ angolo è magiore del retto, quella portione sirá<lb></lb> |
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4268 minore di mezzo cerchio. E, se l’ angolo è minore che ’l retto, la portione sia magiore del mezzo<lb></lb> |
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4269 cerchio. Comme sia il cerchio .abc., del quale il centro .d. e il diametro .adc. e faciase nel mez-<lb></lb> |
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4270 zo cerchio .abc., in sula circonferentia l’ angolo .abc., menate le linee .ab. e .bc. Dico l’ angolo<lb></lb> |
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4271 .abc. essere retto, che in questo modo lo proveró. Menise dal centro .d. la linea .db., siran, per<lb></lb> |
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4272 la .5a. del primo, l’ angolo .abd. e .dab. infra loro iguali. E, perché .bdc. è iguale a’ .2. angoli<lb></lb> |
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4273 .dba. e .dab., per la .32a. del primo e l’ angolo .adb. è doppio al’ angolo .dbc., per la .32a. del<lb></lb> |
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4274 primo, imperoché l’ angolo .dbc. è iguale al’ angolo .dcb., adunque e .2. angoli .cdb. e .adb.<lb></lb> |
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4275 sonno doppi a tutto l’ angolo .abc. Ma quelli .2. angoli, per la .13a. del primo, sonno iguali a<lb></lb> |
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4276 .2. retti. Adunque l’ angolo .abc. è il .1/2. di .2. angoli retti. E peró e gli é retto, che è il primo pro-<lb></lb> |
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4277 posito. Ancora altramente menise .cb. infino al .e., sirá, per la .32a. del primo, l’ angolo .abe.<lb></lb> |
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4278 iguale a’ .2. angoli .a. e .c. E, perché l’ angolo .a. è iguale al’ angolo .abd. e l’ angolo .abd. e l’ ango-<lb></lb> |
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4279 lo .c. è iguali al’ angolo .cbd., sirá l’ angolo .abe. iguale al’ angolo .abc. Adunque l’ angolo<lb></lb> |
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4280 .aeb. e l’ angolo .abc. ciascuno è retto per la diffinitione. E ’l secondo proposito cosí è manife-<lb></lb> |
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4281 sto. Sia nel cerchio .abc., del quale nel centro .d. la portione .abc. magiore del mezzo cerchio<lb></lb> |
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4282 e faciase sopra la circunferentia l’ angolo .abc., menate le linee .ba. e .bc. Dico quello angolo es-<lb></lb> |
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4283 sere minore d’ uno angolo retto. Menise il diametro .ade. e la linea .eb. Sirá, per la prima<lb></lb> |
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4284 parte di questa, l’ angolo .b. tutto retto. Per la qual cosa l’ angolo .abc. è minore per la comu-<lb></lb> |
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4285 ne scientia, perché e gli é parte di quello. Onde seguita il proposito. E ’l terzo si manifesta cosí.<lb></lb> |
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4286 Sia un’ altra volta nel cerchio .abc., del quale il centro .d., una portione di cerchio minore<lb></lb> |
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4287 del .1/2. cerchio. Facciasi sopra la circonferentia l’ angolo .abc., menate le linee .ba. e .bc. Dico que-<lb></lb> |
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4288 sto angolo essere magiore che ’l retto in questo modo. Menise il diametro .ade. e la linea<lb></lb> |
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4289 .be. Sirá, per la prima parte di questa, l’ angolo .abe. retto. Onde l’ angolo .abc. è magiore<lb></lb> |
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4290 che ’l retto, che è il terzo proposito. Gli altri propositi, per loro medesimi, si dichiarano.] 31<lb></lb> |
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4291 S e una linea è contingente a uno cerchio e dal ponto del contato nel cerchio si<lb></lb> |
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4292 meni una linea la quale non passi pel centro del cerchio, dico che l’ angolo, fatto<lb></lb> |
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4293 nela portione che fa la detta linea, sia iguale al’ angolo de fuori fatto dala linea<lb></lb> |
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4294 contingente e dala linea menata nel cerchio. Comme sia il cerchio .cdef., al qua-<lb></lb> |
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4295 le sia la linea .ab. contingente. E sia .g. il centro del detto cerchio. E .d. sia il ponto del contato.<lb></lb> |
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4296 E menise la linea .df. che non passi per lo centro .g. Dico che, facendo un angolo in sula por-<lb></lb> |
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4297 tione del .fed., comme l’ angolo .def., sia iguale al’ angolo .fda. E ancora si faccia un angolo<lb></lb> |
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4298 in sula portione .fcd. e sia l’ angolo .fcd., el quale dico essere iguali al’ angolo .fdb., che cosí te ’l<lb></lb> |
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4299 proveró. Menise il diametro .dgh., sirá, per la .17a. di questo, .bd. perpendiculare sopra la li-<lb></lb> |
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4300 nea .ab. E faciase .fh., sirá, per la prima parte dela passata, l’ angolo .dfh. retto. Adunque e .2.<lb></lb> |
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4301 l’ angoli .adh. e .dfh. sonno retti ciascuno, adunque sonno iguali. Onde, agionto a ciascuno<lb></lb> |
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4302 l’ angolo .hdf., sia l’ angolo .adf. iguale a .2. angoli, che sonno .dfh. e .hdf. Ma questi .2., con<lb></lb> |
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4303 l’ angolo .dhf. sonno iguali a .2. retti, imperoché, per la .32a. del primo, li .3. angoli del trian-<lb></lb> |
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4304 golo .dhf. sonno iguali a .2. retti. Ma l’ angolo .adf., con l’ angolo .bdf., è iguali a .2. retti, per<lb></lb> |
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4305 la .13a. del primo. Adunque l’ angolo .bdf. è iguale al’ angolo .dhf. E l’ angolo .dhf. è iguale<lb></lb> |
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4306 al’ angolo .c., imperoché in una medesima portione di cerchio sonno collocati, comme appare<lb></lb> |
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4307 <lb></lb> |
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4308 <pb></pb></p> |
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4309 <p class="folio"> |
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4310 folio 29r |
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4311 </p> |
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4312 <p class="main"> |
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4313 <lb></lb> |
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4314 </p> |
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4315 <p class="runhead"> |
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4316 Distinctio quarta. Capitulum secundum. 29 |
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4317 </p> |
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4318 <p class="main"> |
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4319 <lb></lb> |
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4320 per la .20a. di questo. E cosí è provato l’ angolo .c. essere iguale al’ angolo .bdf. Ora ci resta a<lb></lb> |
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4321 provare comme l’ angolo .ced. è iguale al’ angolo .fda. Per la .21a. di questo, e .2. angoli .e. e .c.<lb></lb> |
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4322 sonno iguali a .2. retti e, per la .13a. del primo, gli angoli bdf. e .fda. sonno iguali a .2. retti. E<lb></lb> |
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4323 provato habiamo che l’ angolo .fdb. è iguale al’ angolo .c. Adunque, per la commune scientia,<lb></lb> |
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4324 l’ angolo .e. è iguali al’ angolo .fda., che è quello volavamo. 32<lb></lb> |
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4325 Sopra una linea data voglio scrivere una portione di cerchio, nel quale voglio<lb></lb> |
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4326 collocare uno angolo iguale a uno angolo dato. Comme sia .ab. la data linea e<lb></lb> |
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4327 .c. il dato angolo. Voglio sopra la linea .ab. fare una portione de cerchio nela<lb></lb> |
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4328 quale io possi fare uno angolo iguale a l’ angolo .c. dato. Puó essere l’ angolo .c. ret-<lb></lb> |
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4329 to, obtuso e acuto. Sia prima retto: dividerai la linea .ab. in due parti iguali sopra il ponto<lb></lb> |
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4330 .d. e faciase .d. centro e scrivase il cerchio secondo la quantitá del .db. e sopra la portione .dab.,<lb></lb> |
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4331 che è mezzo cerchio, farasse l’ angolo .h., lo quale, per la .30a. di questo, è retto. E peró seguita il<lb></lb> |
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4332 proposito. Ma, se l’ angolo .c. è obtuso, meneró la linea .da. con la linea .ba. causante l’ angolo<lb></lb> |
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4333 .dab. iguale al’ angolo .c. E, dal ponto .a., meneró la linea .ae. perpendiculare sopra la linea<lb></lb> |
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4334 .ad. E, sopra il ponto .b., faró l’ angolo iguale al’ angolo .eab. El quale angolo è quello che lo<lb></lb> |
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4335 obtuso avanza al retto. Menato la linea .bf. infino ala perpendiculare .ae., saran, per la .6a.<lb></lb> |
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4336 del primo, .fb. e .fa. iguali. Fatto adunque il ponto .f. centro, faró il cerchio secondo la quan-<lb></lb> |
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4337 titá del .fa. e sia cerchio .akb. Dove, per la .15a. di questo, la linea .gad. è contingente al cer-<lb></lb> |
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4338 chio detto. Onde l’ angolo fatto nella proportione .akb. è iguale al’ angolo .bad., dove è igua-<lb></lb> |
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4339 le al’ ango[lo] .c. dato. E, se l’ angolo .c. fosse acuto: faró la linea .ag. causante, con la linea .ab., ango-<lb></lb> |
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4340 lo iguale al’ angolo dato, cioé al .c. E, dal ponto .a., meneró .ae. perpendiculare ala linea .ag.<lb></lb> |
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4341 E, sopra il ponto .b., faró l’ angolo iguale al’ angolo .eab., nel quale il retto avanza l’ angolo a-<lb></lb> |
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4342 cuto dato. E produrró la linea .bf. infino ala perpendiculare .ae. Siran, per la .6a. del primo,<lb></lb> |
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4343 .fa. e .fb. iguali. Fatto adunque .f. centro del cerchio, descriveró secondo la quantitá dela li-<lb></lb> |
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4344 nea .fa. el cerchio .abk. Sirá, per lo correlario di questa e per la .15a. di questo, la linea .ag. con-<lb></lb> |
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4345 tingente el cerchio. Per la qual cosa, per la passata, l’ angolo fatto nela portione .akb. è igua-<lb></lb> |
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4346 le al’ angolo .gab. Per la qual cosa è al’ angolo .c. che è il proposito. 33<lb></lb> |
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4347 Sia dato uno cerchio del quale voglio pigliare una portione. Nela quale si faccia<lb></lb> |
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4348 un angolo iguale a un angolo dato. Sia .c. il dato angolo e .ab. il dato cerchio.<lb></lb> |
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4349 Voglio del cerchio .ab. tagliare una portione nela quale si possi fare un angolo<lb></lb> |
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4350 iguale al’ angolo .c. Produrró la linea .dae. contingente il dato cerchio nel ponto<lb></lb> |
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4351 .a. Dal qual meneró nel cerchio la linea .ab. continente, con la linea .ae., un angolo iguale al’ an-<lb></lb> |
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4352 golo .c. Sirá, per la .31a. di questo, la portione .apb. atta a ffarvi uno angolo iguale al’ ango-<lb></lb> |
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4353 lo dato, che è il proposito. 34<lb></lb> |
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4354 Se in uno cerchio .2. rette linee si segano, quello ch’ é fatto d’ una parte d’ una linea<lb></lb> |
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4355 nel’ altra parte de medesima linea é iguale a quello che è fatto dela parte del’ al-<lb></lb> |
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4356 tra linea nel’ altra parte di quella. Voglio, lettore, che questa conclusione, fra l’ altre, te sia<lb></lb> |
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4357 molto familiare, peroché l’ é quella che a molti casi de grandissima importanza ci aiuta, e<lb></lb> |
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4358 in pratica e anche in theorica. E dali antichi philosophy è stata molto adoperata maxime da Ptolomeo,<lb></lb> |
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4359 per tutto il suo almegesto, e in le formationi dele tavole de corda et arcu e ale pratiche vulgari,<lb></lb> |
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4360 per li scemi de ciascuna portione de cerchio, in sapere finire el suo totale diametro. E, da questa nasci<lb></lb> |
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4361 la regola comuna che, data la corda e la saetta, per quella si pó facilmente, con lo presuposito di questa<lb></lb> |
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4362 .34a., trovare l’ avanzo del diametro del cerchio di quella tal portione, o sia magiore o sia minore de<lb></lb> |
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4363 mezzo cerchio. E peró notala bene. Or torniamo ala sua expositione. Comme sienno .2. linee .c. e<lb></lb> |
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4364 .bd., nel cerchio .abcd. sopra il ponto .e. Dico che quello rettangolo che è fatto dal .ae. in .ec. è<lb></lb> |
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4365 iguale a quello ch’ é fatto dal .be. in .ed. Overamente amendui le linee passeranno per lo cen-<lb></lb> |
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4366 tro del detto cerchio overo una over niuna. Prima passino tutte .2., comme nela figura<lb></lb> |
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4367 prima è manifesto. Adunque il ponto .e. sia centro. Adunque tutte .4. le dette linee fienno igua-<lb></lb> |
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4368 li per la diffinitione del cerchio, adunque il proposito seguita. E, se una sola passerá per lo cen-<lb></lb> |
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4369 tro, e sia quella .bd. e il centro del cerchio sia .f. Overamente .bd. segherá la linea .ac. in .2.<lb></lb> |
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4370 parti iguali overo in .2. parti non iguali. Prima sega in .2. parti iguali. Sirá adunque, per<lb></lb> |
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4371 la prima di questo, segata ortogonalmente. Onde meneró .fc., sirá, per la quinta del secondo,<lb></lb> |
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4372 quello ch’ é fatto del .be. in .ed., col quadrato .ef., iguale al quadrato .fd., cioé al quadrato .fc.<lb></lb> |
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4373 Adunque, quello ch’ é fatto del .eb. in .ed. col quadrato del .fe. sia iguali al quadrato .fc. Adon-<lb></lb> |
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4374 que quello che fatto del .fe. in sé e .ec. in sé, per la penultima del primo, è iguali al quadrato<lb></lb> |
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4375 del .fc. Onde quello ch’ é fatto del .eb. in .ed., col quadrato .fe., è iguali al quadrato .fc., cioé a’ .2.<lb></lb> |
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4376 <lb></lb> |
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4377 <pb></pb></p> |
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4378 <p class="folio"> |
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4379 folio 29v |
|
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4380 </p> |
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4381 <p class="main"> |
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4382 <lb></lb> |
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4383 </p> |
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4384 <p class="runhead"> |
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4385 Distinctio quarta. Capitulum secundum. |
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4386 </p> |
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4387 <p class="main"> |
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4388 <lb></lb> |
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4389 quadrati .fe. e .ec. Tratto adunque di ciascuno lo quadrato .fe., rimane quello che è<lb></lb> |
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4390 fatto dai .be. in .ed. iguali al quadrato .ec. E, perché .ec. è iguali al .ea., sirá adunque<lb></lb> |
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4391 il proposito. E, se il .bd. passerá per lo centro e segherá .ac. per parti non iguali nel ponto .e.,<lb></lb> |
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4392 menerai dal centro .f.fg., perpendiculare al .ca. Sirá, per la seconda parte dela .3 a. di questo,<lb></lb> |
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4393 .ag. iguali al .cg. E menise la linea .fc., sirá, per la .5 a. del secondo, quello che è fatto del .be. nel<lb></lb> |
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4394 .ed., col quadrato dela linea .ef., iguale al quadrato .df. e, per la penultima del primo, a’ .2. qua-<lb></lb> |
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4395 drati .fg. e .gc., perché l’ angolo .fgc. è retto. E ancora l’ angolo .fge. è retto. Adunque quel<lb></lb> |
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4396 ch’ é fatto del .be. in .ed., coi quadrati .fg. e .ge., è iguali al quadrato .fd., cioé al quadrato .fc. E<lb></lb> |
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4397 il quadrato dela linea .fc. é quanto e .2. quadrati .fg. e .gc. Onde, tratto di ciascuna parte el qua-<lb></lb> |
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4398 drato .fg., rimarrá quello che è fatto del .be. in .ed., col quadrato .eg., iguali al quadrato .gc.<lb></lb> |
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4399 Ma, per la .5a. del secondo, quello ch’ é fatto del .ae. in .ec., col quadrato dela linea .ge., è igua-<lb></lb> |
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4400 li al quadrato dela linea .gc. Adunque, quel ch’ é fatto del .be. in .ed., col quadrato .ge., è igua-<lb></lb> |
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4401 li a quello ch’ é fatto del .ce. in .ea., col quadrato dela linea .ge. Tratto adunque di ciascuno<lb></lb> |
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4402 il quadrato .ge., sia quello ch’ é fatto del .be. in .ed. iguali a quello ch’ é fatto del .ae. in .ec., che è<lb></lb> |
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4403 il proposito. E, se niuna delle dette linee passerá per lo centro over l’ una dividerá l’ altra per<lb></lb> |
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4404 parti iguali overo per parti non iguali. Prima dividi la .bd. linea .ac. nel ponto .e. E sia<lb></lb> |
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4405 .ae. quanto .ec.; produrró .gfeh., diametro del cerchio passante per lo ponto .e. E divideró an-<lb></lb> |
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4406 cora .ac. per igual parti. Onde lo dividerá ortogonalmente, per la terza di questo. Adunque,<lb></lb> |
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4407 per lo secondo modo di questa conclusione, quello che è fatto del .ge. in .eh. è igual a quel<lb></lb> |
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4408 ch’ é fatto del .ae. in .ec. E, per lo terzo modo di questa, quello ch’ é fatto del .ge. in .eh. è iguali<lb></lb> |
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4409 a quello ch’ é fato del .be. in .ed. Adunque, quello che è fatto del .ae. in .ec. è iguale a quello<lb></lb> |
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4410 ch’ é fatto del .be. in .ed., per la conceptione prima, che è il proposito. E, se niuna dividerá l’ altra<lb></lb> |
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4411 per igual parti, comme sia .ae. divisa per .bd. nel ponto .e. per parti non iguali, meneró<lb></lb> |
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4412 .gfeh., diametro, sirá, per lo terzo modo di questa, quello che è fatto del .ae. in .ec. iguali a quel-<lb></lb> |
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4413 lo ch’ é fatto del .ge. in .eh. E ancora quel ch’ é fatto del .be. in .ed. è iguali, per lo terzo modo<lb></lb> |
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4414 di questa, a quello ch’ é fatto del .ge. in .eh. Onde seguita, quando di .2.cose. ciascuna è iguali<lb></lb> |
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4415 a una ch’ é infra loro, sienno iguali per la conceptione. E, peró, tanto è il fatto del .ae. in .ec., quan-<lb></lb> |
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4416 to il fatto del .be. in .ed., che è il proposito. 35<lb></lb> |
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4417 Se si segnerá fuor d’ un cerchio un ponto e, da quello al cerchio, .2. linee rette si me-<lb></lb> |
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4418 nino, dele quali l’ una seghi il cerchio, l’ altra sia contingente al detto cerchio, quel-<lb></lb> |
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4419 lo che è fatto di tutta la linea segante nela parte di fuori è iguali al quadrato de-<lb></lb> |
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4420 la linea contingente. Comme sia il cerchio .bcd., fuor del quale é il ponto .a. Dal qua-<lb></lb> |
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4421 le si menino .2. linee: .ab. contingente e .adc. segante. Dico che quel ch’ é fatto del .ac. in .ad. è<lb></lb> |
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4422 iguali al quadrato del .ab. Overamente .adc. passerá per lo centro o non. Passi prima per<lb></lb> |
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4423 lo centro che sia .e. e faciase la linea .eb. la qual, per la .27a. di questo, è perpendiculare sopra la<lb></lb> |
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4424 linea .ab. E, perché la linea .dc. è divisa per igual parti nel ponto .e. e a quella è agionta la linea<lb></lb> |
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4425 .da., sirá, per la .6a. del secondo, quello ch’ é fatto del .ca. in .ad., col quadrato dela linea .ed.,<lb></lb> |
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4426 iguali al quadrato dela linea .ea. E il quadrato dela linea .ea., per la penultima del primo, é<lb></lb> |
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4427 quanto e .2. quadrati .ab. e .eb., imperoché l’ angolo .abe. è retto. Adunque, a multiplicare<lb></lb> |
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4428 .ac. in .ad., col quadrato .ed., che è iguali al quadrato .eb., é quanto il quadrato del .eb. e del<lb></lb> |
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4429 .ba. Onde, tratto de ciascuna parte el quadrato .eb., rimarrá quello ch’ é fatto del .ca. in .ad.<lb></lb> |
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4430 iguale al quadrato .ab., che è il proposito. E, se lla linea .ac. non passa per lo centro, piglise<lb></lb> |
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4431 .afeg. passante per lo centro. E menise la linea dal centro al ponto dove la linea .ac. sega il cer-<lb></lb> |
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4432 chio, che sia la linea .ed. E menise .eh. perpendiculare al .ca. Sirá, per la .3a. di questo, .dh.<lb></lb> |
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4433 iguale al .hc. E, perché la linea .dc. è divisa per igual parti nel ponto .h. e a quella è agionto<lb></lb> |
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4434 la linea .ad., sirá, per la .6a. del secondo, quello ch’ é fatto del .ca. in .ad., col quadrato .dh., iguali<lb></lb> |
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4435 al quadrato .ah. Onde, agionto a ciascuno el quadrato .he., sirá quello ch’ é fatto del .ca. in<lb></lb> |
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4436 .ad., coi quadrati di .2. linee .dh. e .he., cioé col quadrato .de., imperoché il quadrato .de. è<lb></lb> |
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4437 quanto li .2. quadrati .dh. e .he., per la penultima del primo (perché l’ angolo .ehd. è retto),<lb></lb> |
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4438 iguali al quadrato .ah. e .he., cioé al quadrato .ae., per la penultima del primo. E il quadrato<lb></lb> |
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4439 .de. è iguali al quadrato .ef., per la diffinitione del cerchio. Adunque, quello ch’ é fatto del<lb></lb> |
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4440 .ca. in .ad., col quadrato .ef., è iguali al quadrato .ea. E ancora (per la sexta del secondo)<lb></lb> |
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4441 quello ch’ é fatto del .ga. in .af., col quadrato dela linea .fe., è iguali al quadrato dela linea .ae.<lb></lb> |
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4442 Per la qual cosa, ciascun di loro ch’ é fatto del .ca. in .ad. e del .ga. in .af., col quadrato dela li-<lb></lb> |
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4443 nea .ef., è iguali al quadrato dela linea .af. E peró e saranno iguali. Tratto adunque di ciascun-<lb></lb> |
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4444 <lb></lb> |
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4445 <pb></pb></p> |
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4446 <p class="folio"> |
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4447 folio 30r |
|
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4448 </p> |
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4449 <p class="main"> |
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4450 <lb></lb> |
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4451 </p> |
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4452 <p class="runhead"> |
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4453 Distinctio quarta. Capitulum secundum 30 |
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4454 </p> |
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4455 <p class="main"> |
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4456 <lb></lb> |
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4457 no el quadrato dela linea .ef., sirá quello ch’ é fato del .ca. in .ad. iguali a quello ch’ é fatto del<lb></lb> |
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4458 .ga. in .af. Ma quello ch’ é fatto del .ga. in .af. è iguale al quadrato dela linea .ab., per lo mo-<lb></lb> |
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4459 do detto di questa. Adunque, quello ch’ é fatto del .ca. in .ad. è iguali a quello che è fatto del<lb></lb> |
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4460 .ab. in sé, che è il proposito. Di questa è da notare che, quando un ponto è dato fuor d’ un<lb></lb> |
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4461 cerchio e, da quello, molte linee si menino nel cerchio segandolo, quello ch’ é fatto di tutte le li-<lb></lb> |
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4462 nee, nela parte di fuora, sia fra loro iguali, imperoché tutte sonno iguali al quadrato dela li-<lb></lb> |
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4463 nea contingente. E, ancora, menando da quel ponto .2. linee contingenti e le fienno infra loro<lb></lb> |
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4464 iguali, imperoché ’l quadrato di ciascuna è iguali a quello che è fatto di tutta la linea segan-<lb></lb> |
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4465 te nela parte di fuora. 36<lb></lb> |
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4466 Se sirá uno ponto fuor del cerchio, dal qual si meni due linee ala circonferentia:<lb></lb> |
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4467 una segante, l’ altra ala circonferentia aplicata e sia il dutto di tutta la linea segan-<lb></lb> |
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4468 te nela parte di fuor iguale al quadrato dela linea aplicata. Di necessitá, quella li-<lb></lb> |
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4469 nea sia contingente, cioé quella aplicata ala circonferentia. Comme sia il ponto<lb></lb> |
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4470 .a., asignato fuor del cerchio .bcd., del quale sia il centro .e. Dal quale si meni al cerchio la linea<lb></lb> |
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4471 .abd. segante quello. E la linea .ac., aplicata ala circonferentia e sia quel ch’ é fatto del dutto del<lb></lb> |
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4472 .da. in .ab. iguali al quadrato .ac. Alora dico la linea .ac. essere contigente. Questa è conver-<lb></lb> |
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4473 sa ala passata. E se .ac. non fosse contingente (per l’ aversario) sia contingente .af. Sirá, per<lb></lb> |
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4474 la passata, quello ch’ é fatto del .da. in .ab. iguale al quadrato .af. Onde il quadrato dela linea<lb></lb> |
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4475 .af. è iguale al quadrato dela linea .ac. Onde .ac. sia iguali al .af., che è impossibile per la .8a.<lb></lb> |
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4476 di questo, adunque .ac. sia contingente, che è il propostio. E questo basti quanto al pri-<lb></lb> |
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|
4477 mo capitolo e, seguendo, diremo del secondo. |
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4478 </p> |
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4479 <p class="main"> |
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4480 De dimensione circulorum eiusque partium. Et tabulis de corda et arcu. Capi-<lb></lb> |
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4481 tulum primum. |
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4482 </p> |
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4483 <p class="main"> |
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4484 Havendo bene indutto a nostro proposito el .3o. de Euclide, hora darasse modo a-<lb></lb> |
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4485 la pratica de mesurare li tondi e ’suoi parti e operaremo tutto con numeri, siché<lb></lb> |
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|
4486 starai atento. Quando adunque del cerchio sai il diametro e vorrai la circonferen-<lb></lb> |
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4487 tia, quello diametro in .3 1/7. multiplica overo quello diametro per .22. multipli-<lb></lb> |
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4488 ca e dividi in .7. e harai quel che è la detta circonferentia. Comme diciammo. E gli é un ton-<lb></lb> |
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4489 do che ’l diametro è .14., quanto è la circonferentia. Multiplicarai .14. per .3 1/7. overo multipli-<lb></lb> |
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4490 carai .14. per .22. e partirai in .7. e harai .44. E .44. dirai giri il detto tondo, benché (com-<lb></lb> |
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4491 me di sotto mostraró) questo non sia pontalmente la veritá ma è molto presso.<lb></lb> |
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4492 E, similmente, per averso, dicendo e gli é un tondo la cui circonferentia è .44., adi-<lb></lb> |
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4493 mando quanto è il diametro. Partirai .44. per 3 1/7. overo .44. per .7. multiplica<lb></lb> |
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4494 e per .22. dividi e harai sempre .14. E .14. sia il detto diametro. Cioé per lo averso<lb></lb> |
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4495 al modo passato, facciendo. |
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4496 </p> |
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4497 <p class="main"> |
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4498 E, volendo trovare l’ area d’ un tondo (comme dichiareró) farai. E diciamo e gli é<lb></lb> |
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4499 un tondo che ’l suo diametro è .14., adimando quanto è quadro. Puoi multiplica-<lb></lb> |
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4500 re la mitá del diametro per la mitá dela sua circonferencia e quello che fanno è l’ a-<lb></lb> |
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4501 rea del detto circulo. Comme multiplicando la mitá del diametro, cioé .7., per la<lb></lb> |
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4502 mitá dela circonferentia, che è .22., fanno .154. e .154. è l’ area dil ditto tondo. Ancora puoi<lb></lb> |
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4503 multiplicare tutto il diametro per la mitá dela circonferentia e partire il produtto in .2. e quel-<lb></lb> |
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4504 lo che ne viene è l’ area detta. Comme nelo exemplo passato: multiplica .14. via .22., fanno .308.<lb></lb> |
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4505 el quale, in .2. partito, vienne .154. per l’ area del detto tondo. Overo ancora multiplicare la<lb></lb> |
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4506 mitá del diametro per tutta la circonferentia e di quel pigliare la mitá, che è quel medesimo.<lb></lb> |
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4507 Ancora poi multiplicare tutto il diametro per tutta la circonferentia e dela somma pigli-<lb></lb> |
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4508 are il .1/4. Comme nelo exempio dato, multiplicarai .14. per .44., fanno .616. Del quale il 1/4. è .154. |
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4509 </p> |
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4510 <p class="main"> |
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4511 E .154. è quadro et cetera. Ancora puoi multiplicare il diametro in se medesimo e di quel pigli-<lb></lb> |
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4512 are gli .11/14. E quello sia l’ area del detto tondo. Comme in detto exemplio: multiplicarai .14. in sé,<lb></lb> |
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4513 fanno .196. del qual gli .11/14. sonno .154. per l’ area del detto tondo. Ancora prendi el .1/4. dela<lb></lb> |
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4514 circonferentia, ch’ é .11., multiplica in sé, fa .121., qual multiplica per lo diametro, che è .14., fa<lb></lb> |
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4515 .1694., qual parti per .11., ne ven .154. per tutta l’ area del tondo che volgesse .44. et cetera. Ancora<lb></lb> |
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4516 puoi multiplicare la circonferentia per sé e la somma partire in .12 4/7. e quello ne viene sirá l’ a-<lb></lb> |
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4517 rea del detto tondo, cioé multiplicare la circonferentia per sé e poi per .7. e dividere in .88. |
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4518 </p> |
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4519 <p class="main"> |
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4520 Comme in detto exemplo: multiplicarai .44. in sé fanno .1936. e questo per .7. fanno .13552.<lb></lb> |
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4521 e questo in .88. dividi, cioé in .8. e .11. e harai .154. per l’ area del detto tondo.<lb></lb> |
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4522 <lb></lb> |
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4523 <pb></pb></p> |
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4524 <p class="folio"> |
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4525 folio 30v |
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4526 </p> |
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4527 <p class="main"> |
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4528 <lb></lb> |
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4529 </p> |
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4530 <p class="runhead"> |
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4531 Distinctio quarta. Capitulum secundum. |
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4532 </p> |
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4533 <p class="main"> |
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4534 <lb></lb> |
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4535 Tutti li detti modi escano del primo modo, cioé di multiplicare la mitá del dia-<lb></lb> |
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4536 metro per la mitá dela circonferentia. E peró disse el nobil geometra Archimede:<lb></lb> |
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4537 el cerchio è iguale a uno triangolo ortogonio fatto, per la basa, di tutta la circon-<lb></lb> |
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4538 ferentia e, per lo catetto, dela mitá del diametro; l’ area del quale s’ á multiplicare<lb></lb> |
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4539 el catetto in mezza la basa, cioé di multiplicare mezzo il diametro in mezza la circonferentia.<lb></lb> |
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4540 E, donde questo procieda, che a multiplicare la mitá del diametro per la mitá dela<lb></lb> |
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4541 circonferentia facia l’ area del tondo, lo voglio dimostrare. sia il cerchio .abgd.<lb></lb> |
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4542 Del quale il centro sia .e. E in quel descriveró una figura rettilinea qual vorró.<lb></lb> |
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4543 E sia uno quadrilatero .abgd. El qual, dal centro .e., risolveró in .4. triangoli:<lb></lb> |
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4544 cioé dal centro a ciascuno angolo tirando le linee. E fienno .eab. e .ebg.egd.eda. E nomi-<lb></lb> |
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4545 nase triangolo equicurio ciascuno di quelli, imperoché le linee .ea.eb.eg.ed. infra loro son-<lb></lb> |
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4546 no iguali, per la diffinitione del cerchio. E, menando a ciascuno dal centro .e. una perpendi-<lb></lb> |
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4547 culare, cadrá ciascuna in sula mitá dela basa del suo triangolo. Onde porremo sopra ala mitá di<lb></lb> |
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4548 dette base li ponti .z.i.t.k. Per li quali produrró, dal centro .e. ala circonferentia, le rette<lb></lb> |
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4549 .el.em.en.eo. E faciase .al.lb.bm.mg.gn.nd.do.oa. e fienno .4. triangoli, sopra le base .ab.<lb></lb> |
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4550 bg.gd.da., fatti. E, perché la retta .ez. è catetto sopra la retta .ab., se multiplicaremo .ez. nela<lb></lb> |
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4551 mitá del .ab., ne perverrá l’ area del triangolo .eab. Similmente, perché .lz. è catetto del trian-<lb></lb> |
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4552 golo .lab., a multiplicare .zl. nela mitá del .ab., ne perviene l’ area del triangolo .lab. Onde,<lb></lb> |
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4553 multiplicando tutta .el., cioé mezzo el diametro del circulo nela mitá del .ab., ne perverrá l’ a-<lb></lb> |
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4554 rea del quadrilatero .ealb. Per lo simil modo, multiplicando .em., cioé .el. nela mitá dela li-<lb></lb> |
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4555 nea .bg., ne perverrá l’ area del quadrilatero .ebmg. E, per lo detto modo, se multiplicaremo<lb></lb> |
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4556 .en. nela mitá del .gd. e .eo. nela mitá del .da., ne perverrá l’ area di quadrilateri .egnd. e .edoa.,<lb></lb> |
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4557 cioé, se multiplicaremo .el., cioé el mezzo diametro del circulo, nela mitá de’ lati de’ quadrilate-<lb></lb> |
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4558 ri .abgd., ne perviene l’ area dela figura di molti lati cadente nel cerchio. Ma l’ area dela det-<lb></lb> |
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4559 ta figura multilatera, che è .al.bm.gn.do., è minore del’ area del circulo. Adonca, dela mul-<lb></lb> |
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4560 tiplicatione dela mitá del diametro del cerchio nela mitá dele rette .ab.bg.gd.da., ne per-<lb></lb> |
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4561 viene meno che l’ area del cerchio. Ma la mitá dele linee .ab.gd.da. e .gb. è meno dela mitá<lb></lb> |
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4562 dela circonferentia del cerchio .abgd. Adunque, a multiplicare la mitá del cerchio, cioé la<lb></lb> |
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4563 mitá dela circonferentia, nela mitá del diametro, fará l’ area del cerchio. Imperoché habia-<lb></lb> |
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4564 mo mostro che, a multiplicare la mitá del diametro nela mitá della detta figura, che è me-<lb></lb> |
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4565 no che la circonferentia del cerchio, fa meno che l’ area del detto circulo. E hora resta a mo-<lb></lb> |
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4566 strare che, a multiplicare la mitá del diametro nel piú che la mitá dela circonferentia, fa piú<lb></lb> |
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4567 che l’ area del detto circulo. E questo per prova é che, a multiplicare .ik., ch’ é mitá del diame-<lb></lb> |
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4568 tro, in mitá del .ab. e .bc. e .cd. e .de. e .ef. e .fg. e .gh. e .ha., fa l’ area dela figura d’ otto angoli pre-<lb></lb> |
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4569 detta. E peró, a multiplicare la mitá del diametro per la mitá dela circonferentia del detto, fa-<lb></lb> |
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4570 rá piú che l’ area del detto circulo, comme appare nela presente figura. E peró adonca, a mul-<lb></lb> |
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4571 tiplicare la mitá del diametro per la mitá dela circonferentia, fará l’ area del detto circulo. E<lb></lb> |
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4572 questo volavamo intendere. |
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4573 </p> |
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4574 <p class="main"> |
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4575 Avendo dichiarato el primo modo di quadrare li circuli, avendo detto che, a mul-<lb></lb> |
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4576 tiplicare tutto il diametro per la mitá dela circonferentia e, quello che fanno,<lb></lb> |
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4577 partendo in .2., haremo l’ area. E questo viene, che, a multiplicare la mitá del dia-<lb></lb> |
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4578 metro per la mitá dela circonferentia, fanno l’ area del circulo. Onde, a multipli-<lb></lb> |
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4579 care tutto il diametro per mezzo la circonferentia, fará .2. cotanti comme tutto el diametro e<lb></lb> |
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4580 a mezzo il diametro .2. cotanti. E cosí ancora, a multiplicare tutta la circonferentia per mezzo<lb></lb> |
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4581 il diametro, fará .2. cotanti del’ area, per la ragione predetta.<lb></lb> |
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4582 E dicemmo che, a multiplicare tutto il diametro per tutta la circonferentia, fanno<lb></lb> |
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4583 .4. cotanti del’ area del cerchio. E questo chiaro appare per la ragione passata.<lb></lb> |
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4584 Imperoché, a multiplicare tutto il diametro per la mezza circonferentia, fará due<lb></lb> |
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4585 cotanti dela detta area. E a multiplicare tutta la circonferentia per mezzo il dia-<lb></lb> |
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4586 metro fa ancora .2. cotanti di detta area. Onde, a multiplicare tutto il diametro per tutta<lb></lb> |
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4587 la circonferentia, fa .4. cotanti di detta area. E questo si deve dichiarare.<lb></lb> |
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4588 Ancora dicemmo che, a multiplicare il diametro per sé e pigliarne .11/14., haremo l’ area del circulo<lb></lb> |
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4589 la quale cosa voglio mostrare. Noi habiamo detto che, a multiplicare il diametro per .3 1/7., s’ á quel-<lb></lb> |
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4590 lo che gira la circonferentia. Adonca, a multiplicare il diametro in sé e poi in .3 1/7. e quello che fanno<lb></lb> |
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4591 <lb></lb> |
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4592 <pb></pb></p> |
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4593 <p class="folio"> |
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4594 folio 31r |
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4595 </p> |
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4596 <p class="main"> |
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4597 <lb></lb> |
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4598 </p> |
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4599 <p class="runhead"> |
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4600 Distinctio quarta. Capitulum secundum 31 |
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4601 </p> |
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4602 <p class="main"> |
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4603 <lb></lb> |
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4604 care il diametro per la circonferentia e, ancora, tanto è a multiplicare il diametro per la<lb></lb> |
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4605 conferentia e partire in .4., quanto a multiplicare il diametro per sé e quello che fan-<lb></lb> |
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4606 no per lo quarto di .3 1/7. E, acioché chiaro appaia. Sienno .2. numeri de’ quali il magiore con-<lb></lb> |
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4607 tenga el minore .8. volte. Dico che tanto è a multiplicare il minore per lo magiore e partire in<lb></lb> |
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4608 .4., quanto a multiplicare il diametro, cioé il minore, in sé e poi per lo quarto del .8., cioé per .2. |
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4609 </p> |
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4610 <p class="main"> |
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4611 E cosí è quanto a multiplicare il diametro in sé e poi per lo quarto di .3 1/7., che è .11/14., e questo vo-<lb></lb> |
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4612 lemmo mostrare. |
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4613 </p> |
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4614 <p class="main"> |
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4615 Quel modo che è de multiplicare la circonferentia in sé et partire in .12 4/7., é quello<lb></lb> |
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4616 che viene l’ area del circulo: per lo passato chiaro appare. Imperoché, se sonno .2.<lb></lb> |
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4617 numeri, e uno sia .8. tanti del’ altro, che tanto è a multiplicare el magiore per lo mi-<lb></lb> |
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4618 nore e pigliare il quarto, quanto a multiplicare il magiore in sé e partirlo per quat-<lb></lb> |
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4619 tro cotanti del .8., cioé in .32. Cosí, a multiplicare la circonferentia in sé e partirla per .4. cotan-<lb></lb> |
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4620 ti di quel che la circonferentia contiene il diametro, che è .12 4/7., é quanto a multiplicare il dia-<lb></lb> |
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4621 metro per la circonferenza e partire in .4. E questo volemmo mostrare.<lb></lb> |
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4622 Ancora e gli é da dimostrare comme e fo trovata da Archimenide la linea circon-<lb></lb> |
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4623 ferentiale essere .3. volte 1/7. del diametro, la quale inventione fo bella e sotile. In que-<lb></lb> |
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4624 sto modo, bene che con brevitá se dica. sia uno cerc[h]io .abgd. del quale il dia-<lb></lb> |
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4625 metro sia .ag. e il centro sia .c. E meneró la linea .ez. contingente il cerchio sopra il<lb></lb> |
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4626 ponto .a. Dove il diametro .ag. é catetto sopra .ez. E faró .zae. lato del’ exagono stante intor-<lb></lb> |
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4627 no al circulo .abgd. E, questo fatto, conciosiacosaché l’ angolo .c. sia .2/3. del retto. E porró .ce.<lb></lb> |
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4628 essere .30., dove .ae. sia .15. E, perché l’ angolo .a. è retto, se del' quadrato del lato .ce. si toglie il<lb></lb> |
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4629 quadrato del lato .ae., cioé .225. di .900., rimane .675. la cui radici è poco meno di .26., cioé .26.<lb></lb> |
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4630 meno .1/52. Di poi dividanse l’ angolo .eca. in .2. mezzi dala linea .cf., che divide l’ arco .ab. sopra il<lb></lb> |
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4631 ponto .y. E, commo s’ á per le demostrationi de Euclide, gli angoli sopra il centro, siando igua-<lb></lb> |
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4632 li, hano iguali archi, onde la periferia .ay. ala periferia .yb. è iguale. Onde .ae. é la mitá del<lb></lb> |
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4633 lato delo exagono. Onde .af. è la mitá d’ una figura di .12. lati contenente il cerchio .abgd. E,<lb></lb> |
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4634 perché l’ angolo .eca. è diviso in .2. parti iguali dala linea .cf., sirá proportionalmente cosí .ec.<lb></lb> |
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4635 al .ca., cosí .ef. al .fa., comme nel sexto de Euclide se dimostra. Onde sia cosí el congionto del<lb></lb> |
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4636 .ca. e .ec. al .ea., cioé comme .56. meno .1/52. é al .ea., cosí el congionto del .ef. e .fa., che è .15., é al .fa.<lb></lb> |
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4637 E, permutati, sia cosí el congionto del .ec. e .ca. al .ea., cioé comme .56. meno 1/52. sonno .a.15., cosí<lb></lb> |
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4638 .ca. e .af. Onde io porró .ca. essere .56. meno .1/52. e .af .15. Onde, congiongnendo li quadrati de-<lb></lb> |
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4639 le linee .af. e .ac., haremo, per lo quadrato dela linea .cf., .3359. o poco meno. La cui radici è a<lb></lb> |
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4640 presso .58. per lo lato .cf. E dipoi divideró l’ angolo .fca. in .2. parti iguali dala linea .ch.e sia<lb></lb> |
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4641 .ah. la mitá del lato d’ una figura equilatera avente .24. lati e scritta intorno al cerchio .abgd.<lb></lb> |
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4642 E, perché l’ angolo .fca. è diviso in .2. parti iguali dala linea .ch., sirá la proportione del con-<lb></lb> |
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4643 gionto del .fc. e .ca. al .ca. comme .fa. al .ha. E, permutati, sia cosí el congionto del .fc. e .ca. al<lb></lb> |
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4644 .fa., cioé comme .114. o poco meno è a .15., cosí .ca. al .ah. Dove porró .ca.114. o poco meno e<lb></lb> |
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4645 .ah. sia .15. Onde, agiongnendo el quadrato di poco meno che .114. e .15. e, di poi, togliendo la ra-<lb></lb> |
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4646 dici haremo .115. o poco meno per la linea .ch. Ancora divideró l’ angolo .hca. in .2. parti<lb></lb> |
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4647 iguali dala linea .ci. e sia .ai. la mitá del lato d’ una figura equilatera avente .48. angoli scrita<lb></lb> |
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4648 intorno al cerchio .abgd. Del quale .ai. la sua proportione è .al.ac. e comme .15. al congion-<lb></lb> |
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4649 to del .ac. e .ch., cioé .229. e poco meno. Dico secondo l’ appressamento. Imperoché le radici<lb></lb> |
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4650 loro sonno sorde e, secondo la propinquitá, diciamo in quel modo. Porró adunque .ca.229.<lb></lb> |
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4651 e poco meno e .ai.15. e agiongo li detti quadrati e di quel piglio la radici e haremo .229.<lb></lb> |
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4652 e .1/3. o poco piú per la linea .ci. E divideró l’ angolo .ica. in .2. igual parti dala linea .ck. Dove<lb></lb> |
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4653 la proportione .ca. al .ak. è comme la proportione del congionto del .ic. e .ca. al .ai. Adunque<lb></lb> |
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4654 la proportione del .ca. al .ak. è quasi comme .458 1/5. a .15. Ma la proportione del .ca. al .ak. è<lb></lb> |
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4655 comme la proportione del diametro .ga. al doppio del .ak. Ma ’l doppio del .ak. è uno lato<lb></lb> |
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4656 d’ una figura avente .96. lati iguali stante intorno al cerchio .abgd. Onde è cosí .458 1/5. a<lb></lb> |
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4657 .15., cosí il diametro .ga. è a uno di deti lati dela figura detta avente .96. lati iguali. Onde, mul-<lb></lb> |
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4658 tiplicando .15. per .9., haremo .1440. per la somma de’ lati di quella figura. Adunque, la pro-<lb></lb> |
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4659 portione di tutti e lati d’ una figura sopra detta al diametro del circulo, el quale cade dentro,<lb></lb> |
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4660 è comme .1440. a .458., che è comme .1. a .3.33/229., che è piccola cosa piú che .3 1/7. E peró disse Archi-<lb></lb> |
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4661 menide il diametro del cerchio essere ala circonferentia comme .1. a .3 1/7. e questo volemmo<lb></lb> |
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4662 mostrare].<lb></lb> |
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4663 <lb></lb> |
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4664 <pb></pb></p> |
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4665 <p class="folio"> |
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4666 folio 31v |
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4667 </p> |
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4668 <p class="main"> |
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4669 <lb></lb> |
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4670 </p> |
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4671 <p class="runhead"> |
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4672 Distinctio quarta. Capitulum secundum. |
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4673 </p> |
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4674 <p class="main"> |
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4675 <lb></lb> |
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4676 A ncora un’ altra volta voglio trovare la detta proportione in una figura, caden-<lb></lb> |
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4677 te dentro al cerchio, che habia .96. lati iguali, in questo modo. Sia il cerchio .abgd.<lb></lb> |
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4678 Dove porró in quello el lato del’ exagono .ad., che è iguali al mezzo diame-<lb></lb> |
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4679 tro .ca. E compiró .gd. e sia il triangolo .gda. ortogonio, imperoché gli é nel<lb></lb> |
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4680 mezzo cerchio .gda., comme nel passato dicemmo. E, perché la linea .ad. è il lato delo exago-<lb></lb> |
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4681 no, sará la periferia .ad. la terza parte dela p[e]riferia .adg. Onde la periferia .dg. è doppia ala periferia<lb></lb> |
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4682 .ad. Onde l’ angolo .gad. è doppio al’ angolo .agd. e sonno amendoi iguali a uno angolo retto. On-<lb></lb> |
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4683 de l’ angolo .agd. è la terza parte d’ un angolo retto. E porró per l’ ordine detto il diametro .ag.30. |
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4684 </p> |
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4685 <p class="main"> |
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4686 Onde la retta .ad. sará .15. E la retta .gd. sia circa .a.26., comme di sopra dicemmo. E divide-<lb></lb> |
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4687 ró l’ angolo .agd. in .2. parti iguali dala linea .gm. E faró la retta .am. e sia la proportione de-<lb></lb> |
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4688 la retta .al. al .ld. comme .ag. al .gd. E, per la congionta proportionalitá, sará cosí .ad. al .ld.<lb></lb> |
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4689 comme el congionto dele rette .ag. e .gd. ala retta .gd. E, per la permutata proportionalitá,<lb></lb> |
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4690 sia cosí .ag. e .gd. ala retta .ad., cioé comme .56. e poco meno é a .15., cosí .gd. al .dl. E, perché<lb></lb> |
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4691 l’ angolo .agd. è diviso in .2. parti iguali dala linea .gm., l’ angolo .amg. è iguali al’ angolo<lb></lb> |
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4692 .gda., imperoché ciascuno è retto, conciosiacosaché fieno nel mezzo cerchio, per la .30a. del<lb></lb> |
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4693 . 3o. L’ altro angolo adunque .gld. al’ altro .gam. equiangoli sonno, adunque, e triangoli<lb></lb> |
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4694 .gdl. e il triangolo .gma. Onde e gli é cosí .dg. al .dl., cosí .gm. al .ma. Onde porró .gm.56.<lb></lb> |
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4695 o poco meno e .ma. sia .15. E sará .am. il lato dela figura di .12. facie iguali, cadente dentro<lb></lb> |
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4696 al cerchio .abgd. Dove .ag. sia .58. o poco meno per lo lato .ag. Ancora divideró l’ angolo<lb></lb> |
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4697 .agm. in .2. parti iguali cola linea .gno. E compiró la retta .ao. e troveró la radici congion-<lb></lb> |
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4698 ta del quadrato .ag., che è (comme ó detto) circa .58. Dove sia cosí .ag. e .gm. al .ma., cioé<lb></lb> |
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4699 comme .114. o pocho meno a .15., cosí .gm. al .mn. Ma cosí .gm. al .mn., cosí .go. al .oa. Son-<lb></lb> |
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4700 no e triangoli .gmn. e .goa. simili e ortogonij. É adunque cosí .114. o poco meno a .15., cosí<lb></lb> |
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4701 .go. al .oa., onde porró .go. essere .114. o poco meno e .oa.15. E torró la radici de’ quadra-<lb></lb> |
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4702 ti dele linee .go. e .oa. E haró, per la linea .ga., .115. meno alcuna cosa. E la linea .oa. é il lato<lb></lb> |
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4703 dela figura aventi .4. lati iguali scritta nel cerchio .abgd. Ancora divideró l’ angolo .ago.<lb></lb> |
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4704 in .2. mezzi dala linea .gp. E compiró .grp. E sia cosí .ga. e .go. al .oa., cosí .go. al .or. Ma<lb></lb> |
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4705 cosí .og. al .or., cosí .gp. al .ra. sia cosí .229. o poco meno a .15., cosí .gp. al .pa. onde porró<lb></lb> |
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4706 .gq. essere .229. o poco meno e, seguendo comme nel’ altre linee, troveró che le facie d’ una fi-<lb></lb> |
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4707 gura avente .96. lati iguali sia al diametro comme .1440. a .458. o poco piú, che è quasi co-<lb></lb> |
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4708 me .3 1/7. a .1o. E, perché gli é poca differentia quella peró posaro’ gli savij phylosophi che ’l dia-<lb></lb> |
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4709 metro alla circonferentia è comme .1o. a .3 1/7. E questo era da mostrare.<lb></lb> |
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4710 E, se uno campo che fosse mezzo cerchio desideri de misurare, l’ area del detto cer-<lb></lb> |
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4711 chio, per uno de’ detti modi truova. E la mitá di quella togli e harai la quadratu-<lb></lb> |
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4712 ra di detto cerchio. Comme sia el semicirculo .abg. Del quale il diametro .ag. sia<lb></lb> |
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4713 .24. e compise il circulo .adgb. E sia il suplemento .adg. ancora mezzo cerchio.<lb></lb> |
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4714 Onde, pigliando la mitá del’ area del circulo .abgd., haremo certamente l’ area dela mitá<lb></lb> |
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4715 del circulo, cioé l’ area del dato semicirculo .abg. Overamente la mitá del diametro, cioé .12.,<lb></lb> |
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4716 in .3 1/7. multiplica e harai .37 1/7. per l’ arco .abg. Del quale piglia la mitá e quella multiplica per la<lb></lb> |
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4717 mitá del diametro. Overo la quarta parte del diametro in tutto l’ arco multiplica e haremo<lb></lb> |
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4718 .226 4/7. per l’ area del mezzo cerchio .abg. Overo del quadrato del diametro togli .11/12. e harai<lb></lb> |
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4719 la detta area. |
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4720 </p> |
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4721 <p class="main"> |
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4722 E, se la notitia del’ arco .abg., che è mita dela linea circonferentiale de tutto cer-<lb></lb> |
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4723 chio, vuoi trovare, dal centro e sopra il diametro .eg., la linea .eb. ortogonalmen-<lb></lb> |
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4724 te riga. E dali ponti .a. e .g. mena le linee .ab. e .gb. e sia l’ angolo .abg. retto. Im-<lb></lb> |
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4725 peroché gli é nel mezzo cerchio .abg. Overo, perché .be. é iguali ala retta .ea. e<lb></lb> |
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4726 .eg., e sará l’ uno e l’ altro triangolo .aeb. e .beg. equicurio. Onde gli angoli .eba. e .eab. e<lb></lb> |
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4727 .ebg. sonno infra loro iguali. E ciascuno di loro è iguali ala mitá del’ angolo retto. Adun-<lb></lb> |
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4728 que ‘.2. angoli che sonno al .b. sonno iguali a uno retto. Retto è adunque l’ angolo .abg. E,<lb></lb> |
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4729 perché .2. rette .ae. e .eb. sonno iguali a .2. rette .be. e .eg. e gli angoli .aeb. e .beg. sonno retti<lb></lb> |
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4730 e iguali e infra loro le rette .ab. e .bg. sonno iguali, onde lo quadrato del diametro .ag. é<lb></lb> |
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4731 doppio al quadrato di ciascuna linea .gb. e .ba. E ancora ciascun de’ quadrati dele linee .gb. e .ba.<lb></lb> |
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4732 a ciascuno de’ quadrati dele linee .ae. e .eb. e ancora .be. e .eg. é doppio. Onde è cosí .ag. al .gb., cosí<lb></lb> |
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4733 .gb. al .be. overo al .eg. Onde, se multiplicaremo .ag. in .be., cioé .24. per .12., haremo .288., per lo quadrato<lb></lb> |
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4734 di ciascuna dele linee .ab. e .bg. overo, se del quadrato del diametro .ga. torreno il mezzo. Overo ra-<lb></lb> |
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4735 <lb></lb> |
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4736 <pb></pb></p> |
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4737 <p class="folio"> |
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4738 folio 32r |
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4739 </p> |
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4740 <p class="main"> |
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4741 <lb></lb> |
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4742 </p> |
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4743 <p class="runhead"> |
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4744 Distinctio quarta. Capitulum secundum. 32 |
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4745 </p> |
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4746 <p class="main"> |
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4747 <lb></lb> |
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4748 doppiarai il quadrato del mezzo diametro .ge. overo .ea., haremo similmente .288. per lo<lb></lb> |
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4749 quadrato di ciascuna linea .gb. e .ba. e é .gb. arco dela mitá del mezzo cerchio .abg. Di poi si<lb></lb> |
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4750 dividiremo la corda .bg. sopra il ponto .c. in .2. parti iguali. Per li ponti .ce. meneremo la linea<lb></lb> |
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4751 df. Sará .df. diametro del cerchio .abgd. e fará gli angoli retti sopra il ponto .c. col la cor-<lb></lb> |
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4752 da .b g. E ancora divida l’ arco .bfg. in .2. parti iguali. Onde, se dal ponto .f. meneremo le linee<lb></lb> |
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4753 .fb.fg., sará ciascuna corda la quarta parte del mezzo cerchio .gba., ala quale notitia verre-<lb></lb> |
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4754 mo cosí. Perché l’ angolo .bce. è retto, se ’l quadrato dela linea, che è .72., cioé la quarta parte<lb></lb> |
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4755 del quadrato dela corda .bg., trarremo del quadrato dela linea .be., che è sotto al’ angolo ret-<lb></lb> |
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4756 to rimarranno .72. per lo quadrato dela linea. Onde tutta .dc. è .12. e radici de .72. E chia-<lb></lb> |
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4757 mase binomio, comme nela parte de sopra, in questa opra del’ arithmetica, fo manifesto. E que-<lb></lb> |
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4758 sto binomio non si puó exprimere con numeri. Onde, se traremo .ec. del .ef. rimarranno .12.<lb></lb> |
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4759 meno radice di .72. per la linea .cf. E chiamase quella linea .cf. residuo overo abscisio: con-<lb></lb> |
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4760 ciosiacosaché sia fatto de numero meno radice. Dipoi piglieremo li quadrati dele linee .fc.<lb></lb> |
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4761 e .cb. e harai el quadrato dela corda .bf. overo .fg. E, comme si pigli el quadrato dele linee<lb></lb> |
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4762 .cf., io te lo voglio dimostrare. Poni e detti nomi, cioé .12. e radice di .72., comme dal lato si<lb></lb> |
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4763 mostra; multiplicherai .12. per .12., fanno .144. E poi multiplica radice di .72. meno in sé, cioé<lb></lb> |
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4764 per radici di .72. meno, fanno .72. che, con .144., fanno .216., del quale tra’ el doppio di .12. mul-<lb></lb> |
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4765 tiplicato in radice di .72., fanno .24. radici di .72. E cosí haremo per lo quadrato dela linea<lb></lb> |
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4766 .cf.216. meno .24. radici di .72., che sonno una radici .41472. Overo, perché la linea .ef. è<lb></lb> |
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4767 divisa comme viene in sul ponto .c., fienno ’.2. quadrati .ef. e .ec. iguali al quadrato .cf. e al dop-<lb></lb> |
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4768 pio di quel ch’ é fatto del .ec. in .ef., per la .7a. del .2o., imperoché li quadrati dele linee .ef. e .ec.<lb></lb> |
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4769 sonno .216. del qual, togliendo el doppio dela multplicatione del .ef. in .ec., cioé di .24. in radi-<lb></lb> |
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4770 ci di .72., fanno .216. meno la radici di .41472. e chiamase el primo reciso, comme sopra nel’ a-<lb></lb> |
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4771 rithmetica mostrai. Al quale quadrato, agionto el quadrato dela linea .bc., cioé .72., haremo<lb></lb> |
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4772 .288. meno la radici de .41472. per lo quadrato dela corda .bf., che si chiama el quarto reci-<lb></lb> |
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4773 so. Del quale la radice è quella linea detta linea minor. E, se secondo l’ appressamento vuoi pro-<lb></lb> |
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4774 cedere a notitia dela linea .bf., la radici di .41472. piglia, che è pocho meno di .203 2/3. che, de<lb></lb> |
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4775 .288. tratta, rimane poco piú di .84 1/3. per lo quadrato de ciascuna delle .4. corde .gf.fb.hb.<lb></lb> |
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4776 .ha. overo, altramente, del quadrato dela linea .ec. truova la radici, ch’ é poco meno di .8 1/2. E<lb></lb> |
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4777 quella tra’ dela linea .ef., cioé di .12., rimane poco piú di .3 1/2. per la linea .cf. Del quale il quadra-<lb></lb> |
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4778 to, che è poco piú di .12 1/3., se l’ agiognamo col quadrato del .bc., haremo poco piú di .84 1/3. per<lb></lb> |
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4779 lo quadrato di ciascuna dele .4. corde. Le quali, multiplicate per lo .4., cioé per lo quadrato<lb></lb> |
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4780 di .4., che è .16., haremo .1350. per lo quadrato dela somma dele .4. corde de’ quali la radici è cir-<lb></lb> |
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4781 ca .36 3/4. Ma l’ arco .abg. è .37 5/7. Onde ancora siamo alquanto de longni al trovare per la noti-<lb></lb> |
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4782 tia di .4. corde. Onde ancora divideró una di quelle .4. corde in .2. corde iguali. E sia la cor-<lb></lb> |
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4783 da .ah. divisa sopra il ponto .i. E meneró li ponti .i.e. e il diametro .kl. che divide l’ arco .akh.<lb></lb> |
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4784 in .2. parti iguali in sul ponto .k. E meneró la corda .ak. La qual sará il lato d’ una figura aven-<lb></lb> |
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4785 ti .16. lati iguali, cadente dentro al cerchio .abg. E verró ala notitia di quello secondo che<lb></lb> |
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4786 è detto di sopra: cioé che del quadrato .ae., cioé di .144., si traga el quadrato dela linea .ai., che<lb></lb> |
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4787 è circa .21 1/11., cioé la quarta parte del quadrato dela linea .ah.122 10/11., per lo quadrato dela li-<lb></lb> |
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4788 nea .ei. Dove quella linea è .11 1/11. Lo quale, se lo togliamo dela linea .ek., rimará .ik. circa .10/11.<lb></lb> |
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4789 Del quale il quadrato è circa 9/11. che, agionto al quadrato dela linea .ai., haremo .21 9/10. per lo<lb></lb> |
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4790 quadrato dela linea .ak., che è la corda del’ ottava parte del’ arco .abg. Onde, se multiplica-<lb></lb> |
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4791 remo .21 10/11. per lo quadrato del .8., haremo circa .1402. per lo quadrato dele .8. corde iguali<lb></lb> |
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4792 cadenti nel mezzo cerchio .abg., dele quali la radici è circa .37 1/2. o poco meno. Ma l’ arco .abg.<lb></lb> |
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4793 é piú di .37 5/7. Per la qual cosa, se per lo detto modo dividerai l’ arco .ak. e di quello truovi<lb></lb> |
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4794 la corda, haremo apresso ala longhezza del’ arco .abg. E cosí, sempre dividendo l’ arco, ha-<lb></lb> |
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4795 remo, in modo che la differentia è cosa insensibile al’ arco. E peró cosí poi venire ala noti-<lb></lb> |
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4796 tia di qual vuoi arco. Nondimeno, a piú tua chiarezza de quello che è ditto, la sequente con<lb></lb> |
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4797 diligentia notarai. Avenga che dificil cosa sia, per queste vie, ala notitia precisa deli archi e<lb></lb> |
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4798 corde pervenire. E, per questo, si manifesta la gran difficultá dela quadratura del cerchio che,<lb></lb> |
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4799 da tutti ‘philosophi, la sua quadratura si conprende essere possibile e dabile, benché finora per nullo sia<lb></lb> |
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4800 trovata, se non quanto per Archimede se dimostra nel suo terzo libro. El qual modo fo per<lb></lb> |
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4801 l’ aproximamento dela proportione del suo diametro ala circunferentia, commo denanze in<lb></lb> |
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4802 questo te dissi e dimostrai et cetera. Or prendi l’ altra. Videlicet.<lb></lb> |
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4803 <lb></lb> |
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4804 <pb></pb></p> |
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4805 <p class="folio"> |
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4806 folio 32v |
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4807 </p> |
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4808 <p class="main"> |
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4809 <lb></lb> |
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4810 </p> |
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4811 <p class="runhead"> |
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4812 Distinctio quarta. Capitulum secundum. |
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4813 </p> |
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4814 <p class="main"> |
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4815 <lb></lb> |
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4816 Questo, acioché piú liquidamente appaia, sia il circulo .abcd. Del quale el diame-<lb></lb> |
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4817 tro .ac. sia .10. E in quello sia data la corda nota .bd. che sia .8. e vogliamo l’ arco<lb></lb> |
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4818 .bad., per la notitia di deta corda. Prima è da trovare la longheza di ciascuna saet-<lb></lb> |
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4819 ta in questo modo. Dividase la corda .bd. in .2. parti iguali sopra il ponto .e. E fa-<lb></lb> |
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4820 ciase .aec. e sia perpendiculare. La quale passerá per lo centro. E, per la .34a. del .3o., tanto fa<lb></lb> |
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4821 .be. in .ed., quanto .ae. in .ec. Dove .be. in .ed. fanno .16. E dirai che tu habia a dividere .10. in .2. parti,<lb></lb> |
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4822 che, l’ una multiplicata nel’ altra facino .16. Che sia l’ una .2., l’ altra .8. Adunque .ae. è .2. e .ec. è .8. |
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4823 </p> |
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4824 <p class="main"> |
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4825 E, similmente, volendo la corda .bd., havendo la saetta .ae. e .ec., multiplicando .2. via .8. e di quel-<lb></lb> |
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4826 lo pigliando la radici che sia .4. per lo .be. e altretanto per lo .ed. E cosí tutta .bd. è .8. Faciase<lb></lb> |
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4827 adunque le .2. corde degli .2. archi .ab. e .ad. per le linee .ab. e .ad. Dele quali, se vuoi la notitia, a-<lb></lb> |
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4828 giongase el quadrato .ae. col quadrato .be. e haremo .20. per lo quadrato d’ una dele linee .ad.<lb></lb> |
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4829 overo .ab. Dipoi dividasi la corda .ad. nel ponto .g., faremo .igfh. diametro e troveremo, per<lb></lb> |
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4830 quelle cose che sonno dette, la notitia dele saette .ig. e .gh., con quai li quadrati dele linee .ag. e<lb></lb> |
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4831 .gi. congiungeremo e haremo el quadrato dela linea .ai., che è la quarta parte di tutto l’ arco .bda.<lb></lb> |
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4832 E cosí faremo frequentemente e haremo apresso a tutto l’ arco .bad. Lo quale dela linea cir-<lb></lb> |
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4833 cunferente, cioé del .adhb., cioé di .31 3/7., trarai e rimarrate l’ arco .bcd. manifesto.<lb></lb> |
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4834 Ancora ci é un altro modo a trovare degli archi le corde de’ mezzi archi de’ qua-<lb></lb> |
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4835 li archi le corde sonno note. El quale Ptolomeo pose nel’ Almagesto. Sia adonca<lb></lb> |
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4836 nel cerchio .abgd. el diametro .bd. noto e ancore la corda .ad. nota. Voglio trova-<lb></lb> |
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4837 re la corda dela mitá del’ arco .ad. Meneró la corda .ab. e sia nota: conciosiacosa-<lb></lb> |
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4838 ché la corda .ad. sia nota e l’ angolo .dab. sia retto per la .30a. del 3o. Onde il quadrato del dia-<lb></lb> |
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4839 metro .db. è iguali a’ .2. quadrati de .2. corde .da. e .ab. E faciase la retta .be. iguale ala retta .ba.<lb></lb> |
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4840 e dividase l’ angolo .abe. in .2. parti iguali dala linea .bz. e comporró le rette .zd.ze.za. E dal<lb></lb> |
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4841 ponto .z., sopra el diametro .bd., meneró il catetto .zi. E, perché e gli è iguale la retta .ab. ala ret-<lb></lb> |
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4842 ta .be., se comunamente si pone la retta .bz., fienno .2. rette .ab. e .bz. a .2. rette .bz. e .be. iguali, im-<lb></lb> |
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4843 peroche infra loro sonno iguali le periferie .az. e .zd. che, comme habiamo mostro, iguali an-<lb></lb> |
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4844 goli in uno cerchio, sopra l’ iguali periferie, sonno fatti overamente al centro overamen-<lb></lb> |
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4845 te sopra la circonferentia sienno fatti. E gli angoli che sonno al .b. sonno iguali infra loro e<lb></lb> |
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4846 sonno sopra le periferie .az.zd. Onde la retta .ze. è iguale ala retta .zd. e ala retta .za. per .4am.<lb></lb> |
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4847 primi. Equicurio é adunque il triangolo .zed. Onde il ponto .i., che è il cadimento del catetto<lb></lb> |
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4848 .zi., é in mezzo dela linea .ed. E, perché ortogonio è il triangolo .bzd., imperoché gli é nel mez-<lb></lb> |
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4849 zo cerchio e sopra le base, dal’ angolo retto, è menato lo catetto a ciascuno triangolo de’ .2.<lb></lb> |
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4850 per li quali ‘l triangolo .bzd. è diviso e ciascuno triangolo á uno angolo retto e uno comune<lb></lb> |
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4851 con tutto el triangolo .bzd. Comme Euclide per la .8a. del .6o. dimostra. Onde sará cosí .bd.<lb></lb> |
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4852 al .dz. cosí .zd. al .di. Onde la multiplicatione del .di. in .bd. è iguale al quadrato dela linea .zd.<lb></lb> |
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4853 E certamente .id. è noto. Conciosiacosaché sia la mitá del .ed., che è noto. Imperoché .be. è<lb></lb> |
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4854 noto che è iguali ala corda .ba. nota. Onde, se si toglie la la corda .ba., cioé .be., del diametro .bd.,<lb></lb> |
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4855 rimarrá .ed. noto. Dove la mitá di quello è .id. che sia noto. Onde, se multiplicaremo el .di. no-<lb></lb> |
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4856 to in .bd. noto, ne perverrá il quadrato dela corda .zd. noto. Onde .zd. sia noto comme habiamo<lb></lb> |
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4857 detto, che ancora, con numeri, porró sia il diametro .bd.10. e la corda .da. sia .8. Dove la<lb></lb> |
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4858 corda .ab. sia .6. E, perché .be. é a quella iguale, sia .be.6. che, tratto del diametro .bd., cioé di<lb></lb> |
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4859 .10., rimane .4. per la retta .ed. De’ quali la mitá, cioé .2., sia tutto .id. E dela multiplicatione del<lb></lb> |
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4860 .id. in .bd. fanno .20. che sonno iguali al quadrato dela corda .zd. Onde la corda .zd. è radici di .20. |
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4861 </p> |
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4862 <p class="main"> |
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4863 Ancora, se traremo el quadrato dela linea .zd. del quadrato del diametro .bd.,<lb></lb> |
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4864 rimarrano .80. per lo quadrato dela corda .bz. De’ quali la radice è .9. meno .1/18.,<lb></lb> |
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4865 la quale se trarremo del diametro .bd., rimarrá .1 1/18. De’ quali se la mitá multipli-<lb></lb> |
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4866 caremo nel diametro .bd. overo se multiplicaremo .1 1/18. per la mitá del diametro,<lb></lb> |
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4867 cioé per .5., fanno .5 5/18. per lo quadrato dela corda, che è corda dela mitá del mezzo del’ arco<lb></lb> |
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4868 .dz. E, secondo questo modo, possiamo trovare le corde dela mitá di qualunque arco dato.<lb></lb> |
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4869 Ma questa tale inventione non è da essere operata da quelli che misurano e campi, che<lb></lb> |
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4870 vogliono procedere secondo un vulgare modo. Imperoché, quando vulgarmente la lon-<lb></lb> |
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4871 ghezza d’ alcuno arco disidereno d’ avere, habiano alcuna misura cognosciuta de .2. o .3.bracia.,<lb></lb> |
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4872 la quale si possa piegare e distendere. E con quella studi intorno misurare gli archi che vo-<lb></lb> |
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4873 gliano misurare. Overamente habino una fune d’ un braccio o piú e con quella studi misurare<lb></lb> |
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4874 intorno gl’ archi de’ cerchi, ficando spesso le canne per lo giro del cerchio, accioche quella fune<lb></lb> |
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4875 non si disvii dala circonferentia del circulo. E cosí potrai la misura di tutti gli archi de’ cerchi<lb></lb> |
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4876 <lb></lb> |
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4877 <pb></pb></p> |
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4878 <p class="folio"> |
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4879 folio 33r |
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4880 </p> |
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4881 <p class="main"> |
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4882 <lb></lb> |
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4883 </p> |
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4884 <p class="runhead"> |
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4885 Distinctio quinta. Capitulum secundum. 40] |
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4886 </p> |
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4887 <p class="main"> |
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4888 <lb></lb> |
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4889 havere. Ma quelli che, secondo la scientia de geometria, vogliano operare piú legiermente, si dará lo-<lb></lb> |
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4890 ro il modo per gli archi a trovare la corda e per la corda gli archi, per le tavole sequenti. Nelle quali {+}<lb></lb> |
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4891 archi noti proposi e, inanzi a ciascuno, la sua corda in pertiche e piedi e oncie e ponti scripti. E la per-<lb></lb> |
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4892 ticha è .6. braccia de terra. E il piede è .18. once. E l’ oncia è .20. ponti. Over la pertica è .108. once<lb></lb> |
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4893 e .2160. ponti. E sonno .66. corde e intendese quelle sienno menate nel mezzo cerchio, del quale il dia-<lb></lb> |
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4894 metro è pertiche .42. E ciascuna corda menata nel cerchio è corda di .2. archi iguali, se la cor-<lb></lb> |
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4895 da será diametro over di .2. archi non iguali, se la corda non será diametro. E peró .2. ar-<lb></lb> |
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4896 chi inanze ala ditta corda ó ordinato, commo per le sequenti tavole se manifesta. E li numeri nel<lb></lb> |
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4897 primo luogo, che son .1.2.3.4., denotano le pertiche deli archi menori. E quelli dela seconda<lb></lb> |
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4898 casella, che son .131.130., representano le pertiche deli archi magiori. E quelli dela terza casella,<lb></lb> |
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4899 che comenza .0.1.2.3.4. et cetera, denotano le pertiche dele corde. E quelli dela quarta deno-<lb></lb> |
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4900 tano li piedi, che comenza .5.5.5.5.4. et cetera. Quelli dela quinta, che comenza .17.17.17. et cetera, deno-<lb></lb> |
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4901 tano once. E li ultimi, nel sexto luogo, dicano li ponti, che comenzano .17.13.4. et cetera. Siche<lb></lb> |
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4902 sempre harai a rimpetto arimpetto l’ uno al’ altro. E sappi commo habia facta questa, cosí ne poi far da te<lb></lb> |
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4903 infinite magiori e minori, commo se elice del ditto Almegesto, sempre proportionando un cer-<lb></lb> |
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4904 chio piccolo a un grande. E, per tal modo, se reggano li pratici agrimensori nel fabricare le ta-<lb></lb> |
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4905 vole deli scemi per le botti, commo quella del .60. e l’ altra del .12. quella del .10., che ala gior-<lb></lb> |
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4906 nata s’ usano, secondo la experienza che, nelli paesi, in alcuna botte, da sé fanno et cetera.<lb></lb> |
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4907 <lb></lb> |
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4908 <lb></lb> |
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4909 1 131 0 5 17 17 34 98 30 2 6 17<lb></lb> |
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4910 2 130 1 5 17 17 35 97 31 0 8 5<lb></lb> |
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4911 3 129 2 5 17 13 36 96 31 4 8 7<lb></lb> |
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4912 4 128 3 5 17 4 37 95 32 2 5 15<lb></lb> |
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4913 5 127 4 4 12 2 38 94 33 0 {+} 9<lb></lb> |
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4914 6 126 5 5 16 10 39 93 34 3 13 0<lb></lb> |
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4915 7 125 6 5 14 5 40 92 35 1 4 15<lb></lb> |
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4916 8 124 7 5 12 9 41 91 35 4 2 10<lb></lb> |
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4917 9 123 8 5 8 16 42 90 36 2 0 0<lb></lb> |
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4918 10 122 9 5 7 8 43 89 36 5 3 5<lb></lb> |
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4919 11 121 10 5 14 2 44 88 37 2 4 6<lb></lb> |
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4920 12 120 11 4 17 18 45 87 37 5 3 2<lb></lb> |
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4921 13 119 12 4 13 6 46 86 38 1 17 15<lb></lb> |
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4922 14 118 13 4 7 16 47 85 38 4 12 13<lb></lb> |
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4923 15 117 14 4 1 0 48 84 38 1 4 0<lb></lb> |
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4924 16 116 15 3 11 18 49 83 39 3 11 15<lb></lb> |
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4925 17 115 16 3 3 12 50 82 39 5 17 2<lb></lb> |
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4926 18 114 17 2 12 8 51 81 40 2 2 1<lb></lb> |
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4927 19 113 18 0 2 15 52 80 40 4 2 10<lb></lb> |
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4928 20 112 19 1 8 12 53 79 41 0 0 11<lb></lb> |
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4929 21 111 20 0 13 18 54 78 40 1 14 5<lb></lb> |
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4930 22 110 21 0 0 0 55 77 40 3 7 8<lb></lb> |
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4931 23 109 21 5 2 16 56 76 40 4 16 2<lb></lb> |
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4932 24 108 22 4 4 5 57 75 41 0 4 12<lb></lb> |
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4933 25 107 23 3 4 8 58 74 41 1 8 1<lb></lb> |
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4934 26 106 24 2 3 2 59 73 41 2 9 0<lb></lb> |
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4935 27 105 25 1 6 6 60 72 41 3 7 14<lb></lb> |
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4936 28 104 25 5 6 2 61 71 41 4 9 2<lb></lb> |
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4937 29 103 26 4 8 0 62 70 41 4 15 10<lb></lb> |
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4938 30 102 27 3 0 3 63 69 41 5 6 9<lb></lb> |
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4939 31 101 28 1 9 7 64 68 41 5 12 17<lb></lb> |
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4940 32 100 28 5 16 4 65 67 41 5 6 14<lb></lb> |
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4941 33 99 29 4 3 9 66 66 42 0 0 0<lb></lb> |
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4942 <lb></lb> |
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4943 <lb></lb> |
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4944 <pb></pb></p> |
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4945 <p class="folio"> |
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4946 folio 33v |
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4947 </p> |
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4948 <p class="main"> |
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4949 <lb></lb> |
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4950 </p> |
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4951 <p class="runhead"> |
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4952 Distinctio quarta. Capitulum secundum. |
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4953 </p> |
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4954 <p class="main"> |
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4955 <lb></lb> |
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4956 Commo, adonca, per queste tavole qual voi arco di cerchi s’ abbia a trovare, lo mo-<lb></lb> |
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4957 straró. E, accioché di quelle migliore doctrina s’ abbia, é da mostrare, se in uno<lb></lb> |
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4958 cerchio .2. archi non iguali fienno, sia la proportione del magiore arco ala sua cor-<lb></lb> |
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4959 da magiore del minore arco ala sua corda, che ancora all’ ochio si puó compren-<lb></lb> |
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4960 dere per le tavole sopra scripte. Imperoché l’ arco del mezzo cerchio ala sua corda (cioé al<lb></lb> |
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4961 diametro) è commo .66. a .42., cioé neli numeri minori, commo .11. a .7. E la proportione de-<lb></lb> |
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4962 l’ arco dela sexta parte del cerchio ala sua corda è commo .22. a .21. Magiore certamente è<lb></lb> |
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4963 la proportione del .11. a .7. che di .22. a .21. Similmente, in tutti gli archi dele sopra scripte ta-<lb></lb> |
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4964 vole, trovarai la proportione del magiore arco ala sua corda avanzare la proportione del mi-<lb></lb> |
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4965 nore arco ala sua corda. Ma, accioché questo appaia. Sia dato uno cerchio .abgd. Nel qua-<lb></lb> |
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4966 le sienno .2. archi non iguali .ab. e .bg. E sia l’ arco .bg. magiore. E menise la corda del’ arco<lb></lb> |
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4967 .abg. che sia la retta .ag. e dividise l’ angolo .abg. in .2. parti iguali dala linea .bd. che seghi la<lb></lb> |
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4968 corda .ag. in sul ponto .e. E compise la retta .ad. e .gd. E, perché l’ angolo che è sotto .abg. é di-<lb></lb> |
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4969 viso in .2. parti iguali dala linea .bd., iguali é l’ angolo .abd. al’ angolo .dbg. Onde iguali é l’ ar-<lb></lb> |
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4970 co .ad. al’ arco .dg., commo per la .25a. del terzo apare. Onde iguale è la retta .ad. ala retta .dg.,<lb></lb> |
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4971 per la .28a. del terzo. Onde, posto de comune la retta .db., fienno .2. rette .ad. e .db. a .2. rette .bd.<lb></lb> |
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4972 e .dg. iguali. Ma la basa .ba. è minore dela basa .bg. Onde l’ angolo .bdg. è magiore del’ an-<lb></lb> |
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4973 golo .bda., per la .25a. del primo. Overo, perché minore è la periferia .ab. dela periferia .bg.,<lb></lb> |
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4974 magiore è l’ angolo che è facto dale rette .gd. e .db. del’ angolo che è facto dale rette .bd. e .da.<lb></lb> |
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4975 Peroché e gli é cosí la periferia .gb. ala periferia .ba., cosí l’ angolo .gdb. al’ angolo .bda. E,<lb></lb> |
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4976 perché gli é iguale .ad. ala retta .dg., se dal ponto .d. si mena el catetto sopra la linea .ag., cade-<lb></lb> |
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4977 rá in mezzo del .ag. Dove caderá infra .eg., cioé sopra la linea .eg. Perché magiore é .ge. del<lb></lb> |
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4978 .ae., conciosiacosaché sia cosí la retta .gb. al .ba., cosí .ge. al .ea. Caggia adonca il catetto sopra<lb></lb> |
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4979 il ponto .h. e sia catetto .dh. E, perché retto è l’ angolo .dhe., magiore è la retta .de. che .dh. e<lb></lb> |
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4980 magiore è .da. che .de. Sienno poste amendoi le linee rette .di. e .df., iguali ala retta .de. E<lb></lb> |
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4981 faró .d. centro con lo spatio dele rette .di. e .de. E faciase l’ arco .ief. magiore e adonca el set-<lb></lb> |
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4982 tore .die. del triangolo rettilineo .dhe. e il settore .dif. é menore del triangolo .dea. On-<lb></lb> |
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4983 de la proportione del settore .die. al settore .def. è magiore dela proportione del triangolo<lb></lb> |
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4984 .dae. al triangolo .dea. Ma la proportione del settore .dei. al settore .def. è commo la pro-<lb></lb> |
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4985 portione del’ angolo .ide. al’ angolo .idf. Magiore è la proportione del triangolo .dhe. al<lb></lb> |
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4986 triangolo rettilineo .dae. Ma la proportione del triangolo .hde. al triangolo .ade. è com-<lb></lb> |
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4987 mo la retta .he. ala .ea., conciosiacosaché amendoi li triangoli sienno sotto una medesima<lb></lb> |
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4988 altezza, per la prima del sexto, che è del .d. in .h., imperoché ’l catetto .de. è perpendiculare ali<lb></lb> |
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4989 triangoli .hde. e .ade. Adonca la proportione del’ angolo .ide. al’ angolo .eda. è magiore de-<lb></lb> |
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4990 la proportione dela retta .ah. ala retta .ae. E, per la congionta proportionalitá, sará la pro-<lb></lb> |
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4991 portione .ida. al’ angolo .eda. magiore dela retta .ah. ala retta .ae. Ma l’ angolo .gdi. è igua-<lb></lb> |
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4992 le al’ angolo .adh. E la retta .gd. é iguale ala retta .ah. Onde la proportione del’ angolo .gdi. al’ ango-<lb></lb> |
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4993 lo .eda. è magiore dela proportione dela retta .gh. ala retta .ae. Ma la proportione del’ angolo .ide.<lb></lb> |
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4994 al’ angolo .ade. è trovata magiore dela proportione .eg. ala retta .ea. Ma la proportione<lb></lb> |
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|
4995 del’ angolo .gde. al’ angolo .ade. è commo la proportione del’ arco .bg. al’ arco .ba. E la pro-<lb></lb> |
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4996 portione del .ge. al .ea. è commo la proportione dela corda .gb. ala corda .ba. Adonca la pro-<lb></lb> |
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4997 portione del’ arco .bg. al’ arco .ba. è magiore proportione dela corda .gb. ala corda .ba. e, per-<lb></lb> |
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4998 mutati, e sará adonca la proportione del’ arco .gb. ala corda .gb. magiore dela proportio-<lb></lb> |
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4999 ne del’ arco .ba. ala corda .ba., ch’ era bisogno mostrare. Ma, per consequire simili noti-<lb></lb> |
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5000 tie, te convien molto bene havere ale mani le .6. specie over modi dele proportionalitá, quali<lb></lb> |
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5001 Euclide mette, con tutta diligentia, nel suo quinto libro. E noi, in questo di sopra, nela parte<lb></lb> |
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5002 de arithmetica, nel tractato dele proportioni e proportionalitá, a tuo documento, habiamo indut-<lb></lb> |
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5003 to chiaramente con exempli palpabili e evidenti, siché lasú recorri a tue occurentie. Peroché, sen-<lb></lb> |
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5004 za ditte proportioni, non è possibile la fabrica de alcuna tavola né de corde e archi né anco<lb></lb> |
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5005 d’ altro. Ma quella per li archi haver la corda e per la corda haver li archi passa el segno in tut-<lb></lb> |
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5006 te difficultá conmo, per tutto l’ almegesto de Ptolomeo, apare e anche in sua cosmografia. Ideo et cetera.<lb></lb> |
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5007 Queste cose intese, se per la corda data d’ alcun cerchio del quale il diametro sia no-<lb></lb> |
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5008 to, e vorrai trovare l’ arco di quella corda, quella corda, per lo diametro dela ta-<lb></lb> |
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5009 vola, multiplica, cioé per .42., e quello che fa dividi per lo diametro del cerchio da-<lb></lb> |
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5010 to e quello ne viene é la corda simile dela corda dela tavola. E di quella piglia l’ ar-<lb></lb> |
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5011 co in ditte tavole. E quello multiplica per lo diametro del dato cerchio e dividi per lo diametro<lb></lb> |
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5012 dele tavole, cioé per .42. E quello ne perverra sirá l’ arco che desideri. E acioché meglo intenda,<lb></lb> |
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5013 <lb></lb> |
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5014 <pb></pb></p> |
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5015 <p class="folio"> |
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5016 folio 34r |
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5017 </p> |
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5018 <p class="main"> |
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5019 <lb></lb> |
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5020 </p> |
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5021 <p class="runhead"> |
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5022 Distinctio quarta. Capitulum secundum. 34 |
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5023 </p> |
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5024 <p class="main"> |
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5025 <lb></lb> |
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5026 sia il cerchio .abg. Del quale il diametro .bg. sia pertiche .10. E in quello sia data la corda .ab.,<lb></lb> |
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5027 che sia pertiche .5. E vogli havere la notitia del’ arco .aeb. Multiplicarai .ab. per .42., cioé<lb></lb> |
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5028 .5. via .42., fanno .210. E dividi per lo diametro .bg., vienne .21. per la corda dele tavole simi-<lb></lb> |
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5029 le ala corda .ab. La qual cerca di trovare infra le tavole e piglia l’ arco che è per lo dritto di<lb></lb> |
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5030 quello. E, perché gli é minore del mezzo cerchio, imperoché dimandi del’ arco .ab. che è an-<lb></lb> |
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5031 cora minore, e sia l’ arco minore .22., el quale per .10. multiplica, cioé per lo diametro .bg., e la<lb></lb> |
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5032 somma dividerai per .42. e verranne .5.5/21. per l’ arco .aeb. E, se vorrai havere la notitia del ma-<lb></lb> |
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5033 giore, truova el magiore arco nele tavole che è per lo diritto della corda ancora trovata. E<lb></lb> |
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5034 troverai essere .110. pertiche. Le quali multiplica per .10. e dividerai per .42. e haremo .26. perti-<lb></lb> |
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5035 che .4/21. per l’ arco .agdb. e cosí hai l’ arco .aeb. essere .5 5/21. e l’ arco .agdb. essere .26 4/21.<lb></lb> |
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5036 Ancora sia la corda .ab.8. pertiche .3. pie’ e .16. oncie .2/3. E il diametro .bg. sia .10., com-<lb></lb> |
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5037 me dicemmo. Multiplica adonca .8. pertiche .3. pie’ .16. oncie .2/3. per .42. e la somma di-<lb></lb> |
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5038 viderai per .10., vienne .36. pertiche e .2. pié, che sonno la corda dele tavole, simile a-<lb></lb> |
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5039 la corda data .bo. Onde l’ arco suo minore, se minore vuoi sapere over magio-<lb></lb> |
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5040 re, se di magiore cerchi, cioé se cerchi del’ arco .ogdb. Dove il minore arco suo è .42. Lo qua-<lb></lb> |
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5041 le multiplica per .10., fanno .420. E, per .42. diviso, ne viene .10. per l’ arco .beo. E, se ’l magiore<lb></lb> |
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5042 arco dela corda, che è pertiche .36. e pie’ .2., multiplicarai per .10., cioé .90. per .10., fanno .900. che,<lb></lb> |
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5043 per .42. diviso, vienne .21 3/7. E .21. pertica .3/7. é l’ arco .ogdb., che è magiore del mezzo cerchio.<lb></lb> |
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5044 Ancora sia il circulo .abgd. Del quale il diametro .ag. sia .12. E la corda .ad. sia .6.<lb></lb> |
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5045 pertiche e uno pie’. Adimandase la notitia del’ arco .afd. che è minore del mezzo<lb></lb> |
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5046 cerchio. Multiplica adonca .6. pertiche e .1o. pie’ per .42., fanno .259. pertiche. Le qua-<lb></lb> |
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5047 li dividi per .12., cioé per .ag., ne perviene .21. pertiche .3. pie’ .2. oncie .16. ponti. E on-<lb></lb> |
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5048 de, a cioché habiamo l’ arco in oncie dela corda, lo vogliamo, per figura geometrica, dimostra-<lb></lb> |
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5049 re. Sia un mezzo cerchio .ezitk. Del quale el diametro .ek. sia pertiche .42., cioé il dia-<lb></lb> |
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5050 metro dele tavole. Del quale si tolga l’ arco .ez. e l’ arco .et. De’ quali .ez. sia .22. e .et. sia .23. E me-<lb></lb> |
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5051 nise le corde .ez.et. E sia .ez.21. e l’ arco .et. sia pertiche .21. e .5. pie’ e .2. oncie e .16. ponti, comme<lb></lb> |
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5052 di sopra trovammo nele tavole. Infra le quali corde cade la corda trovata, dela quale cerchia-<lb></lb> |
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5053 mo l’ arco nele tavole. Onde sappiamo el detto arco cadere infra ’l ponto .z. e il ponto .t. Dove<lb></lb> |
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5054 caggia nel ponto .i. e menise la corda .ie., che sia pertiche .21. e .3. pie’ e oncie .9., comme di sopra mo-<lb></lb> |
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5055 strammo. Onde vogliamo trovare l’ arco .ie. Ma noi sappiamo, per le cose dette, che la pro-<lb></lb> |
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5056 portione del’ arco .ez. ala corda .ez. è minore dela proportione del’ arco .ei. ala corda .ei. Ma,<lb></lb> |
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5057 se noi poniamo la proportione del’ arco .ei. ala corda .ei. quella medesima che á l’ arco .ez.<lb></lb> |
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5058 ala corda .ez. sirá l’ arco .ei.22. pertiche .3. pie’ e oncie .12., che pervengono dela divisione dela<lb></lb> |
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5059 multiplicatione dela corda .ez. nela corda .ei. per l’ arco .ez. Ma la proportione del’ arco .ei.<lb></lb> |
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5060 é in magiore proportione che l’ arco .ez. ala corda .ez. Adonca l’ arco .ei. è piú dele trovate per-<lb></lb> |
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5061 tiche .22. e .3. pie’ e .12. oncie. Ancora, se poniamo l’ arco ala corda .ei. nela proportione del’ ar-<lb></lb> |
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5062 co .et. ala corda .et., sirá l’ arco .et.22. pertiche .4. pie’ .4. oncie .13. ponti. E di sopra trovammo<lb></lb> |
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5063 che ’l circulo .ei. è piú de pertiche .22. pie’ .3. e oncie .12. Onde, se dimezzaremo la differentia che<lb></lb> |
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5064 è da pertiche .22. e .3. pie’ e .12. oncie infino in pertiche .22. e piedi .4. e oncie .4. e ponti .13. e quel-<lb></lb> |
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5065 la mitá che viene agiogneremo sopra .22. pertiche .3. pie’, haremo, secondo l’ apresamento,<lb></lb> |
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5066 la quantité del’ arco .ei. Overo altramente agiongneremo la corda .ez. e la corda .et. e hare-<lb></lb> |
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5067 mo pertiche .45. pie’ .1o. oncie .16. ponti .13. De’ quali la mitá è l’ arco .ei. Overo altramente, to-<lb></lb> |
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5068 gliamo dele corde .ei. e .et. la quantitá dela corda .ez., rimarrá dela corda .oi. la quantitá .ei.<lb></lb> |
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5069 che è pie .3 1/2. E, dela corda .et., rimane la quantita .nt., che è .5. pié, oncie .12. e ponti .16. E po-<lb></lb> |
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5070 niamo l’ arco .zi. al’ arco .zt., cioé a una perticha, comme .oi. al .nt. E questo è che multiplica-<lb></lb> |
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5071 remo l’ arco .tz., che è una pertica, cioé .1260. ponti, e divideremo la somma per .nt., cioé per<lb></lb> |
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5072 ponti .1856. E haremo per l’ arco .zi. pie’ .4. e oncie una e ponti .6. Li quali, agionti al’ arco .ez.<lb></lb> |
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5073 (che è pertiche .22.) haremo pertiche .22. e pie’ .4. e oncie .1a. e ponti .6. per l’ arco .ezi., che è si-<lb></lb> |
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5074 mile al’ arco adimandato .ad. del’ altra figura. Onde, se multiplicaremo quello per la sexta del<lb></lb> |
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5075 diametro .ag., cioé per .2., e divideremo per lo sexto del diametro dele tavole, cioé per .7., ha-<lb></lb> |
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5076 remo pertiche .6. e .2. pie’ e .15. oncie e .15. ponti per l’ arco .ad. E, se l’ arco .abd., che è magiore<lb></lb> |
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5077 del mezzo cerchio, per la corda .ad. vuoi havere, tra’ la corda .ei. della corda .et., rimarranno<lb></lb> |
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5078 ponti .596. E questo multiplica per l’ arco .tz., cioé per ponti .2160. E quello vienne dividi per<lb></lb> |
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5079 .nt., cioé per .1856. e tranne ponti .694. che sonno .1o. pie’ e oncie .16. e ponti .14. per l’ arco .ti.<lb></lb> |
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5080 che, agionto con l’ arco magiore dele tavole, del quale la corda è la linea .et., quel arco è .109. per-<lb></lb> |
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5081 tiche. Haremo .109. pertiche .1o. pie’ e oncie .16. e ponti .14. per l’ arco dele tavole. El quale,<lb></lb> |
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5082 <lb></lb> |
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5083 <pb></pb></p> |
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5084 <p class="folio"> |
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5085 folio 34v |
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5086 </p> |
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5087 <p class="main"> |
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5088 <lb></lb> |
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5089 </p> |
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5090 <p class="runhead"> |
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5091 Distinctio quarta. Capitulum secundum. |
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5092 </p> |
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5093 <p class="main"> |
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5094 <lb></lb> |
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5095 multiplicato per lo sexto di .12. e la somma dividerai per lo sexto di .42., harai pertiche .31. pie’<lb></lb> |
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5096 .1o. oncie .7. e ponti .6 6/7. per l’ arco .abd. |
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5097 </p> |
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5098 <p class="main"> |
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5099 E, se sopra l’ arco .afd. noto la corda .ad. non saputa vuoi havere, multiplica l’ ar-<lb></lb> |
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5100 co .afd. per lo diametro dele tavole e la somma dividi per lo diametro .ag. e ha-<lb></lb> |
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5101 rai pertiche .22. pie’ .4. oncie .1a. e ponti .6. per l’ arco detto .ezi., che è l’ arco cadente<lb></lb> |
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5102 nele tavole infra l’ arco .ez. e l’ arco .zt. E, acioché troviamo la corda del .ci., che è<lb></lb> |
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5103 simile ala corda .ad. del dato cerchio .abgd., del trovato arco .ezi. tra’ l’ arco .ez. Rimarran-<lb></lb> |
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5104 no .4. pie’ e .1a. oncia e .6. ponti, che in tuto sonno ponti .1466. Li quali multiplica per la diffe-<lb></lb> |
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5105 rentia che è infra la corda .ez. e la corda .et., che sonno note dele tavole. É quella differentia<lb></lb> |
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5106 .nt., che è .1856. ponti. E dividi la somma per li ponti d’ una pertica. Cioé per l’ arco .tiz., vienne<lb></lb> |
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5107 .3 1/2. pie’ per la linea .oi. Li quali agiongni ala linea .eo., che è iguale .ez. e harai pertiche .22.<lb></lb> |
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5108 e pie’ .3 1/2. per la corda .ei., che è simile ala corda .ad. E cosí, secondo quelle cosse che habiamo<lb></lb> |
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5109 detto, multiplicarai quella per lo sexto di .12. e dividerai per la sexta di .42., vienne pertiche<lb></lb> |
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5110 .6. e pie’ .1o. per la corda .ad. E cosí, quando li diametri de’ cerchi fienno noti potremo per le da-<lb></lb> |
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5111 te corde note, gli archi non saputi trovare. |
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5112 </p> |
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5113 <p class="main"> |
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5114 Quando adunque gli archi per le corde e le corde per gli archi, per quello che<lb></lb> |
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5115 habiamo detto, saprai trovare e vorrai l’ area d’ alcuna settione di cerchio trova-<lb></lb> |
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5116 re, l’ arco studierai di trovare e la mitá di quello, per la mitá del diametro del cer-<lb></lb> |
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5117 chio, multiplica. E quello che fa sirá l’ area dela detta settione. Verbi gratia. Sia<lb></lb> |
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5118 il settore .abgd., contento sopra le rette .ab. e .ad. e dal’ arco .bgd. Perché el settore è una fi-<lb></lb> |
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5119 gura sotto li ponti .abgd., sirá ciascuna retta .ab. e .ad. mezzo diametro del cerchio. Onde<lb></lb> |
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5120 il ponto .a. sia centro del circulo del quale è tagliato il detto settore di cerchio .abgd. E compise<lb></lb> |
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5121 adunque il cerchio che sia il cerchio .gbed. E sia l’ arco .be. e .ez. iguale al’ arco .bgd. E compi-<lb></lb> |
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5122 se la retta .ae. e .az. Sirá adunque ciascun settore .abe. e .aez. iguali al settore .abgd. Onde<lb></lb> |
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5123 e gli é cosí l’ arco .db. al’ arco .be. cosí el settore .abgd. al settore .abe. Similmente è il settore<lb></lb> |
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5124 .aez. equale al settore .adb. Onde l’ arco .db. é al’ arco .ez. cosí el settore .abe. al settore .aez.<lb></lb> |
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5125 E .3. settori, adunque, .abgd. e .abe. e .aez. sono infra loro iguali. E gli e .2. di quelli al’ altro<lb></lb> |
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5126 sonno .2. cotanti. Onde il settore .adbe. è doppio al settore .aez. E l’ arco .dbe. è dopoio al’ ar-<lb></lb> |
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5127 co .ez. Perché e gli é cosí l’ arco .dbe. al’ arco .ez., cosí il settore .adbe. al settore .aze. Onde, se<lb></lb> |
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5128 tutto il cerchio divideremo in settori, trovaremo che la proportione d’ un di loro a tutto il cir-<lb></lb> |
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5129 culo è comme l’ arco suo ala mitá dela linea circonferente. Imperoché la proportione del nu-<lb></lb> |
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5130 mero fatto del mezzo diametro nela mitá dela linea circonferente del’ arco del settore é al<lb></lb> |
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5131 numero fatto del mezzo il diametro nela mitá dela linea circonferente. Ma quello ch’ é fat-<lb></lb> |
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5132 to dal mezzo il diametro dela mitá dela circonferentia fa l’ area del circulo. Adonca è cosí la multiplica-<lb></lb> |
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5133 tione del mezzo diametro del circulo nela mitá del’ arco del settore al’ area del circulo, cosí el se-<lb></lb> |
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5134 tore al’ area del circulo. Adonca l’ area di tutti li settori sonno fatte dela multiplicatione di .1/2. il diametro<lb></lb> |
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5135 de’ soi cerchi nela mitá degli archi loro. E questo volsi mostrare. E, acioché piú chiaro apaia, per<lb></lb> |
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5136 numeri, sia ciascuna dele rette .ab. e .ad.5.bracia. E l’ arco .bgd. sia .8. Sirá adunque il diame-<lb></lb> |
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5137 tro tutto .10.bracia. Multiplicarai adunque mezzo il diametro .ad. nela mitá del’ arco .bed.<lb></lb> |
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5138 vengano .20. per l’ area del settore .abd. E, se l’ area del settore .abez. vuoi havere, multiplica-<lb></lb> |
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5139 rai mezzo il diametro .ae. nela mitá del’ arco .bez. Vengono .40. per l’ area del settore .abez.<lb></lb> |
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5140 E, se l’ area d’ alcuna portione di cerchio vuoi trovare, minore del mezzo cerchio,<lb></lb> |
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5141 comme l’ area dela settione overo portione di cerchio .abg., del quale la corda<lb></lb> |
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5142 .ag. sia bracia .16. e la saetta .bd. sia bracia .4., el diametro del cerchio onde tale por-<lb></lb> |
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5143 tione viene studia di trovare in questo modo: multiplicando la mitá dela corda<lb></lb> |
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5144 in sé, cioé .8. in sé, fa .64. E partendo in .4., cioé per la saetta, haremo .16. per la linea .de. Onde<lb></lb> |
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5145 il diametro è .20. E piglise il centro del cerchio che sia .f. e faciase .fa. e .fg. E sia il settore .fabg.<lb></lb> |
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5146 Onde multiplicaremo .fb. nela mitá del’ arco .abg., cioé nel’ arco .bg., verranne l’ area del set-<lb></lb> |
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5147 tore .abg. dela quale, se nne torremo l’ area del triangolo .fag., la quale è fatta dela multiplica-<lb></lb> |
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5148 tione del .fd. in .dg., rimaranno l’ area dela settione overo portione di cerchio .abd. E, se<lb></lb> |
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5149 l’ area del rimanente del cerchio, cioé la portione fatta dala retta .ag. e dal’ arco .aeg.<lb></lb> |
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5150 vuoi, la mitá del diametro .fe. nela mitá del’ arco .aeg. multiplica e a quello agiongni<lb></lb> |
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5151 l’ area del triangolo .fag. E harai l’ area dela portione .aeg. E cosí studia de operare in simili et cetera.<lb></lb> |
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5152 <lb></lb> |
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5153 <pb></pb></p> |
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5154 <p class="folio"> |
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5155 folio 35r |
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5156 </p> |
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5157 <p class="main"> |
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5158 <lb></lb> |
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5159 </p> |
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5160 <p class="runhead"> |
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5161 Distinctio quarta. Capitulum tertium. 35 |
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5162 </p> |
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5163 <p class="main"> |
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5164 <lb></lb> |
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5165 E, se l’ area d’ una portione di cerchio, composta d’ un triangolo e d’ una portione,<lb></lb> |
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5166 comme è la figura .abgd., la quale si conosci non essere settore di cerchio per una<lb></lb> |
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5167 di .2. cagioni: overo perché .ab. e .ad. non sonno iguali overo perché .ag. è<lb></lb> |
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5168 minore o magiore del .ab. Prima misura la portione .bgd. E, dipoi, truova<lb></lb> |
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5169 l’ area del triangolo .bda. e agiongni al’ area dela detta portione .bgd. e harai l’ area de-<lb></lb> |
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5170 la figura .abgd. E cosí, in simili, studia di fare et cetera. |
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5171 </p> |
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5172 <p class="main"> |
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5173 E, volendo misurare una figura a modo de’ pesci, cioé composta di due portioni<lb></lb> |
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5174 di cerchio, comme sia la figura .etzi., ch’ é composta di .2. portioni di cerchio, cioé<lb></lb> |
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5175 .eti. e .ezi., prima trova l’ area d’ una portione e poi del’ altra e quelle .2. aree in<lb></lb> |
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5176 una somma agiongni. E harai l’ area dela detta figura .etiz. E cosí ogni figu-<lb></lb> |
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5177 ra che havesse forma di tondo o compositione di tondo studia di misurare. E peró faremo fi-<lb></lb> |
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5178 ne a questo capitolo e delo terzo e ultimo diremo. |
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5179 </p> |
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5180 <p class="main"> |
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5181 Qualiter metiantur superficies in montibus vallibusque existentes. Capitulum tertium.<lb></lb> |
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5182 Quando adunque alcuno campo, che sia nela salita d’ un monte overo inn ascien-<lb></lb> |
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5183 sione d’ alcuna costa overo ripa, vuoi misurare, lo lato dela superficie predetta che<lb></lb> |
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5184 sia in piano studia di trovare e, de’ detti lati, l’ area cerca e quello sia l’ area dela<lb></lb> |
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5185 sopradetta figura overo campo. Non si misurano li monti, ma le superficie ap-<lb></lb> |
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5186 parenti in quello, comme case o dificij, arbori e ancora escemi. Imperoché non secondo<lb></lb> |
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5187 l’ angolo retto stanno elevati, onde si cercha l’ area de’ detti piani sopra’ quali le apparenti su-<lb></lb> |
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5188 perficie de’ monti giaciano. E, sopra quelli piani predetti, d’ ogni cosa, secondo retto angolo è<lb></lb> |
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5189 elevato e comme, per li lati declinanti, queste notitie deli lati dele predette superficie stanti<lb></lb> |
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5190 sotto quelle, voglio dimostrare. Sia dato adunque una linea .ab. per lo lato d’ alcuna declina-<lb></lb> |
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5191 tione di superficie giacente in monte. E sia, sopra .bg., perpendiculare la retta .ag., constituen-<lb></lb> |
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5192 te l’ angolo .agb. retto. É de bisogno, per la notitia delo lato apparente .ab., lo lato oculto .bg.<lb></lb> |
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5193 e ancora la perpendiculare .ag. verissimamente trovare, che in .2. modi si puó fare.<lb></lb> |
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5194 Lo primo modo e quello che li savij misuratori da terre é che, dal ponto .a. capo,<lb></lb> |
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5195 quello che misura ponga la pertica. La quale si stenda inverso il .b., sopra la linea<lb></lb> |
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5196 .ab. E il capo della pertica che è di sopra immobile tieni e l’ altro alza su insino a<lb></lb> |
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5197 tanto che la detta pertica stia equedistante ala linea .gb. El quale saprai per al-<lb></lb> |
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5198 cuno strumento detto archipendolo e, per Vitruvio e Frontino e gli altri architettori, ditto li-<lb></lb> |
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5199 bella, che di sotto te ’l mostraró. E alora, per lo capo dela pertica di sotto, cioé per lo capo ove-<lb></lb> |
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5200 ro ponto .e. poni il filo col piombo. E lascia andare sopra la linea .ab. E, dove il piombo ca-<lb></lb> |
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5201 de, quivi, con quella pertica, incomencia a misurare col detto ordine. E cosí farai infino a tan-<lb></lb> |
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5202 to che harai compiuta tutta la declinatione de .ab. Verbi gratia: sia la prima pertica posta che<lb></lb> |
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5203 sia equedistante alla linea .bg. la pertica .ae. E, per lo ponto .e., caggia il filo sopra il ponto .c.<lb></lb> |
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5204 Sopra il quale poni ancora il capo dela pertica tenendola collo archipendolo equedistan-<lb></lb> |
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5205 te con la linea .bg. E sia la pertica .cf. E dal ponto .f. caggia il filo sopra il ponto .h. E, dal pon-<lb></lb> |
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5206 to .h. verso il .b., poni un’ altra volta la pertica equedistante ala linea .bg., che sia .hi. e, per lo pon-<lb></lb> |
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5207 to .i., caggia il filo sopra il ponto .b. Imperoché tante volte la pertica sia posta cadente sopra<lb></lb> |
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5208 .ab. quante volte una pertica sia nel lato .gb. Imperoché la pertica .ae. è equedistante al lato<lb></lb> |
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5209 .gb., sia l’ angolo .eag. retto. Conciosiacosaché l’ angolo .agb. sia retto per la .29. del po. E,<lb></lb> |
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5210 perché .ec. è il cadimento del filo, se mettaremo la linea per lo .c. al ponto .e., sopra la linea .gb.,<lb></lb> |
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5211 che sia la linea .ek., sia la retta .ek. catetto sopra la retta .gb. Onde la retta .ek. è equedistante e<lb></lb> |
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5212 iguale ala retta .ag. Onde .gk. è iguale ala longhezza della pertica .ae. Similmente, sopra<lb></lb> |
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5213 .h., meneremo la linea .fl., venente sopra la linea .kl., per le cose dette. Onde .kl. sia iguale ala li-<lb></lb> |
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5214 nea overo perticha .cf. Ancora, sopra il caso del filo, cadendo al ponto .i., nel ponto .b. mene-<lb></lb> |
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5215 remo la linea .ib. Sirá quella equedistante e iguale ala linea .hl. Dove .lb. è iguale ala perti-<lb></lb> |
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5216 ca .hi. Adunque, quante volte la pertica fo presa equedistante ala linea .gb. sopra la linea .ab.,<lb></lb> |
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5217 tante volte una pertica fo presa per la longhezza iguale al .gb., comme dicemmo.<lb></lb> |
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5218 L archipendolo adunque è uno strumento di lengno avente forma d’ uno trian-<lb></lb> |
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5219 golo equicurio. E dal’ uno deli angoli pende il filo col piombo. Conciosiacosaché<lb></lb> |
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5220 la basa habia un ponto nel mezzo. E, quando porrai l’ archipendolo e dal’ ango-<lb></lb> |
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5221 lo di sopra penda il filo col piombo e, quando passerá per lo ponto del mezzo la<lb></lb> |
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5222 basa di quello, alora la pertica stará equedistante a quel piano che vuoi misurare, che al’ ochio<lb></lb> |
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5223 si puó vedere, commo nela figura di sotto. Dove la pertica sia la linea .op. sopra la quale è eretto<lb></lb> |
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5224 <lb></lb> |
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5225 <pb></pb></p> |
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5226 <p class="folio"> |
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5227 folio 35v |
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5228 </p> |
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5229 <p class="main"> |
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5230 <lb></lb> |
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5231 </p> |
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5232 <p class="runhead"> |
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5233 Distinctio quinta. Capitulum primum. |
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5234 </p> |
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5235 <p class="main"> |
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5236 <lb></lb> |
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5237 l’ archipendolo .aebgf. E dal ponto .a. cade il filo col piombo .ad. sopra il ponto dela perti-<lb></lb> |
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5238 ca .c. E il detto filo passa per lo ponto .h., che é nela mitá dela linea .ef., equedistante ala ba-<lb></lb> |
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5239 sa del detto archipendolo. Dico alora la pertica .op. essere equedistante ala longhezza del pia-<lb></lb> |
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5240 no. E, quando l’ ascesa andasse per una linea senza salire o sciendere, cioé fosse una comme<lb></lb> |
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5241 nela passata figura, alora nonn’ é de bisogno se none una volta porre l’ archipendolo, impe-<lb></lb> |
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5242 roché, la seconda volta overo l’ altre volte comporrai uno triangolo simile al primo, cioé si-<lb></lb> |
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5243 mile al triangolo .aec., imperoché fienno sempre in equedistanti linee. E cosí haresti la lon-<lb></lb> |
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5244 ghezza del piano .gb. e l’ altezza del monte .ag. |
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5245 </p> |
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5246 <p class="main"> |
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5247 Li antichi ordinavano con le canne uno triangolo simile in questo modo. Descrit-<lb></lb> |
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5248 ta la figura .agb., dela quale il lato .ab. giace nell’ aparentia del monte per lo quale<lb></lb> |
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5249 volevano havere la longhezza del piano .bg. e l’ altitudine .ag., dirizavano adun-<lb></lb> |
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5250 que, sopra la radice del monte, la canna .bc., ortogonalmente posta. Sopra la qua-<lb></lb> |
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5251 le ponevano l’ altra canna causante l’ angolo .c. retto. Del quale l’ altro capo giacerá sopra la li-<lb></lb> |
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5252 nea .ab. La quale canna sia .ce. E sia il triangolo .cbe. simile al triangolo .abg. Onde sia cosí<lb></lb> |
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5253 .be. al .ba., cosí .ec al .bg., per la .2a. del .6o. Multiplicavano adunque .ab. per .ce. e divideva-<lb></lb> |
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5254 no per .be. E cosí havevano notitia del lato .bg. Similmente, perche e gli é cosí .be. al .ba., co-<lb></lb> |
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5255 sí .cb. al .ag., multiplicavano .ab. per .bc. e dividevano per .bc. e cosí havevano l’ altezza .ag.<lb></lb> |
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5256 Molte figure sonno quelle situate e poste ne’ monti, cioé nell’ altezze de’ monti o-<lb></lb> |
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5257 vero descensioni di ripe. Le quali, in diversi modi, sonno erette, le quali, a volerle<lb></lb> |
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5258 misurare, ti bisogna quello salire arrecare a ppiano overo quel scendere recare<lb></lb> |
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5259 a ppiano. E quelle, dipoi, misurare comme l’ altre figure superficiali. Imperoché’ l<lb></lb> |
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5260 piano é quello che è bisogno missurare e non l’ altezza. E molto difficilmente si puó mostra-<lb></lb> |
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5261 re tali figure se nno in sul fatto del misurare. E peró a questo capitolo e distintione faremo fine.<lb></lb> |
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5262 E, seguendo, diremo del dividere le superficie im parti ergo et cetera.<lb></lb> |
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5263 <lb></lb> |
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5264 </p> |
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5265 <p class="head"> |
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5266 Distinctio quinta eiusque divisio. |
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5267 </p> |
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5268 <p class="main"> |
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|
5269 <lb></lb> |
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5270 Lo dire passato, assai evidentemente, ha mostro il modo a trovare l’ area di ciascu-<lb></lb> |
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5271 na superficie. Onde, hora, assai competentemente, mi pare de dimostrare comme<lb></lb> |
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5272 quelle tali superficie in parti overo in sorte si debiano dividere. E peró starai<lb></lb> |
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|
5273 attento. Questa distintione adunque in .4. capitoli la divideremo. Nel pri-<lb></lb> |
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5274 mo mostraremo el dividero de’ triangoli. Nel secondo e quadrilateri. Nel terzo le fi-<lb></lb> |
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5275 gure di molti lati. Nel quarto de’ cerchi e loro parti mostraremo il modo a dividergli.<lb></lb> |
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|
5276 <lb></lb> |
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|
5277 </p> |
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|
5278 <p class="head"> |
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|
5279 De modo dividendi triangulares formas in partes plures proportionabiliter. |
|
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|
5280 </p> |
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|
5281 <p class="main"> |
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|
5282 <lb></lb> |
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|
5283 </p> |
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|
5284 <p class="head"> |
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|
5285 Capitulum primum. |
|
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|
5286 </p> |
|
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|
5287 <p class="main"> |
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|
5288 <lb></lb> |
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5289 Quando adunque uno triangolo in due parti iguali, da uno degli angoli, vuoi di-<lb></lb> |
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5290 videre, é de bisogno, da quel‘ angolo ala mitá dela basa, produrre una linea e ha-<lb></lb> |
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5291 rai lo intento. Verbi gratia. Vogliamo el triangolo .abg., dal ponto .a., in .2. par-<lb></lb> |
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5292 ti iguali dividere. Dividase adunque el lato .bg. in .2. parti iguali, sopra il ponto<lb></lb> |
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5293 .d. e compise la retta .ad. Dico adunque il triangolo .abg. in .2. triangoli iguali essere diviso.<lb></lb> |
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5294 Sonno adunque i triangoli .abd. e adg. simili infra loro e sotto una medesima altitudine.<lb></lb> |
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5295 Per la prima del .6o. e sonno iguali, imperoché gli é cosí .bd. al .dg., cosí il triangolo .abd. al<lb></lb> |
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5296 triangolo .adg., per la ditta pa. É certamente la basa .bd. iguale ala basa .dg. E peró li trian-<lb></lb> |
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5297 goli .abd. e .adg. sonno infra loro iguali, comme dicemmo. Overo, se meneremo il catetto<lb></lb> |
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5298 dal .a. sopra la linea .bg., sirá quel catetto a ciascun triangolo .abd. e .adg. Del quale, se multi-<lb></lb> |
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5299 plicaremo la mitá nela basa .bd., sirá iguale ala multiplicatione dela mitá del detto catetto<lb></lb> |
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5300 nela basa .dg. E, a multiplicare la mitá del catetto nele base .bd. e .dg., fanno l’ area di detti trian-<lb></lb> |
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5301 goli overo l’ area del gran triangolo. E peró dirai il triangolo .adg. essere iguale .adb.<lb></lb> |
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5302 Li triangoli aventi uno angolo iguale hano proportione infra loro composte<lb></lb> |
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5303 de’ lati continenti quel angolo iguale. E acioché questo se chiaresca, sienno li<lb></lb> |
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5304 triangoli .abg. e .gez. aventi iguali gli angoli che sonno al .g. Dico certamente<lb></lb> |
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5305 li detti triangoli essere nella proportione composta di queste proportioni che son-<lb></lb> |
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5306 no fatte da’ lati continenti gli angoli eguali, cioé di quella che á lo lato .bg. al lato .eg. e di quel-<lb></lb> |
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5307 la che á lo lato .ag. al lato .gz., che cosí te ’l proveró. Compise la retta .ae. E pongase nel trian-<lb></lb> |
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5308 golo .abg. e .gez. el triangolo .age. Adunque sirá la proportione del triangolo .abg. al trian-<lb></lb> |
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5309 golo .gez. fatta dele .2. proportioni, cioé di quella che á il triangolo .abg. al triangolo .agf. e<lb></lb> |
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5310 di quella che á il triangolo .age. al triangolo .gez. Ma la proportione del triangolo .abg.<lb></lb> |
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5311 <lb></lb> |
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5312 <pb></pb></p> |
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5313 <p class="folio"> |
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5314 folio 36r |
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5315 </p> |
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5316 <p class="main"> |
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5317 <lb></lb> |
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5318 </p> |
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5319 <p class="runhead"> |
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5320 Distinctio quinta. Capitulum primum. 36 |
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5321 </p> |
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5322 <p class="main"> |
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5323 <lb></lb> |
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5324 al triangolo .age. è comme la basa .bg. ala basa .ge. Conciosiacosa sienno sotto una medesi-<lb></lb> |
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5325 ma altitudine, per la .pa. del .6o. E ancora la proportione del triangolo .eag. al triangolo .gez.<lb></lb> |
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5326 è comme il lato .ag. alo lato .gz. E peró la proportione del triangolo .abg. al triangolo .gez.<lb></lb> |
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5327 é fatta dele proportioni de’ lati .ag. al .ge. e .bg. al .gz., continenti gli angoli iguali, ch’ era de bi-<lb></lb> |
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5328 sogno mostrare. Componsi la proportione del triangolo .abg. al triangolo .gez. dele propor-<lb></lb> |
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5329 tioni del .bg. al .gz. e .ga. al .ge. Imperoché li lati .ag. e .bg. fienno antecedenti e li lati .ge. e<lb></lb> |
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5330 .gz. fienno consequenti. E, benché nel tratatto degli angoli ne dicessimo alcuna cosa, cioé che<lb></lb> |
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5331 la proportione composta era dela multiplicatione di tutti gl’ antecedenti al fatto di tutti li con-<lb></lb> |
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5332 sequenti. E peró la proportione del triangolo .abg. al triangolo gez. è comme il fatto del .bg.<lb></lb> |
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5333 in .ga. al fatto del .eg. in .gz. Onde, se iguale sia la multiplicatione del lato .eg. nello lato .gz.<lb></lb> |
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5334 ala multiplicatione del lato .bg. in .ga., alora sirá iguale il triangolo .egz. al triangolo .abg.<lb></lb> |
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5335 E, se minore, minore e, se magiore, magiore. E questo volsi demostrare.<lb></lb> |
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5336 E, se da un triangolo sia menata una linea retta segante e .2. lati del triangolo che,<lb></lb> |
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5337 con detti due lati, facia triangolo avente uno angolo comune col detto triango-<lb></lb> |
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5338 lo, alora sia la proportione del’ un triangolo al’ altro comme el fatto de’ lati conti-<lb></lb> |
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5339 nenti quel angolo. Onde sia il triangolo .abc. E in quel si meni la linea .de., segan-<lb></lb> |
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5340 te e lati .ca. e .cb. sopra li ponti .e. e .d. Dico il triangolo .abc. havere proportione al triangolo<lb></lb> |
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5341 .dec. comme el fatto del .ac. in .bc. al fatto del .dc. al .ce. Che cosí te ’l proveró. Sopra il lato .ac.<lb></lb> |
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5342 appicheró el triangolo .acf., iguale al triangolo .dec. E, perché li triangoli .abc. e .afc. son-<lb></lb> |
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5343 no sotto una altezza, é cosí .bc. al .fc., cosí il triangolo .abc. al triangolo .afc., per la .pa. del .6o.<lb></lb> |
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5344 Ma la proportione del .ac. al .fc. è comme el fatto del .ac. in .cb. al fatto del .ac. in .cf. Adun-<lb></lb> |
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5345 que la proportione del triangolo .abc. al triangolo .afc. è comme il fatto del .ac. in .bc. al fat-<lb></lb> |
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5346 to del .ac. in .cf. E, perché el triangolo .dec. è iguali al triangolo .acf., é la proportione del tri-<lb></lb> |
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5347 angolo .acb. al triangolo .dce. comme el fatto del .ac. in .bc. al fatto del .ca. in .cf. E, perché<lb></lb> |
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5348 li triangoli .acf. e .dce. sonno infra loro iguale e hano uno angolo comune e gli lati posti in-<lb></lb> |
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5349 torno al’ angolo comune sonno in proportione mutua, comme nela .15a. del .6o. de Euclide si<lb></lb> |
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5350 manifesta, adunque è cosí .ac. al .dc., cosí .ce. al .cf. Onde il fatto del .dc. in .ce. è iguale al fat-<lb></lb> |
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5351 to del .ac. in .cf. Adunque la proportione del triangolo .abc. al triangolo .dec. é comme il<lb></lb> |
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5352 fatto del .ac. in .cb. al fatto del .dc. in .ce., ch’ era bisogno mostrare.<lb></lb> |
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5353 Quando vorrai dividere uno triangolo in .2. parti iguali per una linea la quale si<lb></lb> |
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5354 muova da uno ponto dato, comme sia il triangolo .bgd. nel quale sia il ponto da-<lb></lb> |
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5355 to .a., nela linea .gd. Dal quale ponto .a. voglio menare una linea la quale divida<lb></lb> |
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5356 el detto triangolo in .2. parti iguali. E sia prima il ponto .a. nel mezzo dela linea .gd.<lb></lb> |
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5357 E meneró .a. infino al .b. Dico che .ab. divide quel triangolo in .2. parti iguali. E questo, per<lb></lb> |
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5358 le cose dette, é manifesto. |
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5359 </p> |
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5360 <p class="main"> |
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5361 E, se ’l ponto dato nonn’ é in sul mezzo dela facia, comme in questo altro triangolo<lb></lb> |
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5362 .abg., nel quale è il ponto dato .d. che è piú presso al .b. che al .g. E sia .ab.13. E il<lb></lb> |
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5363 .bg.14. e .ag.15. Adimandase quanto è la linea .gz., cioé in che luogo viene la linea<lb></lb> |
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5364 .dz. Conciosiacosaché .gd. sia .9. Dovemo multiplicare lo lato .ag. per .bg. e di quel-<lb></lb> |
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5365 lo pigliare la mitá, che è .105. E questo divideremo per .dg., cioé per .9., vienne .11 2/3. E .11 2/3. è<lb></lb> |
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5366 dal .g. al .z. E dal .a. al .z. è l’ avanzo infino in .15., che v’ é .3 1/3. E, se vuoi sapere quanto è .dz. mul-<lb></lb> |
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5367 tiplicarai .3 1/3. in sé, cioé .az. e .de. in sé. Imperoché .ad. è ponto il catetto fanno .11 1/9. e .4. e trai<lb></lb> |
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5368 .4. del .11 1/9., rimane .7 1/9. Del quale la radici è .2 2/3. per la linea .pa. e .pd. sia .9 1/3. El quale in sé mul-<lb></lb> |
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5369 tiplica, fanno .87 1/9., ai quali agiongni .4., cioé il quadrato del .de., fanno .91 1/9 per la linea .dz. E<lb></lb> |
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5370 cosí hai fatto quanto .dg. fusse .9. el .zg. sirá .11 2/3. E .dz. sia la radici de .91 1/9. E questo volavamo<lb></lb> |
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5371 dire. |
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5372 </p> |
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5373 <p class="main"> |
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5374 Se, da .2. angoli d’ uno triangolo ala mitá de’ .2. lati sottoposti a quelli angoli, .2. li-<lb></lb> |
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5375 nee rette si menino, e lle si segheranno proportionalmente in tal modo che quel-<lb></lb> |
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5376 la parte che è infra l’ angolo e il ponto del segamento al’ altra parte è .2. cotanti. E,<lb></lb> |
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5377 se, dal’ altro angolo sopra l’ altro lato, per lo ponto della settione, si mena una linea retta, e lla di-<lb></lb> |
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5378 viderá quello lato per .2. parti iguali. Comme sia nel triangolo .abg.: dagli angoli .abg. e<lb></lb> |
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5379 .bag., sopra la mitá de’ lati .ag. e .bg., si menino le rette .ae. .bz., segantise infra loro nel ponto .d.<lb></lb> |
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5380 Dico che la proportione .ad. al .de. è comme .bd. al .dz. e ciascuna di loro, cioé .ad. e .bd., al’ a-<lb></lb> |
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5381 vanzo sonno doppi, che cosí te ’l proveró. Dal ponto .a. meneró la retta .ai. equedistanti ala<lb></lb> |
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5382 linea .bg.E meneró la retta .bz. infino al .i., cioé infino concorra col ponto .i. E fienno i trian-<lb></lb> |
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5383 <lb></lb> |
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5384 <pb></pb></p> |
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5385 <p class="folio"> |
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|
5386 folio 36v |
|
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5387 </p> |
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5388 <p class="main"> |
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5389 <lb></lb> |
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5390 </p> |
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5391 <p class="runhead"> |
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5392 Distinctio quinta. Capitulum primum. |
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5393 </p> |
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5394 <p class="main"> |
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5395 <lb></lb> |
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5396 goli .azi. e .gbz. infra loro simili. Onde e gli é cosí .az. al .zg., cosí .iz. al .zb. e .ia. al .bg. iguali. E<lb></lb> |
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5397 certamente .az. del .gz. iguali, adonca saranno .zi. del .zb. e .ia. de .bg. Onde è cosí .bg. al .be.,<lb></lb> |
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5398 cosí .ad. al .de. e .id. al .db. doppia e adonca .bg. del .be. Onde doppia è .ad. del .de. e .id. del<lb></lb> |
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5399 .db. Perché è iguale .iz. del .zb. Onde communamente, agiongnendo .zd. sia tutta .id. iguale a .2.<lb></lb> |
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5400 rette .bz. e .zd. Ma .id. è mostro essere doppia del .db. Onde doi rette .bz. e .zd. sonno dop-<lb></lb> |
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5401 pie ale rette .bd. Onde, se da ogni parte si toglie .bd., rimarrá .bd. iguali a .2. volte .dz. On-<lb></lb> |
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5402 de .db. del .dz. è doppio e cosí si mostra .ad. essere doppia al .de. E peró è cosí .ad. al .de. com-<lb></lb> |
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|
5403 mo .bd. al .dz., ch’ era bisogno mostrare. E, se dal’ angolo .g., per lo ponto .d., passerá la linea .gt.,<lb></lb> |
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5404 dico che lo lato .ab. é diviso in .2. parti iguali sopra il ponto .t. Meneró adonca .gt. fuore del<lb></lb> |
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|
5405 triangolo .abg. infino a tanto che concorrerá col ponto .k. dela linea .ik. e fienno li triangoli<lb></lb> |
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5406 .adk. e .edg. simili. Onde è cosí .ad. al .de., cosí .ak. al .ge. Ma .ad. del .de. è doppio. Onde la<lb></lb> |
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5407 retta .ak. è doppia ala retta .ge. Ancora, perché simili sonno e triangoli .atk. e .btg., é cosí .ak. al bg.,<lb></lb> |
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5408 cosí .at. al .tb. É adonca .ak. iguale ala retta .bg. e .at. iguale ala retta .tb. Adonca è diviso il<lb></lb> |
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|
5409 lato .ab. nel ponto .t., dala linea .gt., in .2. parti iguali. E questo era bisogno mostrare. E anco-<lb></lb> |
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5410 ra è da sapere che la retta .gt. passa per lo ponto .d. e questo, per quello che s’ é ditto, si manife-<lb></lb> |
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5411 sta. E ancora, per questo, è da sapere che nel triangolo nonn’ é se non un ponto per lo quale<lb></lb> |
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5412 le linee, che si muovano dagli angoli e vanno ala mitá dele facie, di sottostanti ali ditti an-<lb></lb> |
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5413 goli, passano. Adonca, quando un ponto dato nel triangolo sia e voi da quello menare una<lb></lb> |
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5414 linea dividente el triangolo in .2. parti iguali e il detto ponto sia quello di che habiamo ragio-<lb></lb> |
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5415 nato, alora dal’ angolo menarai la linea, infino ala mitá del lato, a quello angolo opposto<lb></lb> |
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|
5416 e harai el desiderio. Commo sia il triangolo .abg. E il ponto dato sia .d., per lo quale passino<lb></lb> |
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5417 le linee menate dagli angoli e menate in sula mitá degli lati opposti, cioé menate .ea. e .bz. e<lb></lb> |
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5418 .gt. Dico che ciascuna di queste linee divide el triangolo in .2. parti iguali. E questo, per quel<lb></lb> |
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|
5419 che s’ é detto, chiaro appare. |
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|
5420 </p> |
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|
5421 <p class="main"> |
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|
5422 E, se ’l ponto dato, dal quale debbia passare la linea che divida il triangolo in .2.<lb></lb> |
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5423 parti iguali, non sia sule linee che si muovano dal’ angolo ala mitá del lato op-<lb></lb> |
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5424 posto, cioé non sia quello dove vien la segatione dele linee, alora dico che, se da-<lb></lb> |
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5425 gli angoli si menano le linee che passino per quel ponto dato, e lle divideranno li<lb></lb> |
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|
5426 lati opposti al’ angolo in .2. parti non iguali, dele quali le .2. parti magiori dela mitá deli la-<lb></lb> |
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|
5427 ti fienno intorno a uno angolo de’ ditti .3. angoli e le .2. minori intorno a un altro angolo e un’ al-<lb></lb> |
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5428 tra parte magiore e l’ altra minore intorno al’ altro angolo. Commo sia il triangolo .dez. Nel<lb></lb> |
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|
5429 quale sia dato il ponto .i. e menisi .dia. e .zig. e .eib., rette che passeranno sopra il ponto .i., el qua-<lb></lb> |
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|
5430 le nonn’ é nella linea menata dall’ angolo ala mitá delo lato opposto. Dico che le .2. parti ma-<lb></lb> |
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|
5431 giori, cioé .gd. e .db., fienno intorno a uno angolo, che sia .d. E le .2. parti minori intorno al’ al-<lb></lb> |
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|
5432 tro angolo, che sia .e. E una dele parti magiori e una dele minori sia intorno al’ altro ango-<lb></lb> |
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5433 lo che è .c. |
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5434 </p> |
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5435 <p class="main"> |
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5436 Sia adonca uno ponto dato nel triangolo, che non sia nelle linee descendenti dal-<lb></lb> |
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5437 l’ angolo ala mitá del lato. E vorrai quello triangolo dividere in doi parti iguali<lb></lb> |
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5438 dala linea passante per lo ditto ponto. Studiarai, per quelle cose che sonno det-<lb></lb> |
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5439 te, l’ angolo contento dale parti non iguali, perché le cose che s’ ánno a dire, per<lb></lb> |
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5440 quelle, si manifestaranno. Verbi gratia. Sia il triangolo .abg. Nel quale sia il ponto dato .d.,<lb></lb> |
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5441 che non sia in alcuna dele linee rette descendenti dal’ angolo e dividenti el lato opposto in<lb></lb> |
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5442 doi parti iguali. E vogliamo dividere il triangolo .abg. in doi mitá dala linea passante<lb></lb> |
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5443 per lo ponto .d. Piglise, prima a ochio, sopra li lati, el cadimento .del., accioché s’ abbia la no-<lb></lb> |
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5444 titia degli angoli contenti dale parti non iguali, che sia l’ angolo che è al .g. E, dal ponto .d., si<lb></lb> |
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5445 meni al lato .ag. la equedistante alo lato .bg., che sia .de. E menise lo lato .ga. infino a tanto<lb></lb> |
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5446 che, multiplicato .de. per quello, faccia la mitá di quanto .ag. in .gb. E sia .gz. Cioé la mitá<lb></lb> |
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5447 dela multiplicatione del .ag. in .gb. sia divisa per .de., che ne venga .gz. Dipoi, ala<lb></lb> |
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5448 linea .gz., appiccarai uno paralello mancante ala figura tetragona, che sia iguali ala<lb></lb> |
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5449 multiplicatione del .ge. in .gz., cioé che si divida .gz. in doi parti. Dele quale una multiplica-<lb></lb> |
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5450 ta per l’ altra facia lo eguale dela multiplicatione del .ge. in .gz., che altramenti non si pó fa-<lb></lb> |
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5451 <lb></lb> |
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5452 <pb></pb></p> |
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5453 <p class="folio"> |
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5454 folio 37r |
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5455 </p> |
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5456 <p class="main"> |
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5457 <lb></lb> |
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5458 </p> |
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5459 <p class="runhead"> |
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5460 Distinctio quinta. Capitulum primum. . 37 |
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5461 </p> |
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5462 <p class="main"> |
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5463 <lb></lb> |
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5464 re, se ’l quadrato dela mitá .gz. avanza ala superficie del .ge. in .gz. overo sia a quella igua-<lb></lb> |
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5465 le, che sia l’ una .gi. e l’ altra .iz. E compise la retta .id. infino al ponto .t. Dico adonca el trian-<lb></lb> |
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5466 golo .abg. esser diviso in .2. parti iguali dala linea .idt., che cosí te ’l proveró. Perché, a mul-<lb></lb> |
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5467 tiplicare del .ge. in .gz. è iguale ala multiplicatione del .gi. in .iz., sará cosí .gz. al .zi. comme .ig.<lb></lb> |
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5468 al .eg. Onde sia cosí .zg. al’ altra parte di sé, cioé al .ig., comme .ig. al’ altra parte di sé, cioé al .ie.<lb></lb> |
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5469 Ma è cosí .gi. al .ie., cosí .gt. al .de. Adunca cosí .gz., al .ig., cosí .gt. al .de. Ma la multiplicatio-<lb></lb> |
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5470 ne del .zg. in .de. è la mitá dela multiplicatione del .ag. in .bg., onde il triangulo .itg. è la mitá<lb></lb> |
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5471 triangulo .abg., comme dicemmo. E, acioché questo s’ abia neli numeri, sia .ab.13. e<lb></lb> |
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5472 .ag.15. e .bg.14. E faciase .ac. catetto, che sia .12. E il cadimemo .bc. sia .5. e .cg. sia .9. e il ponto<lb></lb> |
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5473 .d. sia infra il catetto e l’ angolo .g. E faciase la linea .dk. equedistante al catetto e sia .3. E me-<lb></lb> |
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5474 nise la linea .edl., e e al lato .bg., e sia .ld. 3/16. d’ uno. Voglio, pel ponto .d., menare una<lb></lb> |
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5475 linea che divida lo triangulo .abg. in .2. parti iguali. In questo modo .kc. è iguale e equedistante<lb></lb> |
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5476 al .ld. e il .dk. sia iguale e equedistante ala linea .lc., dove .lc. è .3. e .kc. sia .3/16. e .al. rimará .9. e la<lb></lb> |
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5477 linea .le. sia .6 1/4., che in questo modo lo troverai. Nel triangulo .acg. la linea .le. è equidistan-<lb></lb> |
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5478 te ala linea .cg. Sará cosí .al. al .ac. comme .le. al .gc., per la .2a. del sexto. Dove, multiplicato .al.,<lb></lb> |
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5479 ch’ é .9., per .gc., che è ancora .9., fanno .81. che, diviso per .ac., ne viene .6 3/4. per la linea .le., dela qua-<lb></lb> |
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5480 le, tratta .ld., cioé .3/16., rimarrá .de.6 9/16., per lo quale numero se divideremo la mitá del fatto del<lb></lb> |
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5481 .ag. in .bg., cioé .105., ne perverrá .16. per la linea .gz. Ancora, perché gli é cosí .al. al .ac., cosí .ae. al .ag.,<lb></lb> |
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5482 sia .ae.11 1/4., cioé diviso la multiplicatione del .al. in .ag. per .ac., dove .eg. fieno .3 3/4., cioé il quar-<lb></lb> |
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5483 to dela linea .ga., per la quale multiplicato la linea .gz., fanno .60. Adunque abiamo a dividere<lb></lb> |
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5484 .16. in .2. parti che, multiplicata l’ una per l’ altra, facino .60. Dove del quadrato dela mitá di .16.<lb></lb> |
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5485 cioé .64., ne trarrai .60., rimane .4., del quale la radici è .2., dove, che tratta di .8., riman .6. per una<lb></lb> |
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5486 parte. L’ altra sia infino in .16., che è .10. Adunque .gi. sia .10. Dipoi, diviso .105. per .gi., cioé per<lb></lb> |
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5487 .10., ne viene .10 1/2. per la linea .gt., che passa per lo ponto .d., e la linea .ti. sia radice di .80 1/4., che<lb></lb> |
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5488 la troverai, se il catetto .in. troverrai che sia .8. e .ng. sia .6., ch’ era da mostrare.<lb></lb> |
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5489 Se’ l ponto datto fosse di fora del triangulo dal quale vogliamo menare la linea che<lb></lb> |
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5490 divida il detto triangulo in .2. parti iguali. Comme sia il ponto dato .d., fuori del tri-<lb></lb> |
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5491 angulo .abg. E voglio dividere il detto triangolo in .2. parti iguali dala linea che<lb></lb> |
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5492 si muova dal ponto .d. Io meneró la linea .ad., segante il lato .bg. nel ponto .e. E,<lb></lb> |
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5493 se la retta .be. è iguale ala retta .eg., noi aremo il proposito, che giá molte volte l’ abiamo det-<lb></lb> |
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5494 to. Adunque la linea .aed. sarebe quella che dividerebe il detto triangulo in .2. parti iguali.<lb></lb> |
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5495 Ma non sia .be. iguali al .eg., ma sia una di loro magiore e sia .be. magiore del .eg.<lb></lb> |
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5496 e dal ponto .d. si meni la retta .dz. equidistante ala linea .be. E menise la retta<lb></lb> |
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5497 .ab. infino tochi lo ponto .z. E, perché la retta .be. è magiore dela mità del lato<lb></lb> |
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5498 .bg., la superficie fatta dal .be. in .ba. é piú che la mitá dela superficie fatta dal .ba.<lb></lb> |
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5499 in .bg. E ancora molto piú è la superficie .ba. in .zd. che la mitá dela superficie del .ba. in .gb.,<lb></lb> |
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5500 imperoché magiore è .zd. che .bc. Piglise adunca la superficie .ib. in .zd., che sia iguale ala<lb></lb> |
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5501 mitá dela superficie del .ab. in .bg. E, perché magiore è la superficie del .ab. in .be. dela super-<lb></lb> |
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5502 ficie del .ib. in .zd., sará la proportione del .zd. al .de. minore dela proportione del .ba. al .bi.<lb></lb> |
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5503 Ma la proportione del .zd. al .be. è iguale ala proportione del .za. al .ab. E, per la disgionta<lb></lb> |
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5504 proportionalitá, sirá la proportione del .zb. al .ba. minore dela proportione de .ai. al .ib. On-<lb></lb> |
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5505 de la superficie del .zb. in .bi. è minore dela superficie .ba. in .ai. Agiungase adunca ala retta<lb></lb> |
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5506 .bi. uno paralello iguale ala superficie del .z.b. in .bi., cioé che ala retta .bi. s’ agiunga alcuna li-<lb></lb> |
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5507 nea che, multiplicata in sé e nel .ib., facia iguali ala multiplicatione del .zb. in .bi., che sia .ti., e com-<lb></lb> |
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5508 pise la retta .tkd. La quale linea divide il triangulo .abg. in .2. parti iguali, commo volava-<lb></lb> |
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5509 mo. E con numeri. Sia .ab.13. e .ag.15. e .gb.14. e .zd.10 2/5. e .bz. 1 3/7. Voglio dal ponto .d. mena-<lb></lb> |
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5510 re la linea che divida il triangulo .abg. in .2. parti iguali. Divideró la mitá dela multiplica-<lb></lb> |
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5511 tione del .ab. in .bg., cioé .91., per .dz., cioé per .10 2/5. e verrane .8 3/4. per la linea .bi. E multiplica-<lb></lb> |
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5512 ró .zb., cioé .1 3/7. per .8 3/4., fanno .12 1/2., a’ quali agiongneró. Il quadrato dela mitá dela linea .bi., cioé<lb></lb> |
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5513 la multiplicatione de .4 3/6. in sé, che fanno .19 9/64. che, con .12 1/2., fanno .31. e .41/64., de’ quali la radici è<lb></lb> |
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5514 .5 5/8. per la linea .lt., ai quali, agionto .lb., che è .4 3/6., fanno .10. per la linea .tb. E, perché gli é cosí el<lb></lb> |
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5515 <lb></lb> |
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5516 <pb></pb></p> |
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5517 <p class="folio"> |
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5518 folio 37v |
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5519 </p> |
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5520 <p class="main"> |
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5521 <lb></lb> |
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5522 </p> |
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5523 <p class="runhead"> |
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5524 Distinctio quinta. Capitulum primum. |
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5525 </p> |
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5526 <p class="main"> |
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5527 <lb></lb> |
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5528 .zt. al .tb., cosí .zd. al .bk., multiplicaró .zd. in .bt., cioé .10 3/5. per .10., fanno .104., lo quale divi-<lb></lb> |
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5529 deró per 11 3/7., cioé per la linea .zt., vienne .9 1/10. per la linea .bk. overo .91., che sonno la mitá<lb></lb> |
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5530 dela superficie del .ab. in .bg. Divideró per .bt., cioé per .10., similmente ne viene .9 1/10.<lb></lb> |
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5531 per la linea .bk. E cosí sia il triangolo .abg. diviso in .2. parti iguali dala linea .tkd., comme era<lb></lb> |
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5532 de bisogno. E questo del dividere el triangolo in .2. parti iguali sia abastanza e daremo mo-<lb></lb> |
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5533 do a dividere quello in .3. o piú parti e peró starai atento. |
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5534 </p> |
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5535 <p class="main"> |
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5536 Sia dato uno triangolo .abg. del quale voglio torne la terza parte per una linea<lb></lb> |
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5537 cadente da uno angolo. Dico che bisogna che quella linea caggia in sul terzo de-<lb></lb> |
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5538 la facia oposta a quello angolo, comme per la linea .ad., che è .bd. il terzo dela linea<lb></lb> |
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5539 overo lato .bg. che, per quel che s’ é detto, chiaro si manifesta. Adunque, essendo<lb></lb> |
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5540 .ab.13. e .bg.15. e .ga.14. lo .bd. sia .5. |
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5541 </p> |
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5542 <p class="main"> |
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5543 E, volendo il detto triangolo diveder per terzo, cioé .3. parti iguali, dale linee .de. e<lb></lb> |
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5544 .if., e voglio sapere quanto è dal .a. al .d. e dal .a. al .e. e quanto è la linea .ed. e anco-<lb></lb> |
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5545 ra quanto .ef. e .ai. e .if. Multiplicarai .ab. in sé, cioé .13. in sé, fanno .169. De’ quali pi-<lb></lb> |
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5546 glia il .1/3., che è .56 1/3., e la radice di .56 1/3. è .ae. E, per lo lato .ag., multiplica .14. in sé, fan-<lb></lb> |
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5547 no .196. De’ quali il 1/3. è .65 1/3. e la radici de .65 1/3. é la linea .ad. E, per .ed., multiplica .bg. in sé, fanno<lb></lb> |
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5548 .225. De’ quali il .1/3. è .75. e radici di .75. è .ed. E, dipoi, piglia e .2/3. di .169., che sonno .112 2/3., e radici<lb></lb> |
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5549 di .112 2/3. é la linea .ai. e piglia e .2/3. di .196., che sonno .130 2/3., e radice di .130 2/3. sia la linea .af. E pi-<lb></lb> |
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5550 glia e .2/3. di .225., che sonno .150., e radice di .150. sia la linea .if. e cosí opera in simili.<lb></lb> |
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5551 E, volendo il detto triangolo dividere in .2. parti iguali per la linea .pq., e vorai sa-<lb></lb> |
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5552 pere quanto è .ap. e .aq. e quanto è .pq. Multiplicarai .13. in sé, fanno .169. e di quel<lb></lb> |
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5553 piglia la mitá, che è .84 1/2., e radici de .84 1/2. sia .ap. E, dipoi, multiplica .14. in sé, fanno<lb></lb> |
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5554 .196. De’ quali la mitá è .98., la cui radici è la linea .pq. E, dipoi, multiplica .15. in sé, fan-<lb></lb> |
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5555 no .225. De’ quali la mitá è .112 1/2. e radici de .112 1/2. é la linea .pq. E cosí, in similianti, è da operare.<lb></lb> |
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5556 E, se si dá un ponto in sul lato del triangolo. Comme nel triangolo .gbd. e sia nel<lb></lb> |
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5557 lato .dg. dato il ponto .z. E voglio dal ponto .z. menare una linea che del triango-<lb></lb> |
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5558 lo ne tolga il .1/3. Adimando in che parte dela linea .bd. tocherá la linea che si muo-<lb></lb> |
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5559 ve dal .z. E questo modo farai. Overo il ponto .z. é in sul terzo dela facia .gd. o non.<lb></lb> |
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5560 Se gli é in sul terzo, alora si muova la linea dal ponto .z. e vada in sul’ angolo oposto. E, com-<lb></lb> |
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5561 me ó detto, la linea .bz. harebbe diviso il triangolo in terza parte. Ma, se ’l ponto .z. non è in sul ter-<lb></lb> |
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5562 zo dela linea overo lato .gd. overo .gz. sia piú che ’l terzo overo meno. Sia prima meno:<lb></lb> |
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5563 dove togli del .gd. la terza parte e sia .ag. e compise la retta .ai. equedistante ala linea .bz. E, di-<lb></lb> |
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5564 poi si facia la linea .zi., la quale divide lo triangolo .bgd. in terza parte, cioé lo quadrilatero<lb></lb> |
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5565 .bgzi. é il terzo del detto triangolo. Dove, con numeri, sia .gz.3., dove .zd. sia .12. E multiplichi-<lb></lb> |
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5566 se .gd. per .db., cioé .15. per .14., fanno .210. De’ quali togli e .2/3., sonno .140., li quali dividi per .12.,<lb></lb> |
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5567 haremo .11 2/3. e .11 2/3. è la linea .di. E questo volavamo mostrare.<lb></lb> |
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5568 E, se ’l ponto .z. è dato in modo che .zg. sia piú che ’l .1/3. e meno che .2/3., faciase dal lato<lb></lb> |
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5569 del’ angolo .d. la terza parte del .dg., che sia .ed. e menise .ei. E, dipoi, si meni .zi. Di-<lb></lb> |
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5570 co .zi. essere quella linea che divide il triangolo in terza parte. Cioé che ’l triangolo<lb></lb> |
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5571 .zid. è il terzo del triangolo .bgd. Che con numeri sia .gz.9. e multiplicarai .gd. per<lb></lb> |
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5572 .db., cioé .15. per .14., fanno .210. Del quale il .1/3. è .70. che, diviso per .9., ne viene .7 7/9. per la linea<lb></lb> |
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5573 .di. E cosí sempre a simigliante é da operare. E, se ’l ponto dato fosse infra ’l .e. e .d., operaresti com-<lb></lb> |
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5574 me in quella fatta innanze ala detta. |
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5575 </p> |
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5576 <p class="main"> |
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5577 Ancora sia il triangolo .abg. il quale voglio dividere in .3. parti iguali. Dele quali<lb></lb> |
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5578 parti ciascuna habia uno angolo e uno lato. Divideró .bg. in .2. parti iguali dal<lb></lb> |
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5579 ponto .d. E compiró la linea .ad. E, dipoi, dela linea .ad. piglieró il terzo, che sia<lb></lb> |
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5580 .dc., cioé dala parte del .d., e faró le linee .bc. e .gc. Dico adunque il triangolo .abg.<lb></lb> |
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5581 essere diviso in .3. parti iguali, de’ quali ciascuna è sopra a uno lato del triangolo .abg., le quali<lb></lb> |
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5582 parti sonno li triangoli .abc. e .bgc. e .gac., che cosí el proveró. Perché .dc. è la terza parte de-<lb></lb> |
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5583 la retta .da., sia .ac. doppio del .cd. Onde el triangolo .abc. è doppio al triangolo .bcd., per<lb></lb> |
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5584 primam sexti. E, per questo, ancora el triangolo .acg. è doppio al triangolo .gcd. Ancora per-<lb></lb> |
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5585 ché e gli é iguale la retta .bd. ala retta .dg., iguali sonno li triangoli .cbd. e .cdg. infra loro, pur<lb></lb> |
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5586 per la pa. del .6o. Onde tutto .cbg. triangolo è iguale a ciascun de’ triangoli .cbd. e .cdg. e quel-<lb></lb> |
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5587 li, che a una medesima cosa sonno doppi, infra loro sonno iguali per la conceptione. Diviso<lb></lb> |
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5588 è adunque il triangolo .abg. in .3. parti iguali.<lb></lb> |
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5589 <lb></lb> |
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5590 <pb></pb></p> |
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5591 <p class="folio"> |
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5592 folio 38r |
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5593 </p> |
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5594 <p class="main"> |
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5595 <lb></lb> |
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5596 </p> |
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5597 <p class="runhead"> |
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5598 Distinctio quinta. Capitulum secundum. 38]. |
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5599 </p> |
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5600 <p class="main"> |
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5601 <lb></lb> |
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5602 Ancora sia uno triangolo .abg. e in quello sia dato uno ponto fortuitu .d., per lo<lb></lb> |
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5603 quale .d. voglio passi la linea che divida el triangolo overo che pigli del triango-<lb></lb> |
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5604 lo una parte, comme a dire il terzo. Adimandase in che modo si debia operare. Dal<lb></lb> |
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5605 ponto .d. meneró la linea .ae. e considereró se la settione .be. overo .eg. sia la terza<lb></lb> |
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5606 parte del lato .bg. Onde, se ’l .be. è la terza parte dela retta .bg., alora el triangolo .abe. è la<lb></lb> |
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5607 terza parte del triangolo .abg. E, se niuno di loro è la terza parte del lato .bg., alora meneró<lb></lb> |
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5608 la linea per gl’ altri angoli, passante per lo ponto .d. Dico che, se alcuna dele dette linee piglia<lb></lb> |
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5609 il terzo del lato, e alora dirai quella linea pigliare del triangolo il terzo. E, si troveró questo<lb></lb> |
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5610 non essere, meneró dal ponto .d. la linea .dz. equedistante ala linea .ge. E porró la superficie<lb></lb> |
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5611 .dz. in .zi. iguale ala terza parte dela superficie .ag. in .bg. E apiccheró ala linea .gi. la superficie<lb></lb> |
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5612 equiangola alla quale manchi la figura tetragona iguale ala superficie del .gz. in .zi., che sia<lb></lb> |
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5613 la superficie .it. in .tg. e, comporró la retta .dt. e producerolla infino al .k. Dico adunque che dal<lb></lb> |
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5614 triangolo .abg. n’ é tolta la terza parte, cioé il triangolo .tkg., dala linea .tk. passante per lo pon-<lb></lb> |
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5615 to .d. che, per le cose dette, chiaro appare. E questo basti quanto alo primo capitolo di que-<lb></lb> |
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5616 sta distintione e, seguendo, diremo delo secondo. |
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5617 </p> |
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5618 <p class="main"> |
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5619 Qualiter figure quadrilatere in partes proportionabiliter dividantur. Capitulum secundum.<lb></lb> |
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5620 Le specie de’ quadrilateri sonno .3., cioé paralelli, caput abscisum e diversilatero.<lb></lb> |
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5621 E paralelli sonno figure che hano e lati oposti equedistanti e gli angoli oposti<lb></lb> |
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5622 iguali, per la .34a. del po., de’ quali le specie sonno .4. Li primi sonno tetragoni, che<lb></lb> |
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5623 hano tutti e lati iguali e gli angoli retti. E li secondi sonno parte altera longiore,<lb></lb> |
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5624 che hanno e .2. lati oposti infra loro iguali e tutti gli angoli retti. Nela terza spetie sonno e rom-<lb></lb> |
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5625 bi, che hano e lati igual e gli angoli non sonno retti e né iguali. Nela quarta specie sonno li<lb></lb> |
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5626 romboidi, li quali hano le faccie overo lati oposti iguali, ma gli angoli non sonno iguali e non<lb></lb> |
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5627 retti. E, perché e gli é uno modo solo al dividerli queste .4. specie di paralelli, tutte le loro figu-<lb></lb> |
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5628 re, sotto le medesime notule, porremo, peroché quello che d’ una se dici, quel medesimo del’ al-<lb></lb> |
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5629 tre si possono dire. Sia adunque uno tetragono e parte altera longiore e rombo e romboi-<lb></lb> |
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5630 de .abcd. De’ quali ciascuno voglio dividere in .2. parti iguali. Li detti paralelli, se per diame-<lb></lb> |
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5631 tro gli dividerai, harai fatti di quelli .2. parti iguali. Onde, se meneremo il diametro .ac. ove-<lb></lb> |
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5632 ro .bd., comme è manifesto nela prima figura, saranno in .2. parti divise. Comme havendo<lb></lb> |
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5633 menato il diametro .ac., sia il triangolo .abc. iguale al triangolo .adc., perché il lato .ad. è iguale<lb></lb> |
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5634 al lato .bc. e il lato .ab. iguale al lato .dc. e la basa .ac. è comune a ogni triangolo.<lb></lb> |
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5635 E, se la divisione vuoi incomenzare d’ alcuno de’ lati, quello lato in .2. parti iguali<lb></lb> |
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5636 segherai e, per lo ponto dela settione, sopra il lato opposto, agli altri lati equedis-<lb></lb> |
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5637 tante, menerai una linea, comme nela figura sola appare, nela quale sopra la mitá<lb></lb> |
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5638 del lato .bc. è segnato uno ponto e da quello è menato la retta .ef. equedistante a-<lb></lb> |
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5639 le rete .ba. e .cd. E cosí tutto il quadrilatero .ac. è diviso in .2. parti iguali che sonno .ae. e .fc.<lb></lb> |
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5640 Sonno adunque sopra le base iguali .be. e .ec. e nele medesime linee equedistanti .ad. e .bc.<lb></lb> |
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5641 dove, comme appare per lo primo de Euclide, e lle sonno infra loro iguali.<lb></lb> |
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5642 Similmente, se meneremo la linea .gk. equedistante ale rette .ad. e .bc., dividente<lb></lb> |
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5643 e lati per lo mezzo, cioé dividente li lati .bc. e .ad., sirá ancora diviso il paralello .abcd.<lb></lb> |
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5644 in .2. parti iguali dala linea .gk., comme nela presente figura appare. Sonno certamente li paralelli<lb></lb> |
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5645 .gd. e .bk. infra loro iguali. Conciosiacosaché sienno infra le base iguali e ’lati equedistanti].<lb></lb> |
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5646 E, se sopra ad alcun lato sirá dato il ponto, comme se manifesta nella presente figu-<lb></lb> |
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5647 ra, nela quale è dato il ponto .h., cadente infra ’l ponto .be., segnerai nello posto<lb></lb> |
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5648 di quello il ponto .i., cadente infra ’l .fd. E sia .fi. iguali ala retta .ch. E compise la<lb></lb> |
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5649 retta .hi. Dico certamente essere il detto quadrilatero in .2. parti iguali diviso dala<lb></lb> |
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5650 linea .hi. Che cosí si pruova. Perché, in equedistanti linee .ad. e .bc., le rette .fe. e .hi. son<lb></lb> |
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5651 menate, sirá l’ angolo .ifk. alo angolo .keh. iguali, per la .29a. del primo, perché son coalterni.<lb></lb> |
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5652 E ancora l’ angolo .fik. al’ angolo .khe. e gli angoli che sonno al .k. sonno infra loro iguali<lb></lb> |
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5653 e la basa .fi. ala basa .eh. è iguale. Adunque e il triangolo .fhi. è iguale al triangolo .khe. E,<lb></lb> |
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5654 comunamente, s’ agionga el pentagono .kfabh., sirá el quadrilatero .iabh. iguale al quadri-<lb></lb> |
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5655 latero .abef., che è la mitá di tutto el quadrilatero. |
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5656 </p> |
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5657 <p class="main"> |
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5658 E, se uno di detti paralelli in .2. parti, iguali per la linea menata dal ponto dato in-<lb></lb> |
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5659 fra quello, vuoi dividere. Comme il parelello .ag., infra ’l quale sia dato el ponto .f.,<lb></lb> |
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5660 per lo quale debba passare la linea che divida il detto quadrilatero in .2. parti igua-<lb></lb> |
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5661 li e sia il detto ponto sopra il .1/2. del diametro .bd. Onde puoi dire che ’l detto diame-<lb></lb> |
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5662 <lb></lb> |
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5663 <pb></pb></p> |
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5664 <p class="folio"> |
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5665 folio 38v |
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5666 </p> |
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5667 <p class="main"> |
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5668 <lb></lb> |
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5669 </p> |
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5670 <p class="runhead"> |
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5671 Distinctio quinta. Capitulum secundum. |
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5672 </p> |
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5673 <p class="main"> |
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5674 <lb></lb> |
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5675 tro lo divida per .2. parti iguali. Overo menerai una linea che si parta da uno de’ lati, comme<lb></lb> |
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5676 dal lato .bg., dal ponto .z. E sia la linea .ze. che passi per lo ponto .f. Dico che la linea .ze. è quel-<lb></lb> |
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5677 la che divide il paralello .ag. in .2. parti iguali che, per quello che s’ é detto, chiaro appare. E pe-<lb></lb> |
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5678 ró lascieremo la dimostratione. |
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5679 </p> |
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5680 <p class="main"> |
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5681 S e lo ponto dato fosse fuori del diametro, comme sia il ponto dato .e. nel quadrila-<lb></lb> |
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5682 tero .abgd. Dico che meni il diametro .bd. e in sun quello segna un ponto che<lb></lb> |
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5683 sia in mezzo del detto diametro e sia ponto .f. e menise dal ponto .f. ala linea .fc., che<lb></lb> |
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5684 termini infino agli lati del detto quadrilatero, e sia la lina .zcfe., la quale linea di-<lb></lb> |
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5685 co che la divide il detto paralello in .2. parti iguale. E questo, per quel che s’ é detto, chiaro<lb></lb> |
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5686 è manifesto. |
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5687 </p> |
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5688 <p class="main"> |
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5689 E, se ’l ponto dato fosse fuore del paralello. Comme sia il paralello .abcd., fuor del<lb></lb> |
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5690 quale sia dato il ponto .e. E da quello voglio menare una linea la quale divida il<lb></lb> |
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5691 detto paralello in .2. parti iguali. Meneró in quel paralello il diametro .bd. e in<lb></lb> |
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5692 quello diametro segneró un ponto, che sia .o., el quale sia nel mezzo del detto dia-<lb></lb> |
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5693 metro. E meneró la linea .ekf., la quale linea passi per lo ponto .o. Dico che la linea .ekof. divi-<lb></lb> |
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5694 de il detto paralello per mitá. E questo chiaro appare per le cose dette.<lb></lb> |
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5695 Ancora sienno le .4. spetie di paralelli .abcd. e vogliamo alcuni di quelli divide-<lb></lb> |
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5696 re in .3. parti iguali sopra e .2. lati dati de quello, che sonno .ad. e .bc. Divideró quel-<lb></lb> |
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5697 li in .3. parti iguali, che fienno .ae.ef.fd. E, per gli ponti .e.f., meneró le rette linee<lb></lb> |
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5698 .eg. e .fh. equedistanti agli lati .ab. e .dc. Dico che qual vuoi di questi paralelli é<lb></lb> |
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5699 diviso in .3. parti iguali dale rette .eg. e .fh., che cosí si prova. Perché equedistanti sonno le rette<lb></lb> |
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5700 .ad. e .bc. e con quelle sonno fatte l’ equedistanti .ab.eg.fh. e .dc. fienno infra loro iguali. On-<lb></lb> |
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5701 de i paralelli sonno quadrilateri .ag.eh.fc. E hano le base iguali infra loro, che sonno .bg. e<lb></lb> |
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5702 .gh. e .hc. e ciascuna di quelle é il terzo del lato .bc. Onde ciascuno di quelli paralelli é il terzo<lb></lb> |
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5703 di tutta .ac. paralello, per la pa. del 6o. E questo era da mostrare.<lb></lb> |
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5704 E, se sopra uno ponto dato sopra uno lato si menerá la retta che dal paralello .abcd.<lb></lb> |
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5705 dato tolga la terza parte e il dato ponto si ponga essere sopra la linea overo<lb></lb> |
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5706 il lato .ad., diviso prima el lato .ad. in .3. parti iguali sopra ponti .ef., comme habia-<lb></lb> |
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5707 mo detto, cioé che ciascuna settione .ag.eh.fc. sia il terzo di tutto .ac. e il dato pon-<lb></lb> |
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5708 to sia .e. e meneró la linea .eg. equedistante ala linea .ab., sirá per lo paralello .ag. il terzo del pa-<lb></lb> |
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5709 ralello .ac. E, similmente, se il dato ponto fosse .f., sirá il paralello .fc. il terzo del paralello .ac. e<lb></lb> |
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|
5710 questo chiaro si vede. |
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5711 </p> |
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5712 <p class="main"> |
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5713 Ma se ’l ponto dato non fosse in sul .e. overo in sul .f., sirá adonque infra ’l .a. e .e. ove-<lb></lb> |
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5714 ro infra ’l .e. e .f. overo infra ’l .f. e .d. Sia prima infra ’l .a. e .e. il ponto dato .i. e voglio<lb></lb> |
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5715 dal ponto .i .menare la linea che divida il paralello in una parte che sia il terzo di<lb></lb> |
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5716 tutto, cioé del paralello .ac. ne tolga el .1/3., che cosí faró. Segnato il ponto .f., che sia<lb></lb> |
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5717 .ef .ancora il terzo del .ad. E dal ponto .f. meneró la linea .fh. equedistante ala linea .ab. overo<lb></lb> |
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5718 .dc. Dove il paralello .abfh. e gli é .2/3. del paralello .ac. Dove dividerai il paralello .abfh. in .2. par-<lb></lb> |
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5719 ti iguali dala linea che si parte dal ponto. Onde, quanto .i. é discosto dal .a., tanto faró un pon-<lb></lb> |
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5720 to discosto dal .h. e sia .k. E meneró la linea .ik. Dico che ’l quadrilatero .aibk. è la terza parte<lb></lb> |
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|
5721 del quadrilatero .ac., imperoché gli é la mitá del pararello .abhf. E cosí ancora el quadrila-<lb></lb> |
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5722 tero .ae. è diviso in .3. parti iguali che sonno e quadrilateri .iabk. e .ikhf. e i paralello .fc., com-<lb></lb> |
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5723 me per la figura apare. |
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5724 </p> |
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5725 <p class="main"> |
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5726 E, se il dato ponto .i. sia infra ’l .fd., se oprerá quel medesimo nella parte aversa, cioé<lb></lb> |
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5727 dal paralello .ac. torró il paralello .ag., che sia la terza parte del paralello .ac. Di-<lb></lb> |
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5728 poi il paralello .ec. divideró in .2. parti iguali per la linea menata dal ponto .i., che<lb></lb> |
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5729 sia .il., che toglie del paralello .ac. la terza parte, cioé il quadrilatero .ilcd., comme<lb></lb> |
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|
5730 per la figura si manifesta. |
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5731 </p> |
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5732 <p class="main"> |
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5733 E, se ’l ponto .i. sia infra ’l .e. e .f., taglieró del paralello .ac. il paralello .fc., che sia il ter-<lb></lb> |
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5734 zo del paralello .ac. e l’ avanzo divideró in .2. parti iguali per la linea menata dal pon-<lb></lb> |
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5735 to .i., che sia la retta .im. E tolglise, per la linea .im., el quadrilatero .iabm. dal para-<lb></lb> |
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5736 lello .ac., ch’ é il terzo del paralello .ac.; l’ altre due parti fienno el quadrilatero .imhf. e il para-<lb></lb> |
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5737 lello .fc., comme nella quarta figura designata appare. |
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5738 </p> |
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5739 <p class="main"> |
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5740 Per lo detto modo si puó ogni paralello dividere in .4. o piú parti iguali, come volendo<lb></lb> |
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5741 dividere uno paralello in .4. parti iguali. Divideralo prima in .2. parti iguali o vuoi per lo dia-<lb></lb> |
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5742 metro suo o voi per la retta equedistante a doi suoi lati. Dipoi ciascuna di quelle<lb></lb> |
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5743 <lb></lb> |
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5744 <pb></pb></p> |
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5745 <p class="folio"> |
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|
5746 folio 39r |
|
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5747 </p> |
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5748 <p class="main"> |
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5749 <lb></lb> |
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|
5750 </p> |
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5751 <p class="runhead"> |
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|
5752 Distinctio quinta. Capitulum secundum. 39 |
|
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5753 </p> |
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5754 <p class="main"> |
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|
5755 <lb></lb> |
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5756 2. parti in .2. parti iguali segherai e sia tutto il paralello in .4. parti diviso comme vorrai.<lb></lb> |
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5757 E, volendo dividere uno paralello in .3. parti non iguali, comme sia il paralello .abgd.<lb></lb> |
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5758 el quale in .3. parti non iguali voglio dividere in tal modo che ’l primo habia la .1/2. e ‘l secondo il<lb></lb> |
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5759 terzo. E il terzo habia il sexto. Divideró prima il paralello .ag. in .2. paralelli igua-<lb></lb> |
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5760 li che sonno .az. e .eg. Dipoi, da uno di quelli, segheró il terzo, cioé la terza parte del<lb></lb> |
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5761 paralello .az. overo .eg. che sirá il paralello .ig. Cioé che la retta .id. sia la terza parte del .ed.<lb></lb> |
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5762 Dico il paralello .ag. essere diviso in .3. parti predette dele quali la mitá è il paralello .az. e il sexto<lb></lb> |
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5763 é il paralello .ig. e l’ avanzo, cioé il paralello .et. sia la terza parte di tutto il paralello .ag. ch’ era<lb></lb> |
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5764 bisogno mostrare. Possiamo adunque qualunque parte vogliamo torre dele predette fi-<lb></lb> |
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5765 gure dala linea menata dal ponto dato fuor o dentro overo sopra uno de’ lati del paralello,<lb></lb> |
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|
5766 per quelle cose dette. E questo basti quanto al dividere de’ paralelli e, seguendo, diremo del<lb></lb> |
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|
5767 dividere dele figure dette caput abcisum. |
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|
5768 </p> |
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5769 <p class="main"> |
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5770 Le figure chiamate caput abscisum, cioé le figure dette capo tagliato, sonno di .4.<lb></lb> |
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5771 spetie, de quali la prima se dice mezzo capo tagliato, l’ altra figura se dice igual ca-<lb></lb> |
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5772 po tagliato, la terza diverso capo tagliato. La quarta capo tagliato declinan-<lb></lb> |
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5773 te, commo di sopra havesti in lor misurare. E, perché el modo a dividere de tutte<lb></lb> |
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5774 queste è uno, quelle per ordine porró sotto medesime notule e termini acioché quello che<lb></lb> |
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|
5775 se dici d’ una di tutte l’ altre s’ intenda. Sienno adunque queste .4. specie di quadrilateri segna-<lb></lb> |
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5776 te .abgd., aventi e lati .ad. e .bg. equedistanti e volse ciascuna di queste dividere per la retta<lb></lb> |
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5777 equedistante ale base loro, le quali base sonno .bg., in .2. parti iguali, che in questo modo lo<lb></lb> |
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|
5778 faremo: perché la equistante .ad. e .bg. non sonno iguali, anzi é magiore .gb. che .ad. se me-<lb></lb> |
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5779 neremo le rette .ba. e .gd. nela parte del .ad. infino dove concorrono nel ponto .e. E sia il qua-<lb></lb> |
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|
5780 drato dela retta .ze. la mitá de’ quadrati delle rette .eb. e .ae. e dal ponto .z. si meni la retta .zi. e-<lb></lb> |
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5781 quedistante ala basa .gb. Dico el trapezzo .abgd. essere diviso in .2. parti iguali dala linea .zi.<lb></lb> |
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5782 equedistante alla linea .bg. che cosí si prova. Perché e quadrati dele linee .eb. e .ae. sonno dop-<lb></lb> |
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5783 pi al quadrato dela linea .ez., fienno ancora doppio li triangoli .ebg. e .ead. al triangol .ezi.<lb></lb> |
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|
5784 conciosiacosaché fra loro sienno simili. Ma, se del triangolo .ebg. lasciamo el triangolo .ezi.,<lb></lb> |
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5785 che è iguali a quello triangolo che è nel triangolo .ebg., rimarrá il quadrilatero .izbg. e ‘l triangolo<lb></lb> |
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5786 .ead. iguali al triangolo .ezi.; onde, d’ ogni parte si traga el triangolo .ead., rimane el quadrilatero .zgi.<lb></lb> |
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|
5787 iguale al quadrilatero .ai. Adonca è diviso el quadrilatero .abgd. in .2. parti iguali dala linea .zi. ch’ era<lb></lb> |
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5788 de bisogno mostrare. Che ancora con numeri il mostraremo. Sia .ab.12. e .bg.12. e .ad. sia .3.<lb></lb> |
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5789 e il lato .gd. sia .15. E, perché nel triangolo .ebg. é menata la retta .da. equedistante ala basa<lb></lb> |
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5790 .bg., sia cosí .ad. al .gb., cioé cosí .3. e a .12., cosí .ea. al .eb., per la .2a. del 6o. Onde comme .3. è a<lb></lb> |
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|
5791 .9. cosí .ea. e .ab. Adunque .ea. é .4., cioé il terzo del .ab., adunque .eb. é .16. Agionti adunque e qua-<lb></lb> |
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5792 drati dele linee .eb. e .ea., cioé .256. e .16., fanno .272., de’ quali la mitá, cioé .136. é il quadrato della<lb></lb> |
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|
5793 linea .ez. E, perché e gli é cosí .ez. al .eb. cosí .zi. al .bg., sirá adunque comme el quadrato de-<lb></lb> |
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|
5794 la linea .ez. al quadrato dela linea .eb., cioé commo .136. a .256. che, ne’ minori numeri, é comme<lb></lb> |
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|
5795 .17. a .32. Cosí il quadrato dela linea .zi. é alo quadrato della linea .bg. Onde, se multiplicare-<lb></lb> |
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5796 mo .17. per .144. e divideremo per .32., verranne .76 1/2. per lo quadrato della linea .zi. E, perché<lb></lb> |
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5797 ortogonio è il triangolo .ezi. e ancora il triangolo .ead., onde, multiplicando la mitá del .ez. in<lb></lb> |
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5798 .zi., haremo l’ area del triangolo .ezi. che è .51. che viene dela multiplicatione della quarta par-<lb></lb> |
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5799 te di .136. nel quadrato dela linea .zi., cioé in .76 1/2. La quale multiplicatione è la radice de .2601. |
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5800 </p> |
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5801 <p class="main"> |
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5802 Del quale, togliendo el triangolo .ead. che è .6., rimangono .45. per l’ area del quadrilatero<lb></lb> |
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5803 .azid., el quale .45. è la mitá de .90. ch’ é l’ area di tutto il quadrilatero mezzo caput abscisum .abgd.<lb></lb> |
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5804 perché multiplicando la mitá del .ab. nel congionto del .ad. e .bg., cioé .6. per .15. haremo el det-<lb></lb> |
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5805 to .90. che è doppio de .45. e peró è provato el quadrilatero .ai. essere la mitá di tutto il quadri-<lb></lb> |
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5806 latero .abgd. overo altramente tratta .ez. del .ed., cioé radice di .136., di .16., rimarranno .16. me-<lb></lb> |
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5807 mo radice di .136. per la linea .bz. del quale il .1/2. è .8. meno radice di .34. che, multiplicato per lo<lb></lb> |
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5808 congionto del .zi. e .bg., cioé per .12. e radice di .76 1/2., haremo .45. per l’ area del quadrato .gz.<lb></lb> |
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5809 e questo era da mostrare. |
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5810 </p> |
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5811 <p class="main"> |
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5812 E, se sopra il lato .ad. alcun ponto sia dato e vorrai da quello menare la linea che divi-<lb></lb> |
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5813 da el detto quadrilatero in .2. parti iguali, e lati .ad. e .bg. in .2. parti iguali dividi sopra li<lb></lb> |
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5814 ponti .k. e .t., che cosí si prova. Io compiró la retta .tk. Dico che la retta .tk. é quella che divide<lb></lb> |
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5815 il detto quadrilatero in .2. parti iguali. Adonca, quando el ponto dato fosse in sula mitá dela linea .ad. com-<lb></lb> |
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5816 me il ponto .t., alora el detto quadrilatero è diviso in .2. parti iguali comme nela presente<lb></lb> |
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5817 <lb></lb> |
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5818 <pb></pb></p> |
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5819 <p class="folio"> |
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5820 folio 39v |
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5821 </p> |
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5822 <p class="main"> |
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5823 <lb></lb> |
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5824 </p> |
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5825 <p class="runhead"> |
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5826 Distinctio quinta. Capitulum secundum. |
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5827 </p> |
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5828 <p class="main"> |
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5829 <lb></lb> |
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5830 figura appare che cosí el proveró. Io comporró le rette .bt. e .gt. e fienno e triangoli .tkb. e<lb></lb> |
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5831 .tkg. infra loro iguali. E conciosiacosaché sienno sotto una altezza e habino le base iguali.<lb></lb> |
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5832 Ancora, perché e triangoli .bat. e .tgd. sonno sotto una altezza, è cosí .at. al .td., cosí il trian-<lb></lb> |
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5833 golo .bat. al triangolo .gtd. E certamente .at. iguale al .cd. dove el triangolo .bat. è iguale al<lb></lb> |
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5834 triangolo .gtd. Dimostrato è adonca el triangolo .tkg. iguale al triangolo .tkb. Onde il qua-<lb></lb> |
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5835 drilatero .ak. è iguale al quadrilatero .tg. Diviso adonca il quadrilatero .abgd. in .2. parti<lb></lb> |
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5836 iguali dala linea .tk. e manifestasi, per quello che habiamo detto che in ogni quadrato aven-<lb></lb> |
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5837 ti .2. lati equedistanti, se una retta linea segherá quello proportionalmente, quelli lati segherá<lb></lb> |
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5838 in detta proportionalmente e conterranno medesima quantità comme dicemo .at. al .td. co-<lb></lb> |
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5839 sí .bk. al .kg. Dove fo cosí .at. al .td., cosí el quadrilatero .ak. al quadrilatero .tg.<lb></lb> |
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5840 E, se ’l ponto dato fosse in sulo minor lato in sul ponto degli angoli. Comme sia il qua-<lb></lb> |
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5841 drilatero .abgd. E sia il ponto dato .a. dal quale voglio menare una linea che di-<lb></lb> |
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5842 vida il detto quadrilatero in .2. parti iguali. Prima meneró la linea .tk., la quale<lb></lb> |
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5843 dividerá il detto quadrilatero in .2. parti iguali e passerá per gli ponti .t. e .k., li quali son-<lb></lb> |
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5844 no fatti in sula mitá deli lati .ad. e .bg. E, sopra la retta .kg., torró la retta .kl. iguale ala retta<lb></lb> |
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5845 .at. E sia .bl. la mitá de’ lati .bg. e .ad. e faró la retta .al. Dico el quadrilatero .abgd. essere divi-<lb></lb> |
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5846 so in .2. parti iguali dala linea .al. che cosí el proveró. Perché equedistante é la retta .at. dela<lb></lb> |
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5847 retta .kl. e la retta .al. le taglia, sirá l’ angolo .atm. iguale al’ angolo .lkm. e l’ angolo .tam.<lb></lb> |
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5848 iguale al’ angolo .klm. E ancora gli angoli che sonno al .m. sonno infra loro iguali, conció-<lb></lb> |
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5849 siacosaché sienno contraposti per la .15a. del primo e fatti dale base iguali. Onde il triangolo .amt.<lb></lb> |
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5850 è iguale al triangolo .lmk. ai quali, agionto a ciascuno el quadrilatero .mabk., sirá il triangolo<lb></lb> |
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5851 .abl. iguale al quadrilatero .abkt. che è la mitá di tutto el quadrilatero .ag. comme è manife-<lb></lb> |
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5852 sto nella .3a. figura. |
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5853 </p> |
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5854 <p class="main"> |
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5855 Ancora per simil modo, faresti quando il dato ponto fosse sopra lo lato .ad. nel pon-<lb></lb> |
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5856 to .d. Imperoché si piglierá lo equale dela retta .td. in sulo lato .kb. che sia .kn.<lb></lb> |
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5857 Dico che tutta .gn. sia la mitá degli lati .ad. e .bg. E compilero la retta .dn. la qua-<lb></lb> |
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5858 le dividerá il quadrilatero .abgd. in .2. parti iguali comme si manifesta in que-<lb></lb> |
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5859 sta figura che si proverá per quello che s’ é detto nela precedente figura.<lb></lb> |
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5860 E, se ’l ponto dato fusse infra ’l .td. overo .at., comme sia il ponto dato .o. infra ’l .td.,<lb></lb> |
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5861 menise prima .tk., el quale divida el quadrilatero in .2. parti iguali sopra li ponti<lb></lb> |
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5862 fatti nel mezzo de’ lati .ad. e .bg. Dipoi si pigli del .kb. lo equale al .to. e sia .kn. E<lb></lb> |
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5863 menise la linea .on. la quale dico che divide el quadrilatero .abgd. in .2. parti igua-<lb></lb> |
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|
5864 li comme volavamo. |
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5865 </p> |
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|
5866 <p class="main"> |
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5867 Similmente è da operare quando li ponti dati fosseno nel lato .bg. in luogo che la<lb></lb> |
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5868 distantia dal ponto .k. al ponto dato sia quanto dal ponto .t. al’angolo oposto a<lb></lb> |
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|
5869 quel ponto. Comme se fosse dal lato del .kg., é de bisogno la distantia sua sia me-<lb></lb> |
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5870 no che dal .t. al .a. E, similmente, se fosse dal lato del .kb., é de bisogno sia meno che<lb></lb> |
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5871 dal .td. E, se ’l ponto dato fosse piú distante dal ponto .k. che non n’ é dal ponto .t. al’angolo opo-<lb></lb> |
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5872 sto, che comme si debba operare al dividere el detto quadrilatero lo mostraro. Sia il quadri-<lb></lb> |
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5873 latero .abgd., nel quale sia il ponto dato .b. dal quale voglio menare una linea che divida lo<lb></lb> |
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5874 detto quadrilatero in .2. parti iguali. Prima faró la linea .dn., la quale dividerá lo detto qua-<lb></lb> |
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5875 drilatero in .2. parti iguali. E, questo fatto, comporró la linea .bd. ala quale linea faró la linea<lb></lb> |
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5876 che si parta dal ponto n. e sia equedistante alla linea .bd. e sia la linea .nc. E comporró la ret-<lb></lb> |
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5877 ta .bc. Dico el quadrilatero .abgd. essere diviso in .2. parti iguali dala linea .bc. che è mena-<lb></lb> |
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5878 ta dal ponto dato .b., che cosí te ’l proveró. Li triangoli .ncb. e .ncd. sonno infra loro iguali:<lb></lb> |
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5879 conciosiacosaché sienno infra le equedistanti .bd. e .nc. e sopra le medesime base .nc. Ai qua-<lb></lb> |
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5880 li agionto di comune el triangolo .cng. sia il triangolo .cgb. iguale el triangolo .dgn. Ma ’l<lb></lb> |
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5881 triangolo .dgn. è la mitá del quadrilatero .ab.gd. Onde el triangolo .cgb. è la mitá del qua-<lb></lb> |
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5882 drilatero .abgd. ch’ era de bisogno mostrare. É ancora a sapere che, se ’l ponto dato sia infra<lb></lb> |
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5883 .bn. overo nel’ altra parte, é de bisogno menare da quello ponto .n. la equedistante alla linea<lb></lb> |
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5884 che se parte dal .d. al ponto dato. E compilare dal ponto dato una linea retta al ponto dove<lb></lb> |
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5885 la linea equedistante cade infra ’l ponto .d. e .c. |
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5886 </p> |
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5887 <p class="main"> |
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5888 Similmente è da operare se ’l ponto dato fosse sopra la linea .bg., infra li ponti .l.g.<lb></lb> |
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5889 Verbi gratia: sia uno quadrilatero .abgd. diviso in .2. parti iguali dala linea .al. e<lb></lb> |
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5890 sia el ponto dato .g. e comporró per lo detto ordine nela retta .ga. E a quello mene-<lb></lb> |
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5891 <lb></lb> |
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5892 <pb></pb></p> |
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5893 <p class="folio"> |
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5894 folio 40r |
|
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5895 </p> |
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5896 <p class="main"> |
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5897 <lb></lb> |
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5898 </p> |
|
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|
5899 <p class="runhead"> |
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5900 Distinctio quinta. Capitulum secundum. 40 |
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|
5901 </p> |
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5902 <p class="main"> |
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5903 <lb></lb> |
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5904 ró la equedistante retta .lf. e comporró .gf. che divida il quadrilatero .abgd. in .2. parti iguali<lb></lb> |
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5905 e partase dal ponto .g. che se perverá per l’ ordine e modo detto. E cosí habiamo mostro in che<lb></lb> |
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5906 modo el quadrilatero .abgd. si divide in .2. parti iguali dagli .2. lati equidistanti dala linea me-<lb></lb> |
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5907 nata dal ponto dato in suli deti lati in che luogo voi. Ora é da mostrare commo si divida da-<lb></lb> |
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5908 la linea che esca dagli altri .2. lati. |
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5909 </p> |
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5910 <p class="main"> |
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5911 Sia adonca uno quadrilatero .abgd. diviso in .2. parti iguali dala linea .zi. la quale<lb></lb> |
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5912 trovo per lo passato modo. E ancora si meni la linea .gf. la quale divida il detto qua-<lb></lb> |
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5913 drilatero in .2. parti iguali per la passata. E il ponto dato sia sopra la linea .ab., in<lb></lb> |
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5914 qual parte voi: cioé o infra ’l .bz. o fra’ l .zf. overo infra ’l .fa. overo uno de’ ditti ponti. Onde,<lb></lb> |
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5915 se il dato ponto fosse .b., si dividerebe il detto quadrilatero da linea .bc. in .2. parti iguali commo<lb></lb> |
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5916 giá habiamo dichiarato. E, similemente, se il dato ponto fosse .z. la retta .zi. lo dividerebe in .2.<lb></lb> |
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5917 parti iguali comme é mostro. E, se ’l ponto dato é .f., ancora .fg. segará il quadrilatero .abgd.<lb></lb> |
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5918 in .2. parti iguali commo dicemmo, essendo el ponto .a., la linea .al. é quella che llo dividerebe. E<lb></lb> |
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5919 secondo quello che s’ é detto. Ma sia il dato ponto .h. infra ’l .bz., meneró la linea .hi. e dal pon-<lb></lb> |
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5920 to .z. faró la linea .zc. equedistante ala linea .hi. e meneró la linea .hc., la quale linea .hc. dico<lb></lb> |
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5921 che la divide el quadrilatero .abgd. in .2. parti iguali che cosí se prova: perché nelle equedistan-<lb></lb> |
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5922 ti e infra quelle cioé .hi. e .zc. e sopra la base .zh. sonno li trianguli .zih. e .hcz., quai triangoli son-<lb></lb> |
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5923 no infra loro iguali. Onde agiungnendo a ciascuno dei triangoli .hiz. e .hic., che sonno iguali, lo quadri-<lb></lb> |
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5924 latero .igbh., será lo quadrilatero .cgbh. iguali al quadrilatero .izbg., che è la mitá di tutto el qua-<lb></lb> |
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5925 drilatero .abgd. commo è mostro. Adunque la linea .hc. divide il quadrilatero in doi parti iguali.<lb></lb> |
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5926 E, se dal ponto .p. posto infra ’l .zf. voi che si parta la linea che divida il detto quadrila-<lb></lb> |
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5927 tero in .2. parti iguali è de bisogno facia la retta .pi. e dal ponto .z. si facia la linea<lb></lb> |
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5928 .zq., la quale sia equedistante ala linea .pi. e dipoi si meni la linea .pq., la quale di-<lb></lb> |
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5929 co che divide il detto quadrilatero in .2. parti iguali, e la prova non bisogna, conciosia-<lb></lb> |
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|
5930 cosaché per quelle cosse che sonno dette chiaro appaia. |
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|
5931 </p> |
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|
5932 <p class="main"> |
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|
5933 E, se ’l ponto dato fosse infra ’l .f. e .a. e sia .r., dico che si facia la linea .rg. e dal ponto .f.<lb></lb> |
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5934 si meni la linea .fs. equedistante ala linea .rg. e, dal ponto .r. al ponto .s., si meni la li-<lb></lb> |
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5935 nea .rs., la quale linea dico che la divide lo quadrilatero .abgd. in .2. parti iguali che<lb></lb> |
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|
5936 ancora chiaro appare per quello s’ é detto. |
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|
5937 </p> |
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|
5938 <p class="main"> |
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5939 Ancora dimostraró per uno altro modo commo un quadrilatero di .2. lati equedi-<lb></lb> |
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5940 stanti si debi dividere dali angoli soi. Sia il quadrilatero .abgd. capo tagliato, del<lb></lb> |
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|
5941 quale .ad. e .bg. soi lati sienno equedistanti e il minore di quelli sia .ad.; meneró in<lb></lb> |
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5942 quello el diametro .ag. e .bd., infra lor segantesi in sul ponto .m. E, perché e doi lati<lb></lb> |
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5943 .ad. e .bg. sonno equedistanti, simili sonno e triangoli .amd. e .bmg. Onde e gli é cosí .bg. al .ad.<lb></lb> |
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5944 commo .bm. al .md. e .gm. al .ma. Magiore é certamente .bg. che .ad., magiore sia .gm. che .ma.,<lb></lb> |
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5945 dividinse adunque li diametri .ag. e .bd. in .2. parti iguali sopra li ponti .e. e .z. e per lo ponto .e.<lb></lb> |
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5946 equedistante al diametro .bd. si meni la retta .ec. e compise la retta .bc. Dico el quadrilatero<lb></lb> |
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5947 .abgd. essere diviso in .2. parti iguali dal’ angolo .b. per la linea .bc. che cosí il prover. Compon-<lb></lb> |
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5948 gase la retta .be. e .ed. E, perché il ponto .e. é in mezzo del diametro .ag. fienno e triangoli .ade.<lb></lb> |
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5949 e. gde. infra loro iguali. E ancora il triangolo .abe. è iguale al triangolo .gbe. Onde il qua-<lb></lb> |
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5950 drilatero .edab. è la mitá di tutto il quadrilatero .abgd. E, perché li triangoli .bdc. e .bde.<lb></lb> |
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5951 sonno infra le equedistanti .bd. e .ec., sopra la basa .bd., sonno infra loro iguali, ai quali agionto<lb></lb> |
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5952 a ciascuno el triangolo .abd., sará el quadrilatero .abcd. iguale al quadrilatero .abed. Ma<lb></lb> |
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5953 il quadrilatero .abcd. è la mitá di tutto el quadrilatero .abgd. Onde ancora el quadrilate-<lb></lb> |
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5954 ro .abcd. sia la mitá di tutto el quadrilatero .abgd., ch’ era bisogno mostrare. Similmente,<lb></lb> |
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5955 se meneremo la linea .zf. equedistante ala linea diametrale .ag. e comporremo la retta .gf.,<lb></lb> |
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5956 sará il quadrilatero .abgd. diviso in .2. parti iguali dala linea .gf. menata dal’ angolo .g. che,<lb></lb> |
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5957 per lo detto modo, lo poi provare. |
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5958 </p> |
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5959 <p class="main"> |
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5960 Diciamo adonca, quando uno ponto è dato for del quadrilatero, in che modo in<lb></lb> |
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5961 .2. parti iguali si deba dividere per una linea passante su per lo ponto dato. On-<lb></lb> |
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5962 de sia un’ altra volta il quadrilatero di .2. lati equidistanti diviso in .2. parti iguali da-<lb></lb> |
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5963 le linee che si parteno dagli angoli: cioé dale linee .al.dn.gf.bc. Le quali, menate<lb></lb> |
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5964 da ogni parte infinitamente nei ponti .e.z.i.t.k.h.o.p., le dimostra adonca non se potere dare al-<lb></lb> |
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5965 cuno ponto se nno infra le dette linee: overo in sulle dette linee. Onde, se alcun ponto será da-<lb></lb> |
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5966 to in sulle dette linee che sonno .4., overo ne’ termini di quelle, sirá quella linea retta menata dal<lb></lb> |
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5967 detto ponto dividente il quadrilatero .abgd. in .2. parti iguali. Commo se ’l dato ponto fosse<lb></lb> |
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5968 <lb></lb> |
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5969 <pb></pb></p> |
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5970 <p class="folio"> |
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5971 folio 40v |
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5972 </p> |
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5973 <p class="main"> |
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5974 <lb></lb> |
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5975 </p> |
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5976 <p class="runhead"> |
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5977 Distinctio quarta. Capitulum secundum. |
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5978 </p> |
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5979 <p class="main"> |
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5980 <lb></lb> |
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5981 e da quello si menerebe la linea .eal. la qual divide il quadrilatero in doi parti iguali e quel<lb></lb> |
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5982 medesimo intenderai degli atri. Overamente, se ’l dato ponto fosse .q., cadente infra le linee .ea.<lb></lb> |
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5983 e .zd. sopra il lato .ad. e voglio che dal ponto .q. esca la retta dividente il quadrilatero .abgd. in<lb></lb> |
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5984 .2. parti iguali, perché equedistanti sonno e llati .ad. e .bg., faró la retta .qm. e menerola infino<lb></lb> |
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5985 al ponto .v. Dico el quadrilatero .abgd. essere divisso in .2. parti iguali dala linea .qrsu. e in-<lb></lb> |
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5986 tendi che ’l ponto .m. sia el ponto dove le linee diametrali s’ intersegano, che in .2. parti sia di-<lb></lb> |
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5987 viso e cosí te ’l proveró. Perché le rette .ad. e .nl. sonno infra loro equedistanti e iguali e in quella<lb></lb> |
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5988 la retta .rs. passa, sia l’ angolo .ars. iguali al’ angolo .rsl. E ancora l’ angolo .sla. al’ angolo .lar.<lb></lb> |
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5989 è iguale e le rette .am. e .md. sonno infra loro iguali. Imperoché le rette .ad. e .ln. sonno infra lo-<lb></lb> |
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5990 ro iguali. Onde .ax. è iguale al .xl., onde perché le tagliate e equedistanti le rette si segano<lb></lb> |
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5991 per igual parti si segano commo in geometria si mostra. Onde iguale è la retta .nx. ala retta<lb></lb> |
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5992 .xd. e .ax. ala retta .xl. comme ó detto. Onde il triangolo .arx. e .xsl. sonno infra loro iguali e<lb></lb> |
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5993 equiangoli e equilateri. Onde, se a ciascuno s’ agiongi el quadrilatero .absx., sará el quadrila-<lb></lb> |
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5994 tero .absr. iguali al triangolo .abl., cioé ala mitá del quadrilatero .abgd. segato. E adonca<lb></lb> |
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5995 il quadrilatero .abgd. in .2. parti iguali diviso dala linea .rs. che esci dal ponto .q. Similmen-<lb></lb> |
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5996 te, se sirá dato il ponto infra la linea .nl., infra le rette .nh. e .lk., per lo detto modo sirebe da ope-<lb></lb> |
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5997 rare, cioé da quel ponto si menase la retta che passasse per lo ponto .x. e facesse quella retta so-<lb></lb> |
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5998 pra il lato .ad. Verbi gratia: se ’l ponto dato sirá .v., didise per la linea .usxr. in .2. parti iguali, com-<lb></lb> |
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5999 me era di bisogno mostrare. |
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6000 </p> |
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6001 <p class="main"> |
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6002 Per simili figure quadrilatere è da notare commo in exponendo Euclide, nel<lb></lb> |
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6003 principio del suo primo libro, solemo dire dove diffinendo le specie dele figure ret-<lb></lb> |
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6004 tilinee, asegna dele quadrilatere solo quelle .4. specie, cioé quadrato equilatero,<lb></lb> |
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6005 tetragono longo, helmuaym e simile helmuaym e l’ altre tutte a queste non simi-<lb></lb> |
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6006 li, de .4. lati existenti, le chiama helmuariffe, cioé a modo nostro irregulari, che sonno molte<lb></lb> |
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6007 sorti e avenga che qui, de simili quadrilateri parlando, de tutte apieno non si tratti. Nondi-<lb></lb> |
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6008 meno tu, per tuo peregrino ingegno, le ritroverai che sonno de grande industria lor divisioni.<lb></lb> |
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6009 Ma pur sempre per viam triangulorum faciliter le dedurai siché bisogna che l’ ingengno aiuti<lb></lb> |
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6010 el libro, che ogni cosa non é possibile scrivere. Ideo et cetera.<lb></lb> |
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6011 <lb></lb> |
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6012 Se alcuna figura detta capo tagliato in .3. parti iguali dale rette equedistati ale<lb></lb> |
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6013 sue base voi dividere, comme sia il quadrilatero .abgd. del quale e .2. lati .ad. e .bg.<lb></lb> |
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6014 sienno equedistanti, gli altri lati .ab. e .gd. dai ponti .ad. concorrino infino al .e. E faciase la linea<lb></lb> |
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6015 .zti. e sia la proportione del .zi. al .it. comme la proportione del quadrato dela linea .eb. al quadrato<lb></lb> |
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6016 dela linea .ae. E divideró .tz. in .3. parti iguali che sienno .tk.kl.lz. e porró il quadrato dela linea .em. al quadra-<lb></lb> |
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6017 to dela linea .eb. comme .zi. al .ik. E ancora porró il quadrato dela linea .en. al quadrato dela<lb></lb> |
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6018 linea .eb. comme .li. al .zi. E, per li ponti .nm., meneró le rette equedistanti ale basi .bg. e produ-<lb></lb> |
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6019 ceró la retta .mo. e .np. Dico el quadrato .ag. essere diviso in .3. parti iguali che sonno quadri-<lb></lb> |
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6020 lateri .adom. e .mopn. e .npgb., che cosí te ‘l proveró. E adonca (comme dissi) come il quadra-<lb></lb> |
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6021 to dela linea .be. al quadrato dela linea .ae., cosí el triangolo .ebg. al triangolo .ead. Anco-<lb></lb> |
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6022 ra, perché e gli é cosí .zi. al .ik., cosí il quadrato dela linea .be. al quadrato dela linea .me. Ma<lb></lb> |
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6023 comme il quadro .be. el quadrato .me., cosí el triangolo .ebg. al triangolo .emo. e ancora cosí<lb></lb> |
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6024 el .zi. al .il., cosí el triangolo .ebg. al triangolo .enp. E, per la disgionta proportionalitá, sia co-<lb></lb> |
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6025 sí .it. al .tk., cosí il triangolo .ead. al quadrilatero .ao. E ancora é cosí .tk. al .kl., cosí il quadri-<lb></lb> |
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6026 latero .ao. a il quadrilatero .mp. É certamente .tk. iguale al .l.kl. Onde e .ao. quadrilatero sa-<lb></lb> |
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6027 rá iguale al quadrilatero .mp. Ancora sirá cosí il .kl. al .lz., cosí il quadrilatero .mp. al quadri-<lb></lb> |
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6028 latero .ng.; è certamente .kl. iguali al .lz. e .mp. quadrilatero sirá iguale al quadrilatero .ng.; igua-<lb></lb> |
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6029 li adonca sonno i quadrilateri infra loro .ao. e .mp. e .ng. comme dicemmo. E, acioché con numeri<lb></lb> |
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6030 l’ abia, sia .ad.6. e .bg.15. e .ab.12. e .dg. 15. E sia l’ angolo .b. retto. Donde tutta .eb. sia .20., che ’l suo<lb></lb> |
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6031 quadrato cioé .400., è al quadrato dela linea .ea. comme .25. a .4. Porró adonca .zi.25. e .ti.4.,<lb></lb> |
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6032 rimane .tz.21., li quali dividi in .3. parti iguali, sirá ciascuna dele parti .tk.kl.lz.7. Onde .ik.<lb></lb> |
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6033 è .11. Multiplicarai adonca .11. per .400. e dividerai per .25. e virranne .176. per lo quadrato de-<lb></lb> |
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6034 la linea .em. Ancora stenderó .li., cioé .18. per .400. e divideró per .25. e haremo .288. per lo qua-<lb></lb> |
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6035 drato dela linea .en. Adonca .am. è radici di .176. meno .8. perché .ea. è .8. per la linea .mn., é ra-<lb></lb> |
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6036 dici de .288. meno .R. di .176. e la linea .nb. è .20. meno .R. de .288. E, perché e gli é cosí el quadrato de-<lb></lb> |
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6037 la linea .ea. al quadrato dela linea .ad., cioé comme .16. a .9., cosí el quadrato dela linea .em. al quadrato .mo. dove,<lb></lb> |
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6038 <lb></lb> |
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6039 <pb></pb></p> |
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6040 <p class="folio"> |
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6041 folio 41r |
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6042 </p> |
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6043 <p class="main"> |
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6044 <lb></lb> |
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6045 </p> |
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6046 <p class="runhead"> |
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6047 Distinctio quinta. Capitulum secundum. 41 |
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6048 </p> |
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6049 <p class="main"> |
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6050 <lb></lb> |
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6051 se multiplicarai .176. per .9. e divideremo per .16., haremo .99. per lo quadrato dela linea .mo.<lb></lb> |
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6052 Ancora se .1/16. de .288. et cetera E questo basti sopra li quadrilateri che si dicano capo tagliato e diremo<lb></lb> |
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6053 de’ quadrilateri detti diversilateri. |
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6054 </p> |
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6055 <p class="main"> |
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6056 Prima voglio dimostrare commo si divida el quadrilatero diversilatero in .2. parti<lb></lb> |
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6057 iguali dal’ angolo dato. Sia il quadrilatero .abcd. el quale voglio dividere in .2.<lb></lb> |
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6058 parti iguali dal’ angolo .a. Meneró prima el diametro .bd. opposto al’ angolo .bad.<lb></lb> |
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6059 e segaró el detto diametro per lo diametro .ac. in nel ponto .e. e fienno le rette .be. e .ed.<lb></lb> |
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6060 iguali o non. Sienno prima iguali: e, perché iguale è la retta .be. del .ed., sará il triangolo .abe.<lb></lb> |
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6061 iguale al triangolo .ade. e il triangolo .ebc. è iguale al triangolo .acd. Onde tutto il triango-<lb></lb> |
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6062 lo .abc. a tutto il triangolo .acd. è iguale. Diviso é adonca il quadrilatero .abcd. in .2. parti igua-<lb></lb> |
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6063 li dal diametro .ac. usciente dal’ angolo dato ch’ era bisogno mostrare. Ma non sia la linea<lb></lb> |
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6064 .be. iguale ala retta .ed., ma sia .bz. iguale ala retta .zd. E meneró la retta .zi. equedistante al dia-<lb></lb> |
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6065 metro .ac., commo in questa figura si manifesta e comporró la retta .ai. Dico adonca il quadrila-<lb></lb> |
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6066 tero .abcd. essere diviso in .2. parti iguali dala linea .ai. che esci dal’ angolo .a. dato, che sonno<lb></lb> |
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6067 li triangoli .abi. e il quadrilatero .aicd., che cosí te’ l proveró. Io comporró la retta .az. e .cz. e fienno<lb></lb> |
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6068 li triangoli .azd. e .dzc. Iguali a’ triangoli .abz. e .bzc. Onde il quadrilato .aicd. é iguale<lb></lb> |
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6069 al triangolo .abi. Onde il quadrilatero .abcd. è diviso in .2. parti iguali commo bisogna.<lb></lb> |
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6070 Ancora, se da uno ponto dato sopra uno de’ lati del quadrilatero diversilatero vor-<lb></lb> |
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6071 rai dividere per una linea retta in .2. parti iguali: commo il quadrilatero .ac. el quale<lb></lb> |
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6072 voglia dividere per una linea usciente dal ponto .e. sopra il lato .ad. Divideró prima il<lb></lb> |
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6073 quadrilatero .ac. in .2. parti iguali dal’ angolo .d. e sia la retta .td. quella linea che<lb></lb> |
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6074 lo divide in parti iguali e faró la linea .et. la quale porró sia equedistante ala retta .cd. Commo<lb></lb> |
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6075 appare in ditta figura. E faciase la retta .ec. Dico certamente la retta .ec. essere quella che divi-<lb></lb> |
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6076 de il quadrilatero .abcd. in .2. parti iguali, la quale linea escí dal ponto .e., che cosí si pruova.<lb></lb> |
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6077 Perché equedistanti sonno le rette .et. e .cd., fienno li triangoli .ecd. e .tde. infra loro iguali, ma<lb></lb> |
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6078 il triangolo .tcd. è la mitá del quadrilatero .ac.; diviso é adonca el quadrilatero .ac. in .2. par-<lb></lb> |
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6079 ti iguali dal ponto .e., dato ch’ era bisogno fare. Ma, se la retta .et. non fosse equedistante ala<lb></lb> |
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6080 retta .cd., faró adonca la retta .dz. equedistante ala retta .te. e comporró .ez. commo appare nel-<lb></lb> |
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6081 la ditta figura. Dico adonca el quadrilatero .ec. essere diviso in .2. parti iguali dala linea .ez.,<lb></lb> |
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6082 che cosí si prova. Sonno certamente li triangoli .dze. e .dzt. infra loro iguali, perché sonno in-<lb></lb> |
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6083 fra medesime base e in linee equedistanti; a’ quali, se s’ agiongni in commune il triangolo .dez., sia<lb></lb> |
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6084 el quadrilatero .ezcd. iguale al triangolo .det., cioé ala mitá di tutto el quadrilatero .ac. Ed é<lb></lb> |
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6085 da notare che, se ’l diametro .bd. divide il quadrilatero .ac. in .2. parti iguali, ne perviene similmente<lb></lb> |
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6086 quello che habiamo detto in queste .2. figure. Ma se la divisione caderá sopra il lato .ab.,<lb></lb> |
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6087 commo la retta .di. che sia dividente il quadrilatero .ac. in .2. parti iguali, altramente si debbia<lb></lb> |
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6088 operare. Divideró .ac. dal’ angolo .b. con la retta .bk. e alora se ’l .k. sirá il dato ponto sopra il<lb></lb> |
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6089 lato .ad., dal quale sia de bisogno partire la retta dividente el quadrilatero .ac. in .2. parti igua-<lb></lb> |
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6090 li e la retta sua quella linea .kb. Ma, se ’l .k. non sirá il dato ponto, sirá allora il dato pon-<lb></lb> |
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6091 to infra ’l .k. e .a. Sia prima il dato ponto .e. infra ’l .dk. e facciase la retta .be. E dal ponto .k. si me-<lb></lb> |
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6092 ni la retta .kl. equedistante ala retta .eb. e facciase la retta .el. dividente il quadrilatero .ac. in .2.<lb></lb> |
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6093 parti iguali che sonno li quadrilateri .al. e .ce. commo è manifesto nella figura passata che an-<lb></lb> |
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6094 cora si prova per l’ ordine ditto. |
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6095 </p> |
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6096 <p class="main"> |
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6097 Ancora, se il dato ponto fosse infra ’l .a. e .k. e compise similmente la retta .eb. e dal<lb></lb> |
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6098 ponto .k. si meni la retta .km. equedistante ala retta .eb. e faciase la retta .em. di-<lb></lb> |
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6099 vidente il quadrilatero .ac. in .2. parti iguali, che sonno li quadrilateri .am. e .ec.,<lb></lb> |
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6100 commo nella sexta figura appare, cioé nella figura presente, che si prova per lo<lb></lb> |
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6101 modo che nell’ altre figure s’ é provato. E peró non á bisogno de dimostratione e questo vo-<lb></lb> |
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6102 levamo mostrare. Insegnaró certamente, per lo modo di sopra detto, commo si divida-<lb></lb> |
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6103 no li quadrilateri diversilateri dal ponto dato sopra uno de’ lati soi. Sia il quadrilatero<lb></lb> |
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6104 .abcd. e il dato ponto sia .e., cadente prima sopra la mitá delo lato .ab. E menise dal ponto<lb></lb> |
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6105 .d. la retta .dz. equedistante ala retta .ab. e dividerolla in .2. parti iguali: sopra il ponto .i. E fa-<lb></lb> |
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6106 ró .ei.ec. e .ic. E faró la linea .it. equedistante ala linea .ec. e faró la linea .et. Dico che la li-<lb></lb> |
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6107 nea .et. è quella che divide il quadrilatero .ac. in .2. parti iguali dala linea .et., che cosí si pro-<lb></lb> |
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6108 va. El quadrilatero .ez. è la mitá del quadrilatero .az. per la .34a. del primo de Euclide. Simil-<lb></lb> |
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6109 mente, perché le base .zi. e .id. sonno infra loro iguali, fienno li triangoli .czi. e .cid. infra lo-<lb></lb> |
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6110 <lb></lb> |
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6111 <pb></pb></p> |
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6112 <p class="folio"> |
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6113 folio 41v |
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6114 </p> |
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6115 <p class="main"> |
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6116 <lb></lb> |
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6117 </p> |
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6118 <p class="runhead"> |
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6119 Distinctio quinta. Capitulum secundum. |
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6120 </p> |
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6121 <p class="main"> |
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6122 <lb></lb> |
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6123 ro iguali. Onde il triangolo .ciz. la mitá è del triangolo .cdz.; sirá ancora el quadrilatero .ez.<lb></lb> |
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6124 la mitá del quadrilatero .az. Adonca el quadrilatero .ebci. è la mitá del quadrilatero .ac.<lb></lb> |
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6125 Sonno certamente li triangoli .eci.ect. infra loro iguali, ai quali, se s’ agiongni in commu-<lb></lb> |
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6126 ne el triangolo .ebc., sirá il quadrilatero .ebct. iguale al quadrilatero .ebci., ma il quadrila-<lb></lb> |
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6127 tero .ebci. è la mitá del quadrilatero .ec. Onde il quadrilatero .bt. è la mitá del quadrilate-<lb></lb> |
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6128 ro .ac. Commo era de bisogno. Le quali cose dette e date e provate, potremo dividere in .2.<lb></lb> |
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6129 parti iguali ogni quadrilatero da ogni ponto dato sopra ad alcuno de’ lati di quelli. E, se ’l pon-<lb></lb> |
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6130 to dato sirá fuor over dentro, per lo quale la linea sia de bisogno passi che divida el quadri-<lb></lb> |
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6131 latero in doi parti iguali, potremo queste cose, coll’ operatione delle linee over fili subtilmen-<lb></lb> |
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6132 te operare. Ora dimostraremo il modo a torre d’ un quadrilatero una parte nominata<lb></lb> |
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6133 dal’ angolo dato. |
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6134 </p> |
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6135 <p class="main"> |
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6136 Sia adonca el quadrilatero del quale voglio dal’ angolo .d. tagliare una parte da-<lb></lb> |
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6137 ta e conosciuta, commo sia la terza parte. Faró il diametro opposto al’ angolo .d.,<lb></lb> |
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6138 che sia il diametro .ac., el quale segaró col diametro .db. sopra il ponto .e. La retta<lb></lb> |
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6139 .ce. overamente é la terza parte overo non cioé la terza parte de tutto el diametro<lb></lb> |
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6140 .ac. Sia prima .ec. il terzo de tutta .ac. Dico che ’l diametro .bd. togli del quadrilatero .abcd.<lb></lb> |
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6141 la terza parte, la quale è il triangolo .acb., che cosí te ’l proveró. Perché el triangolo .dce. e .dca<lb></lb> |
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6142 sonno sopra una altitudine saranno im proportione infra loro commo la basa .ce. al .ca. e per<lb></lb> |
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6143 quel medesimo, li triangoli .bce. e .abc. sonno in ditta proportione. Onde è cosí .ce. al .ac. co-<lb></lb> |
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6144 sí il triangolo .bcd. al quadrilatero .ac. e certamente la retta .ce. il terzo del .ac. Onde il trian-<lb></lb> |
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6145 golo .bcd. è il terzo del quadrilatero .ac. tagliato e adonca la parte data: cioé la terza par-<lb></lb> |
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6146 te del quadrilatero .ac. per la linea .bd. che esci dal ponto .d. ch’ é angolo del detto quadrilate-<lb></lb> |
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6147 ro ch’ era de bisogno fare. Ma non sia .ce. la terza parte del .ac. Porró adonca .cg. il terzo<lb></lb> |
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6148 del .ac. e dal ponto .g. meneró la retta .gf. equedistante al diametro .bd. e comporró la retta<lb></lb> |
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6149 .df. la quale toglie del quadrilatero .ac. la terza parte, la quale dimostraró essere de tutto el<lb></lb> |
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6150 quadrilatero .ac. il terzo per prova chiara. Imperoché .cg. è il terzo del .ca., siran li triangoli<lb></lb> |
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6151 .bgc. e .cdg., cioé il quadrilatero .bcdg., la terza parte de’ triangoli .abc. e .acd., cioé del qua-<lb></lb> |
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6152 drilatero .abcd. el quale, per quello che s’ é detto, se dimostra chiaro essere iguale al quadri-<lb></lb> |
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6153 latero .fbdc. et cetera. E questo basti quanto al secondo capitolo dela presente distinctione e se-<lb></lb> |
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6154 quendo diremo del terzo.<lb></lb> |
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6155 <lb></lb> |
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6156 Modus dividendi multi lateras formas: videlicet pentagonas, exagonas et cetera in par-<lb></lb> |
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6157 tes proportionabiliter plures: capitulum tertium. |
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6158 </p> |
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6159 <p class="main"> |
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6160 Se un pentagono equilatero e equiangolo da uno de’ soi angoli in .2. parti igua-<lb></lb> |
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6161 li voi dividere, la linea retta da quello angolo sopra la mitá del lato a lui opposto<lb></lb> |
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6162 mena. Verbi gratia: sia il dato pentagono equilatero e equiangolo .abcde. e il<lb></lb> |
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6163 dato angolo sia .a. e sopra la mitá del lato .cd. la linea .af. mena. Dico el pentago-<lb></lb> |
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6164 no detto essere diviso in .2. parti iguali dala linea .af., che cosí te ’l proveró. Comporró le ret-<lb></lb> |
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6165 te .ac. e .ad.; seranno li triangoli .abc. e .aed. infra loro iguali e equiangoli e gli triangoli .afc.<lb></lb> |
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6166 e .afd. iguali. Onde il quadrilatero .aefd. è iguale al quadrilatero .abcf. e diviso adonca il<lb></lb> |
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6167 pentagono detto in .2. parti iguali dal’ angolo .a. per la linea .af. ch’ era bisogno mostrare.<lb></lb> |
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6168 E, per questo, se manifesta che quando in un pentagono equilatero e equiangolo sopra<lb></lb> |
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6169 la mitá di qual voi lato si dirizza una linea al’ angolo, la quale pigli el detto lato, che quella li-<lb></lb> |
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6170 nea divide il detto pentagono in doi parti iguali. |
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6171 </p> |
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6172 <p class="main"> |
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6173 E, se dal ponto dato .g. sopra uno lato del medesimo pentagono e da quello voi<lb></lb> |
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6174 si parta la linea la quale divida il detto pentagono in .2. parti iguali, é da consi-<lb></lb> |
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6175 derare se ’l ponto .g. è in mezzo del lato .ab., perché alora è da menare una linea<lb></lb> |
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6176 dal dato ponto .g. infino al’ angolo opposto .d. e sia la retta .gd. dividente il pen-<lb></lb> |
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6177 tagono .abcde. in doi parti iguali. E, se ’l ponto dato sirá infra ’l .a. e .g., che sia .h., allora di-<lb></lb> |
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6178 videró il pentagono dala linea .af. in .2. parti iguali. E compileró la retta .hf. e dal ponto .a.<lb></lb> |
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6179 comporró la linea .ai. equedistante alla linea .hf. e faró la linea .hi., la quale divide il pentago-<lb></lb> |
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6180 no .abcde. in .2. parti iguali, che cosí si prova. Perché li triangoli .hfa. e .hfi. sonno infra .2.<lb></lb> |
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6181 linee equedistanti e sopra la medesima basa .hf. e peró infra loro sonno iguali per la .36a. del<lb></lb> |
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6182 primo, se a ogni parte s’ agiongni il quadrilatero .hbcf., sirá il quadrilatero .hbci. iguale al<lb></lb> |
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6183 quadrilatero .abcf. Ma il quadrilatero .abcf. è la mitá del pentagono .abcde. Onde il<lb></lb> |
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6184 <lb></lb> |
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6185 <pb></pb></p> |
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6186 <p class="folio"> |
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6187 folio 42r |
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6188 </p> |
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6189 <p class="main"> |
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6190 <lb></lb> |
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6191 </p> |
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6192 <p class="runhead"> |
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6193 Distinctio quinta. Capitulum tertium. 42 |
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6194 </p> |
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6195 <p class="main"> |
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6196 <lb></lb> |
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6197 quadrilatero .hbci. sia ancora la mitá del pentagono .abcde. commo volavamo. Ma<lb></lb> |
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6198 non sia el dato pentagono equilatero e equiangolo, commo il presente pentagono segnato<lb></lb> |
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6199 .abgde. el quale voglio dividere in .2. parti iguali. Produrró la retta .eg. e, sopra la mitá di<lb></lb> |
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6200 quella, segneró il ponto .z. dal quale meneró la retta linea che divida el quadrilatero .abeg.<lb></lb> |
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6201 in .2. parti iguali, la quale sirá la linea .zi. e comporró la retta .zd. la quale ancora divide il trian-<lb></lb> |
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6202 golo .edg. in .2. parti iguali. E la retta .zd. overo é una colla linea .zi. overo non. E sia prima<lb></lb> |
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6203 l’ una e l’ altra per lo diritto: cioé sia la linea .di. una linea, commo in questa figura è manife-<lb></lb> |
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6204 sto. Dico la linea .id. essere quella che divide el ditto pentagono in .2. parti iguali commo era<lb></lb> |
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6205 de bisogno. Ma non sia la retta .zd. per lo diritto colla retta .zi. Comporró la retta .id. e<lb></lb> |
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6206 porró a quella la equedistante la retta .zl. e comporró la retta .il., la quale ancora dividerá il<lb></lb> |
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6207 pentagono .abgde. in .2. parti iguali, che cosí si prova. La retta .zd. e .iz. dividono il pentago-<lb></lb> |
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6208 no .abgde. in .2. parti iguali e sonno il triangolo .idz. e .idl., infra loro iguali, ai quali, agio-<lb></lb> |
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6209 gnendo il quadrilatero .ibgd., sirá il pentagono .ibgdl. iguale al pentagono .ibgdz., cioé<lb></lb> |
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6210 ala mitá del pentagono .abgde., commo era de bisogno. E cosí, secondo questo modo, pos-<lb></lb> |
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6211 siamo produre la divisione sopra ogni lato del pentagono da ogni ponto dato sopra quel-<lb></lb> |
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6212 lo e ancora lo potremo in piú parti dividere per quello che s’ é ditto. Imperoché, se la retta<lb></lb> |
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6213 .eg. in .3. parti iguali dividerai e, per li ponti della divisione, in parti iguali dividerai il qua-<lb></lb> |
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6214 drilatero .abgc. e il triangolo .egd. e dapoi, procedendo per l’ ordine dimostrato, haremo<lb></lb> |
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6215 quello vogliamo. |
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6216 </p> |
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6217 <p class="main"> |
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6218 E, se ’l pentagono hará forma dimitra, commo il pentagono .bcdef., meneremo<lb></lb> |
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6219 a retti angoli, sopra il lato del pentagono .cd., la linea .fa. e investigaró l’ area del<lb></lb> |
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6220 .bcaf. e ancora l’ area del quadrilatero .deaf. che, se le trovi iguali, sará il ditto pen-<lb></lb> |
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6221 tagono diviso in li ditti doi quadrilateri per doi parti iguali dala linea .af. E, se<lb></lb> |
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6222 l’ area de’ detti .2. quadrilateri non è iguale e sia el magiore di loro l’ area del quadrilatero .fabe.,<lb></lb> |
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6223 secondo la quantitá d’ uno numero e porró el ponto .z. tanto discosto dal ponto .a. in verso<lb></lb> |
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6224 .c. che, multiplicato .za. in .af. fará il detto numero e comporró la retta .fz. e sará l’ area del trian-<lb></lb> |
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6225 golo .faz. la mitá del detto numero. Onde la linea .fz. divide il pentagono in .2. parti iguali,<lb></lb> |
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6226 cioé in .2. quadrilateri che sonno .bcdf. e .fzdc. e questo è fatto secondo un vulgar modo.<lb></lb> |
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6227 Ma, se secondo geometria vorrai questo adoperare, comporremo .be. e divideremo .be. in<lb></lb> |
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6228 .2. parti iguali in sul ponto .i., commo nella figura appare, e divideremo el quadrilatero .bcde.<lb></lb> |
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6229 in .2. parti iguali dala linea .ik., la quale passerá per lo ponto .f.; sirá il detto pentagono diviso<lb></lb> |
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6230 in .2. parti iguali per la linea .ik. che dico fosse .ifk., perché del quadrato .bkci. togliamo il<lb></lb> |
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6231 triangolo .bif. e del quadrilatero .ikde. auferamus triangulum .ief.; rimaranno li quadrati-<lb></lb> |
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6232 lateri .fc. e .fd. infra loro iguali. Adonca la linea .ik. divide il pentagono in .2. parti iguali.<lb></lb> |
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6233 E, se la retta ik. sega il lato .ef. nel ponto .l., commo nella figura appare, porró .ml. al .lf. com-<lb></lb> |
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6234 mo .il. al .lk. e comporró la linea .km. La quale dico che la divide il pentagono in .2. parti igua-<lb></lb> |
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6235 li dele quali una parte sirá il quadrilatero .mkde. e l’altra il pentagono .bokmf. La pro-<lb></lb> |
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6236 va: faciase la retta .if., fienno li triangoli .ifl. e .kml. infra loro iguali, conciosiacosaché sienno<lb></lb> |
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6237 la proportione del ml. al .fl. commo .il. al .kl., onde il triangolo .ife. è iguale al triangolo .ile.<lb></lb> |
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6238 e .klm. è certamente e il triangolo .ife. iguale al triangolo .ibf. Onde il triangolo .bif. é igua-<lb></lb> |
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6239 le al triangolo .iel. e .klm. Onde se dal quadrilatero .bcki. togliamo .bfi. e dal quadrilatero<lb></lb> |
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6240 .ikde. togliamo li triangoli .ile. e .klm., rimarranno li quadrilateri .bckl. e il triangolo .ifl.<lb></lb> |
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6241 iguali al quadrilatero .kdem. é certamente il triangolo .klm. eguale al triangolo .ilf. On-<lb></lb> |
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6242 de il pentagono .bcfmf. è iguale al quadrilatero .mkde., commo ó detto. La quale divisione<lb></lb> |
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6243 produrremo dal ponto .f. componendo .fk. e a quello porremo la equedistante .nm. e fare-<lb></lb> |
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6244 mo .nf., la quale divide il detto pentagono in .2. parti iguali. E ancora in .3. parti iguali si<lb></lb> |
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6245 possono, per le cose dette, dividere. |
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6246 </p> |
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6247 <p class="main"> |
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6248 E volendo dividire uno exagono, commo lo exagono .abcdef. in .2. parti igua-<lb></lb> |
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6249 li, dove menerai la linea .ad. la quale dividerá il detto exagomo in .2. parti che son-<lb></lb> |
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6250 no .2. quadrilateri .adef. e .adcb. E ponendo il ponto .g. in sula linea .ab. mene-<lb></lb> |
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6251 rai la linea .gps., la quale passi per lo ponto .q. segnato in sul mezzo dela linea .ad.<lb></lb> |
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6252 Dico la linea .gs. dividere lo exagono in .2. parti iguali, commo era de bisogno. E cosí potre-<lb></lb> |
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6253 mo mutare le divisione mutando li ponti e cosí, commo habiamo detto nello exagono e del<lb></lb> |
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6254 pentagono possiamo ancora d’ altre figure de piú lati procedere e in piú parti dividerli, com-<lb></lb> |
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6255 mo sarebbe in .3. o piú parti. E questo basti quanto al terzo capitolo dela detta distincio-<lb></lb> |
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6256 <lb></lb> |
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6257 <pb></pb></p> |
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6258 <p class="folio"> |
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6259 folio 42v |
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6260 </p> |
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6261 <p class="main"> |
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6262 <lb></lb> |
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6263 </p> |
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6264 <p class="runhead"> |
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6265 Distinctio quinta. Capitulum quartum. |
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6266 </p> |
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6267 <p class="main"> |
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6268 <lb></lb> |
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6269 ne e seguendo diremo del quarto e ultimo.<lb></lb> |
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6270 <lb></lb> |
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6271 </p> |
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6272 <p class="head"> |
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6273 De divisione circulorum in partes ad invicem proportionabiles. Capitulum quartum. |
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6274 </p> |
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6275 <p class="main"> |
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6276 <lb></lb> |
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6277 Se alcuno cerchio in .2. parti iguali voi segare per una linea menata da uno pon-<lb></lb> |
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6278 to dato in sula perifera sua overo fuor del cerchio, quel ponto, col centro del cer-<lb></lb> |
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6279 chio, componi e quella passi per infino ala perifera dal’ altro lato e harai quello<lb></lb> |
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6280 voi. Verbi gratia: Sia el dato circulo .abgd. e fuor di quello sia dato il ponto .e. e<lb></lb> |
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6281 sia .z. centro del cerchio. Compila .ez. infino al ponto .g., sia tutta .ag. diametro di cerchio<lb></lb> |
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6282 .abgd. Imperoché ’l diametro del cerchio è la retta menata pe ’l centro e terminata da ogni<lb></lb> |
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6283 parte dela periferia e la retta .ezg. è diametro del cerchio, adonca ciascuna parte .eazg. e<lb></lb> |
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6284 .ebzg. è mitá de ditto cerchio. E, se infra ’l cerchio il ponto dato sará .i. comporró .iz. e me-<lb></lb> |
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6285 nerolla in ogni parte neli ponti .b. e .d. e sia la retta .bd. diametro del cerchio che sega .abgd.<lb></lb> |
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6286 similmente in .2. parti iguali per la linea .bizd., commo da lato appare nella figura presen-<lb></lb> |
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6287 te. E, se uno cerchio dato vorrai dividere in tre parti iguali constituisce in quello uno trian-<lb></lb> |
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6288 golo equilatero. Commo sia il cerchio .abg. nel quale sia el centro .d. e in quello compilere-<lb></lb> |
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6289 mo uno triangolo equilatero e equiangolo e comporemo le rette .da.db.dg. che divideran-<lb></lb> |
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6290 no il cerchio in tre parti iguali, dele quali l’ una sia il settore .dab., la seconda il settore .dbg.,<lb></lb> |
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6291 la terza il settore .dga. Verbi gratia. Perché iguali sonno le rette .ab.bg.ga., perché le ret-<lb></lb> |
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6292 te iguali nelli cerchi iguali hano archi iguali e dal centro sonno infra loro igualmente disco-<lb></lb> |
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6293 ste. E peró le parti infra loro fienno iguali. Adonca le rette .da.db.dg. in tre settioni iguali<lb></lb> |
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6294 dividono el cerchio, commo era de bisogno. E, se in piú parti quelli dal centro voi divi-<lb></lb> |
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6295 dere, in quante parti quante voi, in tante dividi la circunferentia di quel cerchio e questo sen-<lb></lb> |
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6296 za fatiga harai se le sexte overo una misura che si possa curvare harai.<lb></lb> |
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6297 Se il cerchio per le linee equedistanti in tre parti voi dividere questo, senza gran<lb></lb> |
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6298 fatiga, non si puó fare. Nientedimeno commo questo, secondo l’ appressamento,<lb></lb> |
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6299 si debbia fare lo mostraró. Porró prima nel cerchio .abg. la retta .ag., che sia il<lb></lb> |
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6300 lato del triangolo equilatero cadente in quello. Onde la periferia .ag. sirá. la ter-<lb></lb> |
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6301 za parte di tutta la periferia del cerchio .abg. E porró dal centro .d. la retta .db. equedistan-<lb></lb> |
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6302 te ala retta .ag. e comporró le rette .da.dg.ab. e .bg. e fienno li triangoli .bag. e .dag. infra<lb></lb> |
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6303 loro iguali. Ali quali agionto l’ augmento commune dela portione del cerchio contenta sot-<lb></lb> |
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6304 to la retta .ag. e l’ arco .ag., sirá la figura contenta sotto le rette .ba. e .bg. e l’ arco .ag. iguale al<lb></lb> |
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6305 settore .dag., che è la terza parte del circulo. Adonca la figura fatta dale rette .ba. e .bg. e l’ar-<lb></lb> |
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6306 co .ag. è la terza parte del cerchio .abg. ali quali, agionto la portione del circolo contenta sot-<lb></lb> |
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6307 to la retta .ab. e la perifera .agb. piú dela terza parte del circolo .abg., secondo la quantitá<lb></lb> |
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6308 dela portione .bg., onde troveró l’ area dela portione fatta e contenta sotto la retta .bg. e la<lb></lb> |
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6309 periferia .bg. e divideró il doppio sopra la retta .ab. e poco piú di quello ne perverá. Trarró<lb></lb> |
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6310 dela periferia .bg. e sia quello l’ arco .be. E comporró la retta .ae. e sirá secondo l’ apressamen-<lb></lb> |
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6311 to la portione .age. il terzo del cerchio .abg. Dipoi, sopra la retta .ae., meneró dal centro .d.<lb></lb> |
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6312 el catetto .dz. e produrollo infino al .i. e sia la retta .di. iguale ala retta .dz. E, per lo ponto .i.<lb></lb> |
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6313 meneró la retta .tk. equedistante ala retta .ae. e sia la portione .thk. iguale ala portione .abe.<lb></lb> |
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6314 L’ avanzo certamente che si contiene sotto le rette .ae. e .tk. e l’ arco .ke. e .at. sirá l’ altra terza parte,<lb></lb> |
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6315 che cosí si prova. Meneró dal ponto .e. sopra la linea .ab. il catetto .ec. e sia quello che pervie-<lb></lb> |
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6316 ne del doppio dell’ embado dela portione del .bg. nella retta .ab. e compongase la retta .eb. e<lb></lb> |
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6317 sia il triangolo .abe. iguale alla portione del cerchio. Onde .bg., quando si togli il triango-<lb></lb> |
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6318 lo .abe. della portione del cerchio .agb., rimarrá la portione .aeg. terza parte del circulo .abg.<lb></lb> |
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6319 e, perché la retta .tk. e .ae. equedistanti sonno, equalmente distante del centro, sonno iguali<lb></lb> |
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6320 e, perché iguale è la retta .tk. ala retta .ae., quelle doi linee rette da uno medesimo cerchio<lb></lb> |
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6321 iguale parte pigliano. Commo per lo terzo de Euclide mostrammo. Iguale é adonca la por-<lb></lb> |
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6322 tione .tkl. ala portione .aeg. Onde la terza parte è la settione .tkl. l’ avanzo che se contie-<lb></lb> |
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6323 ne infra le equedistanti .tk. e .ae. rette. E l’ arco .ik. e .at. sirá l’ avanzo, cioé l’ altra terza parte, com-<lb></lb> |
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6324 mo bisogna. E, accioché questo s’ abia piú presto, quanto piú sotilmente poddi trovar la pro-<lb></lb> |
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6325 portione dela retta .dz. al mezzo diametro del circolo e trovai quella essere secondo l’ apres-<lb></lb> |
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6326 samento, commo .9. a .34. Onde, quando alcuno cerchio in tre iguali parti colle rette equedi-<lb></lb> |
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6327 stanti vorrai dividere, el diametro del cerchio studiamo di trovare e la mitá di quello nella<lb></lb> |
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6328 <lb></lb> |
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6329 <pb></pb></p> |
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6330 <p class="folio"> |
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6331 folio 43r |
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6332 </p> |
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6333 <p class="main"> |
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6334 <lb></lb> |
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6335 </p> |
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6336 <p class="runhead"> |
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6337 Distinctio quinta. Capitulum quartum. 43 |
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6338 </p> |
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6339 <p class="main"> |
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6340 <lb></lb> |
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6341 detta proportione da ogni parte sirá diviso. |
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6342 </p> |
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6343 <p class="main"> |
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6344 Similmente, se ’l circulo .abgde. in quatro parti iguali vorrai dividere, li diametri<lb></lb> |
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6345 .az. e .ge., sopra il centro del cerchio .c., a retti angoli, faró segare. Dapoi el mezzo<lb></lb> |
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6346 diametro .ez., sopra il ponto .k., divideró e sia la proportione .ck. al .ez. commo .611.<lb></lb> |
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6347 a .1522. E porró la retta .cl. iguale ala retta .ck. e, per li ponti .lk., produrró le ret-<lb></lb> |
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6348 te .bh. e .df. seganti el diametro .az. ad angoli retti. Sirá ciascuno de’ mezzi cerchi, secondo l’ a-<lb></lb> |
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6349 pressamento, diviso in doi parti iguali. |
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6350 </p> |
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6351 <p class="main"> |
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6352 E, se in doi linee equedistanti voi porre il terzo d’ uno cerchio. Commo sia il cer-<lb></lb> |
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6353 chio .abg. del quale il centro sia .d. Porrola equedistame la retta .dg. ala ret-<lb></lb> |
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6354 ta .ab. E divideró la periferia .ab. in doi parti iguali sopra il ponto .e. e meneró<lb></lb> |
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6355 la retta .ez. equedistante ala retta .bg. Dico che la figura contenta infra le rette eque-<lb></lb> |
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6356 distantti .ez. e .bg. e dagli archi .cg. e .eb. è il terzo di tutto il cerchio .abg. La prova. Meneró<lb></lb> |
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6357 le rette .da. e .db. e .ag. fienno li triangoli .gab. e .dab. iguali, ai quali, quando s’ agiongnerá<lb></lb> |
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6358 la portione .abe., sirá la figura contenta sotto le rette .ga. e .gb. e l’ arco .aeb. iguale al set-<lb></lb> |
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6359 tore .da. e .eb., che è la terza parte del circulo .abg. Adonca la figura contenta dale rette .ga.<lb></lb> |
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6360 e .gb. e l’ arco .aeb. é il terzo del circulo .agb. E, perché la retta .bg. e .ez. sonno equedistanti,<lb></lb> |
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6361 sia l’ arco .eb. e .gz. iguale. Ma l’ arco .eb. è iguale al’ arco .ae. Adonca l’ arco .ae. è iguale al’ ar-<lb></lb> |
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6362 co .gz. al quale, agionto l’ arco .bg., sirá certamente l’ arco .aebg. iguale al’ arco .ebgz. Onde<lb></lb> |
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6363 la portione del cerchio .ezgb. è iguale ala portione del circulo .agbe. Onde d’ ogni parte si<lb></lb> |
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6364 tolga la portione contenta dala retta .bg. e l’ arco .gb. rimarrá la figura contenta dale rette<lb></lb> |
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6365 .gb. e .eb. e dagli archi .be. e .gz. iguale ala terza parte del cerchio, cioé ale figure contente dala<lb></lb> |
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6366 rette .ga. e .gb. e dal’ arco .aeb., ch’ era bisogno mostrare. |
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6367 </p> |
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6368 <p class="main"> |
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6369 Se il mezzo cerchio .abg. in doi parti iguali voi segare, dividi la retta .ag. in doi<lb></lb> |
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6370 parti iguali dala linea .db. e la prova è questa: productte le rette .ba. e .bg., fien-<lb></lb> |
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6371 no li triangoli .bdg. e .bda. infra loro iguali, é certamente il lato .ad. del lato .dg.<lb></lb> |
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6372 iguali e la retta .bd. è commune e ancora gli angoli che sonno al .d. sonno retti.<lb></lb> |
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6373 E, perché la retta .bg. è iguale ala retta .ba. e le rette iguali nel cerchio hano iguali portio-<lb></lb> |
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6374 ni e perché la settione .bg. è iguale ala settione .ba. e li triangoli .bdg. e .bda. sonno iguali.<lb></lb> |
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6375 Onde la settione .bdai. ala settione .bgdi. è iguale. Diviso é adonca el mezzo cerchio .abg.<lb></lb> |
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6376 in doi parti iguali, ch’ era de bisogno fare. E, se colla retta equedistante ala basa .ag. voi<lb></lb> |
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6377 dividere in doi parti iguali. El mezzo diametro .bd. sopra il ponto .z. dividi e sia .dz. al .zb.<lb></lb> |
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6378 commo .611. a .1512. e per lo ponto .z. mena la retta .ei., la quale dividerá similmente el mezzo<lb></lb> |
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6379 cerchio .abg. in doi parti iguali, che saranno, per quelle cose ditte nella divisione de’ cerchi<lb></lb> |
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6380 in quatro parti. |
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6381 </p> |
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6382 <p class="main"> |
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6383 E, se la portione del cerchio, overo sia menore overo magiore del mezzo cerchio,<lb></lb> |
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6384 in doi parti iguali voi dividere, in detto modo, sopra la mitá dela corda sua la<lb></lb> |
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6385 saetta menarai. Verbi gratia. Sia data la portione magiore che ’l mezzo cerchio<lb></lb> |
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6386 .abg. e sopra la mitá dela sua corda sia menata la saetta .da., la quale dico che di-<lb></lb> |
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6387 vide la detta portione .abg. in doi parti iguali che sonno .abd. e .adg., che se proverrebbe<lb></lb> |
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6388 per quelle cose che sonno ditte nel mezzo cerchio. E, se dal ponto .d. menaremo la retta .dc.,<lb></lb> |
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6389 a retti angoli sopra la corda .bg., la quale dividerá ancora la ditta portione in doi parti igua-<lb></lb> |
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6390 li, commo volavamo e haremo diviso la portione magiore e la portione minore.<lb></lb> |
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6391 E, se ’l mezzo cerchio .abgc. in tre parti iguali voi dividere, la retta .bc. in doi par-<lb></lb> |
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6392 ti iguali dividi al ponto .d. Dipoi l’ arco .bac. in tre parti iguali dividi in sugli pon-<lb></lb> |
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6393 ti .a.g. e compilerai .ad. e .dg. che dico che dividono el mezzo cerchio in tre par-<lb></lb> |
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6394 ti iguali, che si prova, perché el mezzo cerchio è la portione .abdc. Sia il .d. cen-<lb></lb> |
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6395 tro del cerchio del quale cerchio la mittá è il detto semicirculo .abdc. Onde le rette .db.dc.<lb></lb> |
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6396 .da.dg. sonno iguali infra loro, sirá ciascuna di quelle figure settore di cerchio contente sot-<lb></lb> |
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6397 to le doi rette e gli archi preditti. Onde quando li sonno sotto medesime linee e medesimi ar-<lb></lb> |
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6398 chi sonno certamete infra loro iguali. Diviso é adonca il mezzo cerchio in tre parti iguali che<lb></lb> |
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6399 sonno .dab.dag.dgc. ch’ era bisogno fare. |
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6400 </p> |
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6401 <p class="main"> |
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6402 E, se una figura contenta sopra doi rette e l’ arco dela periferia in doi parti igua-<lb></lb> |
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6403 li voi dividere, commo sia el settore di cerchio .abcd. Dico che divida l’ arco .ac.<lb></lb> |
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6404 in doi parti iguali in sul ponto .b. e dal ponto .d. si meni la retta .db. la quale dico<lb></lb> |
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6405 che la divide la detta figura in doi parti iguali commo era bisogno. E questo é<lb></lb> |
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6406 <lb></lb> |
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6407 <pb></pb></p> |
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6408 <p class="folio"> |
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6409 folio 43v |
|
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6410 </p> |
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6411 <p class="main"> |
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6412 <lb></lb> |
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6413 </p> |
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6414 <p class="runhead"> |
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6415 Distinctio sexta. Capitulum primum. |
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6416 </p> |
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6417 <p class="main"> |
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6418 <lb></lb> |
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6419 bene da considerare imperoché bisogna che sia settore di cerchio. Potrei sopra questa<lb></lb> |
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6420 parte assai cose dire, ma le ditte bastino. Onde adonca faremo fine generalmente a tuta la<lb></lb> |
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6421 distintione. E peró diremo deo gratias.<lb></lb> |
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6422 <lb></lb> |
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6423 Distinctio sexta. De his que ad corporum dimensiones spectante ex .11o. Euclidis.<lb></lb> |
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6424 Capitulum primum.<lb></lb> |
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6425 <lb></lb> |
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6426 La presente distintione non voglio dividerla in alcune parti, conciosiaché com<lb></lb> |
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6427 brevitá sia de bisogno dire. E peró adonca starai atento a quello che sia necessa-<lb></lb> |
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6428 rio. Dico adonca le genere de’ corpi essere molti. Corpo è quello che ha longhez-<lb></lb> |
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6429 za, altezza e larghezza. Sonno adonca li corpi questi. E solidi. Serratili. Pi-<lb></lb> |
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6430 ramide. Colonne. Spere. E loro parti. E ancora corpi di molte base, commo de sotto de’ cor-<lb></lb> |
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6431 pi regulari intenderai. El solido è quello che ha longhezza larghezza e altezza e fasse di .6. su-<lb></lb> |
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6432 perficie, commo sonno e dadi, cassi, le citerne e simili. El serratile è la mitá del solido, che è fa-<lb></lb> |
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6433 tto di .3. paralelli e doi triangoli, che sonno quando le superficie segha el solido di .6. base so-<lb></lb> |
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6434 pra il diametro suo risolvendolo in doi parti iguali. De quali ciascuna se dici serratil. La pir-<lb></lb> |
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6435 ramide è una figura di che basa sia d’ uno solo ponto dedutta. La colonna è una figura or-<lb></lb> |
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6436 togonalmente elevata sopra la basa circulare e terminata nel cerchio, che è iguale ala sua ba-<lb></lb> |
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6437 sa. Overo è quella che á per basa uno paralello e simile per capo. La piramide colonnale è<lb></lb> |
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6438 quella che ’l triangolo rettangolo segue il lato che tiene il retto angolo a quel ponto mena-<lb></lb> |
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6439 ta. La spera è uno corpo ritondo che vulgarmete noi la diciamo palla et é quella che fa il<lb></lb> |
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6440 mezzo cerchio intorno al diametro che sta fixo e non si muove e a quel ponto menato don-<lb></lb> |
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6441 de si mosse, dela quale il suo centro é il centro del mezzo cerchio, dal quale tutte le linee rette<lb></lb> |
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6442 che vanno ala superficie dela spera sonno iguali infra loro. Mezza spera é quella che la mi-<lb></lb> |
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6443 tá contiene. Dela quale la basa è uno cerchio, il magiore che nella spera si puó circinare. Del<lb></lb> |
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6444 quale il diametro è il diametro dela spera de’ quali i soi termini sonno ditti poli dela spera.<lb></lb> |
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6445 La portione dela spera è quella che piú overo meno dela mezza spera consiste. E solidi di<lb></lb> |
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6446 molte base sonno di molti modi, de’ quali sonno solidi di .8. base e .12. base e di .20. base eguali,<lb></lb> |
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6447 li quali Euclide insengna mettere nella spera nel .14o. libro. Sonno ancora molte altre infini-<lb></lb> |
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6448 te figure corporee dele quali la misura s’ á per quello che habiamo detto. E debbi notare che,<lb></lb> |
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6449 quando diciamo questo corpo è tanto quadro, commo a dire .20. braccia quadro, intendia-<lb></lb> |
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6450 mo che in quello entrarebbe uno cubo, che fosse per faccia uno braccio, .20. volte: cioé uno brac-<lb></lb> |
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6451 cio quadro corporale é uno braccio longo, uno braccio grosso, un braccio alto e á tutti gli an-<lb></lb> |
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6452 goli retti. E cosí intendi degli altri nomi de misure. E, accioché quello che dire debbio sia be-<lb></lb> |
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6453 ne inteso, alcune cose che son nel .11o. de Euclide narraremo, delle quali queste sonno. Cioé<lb></lb> |
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6454 una linea retta, per niuno modo, puó essere in piano e in alto. E, quando .2. linee in-<lb></lb> |
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6455 fra loro si segano, amendoi sonno in una superficie. E in ogni triangolo é in una superfi-<lb></lb> |
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|
6456 cie. E, quando .2. piane superficie infra loro si segano, quella settione è commune. E, quan-<lb></lb> |
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6457 do stará una linea retta sopra una commune settione di .3. linee che con ciascuna dele ret-<lb></lb> |
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6458 te faccia angolo, queste tre linee sonno in una superficie. Ogni superficie stante ortogo-<lb></lb> |
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6459 gonalmente sopra una superficie infra loro si segano e commune é la loro settione ala perpen-<lb></lb> |
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6460 diculare dela subjecta superficie. L’ angolo solido, se tre angoli contiene, li .2. insiemi presi<lb></lb> |
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|
6461 sonno magiori che l’ altro. Ogni angolo piano che è contenuto dal’ angolo solido sonno<lb></lb> |
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6462 minori di .4. angoli retti. Se fra iguali linee tre angoli piani sonno fatti, de’ quali .2. pre-<lb></lb> |
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6463 si sonno magiori del’ altro, si puó con quelli collocare per corde uno triangolo. El soli-<lb></lb> |
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6464 do de equedistanti lati, se la superficie lo sega per lo diametro, le .2. superficie opposite è de<lb></lb> |
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6465 necessitá segarse per lo mezzo. E solidi de lati equedistanti sopra una medesima basa so-<lb></lb> |
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6466 pra una linea di medesima altezza sonno iguali. E solidi de lati equedistanti sopra una<lb></lb> |
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6467 medesima basa e non sienno sopra una linea non sonno iguali. E solidi de equedistanti<lb></lb> |
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6468 lati sopra le base eguali e ortogonalmente levate con iguali linee sonno iguali. E solidi si-<lb></lb> |
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6469 mili si dicano quelli che hano gli angoli iguali e intorno agli angoli iguali e lati proportio-<lb></lb> |
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6470 nali. Tutte le colonne de quali sia una base overo sopra medesime base é la loro propor-<lb></lb> |
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6471 tione commo l’ axe al’ axe. E questo basti a notitia de’ corpi.<lb></lb> |
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6472 <lb></lb> |
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6473 </p> |
|
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6474 <p class="head"> |
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6475 Qualiter solida rectangula et cubi mensurentur. Capitulum secundum. |
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6476 </p> |
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6477 <p class="main"> |
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6478 <lb></lb> |
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6479 <pb></pb></p> |
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6480 <p class="folio"> |
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|
6481 folio 44r |
|
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6482 </p> |
|
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6483 <p class="main"> |
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|
6484 <lb></lb> |
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|
6485 </p> |
|
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|
6486 <p class="runhead"> |
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|
6487 Distinctio sexta. Capitulum secundum. 44 |
|
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6488 </p> |
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6489 <p class="main"> |
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|
6490 <lb></lb> |
|
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6491 Se uno solido voi mesurare, prima trova l’ area della superficie dela basa e quel-<lb></lb> |
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6492 la, per l’ altezza, multiplica e quella multiplicatione sia l’ area corporale di ditto so-<lb></lb> |
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6493 lido. Sonno i solidi nominati in diversi nomi commo chiare appare per lo<lb></lb> |
|
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6494 sequente dire. Ma in questo nome s’ inchiude che lo lato del’ altezza sia perpendi-<lb></lb> |
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6495 culare ala superficie dela basa]. |
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|
6496 </p> |
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6497 <p class="main"> |
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6498 E gli é un cubo o vogliamo dire dado, che per ogni faccia é .10. braccia. Adiman-<lb></lb> |
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6499 do quanto è l’ area corporale di ditto cubo. Prima trova l’ area dela basa cioé<lb></lb> |
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6500 l’ area superficiale, dove multiplicarai .10., ch’ é longo, per .10., ch’ é largo, fanno .100.<lb></lb> |
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|
6501 e .100. braccia quadre è l’ area e queste multiplica per l’ altezza, cioé per .10., fanno .1000. e .1000.<lb></lb> |
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6502 braccia corporali è la detta figura. E, se voi il diametro di detto cubo, el quale diciamo il cu-<lb></lb> |
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6503 bo .aeg. E questo s’ ará se li quadrati .ab. e .bc. insiemi agiongni e haremo .200. per lo quadra-<lb></lb> |
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6504 to dela linea .bd., al quadrato dela quale linea, agiongnendo il quadrato dela linea .dh., che è<lb></lb> |
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6505 .100., haremo .300. per lo quadrato del diametro .bh. Adonca .bh. è radici de .300. E debbi<lb></lb> |
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6506 sapere che la retta .bh. è diametro nella spera nella quale cade el cubo .aeg. E, se solamente<lb></lb> |
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6507 voi per la notitia del diametro .bh. havere e lati del cubo e l’ area sua; prima desidera d’ avere<lb></lb> |
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6508 il quadrato del diametro e di quello che fanno piglia il .1/3. e harai il quadrato delo lato del<lb></lb> |
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6509 cubo el quale tetragono, se per la radici sua il multiplicarai, certamente l’ area del detto cubo<lb></lb> |
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6510 haremo. |
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|
6511 </p> |
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6512 <p class="main"> |
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6513 E, se alcuno dirá io agionsi il quadrato del diametro con li lati soi e pervennene<lb></lb> |
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6514 .310. Adimandase quanto è il lato del cubo. Porrai il lato del cubo essere una<lb></lb> |
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6515 cosa, dove il quadrato suo è uno censo e .3. cotanti sonno .3.censi., ai quali agionto<lb></lb> |
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6516 una cosa, sonno tre censi e una cosa iguali a .310. Dove areca a uno censo e harai<lb></lb> |
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6517 uno censo e .1/3. cosa iguali a .103. Dimezza le cose, sonno .1/6. e in sé multiplica, sonno .1/36., agiongni<lb></lb> |
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|
6518 a .103 1/3., fanno .103 13/36., de’ quali togli la radici che è .10 1/6. De’ quali tra .1/6., rimane .10. per lo lato .ab.,<lb></lb> |
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|
6519 cioé per lo lato del cubo. |
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|
6520 </p> |
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|
6521 <p class="main"> |
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6522 E, se alcuno dicesse io ó agionto el quadrato del diametro del cubo col quadra-<lb></lb> |
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|
6523 to del lato di detto cubo e feciono .400. Adimando quanto è il lato del cubo.<lb></lb> |
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6524 Poni il lato del cubo una cosa che, in sé multiplicata, fa un censo, la quale agion-<lb></lb> |
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6525 gni a .3. cotanti, cioé col quadrato del diametro .bh., fanno .4.censi. che sonno igua-<lb></lb> |
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|
6526 li a .400., dividi adonca .400. per quatro e haremo .100. per lo quadrato delo lato del cubo<lb></lb> |
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6527 ch’ é .ab. onde .ab. é .10. |
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6528 </p> |
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|
6529 <p class="main"> |
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6530 Ancora e gli é un solido del quale la basa sia el tetragono havente in ciascuno la-<lb></lb> |
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|
6531 to .10. E la longhezza, o vogliamo dire l’altezza sua, è .12. Adimando quanto cor-<lb></lb> |
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|
6532 porali è la detta figura. Prima trova l’ area dela basa, ch’ é .100. e questo multi-<lb></lb> |
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|
6533 plica per .12., fanno .1200., cioé multiplica per l’ altezza e .1200. corporali é la detta<lb></lb> |
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|
6534 figura di detto solido. E, se ’l suo diametro voi, multiplica .10. in sé e radoppia, fanno .200.<lb></lb> |
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6535 per lo quadrato dela linea .bd. e questo agiongni al quadrato dela linea .ie., cioé .a.144., fan-<lb></lb> |
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6536 no .344. per lo quadrato del diametro .be. |
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6537 </p> |
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6538 <p class="main"> |
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6539 E, se si propone el diametro del solido de equedistami lati é radici di .344. e de-<lb></lb> |
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6540 la basa tetragona e la sua altezza avanca el lato dela basa .2. e adimandasi quan-<lb></lb> |
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6541 to è il lato suo dela basa e la sua altezza. Multiplicarai una cosa che porrai<lb></lb> |
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6542 sia il lato dela basa e radoppia, haremo .2. censi. E dipoi multiplica l’ altezza in sé, cioé una co-<lb></lb> |
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6543 sa e .2., fanno .1. censo e .4.cose. e .4. per numero. Agiongni a’ .2.censi., haremo .3.censi.4.cose. e<lb></lb> |
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6544 .4. E questo è iguali a .344., dove tragase da ogni parte .4., haremo .3.censi.4.cose. iguali a .340.,<lb></lb> |
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6545 dove secondo l’ algebra opera e harai la cosa valere .10. e cosí l’ altezza troverai essere .12. e co-<lb></lb> |
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6546 sí farai le simiglianti. |
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6547 </p> |
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6548 <p class="main"> |
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6549 Ancora e gli é un solido che la basa è tetragona e la sua altezza è .2. piú che lo lato<lb></lb> |
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6550 dela basa e agiongnendo insiemi il quadrato dela faccia dela basa e il quadrato del dia-<lb></lb> |
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6551 metro dela basa e il quadrato del diametro del solido fanno .688. E adimandasi quan-<lb></lb> |
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6552 to è il lato dela basa e quanto la sua altezza. Perché il quadrato del diametro .eb. fa dello<lb></lb> |
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6553 agiognimento de’ quadrati dele linee .ab. e .bg. e .ie., se dimezzaremo adonca .688., che è .344., hare-<lb></lb> |
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6554 mo el quadrato dela linea .be. Onde operarai commo nella passata. |
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6555 </p> |
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6556 <p class="main"> |
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6557 Ancora agionsi lo lato dela basa collo lato del’ altezza e col quadrato del diametro<lb></lb> |
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6558 suo e feciono .366. Porró adonca el lato dela basa una cosa, dove l’agregatione<lb></lb> |
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6559 preditta è .3.censi. e .2.cose. e .2., dove operarai secondo l’ algebra e harai lo lato dela<lb></lb> |
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6560 basa essere .10. e l’ altezza .12. Intendi che si proponga che l’ altezza sia .2. piú che ’l lato dela basa.<lb></lb> |
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6561 <lb></lb> |
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6562 <pb></pb></p> |
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6563 <p class="folio"> |
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|
6564 folio 44v |
|
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6565 </p> |
|
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6566 <p class="main"> |
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6567 <lb></lb> |
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|
6568 </p> |
|
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|
6569 <p class="runhead"> |
|
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|
6570 Distinctio sexta. Capitulum secundum. |
|
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|
6571 </p> |
|
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|
6572 <p class="main"> |
|
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|
6573 <lb></lb> |
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|
6574 </p> |
|
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|
6575 <p class="head"> |
|
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|
6576 Possonsi in questi modi proporre molte questioni le quali per questi modi s’ asolvono. |
|
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|
6577 </p> |
|
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|
6578 <p class="main"> |
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|
6579 <lb></lb> |
|
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6580 Ancora sia uno solido de equedistanti lati .aeg., del quale l’altezza .hd. è .12. e lla lon-<lb></lb> |
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6581 ghezza dela basa, cioé .ad. sia .11. e la larghezza dela detta basa sia .10. e sia ortogo-<lb></lb> |
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|
6582 nalmnente eretto. Onde volendo l’ area corporale de ditta figura multiplicarai la lon-<lb></lb> |
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6583 ghezza dela basa per la larghezza e haremo .110. e questo multiplica per la sua altezza, cioé<lb></lb> |
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|
6584 per .12., fanno .1320. per l’ area corporale de ditto solido. E, volendo il suo diametro, cioé .hb., multiplica .10.<lb></lb> |
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6585 in sé e .11. in sé e .12. in sé e agiongni insiemi e haremo .365. per lo quadrato del diametro del ditto soli-<lb></lb> |
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6586 do, cioé per lo quadrato dela linea .bh. Adonca .bh. sia la radici de .365. |
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6587 </p> |
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6588 <p class="main"> |
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6589 E per simil modo diremo adonca: e gli é un solido de equedistanti lati che l’uno lato dela<lb></lb> |
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6590 basa avanza al’altro uno e l’altezza del detto solido è uno piú che ’l magiore lato, cioé<lb></lb> |
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6591 .2. piú che ’l minore lato dela basa e il suo diametro è radici di .365. Adimandase et cetera.<lb></lb> |
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6592 Porrai per lo minore lato una cosa, per lo magiore .1a. cosa e .1o., per l’ altezza .1a. cosa e .2. Dove li<lb></lb> |
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6593 quadrati, agionti insiemi, sonno .3.censi.6.cose.5. e questo è iguali a .365. Dove opera secondo l’ algebra, ha-<lb></lb> |
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|
6594 rai lo lato minore essere .10. e il magiore .11. e l’ altezza sua è .12., commo volevi. Similmente, se ’l<lb></lb> |
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6595 si dicesse io ó agionto li quadrati del .ab. e del .ad. e del .dh. col quadrato del .bh. e fo tutto quello<lb></lb> |
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6596 .730. Adimandase ciascuno lato. Dimezzaró .730., del quale la mitá è .365., per lo quadrato .hb., dove<lb></lb> |
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6597 opera commo di sopra e virrá. E, se sia detto io agionsi l’ altezza .dh. col quadrato del diametro .hb. e fe-<lb></lb> |
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6598 ciono .377. e .fo.dh. 1o. piú che .da. e .da. 1o. piú che .ab., porró similmente .ba. 1a. cosa, onde .ad. sia<lb></lb> |
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6599 una cosa e .1. e .hd. sia .1a. cosa e .2. e gli quadrati dele .3. linee sienno .3.censi.6.cose.5., che sonno iguali al<lb></lb> |
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6600 quadrato del diametro .hb. al quale agionto l’ altezza .dh., che è radici e .2., cioé una cosa e .2., fanno .3.<lb></lb> |
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6601 censi .7.cose. e .7. iguali a .377. Dove opera secondo la regola e harai .dh.12. e .da.11. e .ab.10., com-<lb></lb> |
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6602 mo volavamo. E questo de’ solidi basti e diremo de’ serratili.<lb></lb> |
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6603 <lb></lb> |
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6604 </p> |
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6605 <p class="head"> |
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6606 De corpore seratili eiusque dimensione necnon columnan cuiuscunque generis earunque pirramidum. |
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6607 </p> |
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6608 <p class="main"> |
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6609 <lb></lb> |
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|
6610 </p> |
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|
6611 <p class="head"> |
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6612 Capitulum tertium. |
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6613 </p> |
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6614 <p class="main"> |
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6615 <lb></lb> |
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6616 Comme ó ditto el serratile è quello che è segato del solido quando el filo segnato<lb></lb> |
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6617 va su per lo diametro. E peró Euclide dici cosí. Corpo serratile è quello che á .5. superficie dele qua-<lb></lb> |
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6618 li .3. sonno paralelle e le .2. triangulari. E sappi che li triangoli sonno iguali. E l’ uno è ba-<lb></lb> |
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6619 sa, l’ altro è capo. E, a volere l’ area corporale d’ uno serratile, prima trova l’ area superficiale dela basa<lb></lb> |
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6620 triangulare. E quella area superficiale multiplica per l’ altezza del ditto seratile e harai l’ area corporale de dit-<lb></lb> |
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6621 to seratile. Commo sia uno seratile .abgdez. del qual li triangoli .abg. e .dez. sonno amendoi equila-<lb></lb> |
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6622 teri o equicurii over diversilateri. E, di che forma sia, prima trova l’ area del triangolo .abg., com-<lb></lb> |
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6623 mo nel trattato de’ triangoli dicemmo. E diciamo el lato .ab. sia .13. e .bg.14. e .ag.15. Onde sia an-<lb></lb> |
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6624 cora .de.13. e .ez.14. e .dz.15. Meneró nel triangolo .abg. el catetto .al. che sia .12. el quale multiplicaró<lb></lb> |
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6625 per la mitá del .bg. e haremo .84. per l’ area del triangolo .abg. che, multiplicato per l’ altezza .ad. che sia .20.,<lb></lb> |
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6626 virranne .1680. per l’ area corporale del seratile .abgdez. E, se la retta .da. non stesse ortogonalmente<lb></lb> |
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6627 sopra il triangolo .abg., ma inchinasse da alcuna parte, alora studia di trovare la perpendicu-<lb></lb> |
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6628 lare del ditto seratile e opera commo di sopra. E, se la basa fosse paralella, commo fosse .eb. e .gz., a-<lb></lb> |
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6629 lora multiplica .eb. per .gz. fanno .280. E questo multiplica per la mitá del catetto .al., cioé per .6., fanno .1680. per l’ a-<lb></lb> |
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6630 rea corporale preditta. E, havendo a trovare l’ area corporale d’ alcun corpo composto del soli-<lb></lb> |
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6631 do e del seratile, prima trova l’ area del solido, intendi corporale, e poi quella del seratile e in uno agiongni e<lb></lb> |
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6632 harai quello voi e di queste figure sonno l’ arche da tenere grano.<lb></lb> |
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6633 E, se voi misurare una colonna dela quale il diametro dela basa sia .7. e l’ altez-<lb></lb> |
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6634 za, cioé l’ asse sia .20., prima trova l’ area superficiale dela basa dicendo: e gli é<lb></lb> |
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6635 un tondo che ’l diametro è .7., quanto é quadro. Dove li modi oprati tenendo harai essere quadro<lb></lb> |
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6636 .38 1/2., li quali per la sua alteza multiplica, cioé per .20., fanno .770. e .770. corporale è la detta colonna.<lb></lb> |
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6637 Ed é da saper che tutti i corpi, in che modo sienno, quando infra loro sienno simili, sia la propor-<lb></lb> |
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6638 tione del’ area corporale del’ uno al’ area corporale del’ altro, commo la proportione de’ lati tripli-<lb></lb> |
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6639 cata, cioé cubicata. Verbi gratia: sia d’ alcun solido la longhezza .2. e del’ altro .3., sia la pro-<lb></lb> |
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6640 portione del minore solido al magiore commo la proportione del .2. al .3. triplicata, cioé<lb></lb> |
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6641 commo il cubo fatto da .2. al cubo fatto da .3., cioé commo .8. a .27. é certamente .8. a .27. infra .3. quantitá<lb></lb> |
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6642 formata, cioé cosí .8. a .12. e cosí .12. a .18. e cosí .18. a .27. E peró la proportione del .8. al .27. è tripli-<lb></lb> |
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6643 cata la proportione del .2. al .3., onde, sapendo del corpo del minore, la quadratura multiplicarai per .27.<lb></lb> |
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6644 e dividerai la summa per .8. e harai l’ area corporale del minore corpo.<lb></lb> |
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6645 Quando alcuna piramide voi misurare, l’ area dela sua basa di che forma sia, per lo<lb></lb> |
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6646 terzo del’ altezza multiplica. E quello che ne pervirrá sirá l’ area dela detta piramide: la quale<lb></lb> |
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6647 regola perviene di quello che è ditto in Euclide, cioé che ’l solido è doppio al seratile e<lb></lb> |
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6648 <lb></lb> |
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6649 <pb></pb></p> |
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6650 <p class="folio"> |
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6651 folio 45r |
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6652 </p> |
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6653 <p class="main"> |
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6654 <lb></lb> |
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6655 </p> |
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6656 <p class="runhead"> |
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6657 Distinctio sexta. Capitulum tertium. 44 |
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6658 </p> |
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6659 <p class="main"> |
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6660 <lb></lb> |
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6661 il seratile è .3. cotanti che ’l suo piramide havente la basa triangulare. E a questo indurremo<lb></lb> |
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6662 la piramide .abgd. havente e lati iguali e la basa triangulare, che è .abg., e la sua altezza quel-<lb></lb> |
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6663 la che cade ortogonalmente dal ponto .d., ortogonalmente sopra le base .abg., che è de bisogno<lb></lb> |
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6664 trovare il cadimento suo nel triangolo .abg. Cosí si fa: intorno al triangolo .abg. scrivi il<lb></lb> |
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6665 cerchio .abc., intorno al centro .e. Dico certamente che ’l ponto .e. et il cadimento dela perpendi-<lb></lb> |
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6666 culare cadente dal ponto .d. sopra la superficie del triangolo .abg. El quale, se non fosse, sia<lb></lb> |
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6667 adonca .z. e faciase .dz.za.zb.zg. E perché .dz. è perpendiculare sopra il piano .abg. e sia ret-<lb></lb> |
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6668 to ciascun degli angoli .dza. e .dzb. e .dzg.; li lati adonca .dz. e .za possono sopra la retta .da.<lb></lb> |
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6669 Similmente li quadrati dele rette .dz. e .zb. sonno iguali al quadrato dela linea .db. e il qua-<lb></lb> |
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6670 drato dele rette .dz. e .zg. sonno iguali al quadrato dela linea .dg., ma li quadrati .da.db.dg.<lb></lb> |
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6671 sonno iguali infra loro, perché le ditte linee sonno iguali. Adonca li .2. quadrati dele linee .dz.<lb></lb> |
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6672 e .za. agli .2. quadrati dele linee .dz. e .zb. sonno iguali. Dove, de ciascuna parte togliendo il<lb></lb> |
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6673 quadrato dela linea .dz., rimarranno li quadrati dele linee .za. e .zb. iguali. Onde .za. è igua-<lb></lb> |
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6674 le al .zb. Similmente si mostrará la retta .zg. essere a ciascuna dele rette .za. e .zb. Adonca dal<lb></lb> |
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6675 ponto .z. ala periferia del cerchio concorrono .3. rette infra loro iguali. Onde il ponto .z. è cen-<lb></lb> |
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6676 tro di cerchio .abg., per la .9a. del terzo de Euclide, che non è in niuno modo vero. Adonca<lb></lb> |
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6677 il cadimeto è in sul centro .e., centro del cerchio .abg. Meneró adonca el catetto .de., del qua-<lb></lb> |
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6678 le la terza parte multiplicaró per l’ area del triangolo .abg. e haremo la capacitá dela pira-<lb></lb> |
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6679 mide .abg. che con numeri la mostraremo. Sia ciascuno de’ lati dela piramide .12. e compi-<lb></lb> |
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6680 se la retta .ae. e .be. nel ponto .i. e .t. Sirá il ponto .i. sopra il lato .bg. aponto nel mezzo di quel-<lb></lb> |
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6681 lo. Imperoché gli angoli .bai. e .gai. sonno infra loro iguali. E, perché .ab. e .ag. sonno infra<lb></lb> |
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6682 loro iguali, segherá ancora .bt. il terzo del catetto .ai., commo monstrammo nel trattato de’ tri-<lb></lb> |
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6683 angoli. Tratto adonca el quadrato del .bi. del quadrato .ab., cioé .36. di .144., rimarranno .108.<lb></lb> |
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6684 per lo quadrato dela linea .ai. E, perché .ei. è il terzo del .ai., sirá el quadrato dela linea .ei. el<lb></lb> |
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6685 nono del quadrato .ai. Adonca el quadrato dela linea .ei. é .12., a’ quali agionto il quadrato<lb></lb> |
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6686 dela linea .ib. fienno .48. per lo quadrato dela linea .eb. Overamente, perché .ae. è gli .2/3.<lb></lb> |
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6687 del .ai., sirá al quadrato dela linea .ae. li .4/9. del quadrato dela linea .ai., cioé li .4/9. di .108., el quale<lb></lb> |
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6688 tratto del quadrato dela linea .db., rimarrá il catetto .de. radici di .96. e, multiplicato .ai. in .bi.,<lb></lb> |
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6689 fanno radici di .3888. per l’ area del ditto triangolo .abg. che, multiplicato nel terzo del .de.,<lb></lb> |
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6690 cioé nel nono del quadrato di .96., cioé in .10 2/3., fanno .41472. per lo quadrato del’ area dela dit-<lb></lb> |
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6691 ta piramide .abgd., del quale la radici è circa .203 3/5., over un poco piú, el quale è meno d’ un .1/20.<lb></lb> |
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6692 et piú di .1/21. E per questo è manifesto che, in ogni piramide equilatera, el quadrato dela sua<lb></lb> |
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6693 altezza é gli .2/3. del quadrato d’ un suo lato, commo Euclide mostra. E ancora el quadrato dela<lb></lb> |
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6694 linea retta, che procede da ciascuno angolo al ponto del cadimento dela sua altezza, é il ter-<lb></lb> |
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6695 zo del quadrato del suo lato, che in questo modo lo possiamo investigare. Il .bi. è la mitá<lb></lb> |
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6696 del .bg. Onde il quadrato .bi. è il quarto del quadrato del .ab. Onde il quadrato del .ai.<lb></lb> |
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6697 è gli .3/4. del quadrato, che fatto dala retta .ab., è certamente .ae. gli .2/3. del .ai., onde il quadrato del<lb></lb> |
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6698 .ae. è gli .4/9. del quadrato dela retta .ai. Adonca il quadrato del .ae., cioé del .be. over del .ge. sia li<lb></lb> |
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6699 .4/9. de .3/4. del quadrato del lato .ab. Ma gli .4/9. de’ .3/4. d’ alcuna cosa é quanto e .3/4. di .4/9. de ditta cosa,<lb></lb> |
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6700 cioé .1/3. Adonca il quadrato di ciascuna dele rette .ac.be.ge. è .1/3. del quadrato del lato .ba., cioé<lb></lb> |
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6701 del lato .da. Onde, se dal quadrato del lato .da. si tolga il quadrato .ae., rimarranno per lo<lb></lb> |
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6702 quadrato del lato .de. li .2/3. del quadrato delo lato .da. commo dissi. E ancora è da notare che,<lb></lb> |
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6703 in ogni piramide dela quale la basa è triangulare havente e lati che si partano dagli ango-<lb></lb> |
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6704 li dela basa infra loro iguali, il cadimento di ditta altezza é sopra il centro del cerchio conte-<lb></lb> |
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6705 nente la basa de ditta piramide. |
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6706 </p> |
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6707 <p class="main"> |
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6708 Ancora e gli é una piramide .abcd. havente li lati .da.db.dc. infra loro iguali. E<lb></lb> |
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6709 la sua basa cioé el triangolo imperoché la pongo triangulare. E sia .ab.13.ac.15.<lb></lb> |
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6710 .bc.14., dove il catetto è .12., cioé .ae., per lo quale diviso la superficie del .ba. in .ac.,<lb></lb> |
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6711 cioé .195. vienne .16 1/4. per lo diametro del cerchio contenente il diametro .ac. del qua-<lb></lb> |
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6712 le la mitá è .8 1/8. e questa è la differentia che è dal cadimento del’ altezza de ditta piramide a ci-<lb></lb> |
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6713 ascuno degli angoli dela ditta basa. E diremo ciascuno de’ lati .da.db.dc. essere .12. dove, se del<lb></lb> |
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6714 quadrato di .12. che è .144. si togli el quadrato del .8 1/8., che è .66 1/64., rimarranno .78. meno .1/64. per<lb></lb> |
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6715 lo quadrato del catetto .dg. del quale, se pigliaremo il .1/9. e multiplicarenlo contra al quadra-<lb></lb> |
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6716 to del’ area dela basa, cioé contra .7056., se hará il quadrato dela piramide.<lb></lb> |
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6717 <lb></lb> |
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6718 <pb></pb></p> |
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6719 <p class="folio"> |
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6720 folio 45v |
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6721 </p> |
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6722 <p class="main"> |
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6723 <lb></lb> |
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6724 </p> |
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6725 <p class="runhead"> |
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6726 Distinctio sexta. Capitulum tertium. |
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6727 </p> |
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6728 <p class="main"> |
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6729 <lb></lb> |
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6730 E debi notare che, se la basa de simile piramide sirá d’ angoli acuti, alora il cadimen-<lb></lb> |
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6731 to del’ altitudine dela ditta basa cadrá dentro al triangolo .abc.; se l’ angolo .abc.<lb></lb> |
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6732 sia retto, alora il cadimento dela ditta piramide sia in sul mezzo diametro .bc. E,<lb></lb> |
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6733 se l’ angolo .abc. sirá obtuso, alora il cadimento de ditta piramide sia fuori del tri-<lb></lb> |
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6734 angolo dala parte .bc. Le qual cose asa’ te fien piú chiare di sotto in questo: nel particular tratta-<lb></lb> |
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6735 to de’ corpi regulari, dove apieno ancora de ditte piramidi se trattará a inteligentia deli .5.<lb></lb> |
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6736 corpi regulari, cioé tetracedron, che è de .4. base triangulari equilatere e equiangole e con-<lb></lb> |
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6737 sta de .4. piramidelle simili al tutto che lo compongano. E delo exacedron, che consta de .6. qua-<lb></lb> |
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6738 drilatere e octocedron, che ne contien .8. triangole e duodecedron, che ne riceve .12., ciascuna<lb></lb> |
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6739 pentagonale e lo icocedron, che è formato de .20. piramidelle, che ognuna è triangula, in le qua-<lb></lb> |
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6740 li piramide tutti li detti corpi sempre se hano a resolvere tanquam in sua componentia, per-<lb></lb> |
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6741 ché, sí commo de triangoli rettilinei tutte le superficie rectelinee piane si compongano, cosí<lb></lb> |
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6742 tutte le specie di corpi rectilinei se hano a componere de piramidi triangulari, avenga che non<lb></lb> |
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6743 sempre abino a essere equilatere ditte piramidi. E cosí poi le piramide, maxime quelle che ha-<lb></lb> |
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6744 veranno la basa trilatera, si pó resolvere in doi equali piramide fra loro e tutta la grande si-<lb></lb> |
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6745 mili. E ognuna de quelle in doi seractili li quali insiemi presi de necessitá siranno magiori che<lb></lb> |
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6746 mitá de ditta piramide, commo nella terza del .12o. libro dimostra Euclide, siché a lui sem-<lb></lb> |
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6747 pre recorri in ditto .12o., che di loro difusamente ne parla et cetera.<lb></lb> |
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6748 Ancora porró una piramide dela quale la basa .abc. sia triangolo equilatero over<lb></lb> |
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6749 equicurio e de’ lati adonca .da.db.dg. doi solamente sonno iguali, che sienno<lb></lb> |
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6750 .db. e .dg. Voglio dimostrare il modo in che modo se habia l’ altezza di quella pi-<lb></lb> |
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6751 ramide. Perché nel triangolo .dbg. ‘2. lati .db. e .dg. sonno iguali, se dal ponto<lb></lb> |
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6752 .a. si menerá la retta da ogni parte in infinito ne’ ponti .f. e .c., dapoi sopra la retta .fc. produr-<lb></lb> |
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6753 remo el catetto dal ponto .d., che è il catetto del’ altezza dela piramide .dabg., dela quale al-<lb></lb> |
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6754 tezza il cadimento alcuna volta cade infra ‘l .ae., alcuna volta fuori del triangolo infra gli pon-<lb></lb> |
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6755 ti .af. over .ec. over alcuna volta nel ponto .a. over nel ponto .e., che ogni cosa mostraremo<lb></lb> |
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6756 con numeri. Sia adonca ciascuno de’ lati .ab.ag.10. e il lato .bg. sia .12. e ciascuno de’ lati .db.<lb></lb> |
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6757 e .dg. sia .14. e il lato .da.12. Troveró prima il catetto .de., che è radici di .160., cioé trahendo el qua-<lb></lb> |
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6758 drato del cadimento .be. sie tratto del quadrato del lato .db. Similmente il catetto .ae. sia .8. |
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6759 </p> |
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6760 <p class="main"> |
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6761 Dapoi troverró el cadimento d’ altri angoli .dae. sopra la basa .ae., cosí del quadrato che è<lb></lb> |
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6762 fatto del .de. togli el quadrato del .da., cioé .144. di .160., rimangono .16., li quali divisi per .at.<lb></lb> |
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6763 vienne .2., che agionti con .ae. fanno .10., de’ quali la mita, cioé .5., è il magiore cadimento dal<lb></lb> |
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6764 lato del magiore .de. Onde porró .h. sopra il ponto .e., che sia .eh.5., rimarrá .ah.3. Adonca el<lb></lb> |
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6765 quadrato dela linea .ah. si togli del quadrato dela linea .da. over del quadrato dela linea<lb></lb> |
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6766 .de. si traga el quadrato dela linea .eh.; sirá quel che rimarrá .135. per lo quadrato del catet-<lb></lb> |
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6767 to .dh. Ed, é da notare che, se l’ angolo .dah. fosse retto, alora .da. sarebbe il catetto discenden-<lb></lb> |
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6768 te dal ponto .d. E, se l’ angolo .dah. fosse magiore che ’l retto, alora el catetto discendente dal pon-<lb></lb> |
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6769 to .d. caderebbe fuori del triangolo sopra la linea .af. Similmente, se l’ angolo .deh. fosse retto, a-<lb></lb> |
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6770 lora .de. sirebbe l’ altezza dela piramide. E, se obtuso, alora il catetto descendente dal ponto<lb></lb> |
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6771 .d. cadrebbe infra li ponti .ec. fuori del triangolo. E, se ciascuno degli angoli .dah. e .deh. fos-<lb></lb> |
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6772 se, acuto, alora la perpendiculare infra ‘ponti .ae. cadrebbe. |
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6773 </p> |
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6774 <p class="main"> |
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6775 Sia ancora un’ altra volta la piramide dela quale la summitá è .a. e la sua basa sia<lb></lb> |
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6776 il triangolo .bcd. diversilatero. Dela quale la basa sia .bc., cioé el lato .bc. sia .13. e<lb></lb> |
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6777 .cd. sia .14. e .db.15., del quale il catetto sia la retta .be.; degli altri lati descendenti<lb></lb> |
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6778 dal ponto .a. al ponto .bcd. sienno e lati .ac. e .ad. iguali e sia ciascuno .10. e l’ altro<lb></lb> |
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6779 lato .ab. sia .15. Voglio adonca il catetto cadente dal ponto .a. sopra il piano nel quale è il trian-<lb></lb> |
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6780 golo .bcd. Meneró la linea .fg. causante retto angolo .gfd. e .gfe., comporró la retta .ag. e nel<lb></lb> |
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6781 triangolo .afg., sopra la linea .fg., il catetto produrró che sia il catetto cadente dal ponto .a.<lb></lb> |
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6782 sopra il lato del triangolo .bcd. el quale catetto per numero cosí si trova per noticia deli an-<lb></lb> |
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6783 goli, commo nel secondo Euclide mostra. Perché retto è l’ angolo .gfd. e peró sia la linea .fg. equi-<lb></lb> |
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6784 distante al catetto .be. Onde e gli é cosí .df. al .de. cosí .dg. al .db. certamete e gli é .de.9. e<lb></lb> |
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6785 .df.7., cioé la mitá del .cd., imperoché equicurio è il triangolo .acd. Adonca è cosí .7. al .9.<lb></lb> |
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6786 cosí .dg. non saputo al .db. saputo che è .15. Onde, multiplicando .7. per .15. e diviso per .9., vienne<lb></lb> |
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6787 .11 2/3. per la linea .dg. non saputa. Similmente, se multiplicaremo .df. per .be., cioé .7. per .12. e dividere-<lb></lb> |
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6788 mo per .de., virranne .9 1/3. per la linea .fg. Dapoi nel triangolo .abd., sopra il lato .bd. meneremo<lb></lb> |
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6789 <lb></lb> |
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6790 <pb></pb></p> |
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6791 <p class="folio"> |
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6792 folio 46r |
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6793 </p> |
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6794 <p class="main"> |
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6795 <lb></lb> |
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6796 </p> |
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6797 <p class="runhead"> |
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6798 Distinctio sexta. Capitulum tertium. 46 |
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6799 </p> |
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6800 <p class="main"> |
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6801 <lb></lb> |
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6802 il catetto .ah. e sia il cadimento piú longo .bh.11 2/3. e il minore sirá .3 1/3. Onde il quadrato del<lb></lb> |
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6803 catetto .ah. sirá .88 8/9., col quale agiongnendo el quadrato dela linea .hg., che è .69 4/9., faranno<lb></lb> |
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6804 .158 1/3. per lo quadrato dela linea .ag. Ancora, se del quadrato dela linea .ad. si togli il quadra-<lb></lb> |
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6805 to del .df., cioé .49. de .100., rimarranno .51. per lo quadrato dela linea .af. Adonca la linea<lb></lb> |
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6806 .af. è radici di .51. e la linea .ag. è radici de .158 1/3. e la linea .fg. è .9 1/3. E, poiché habiamo e lati del<lb></lb> |
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6807 triangolo .afg. noti, possiamo la noticia del catetto descendente dal .a. sopra la linea .fg. (per<lb></lb> |
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6808 la dottrina che insengnanmo nel trovare l’ area de’ triangoli) havere.<lb></lb> |
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6809 Ancora sia una piramide .abgd. dela quale la basa sia il triangolo .abg. diversi-<lb></lb> |
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6810 latero e gli lati discendenti dal ponto .d. ne’ ponti .a.b.g. sienno similmente di-<lb></lb> |
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6811 versi e non iguali, che alcuna volta ne sia uno di loro ortogonalmente ritto so-<lb></lb> |
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6812 pra il triangolo .abg. E alcuna volta fienno tutti e lati declinanti. Verbi gratia:<lb></lb> |
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6813 sia il lato .ab.10. e .ag.9. e .bg.5. e .da.15. e il lato .db. sia .13. e il .dg. sia .12. In questa piramide rit-<lb></lb> |
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6814 ta .dg. catetto imperoché li quadrati dele linee .bg. e .gd. sonno iguali al quadrato dela li-<lb></lb> |
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6815 nea .bd. e ancora li quadrati dele linee .dg. e .ga. sonno iguali al quadrato dela linea .ad.<lb></lb> |
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6816 Adonca, se ’l terzo del .gd. si meni per l’ embado, over area, del triangolo .abg., sará l’ embado o-<lb></lb> |
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6817 ver capacitá dela piramide .abgd. Ma non sia il catetto dela piramide alcuna dele linee<lb></lb> |
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6818 discendenti dale summitá sua, commo nela piramide .abcd. dela quale il lato .ab. sia .13., bc.<lb></lb> |
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6819 .14. e .ac.15. e il lato .bd. sia .15., .dc. sia .13., .da.14. Prima nelo triangolo .abc. meneró il catet-<lb></lb> |
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6820 to .ae. e nel triangolo .dbc. meneró il catetto .df. e per lo ponto .f. meneró lo catetto .fh. eque-<lb></lb> |
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6821 distante al catetto .ae. e sia l’ angolo .hfc. retto. Dapoi nel triangolo .dac. meneró il catetto<lb></lb> |
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6822 .dg. sopra la retta .ac. De’ quali tutti siranno manifesti: la linea .dh., ‘lati del triangolo .dfh. sia<lb></lb> |
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6823 noti. E peró il catetto cadente in quello dal ponto .d. sopra la linea .fh. sia nota che sirá l’altez-<lb></lb> |
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6824 za dela piramide. Verbi gratia: el catetto .ae. è .12., il cadimento .be. è .5. e il cadimento .ec.9. |
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6825 </p> |
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6826 <p class="main"> |
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6827 Similmente il catetto .df. nella superficie del triangolo .dbc. è .12. e il diametro .fc. è .5. E, per-<lb></lb> |
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6828 ché .fh. è equedistante al catetto .ae., sia cosí .cf. al .ce. cosí .fh. al .ae. e .ch. al .ca. Onde .fh. è .6 2/3.<lb></lb> |
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6829 e .ch. è .8 1/3. Dapoi, acioché troviamo el catetto .dg., trarró el quadrato .dc. del quadrato del<lb></lb> |
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6830 lato .da., cioé .169. di .196., rimane .27. e quali, divisi per .ac., vienne .1 4/5. che, agionti con .ac., fan-<lb></lb> |
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6831 no .16 4/5., de’ quali la mitá è .8 2/5., che è il cadimento del magiore .ag. Onde il caso minore .gc. è<lb></lb> |
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6832 .6 3/5. e per questo s’ ánno .11 1/5. per lo catetto .dg., del quale el quadrato, se s’ agiongni al quadra-<lb></lb> |
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6833 to del .gh., fienno .128 4/9. per lo quadrato del .dh. che è .11 1/3. Resta che nel triangolo .dfh. trovia-<lb></lb> |
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6834 mo el catetto cadente dal ponto .d. sopra la linea .fh. e sirá il caso magiore .fi. 4 1/2. Onde il ca-<lb></lb> |
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6835 tetto .di., che è l’ altezza dela piramide sia radici di .123 3/4., per la quale se multiplicaremo per lo<lb></lb> |
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6836 terzo del’ area del triangolo .abc., cioé .28., haremo radici di .97020. per l’ area corporale dela<lb></lb> |
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6837 piramide .abcd. E debbi notare, se l’ angolo .dcb. del triangolo .dbc. sirá obtuso, alora el catet-<lb></lb> |
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6838 to .df. caderá di fuori del triangolo .dbc., commo in questa altra figura si manifesta, nela qua-<lb></lb> |
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6839 le poniamo el lato .ac.13. e .bc.9. e il lato .ab. radici di .160 e il lato .da. sia .19. e il lato .db. sia .17.<lb></lb> |
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6840 e il lato .dc. sia .10. Onde il catetto .ae. sia .12. e il caso .be. sia .4. et .ec. sia .5. E dapoi trarremo<lb></lb> |
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6841 el quadrato del lato .dc. del quadrato del lato .db., rimarranno .189., lo quale dividerai per .bc., cioé per .9., vien-<lb></lb> |
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6842 ne .21. che, agionto a .9., fanno .30. del quale la mitá è .15. per lo cadimento .bf. Onde il ponto .f. cade di<lb></lb> |
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6843 fuora del .bc. e sia .cf.36. el quale quadrato, tratto del quadrato del lato .dc. rimane .64., cioé<lb></lb> |
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6844 tratto .36. di .100., rimane .64. per lo quadrato del catetto .df. Onde .df. è .8. Dapoi meneró<lb></lb> |
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6845 la linea .fh. equedistante al catetto .ae. e farollo concurrere con la linea .ba. nel ponto .h. e<lb></lb> |
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6846 comporró .dh. e fie il triangolo .hbf. ortogonio simile al triangolo .abc. Onde sia cosí .be.<lb></lb> |
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6847 al .ea. cosí .bf. al .fh. e .hf. è .45., del quale el quadrato è .2025. al quale, agionto el quadrato del-<lb></lb> |
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6848 la basa .bf., che è .225., vienne .2250. per lo quadrato dela linea .bh. Overo, perché e gli é cosí .be.<lb></lb> |
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6849 al .bf. cosí .ba. al .bh., se multiplicaremo el quadrato del .bf. per lo quadrato del .ba., cioé .225.<lb></lb> |
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6850 per .160. e divideremo la summa per lo quadrato del .be., che è .16., vienne similmente .2250.<lb></lb> |
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6851 per lo quadrato dela linea .bh. Dapoi, acioché veniamo ala noticia dela linea .hd., troveró<lb></lb> |
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6852 il catetto cadente dal ponto .d. in sul lato .ba., che cosí si fa. Del quadrato fatto del .da. si<lb></lb> |
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6853 traga el quadrato del .db., cioé .289. di .361., rimane .72. Dapoi, per lo lato .ba., che è radici di<lb></lb> |
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6854 .160., dividi. Onde il quadrato di .72., che è .5184., dividi per .160., vienne .192 2/5. del quale, se si<lb></lb> |
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6855 trae el doppio dela radici dela multiplicatione de .32 2/5. in .160., che è doppio .144., rimarran-<lb></lb> |
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6856 no .48 2/5. per lo quadrato del doppio minore cadimento che sia .bk. Onde il quadrato del<lb></lb> |
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6857 .bk. sia .12 1/10., cioé il .1/4. di .48 2/5., e quali .12 1/10., tratti del quadrato dela linea .db., rimarranno<lb></lb> |
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6858 .176 9/10. per lo quadrato del catetto .dk. Dapoi cosí, acioché habiamo noticia del .bk., agiongneró il<lb></lb> |
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6859 <lb></lb> |
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6860 <pb></pb></p> |
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6861 <p class="folio"> |
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6862 folio 46v |
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6863 </p> |
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6864 <p class="main"> |
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6865 <lb></lb> |
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6866 </p> |
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6867 <p class="runhead"> |
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6868 Distinctio sexta. Capitulum tertium. |
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6869 </p> |
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6870 <p class="main"> |
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6871 <lb></lb> |
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6872 quadrato .bk. col quadrato che è fatto dal .bh., cioé .12 1/10. con .2250., fanno .2262 1/10., del quale tor-<lb></lb> |
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6873 ró el doppio dele radici, ch’ é fatto del .12 1/10. in .2250., el quale doppio è .330., rimarranno .1932 1/10.<lb></lb> |
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6874 per lo quadrato dela linea .kh. al quale, agionto el quadrato del .dk., cioé .276 4/19., virranno<lb></lb> |
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6875 .2209. per lo quadrato dela linea .dh. Onde .dh. è .47. Dapoi, acioché troviamo il catettto del<lb></lb> |
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6876 triangolo .dfh. cadente dal ponto .d. sopra il lato .fh. che è .45. e il lato .df. è .8., operaremo al<lb></lb> |
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6877 modo detto e trovaremo quel catetto essere cadente fuori dela linea .hf., secondo la quantitá<lb></lb> |
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6878 del .1 1/3. el quale cadimento è .fl. E tragase el quadrato .fl. del quadrato .df., che è .64., rimarran-<lb></lb> |
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6879 no .62 2/9. per lo quadrato dela linea .dl., cioé il catetto dela piramide .abcd., che quel medesi-<lb></lb> |
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6880 mo troverai, se ’l quadrato delo lato .dh. si trae del quadrato del cadimento .hl. over se del<lb></lb> |
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6881 quadrato dela linea .d. ci si togli il quadrato dela linea .cl., é il quadrato dela linea .cl. igua-<lb></lb> |
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6882 le a’ .2. quadrati dele linee .cf. e .fl. Over altramente io meneró la linea .ae. ne’ ponti .m. e sia<lb></lb> |
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6883 .em. iguale al .fl. Onde tutta .am. sirá .13 1/3. e compileró .lm. che sia iguale del .ef., che è .11., e agion-<lb></lb> |
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6884 gneró li quadrati dele linee .am. e .ml. e haró il quadrato dela linea .al. el quale, se io lo tra-<lb></lb> |
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6885 go del quadrato dela linea .ca., cioé di .361., rimarrá similmente .62 2/9. per lo quadrato del ca-<lb></lb> |
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6886 tetto .dl. E per questo è manifesto che la retta .dl. è ortogonalmente sopra il piano .flh., con-<lb></lb> |
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6887 ciosiacosaché fanno retti angoli con le linee .lh.la.le. Dapoi, se ’l catetto .dl. multiplicaremo per lo .3o. del<lb></lb> |
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6888 triangolo .abc., che è .18., haremo .20160. per lo quadrato del’ area corporale de ditta piramide .abcd.<lb></lb> |
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6889 E cosí in tutte le piramide studia di fare e potresti ancora ala noticia di ciascuna dele ditte pi-<lb></lb> |
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6890 ramide procedere secondo gli strumenti sopra ditti, cioé canne over con filo con biombo e<lb></lb> |
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6891 verissimamente harai tutto. E, se la basa d’ alcuna piramide sirá quadrilatera over di molti<lb></lb> |
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6892 lati, sempre il terzo dela sua altezza nel’ area dela basa multiplica, imperoché chiaro appa-<lb></lb> |
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6893 re che la piramide triangulare è il terzo del serratile. E cosí ciascuna piramide è subtripla al<lb></lb> |
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6894 suo chelyndro over colonna, per la .9a. del .12o. de Euclide. Dove, quando la piramide á per<lb></lb> |
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6895 basa uno quadrilatero over multilatero, lo puoi risolvere in triangoli e harai la prova resol-<lb></lb> |
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6896 vendo la piramide in tante piramide quanti triangoli hai fatto dela basa e ciascuna pirami-<lb></lb> |
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6897 de hará per basa uno triangolo e l’ altezza di ciascuna sirá una medesima. E peró, multiplican-<lb></lb> |
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6898 do il terzo di ciascuna nel’ area dela basa, harai quello e tutta la piramide.<lb></lb> |
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6899 E, se d’ alcuna piramide si togli una piramide per la superficie equedistante ala sua<lb></lb> |
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6900 basa e vorrai sapere l’ area di quello rimane dela piramide dicurtata, che cosí si chia-<lb></lb> |
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6901 ma, del’ area corporale di tutta la piramide trai l’ area dela piramide picola e quel-<lb></lb> |
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6902 lo che rimane sirá l’ area dela piramide decurtata. Verbi gratia: dela piramide<lb></lb> |
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6903 .dabg., dela quale la basa è uno triangolo .abg. e sia equidistante ala basa .ezi. dela piramide .dezi.<lb></lb> |
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6904 dela quale é da sapere l’ area. Dico che del’ area dela piramide .abgd. si traga l’ area dela<lb></lb> |
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6905 piramide .dezi. e quello che rimane sia l’ area dela piramide decurtata .abgezi. Overo<lb></lb> |
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6906 altramente, perché li triangoli .czi. equedistante è al triangolo .abg. e sonno li lati del trian-<lb></lb> |
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6907 golo .ezi. seganti le superficie .dab. e .dbg. e .dag., fienno ancora e lati di quelli triangoli eque-<lb></lb> |
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6908 distanti: cioé il lato .ez. al lato .ab. e il lato .ei. al lato .ag. e il lato .zi. al lato .bg. Perché ne’ trian-<lb></lb> |
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6909 goli .dab. e .dag. e .dbg. menate sonno le rette .ez.ei. et .zi. equedistanti alle base .ab.ag.bg.,<lb></lb> |
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6910 sirá uno gli angoli di fuori iguali agli oposti e dentro. Certamente .dez. angolo al’ angolo .dab.<lb></lb> |
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6911 e ancora gli angoli .dze. e .dzi. e .diz. e .die. e .dei. agli angoli .dba.dbg.dgb.dga.dag.<lb></lb> |
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6912 sonno iguali. Sonno ancora gli angoli .zei. et .eiz. et .ize. iguali agli angoli .bag. et .agb. et .gba.,<lb></lb> |
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6913 perché la superficie .ezi. è equedistante ala superficie .abg. Onde gli angoli solidi che son-<lb></lb> |
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6914 no al .e. et al .z. e .i. dela piramide .dezi. sonno iguali agli angoli solidi che sonno al .a. e .b. e .g.<lb></lb> |
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6915 Onde è manifesto la piramide .dezi. essere simile a tutta la piramide .dabg. E le pirami-<lb></lb> |
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6916 de simili infra loro sonno nela proportione cubicata de’ lati simili. Onde la proportione<lb></lb> |
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6917 dela piramide .dabg. ala piramide .dezi. è proportione triplicata del lato .bg. al lato .zi.<lb></lb> |
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6918 Onde poniamo essere cosí .bg. al .zi., cosí la quantitá del .c. al .f. e cosí .c. al .f. cosí .f. al .x. cosí .x. al<lb></lb> |
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6919 .y. sia il .c. al .y. proportione triplicata del lato .bg. al lato .zi. Onde é cosí .c. al .y. cosí la pira-<lb></lb> |
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6920 mide .dagb. ala piramide .dezi. Tolgase adonca dela quantitá .c. la quantitá .y. e rimanga<lb></lb> |
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6921 la quantitá .h. Sirá ancora commo il .c. al .y. cosí la quantitá .hx. ala quantitá .y. Adonca é co-<lb></lb> |
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6922 sí la quantitá .hy. ala quantitá .y. cosí la piramide .dagb. ala piramide .dezi. Sia adonca dis-<lb></lb> |
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6923 iunctim cosí lo .h. al .y. cosí la corta piramide .abgezi. ala piramide .dezi. Onde, se gli lati de’ trian-<lb></lb> |
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6924 goli .abg. et .ezi. e l’ altezza dela piramide .dabg. fienno saputi, sirá ancora manifesto, per quello che s’ é<lb></lb> |
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6925 ditto, l’ area dela corta piramide .abgezi., che si mostrará con numeri. Sia il lato .ab.13.ag.15.<lb></lb> |
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6926 .bg.14. e l’ altezza dela piramide .dabg., che sia la retta .dk. e sia .24. E sopra la .1/2. de’ lati .da.db.dg.<lb></lb> |
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6927 <lb></lb> |
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6928 <pb></pb></p> |
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6929 <p class="folio"> |
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6930 folio 47r |
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6931 </p> |
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6932 <p class="main"> |
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6933 <lb></lb> |
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6934 </p> |
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6935 <p class="runhead"> |
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6936 Distinctio sexta. Capitulum tertium. 47 |
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6937 </p> |
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6938 <p class="main"> |
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6939 <lb></lb> |
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6940 passi la superficie del triangolo .ezi. e sia il lato .ez.6 1/2. e il lato .ei.7 1/2. e il lato .zi.7. e il catetto .dk.<lb></lb> |
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6941 ponemmo .24., che è segato dala superficie .ezi. in .2. mezzi nel ponto .t. E poniamo la quanti-<lb></lb> |
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6942 tá .c. essere .8. e, perché lo lato .bg. é doppio al lato .zi. sia la quantitá .c. doppia ala quantitá .f.<lb></lb> |
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6943 dove .f. sia .4. e l’ .x. sia .2. e l’ .y. sia uno e tolgase dela quantitá .c. iguali ala quantitá .y., cioé uno<lb></lb> |
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6944 del .8., rimarranno .7. per la quantitá .h. E, perché troviamo essere cosí .h. al .y. cosí la piramide<lb></lb> |
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6945 corta .abgezi. ala piramide .dezi., é certamente .h. sette cotanti del .y., onde la piramide<lb></lb> |
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6946 .abgezi. corta è .7. cotanti dela piramide .dezi. Ma la piramide .dezi. è .84., che pervengono<lb></lb> |
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6947 del multiplicamento dela terza parte del’ altezza dela ditta piramide cioé .dt., che è .4., nel<lb></lb> |
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6948 triangolo .ezi., che è .21. Onde, multiplicando .84. per .7., fanno .588. per l’ area dela piramide<lb></lb> |
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6949 corta .abgezi. Overo, se di tutta la piramide .dabg., che è .672., che vengono del multiplica-<lb></lb> |
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6950 mento dela terza parte del’ altezza del .dk., che è .8., nel triangolo .abg., che è .84. E trarremo<lb></lb> |
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6951 .84., cioé la piramide .dabg., che è .672., che pervengono commo ó ditto del .8. in .84., rimar-<lb></lb> |
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6952 ranno .588. per la piramide corta .abgezi., commo volavamo. |
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6953 </p> |
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6954 <p class="main"> |
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6955 Ancora sia una piramide .dabg. dela quale la summitá sia .d. e tolghise dala pi-<lb></lb> |
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6956 ramide .dabg. la piramide .dezi. Dico che l’ area dela corta piramide .abgezi.<lb></lb> |
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6957 perviene del dutto dela terza parte e l’ altezza sua, che è .ek., nela summa del’ area<lb></lb> |
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6958 dela basa e del capo di quella e dela superficie che è nela proportione media in-<lb></lb> |
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6959 fra la superficie dela basa e del capo, che cosí si prova. Sopra la retta .bg. ordineró la super-<lb></lb> |
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6960 ficie di retti angoli .bm. iguale ala superficie del triangolo .abg. e porró la linea .ng. iguale al<lb></lb> |
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6961 .zi. e aplicaró sopra la retta .ng. la superficie di retti angoli .nopg. iguale ala superficie del<lb></lb> |
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6962 triangolo .ezi. e meneró la retta .po. infino al .q. Dico prima la superficie .np. essere simile a-<lb></lb> |
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6963 la superficie .bm., che cosí lo proveró. Perché commo nela passata figura triangulare pira-<lb></lb> |
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6964 mide e triangoli .abg. et .ezi. sonno dimostrati essere equiangoli, siranno ancora per questo<lb></lb> |
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6965 li presenti infra loro simili. E gli triangoli simili sonno nela doppia proportione de’ lati simi-<lb></lb> |
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6966 li, commo per la .17a. e .18a. del sexto libro Euclide prova. E lati certamente .bg. et .zi. infra lo-<lb></lb> |
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6967 ro sonno simili. Onde la proportione del triangolo .abg. al triangolo .ezi. è duplicata del<lb></lb> |
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6968 lato .bg. al lato .zi. Sia adonca cosí .bg. al .zi. cosí .zi. al .u. Onde è cosí .bg. al .u. cosí il trian-<lb></lb> |
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6969 golo .abg. al triangolo .ezi., ma al triangolo .abg. è iguale la rettangula .bm. e al triangolo<lb></lb> |
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6970 .ezi. è iguali la superficie .up. Onde è cosí .bg. al .u. cosí la superficie .bm. ala superficie .np. et è<lb></lb> |
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6971 .ng. iguale ala retta .ci. Onde è cosí .bg. al .ng. cosí .ng. al .u. E, perché dela continua propor-<lb></lb> |
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6972 tione cosí è la prima ala seconda cosí la seconda ala terza cosí la figura che è ala prima ala fi-<lb></lb> |
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6973 gura che è ala seconda simelmente descritta. Onde è cosí .bg. al .u. cosí el quadrilatero descrit-<lb></lb> |
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6974 to .bm. al quadrilatero descritto dala linea .gn. simile al quadrilatero .bm. Onde se ’l quadrila-<lb></lb> |
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6975 tero .np. nonn’ é simile al quadrilatero .bm., discrivase sopra la linea .ng. e nel’ angolo .ngm.<lb></lb> |
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6976 un altro quadrilatero che sia magiore over minore .np. al quadrilatero .bm. che è descritto<lb></lb> |
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6977 dala linea .bg. harebbe la proportione quella medesima che .bg. al .u. Cosí è dimostrato el<lb></lb> |
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6978 quadrilatero .mb. al quadrilatero .np. Adonca al quadrilatero .bm. á doi diverse a quella<lb></lb> |
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6979 medesima proportione che è inconveniente. Similmente adonca el quadrilatero .np. al<lb></lb> |
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6980 quadrilatero .bm., commo ho ditto, le superficie simili intorno agli angoli hano e lati proportionali.<lb></lb> |
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6981 Onde è cosí .bg. al .gn. cosí .ng. al .gp. e per la permutata proportione è cosí .bg. al .gn. cosí<lb></lb> |
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6982 .mg. al .gp., ma commo .mg. al .pg. cosí la superficie .bm. ala superficie .bp. E ancora cosí .bg.<lb></lb> |
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6983 al .qg., cioé commo .mg. al .pm. cosí la superficie .bp. ala superficie .pn. Adonca è cosí la su-<lb></lb> |
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6984 perficie .bm. ala superficie .bp. commo la superficie .bp. ala superficie .pn. Onde la superfi-<lb></lb> |
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6985 cie .bp. è mezzana nela proportione infra la superficie .bm. e la superficie .pn. E questo infra<lb></lb> |
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6986 la superficie .abg. e il triangolo .ezi. Onde è mostro che la multiplicatione dela terza par-<lb></lb> |
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6987 te del .tk. nel congionto dele superficie de’ triangoli .abg. et .ezi. e dela superficie .bp., cioé nel<lb></lb> |
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6988 congionto dele superficie .bm. e .bp. e .pn. fará l’ area del tagliamento dela piramide .abgezi.<lb></lb> |
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6989 prima è manifesto che l’ area di tutta la piramide .dabg. se ha dela multiplicatione dela ter-<lb></lb> |
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6990 za parte del’ alteza del .dk. nella superficie del triangolo .abg., cioé nella superficie .bm. Adon-<lb></lb> |
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6991 ca, a multiplicare .dk. nela superficie .bm., ne perviene .3. cotanti del’ area dela piramide .dabg.<lb></lb> |
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6992 Ma la multiplicatione del .dk. nela superficie .bm. sonno iguali ale multiplicatio-<lb></lb> |
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6993 ni del .dt. et .tk. nela superficie .bm. Adonca dele multiplicationi del .dt. in .bm. e del .tk. in .bm.<lb></lb> |
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6994 ne perviene .3. cotanti del’ area dela piramide .dabg., dela qual summa, se se ne togli .3. cotanti<lb></lb> |
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6995 del’ area dela piramide .dezi., la qual summa s’ á del multiplicare del .dt. nela superficie del<lb></lb> |
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6996 triangolo .ezi. cioé nela superficie .np., rimarrá la multiplicatione del .tk. nela superficie<lb></lb> |
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6997 <lb></lb> |
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6998 <pb></pb></p> |
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6999 <p class="folio"> |
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7000 folio 47v |
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7001 </p> |
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7002 <p class="main"> |
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7003 <lb></lb> |
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7004 </p> |
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7005 <p class="runhead"> |
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7006 Distinctio sexta. Capitulum tertium. |
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7007 </p> |
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7008 <p class="main"> |
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7009 <lb></lb> |
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7010 .bm. e del .dt. nela superficie .r. e .s. per l’ area dela corta piramide .eg. Queste cose dimostrate,<lb></lb> |
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7011 io comporró le rette .bk. e .zt., che sonno nela superficie del triangulo .dkb. e sian le rette .bk.<lb></lb> |
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7012 e .tz. equedistanti. Onde è cosí .bd. al .zd. cosí .kd. al .tk. Ma commo .bd. al .zd. cosí .bg. al<lb></lb> |
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7013 .ng. e .mg. al .pg. dove, per la disiuncta proportionalitá sia cosí .bn. al .ng. over commo .mp.<lb></lb> |
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7014 al .pg. cosí .kt. al .dr., ma commo .mp. al .pg. cosí la superficie .qm. ala superficie .bp. Onde<lb></lb> |
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7015 la multiplicatione del .tk. nela superficie .bp. è iguale ala multiplicatione del .de. in superfi-<lb></lb> |
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7016 cie .r., cioé nela superficie .q.m. Ancora, perché e gli é cosí .bn. al .ng. cosí la superficie .s. cioé la<lb></lb> |
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7017 superficie .bc. ala superficie .np. Onde la multiplicatione del .tk. nela superficie .np. è igua-<lb></lb> |
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7018 le ala multiplicatione del .rd. nela superficie .s. Adonca la multiplicatione del .tk. nel con-<lb></lb> |
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7019 gionto dele superficie .bp. e .pn. è iguale ala multiplicatione del .de. nela superficie .rs. Com-<lb></lb> |
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7020 munamente si ponga la superficie del .tk. nela superficie del .bm., sirá la multiplicatione del<lb></lb> |
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7021 .tk. nele superficie .bm.bp.pk. iguale ala multiplicatione del .tk. in .bm. e del .de. nela super-<lb></lb> |
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7022 ficie .rs. Ma la multiplicatione del .tk. nela superficie .bm. e del .de. nela superficie .rs. ne<lb></lb> |
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7023 perviene .3. cotanti del’ area dela corta piramide .eg. Adonca el dutto del .tk. nela superficie<lb></lb> |
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7024 .bm.bp.pn., cioé nela superficie de’ triangoli .abg.ezi. e nela superficie .bp., che è mezzana in-<lb></lb> |
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7025 fra li triangoli ditti, ne perviene .3. cotanti del’ area dela tagliata piramide .eg. Onde, a mul-<lb></lb> |
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7026 tiplicare la terza parte del .tk. nela superficie de’ triangoli .abg. et .ezi. nela superficie .bp., ne<lb></lb> |
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7027 perviene l’ area ditta, cioé dela tagliata piramide e questo é quello che io volia dimostrare.<lb></lb> |
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7028 E con numeri sia il lato .bg.12. e il catetto cadente nel triangolo .abg. dal ponto .a. in sul la-<lb></lb> |
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7029 to .bg. sia .15. e il lato .zi. sia .4. e l’ altezza del .tk. sia .12. Onde sia cosí .3. a uno cosí .bg. al .zi.3.<lb></lb> |
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7030 cotanti. E adonca il lato .bg. del lato .zi., cioé cosí .bg. al .zi. cosí .zi. al .u. Onde .bg. é nove cotan-<lb></lb> |
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7031 ti del .u. E, perché e gli é cosí .bg. al .u., cosí la superficie del triangolo .abg. ala superficie del<lb></lb> |
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7032 triangolo .ezi. La superficie adonca del triangolo .abg. è .9. cotanti del triangolo .ezi., impero-<lb></lb> |
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7033 ché la superficie del triangulo .abg. è .90. quadro, che se hano dela multiplicatione del ditto<lb></lb> |
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7034 catetto nela mitá del lato .bg., dove il nono di .90. sonno .10. quadre per l’ area del triangolo<lb></lb> |
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7035 .ezi. E la superficie mezzana infra ’l triangolo .ezi. e il triangolo .abg. è .30., imperoché tal parte<lb></lb> |
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7036 é .30. di .90. commo .10. di .30. Agionte queste .3. superficie in una summa, cioé .90.30.10., fanno<lb></lb> |
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7037 .130. li quali, multiplicati per lo terzo del .kt., cioé per .4. fanno .520. per l’ area dela corta pira-<lb></lb> |
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7038 mide .eg. ala quale summa ancora virremo se dela quadratura di tutta la piramide .dabg.<lb></lb> |
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7039 torremo la piramide .dezi. che cosí si fa. Perché e gli é cosí .bg. al .zi. cioé commo il .bg. al .ng.<lb></lb> |
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7040 cosí .dk. al .dt., per la disiuncta proportionalitá, sirá cosí .bn. al .ng. cosí .kt. al .td. Onde, se<lb></lb> |
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7041 multiplicaremo .ag. per .bt., cioé .4. per .12., e divideremo per .ba., cioé per .8., vienne .6. per lo<lb></lb> |
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7042 catetto .dt. Onde tutta .dk. è .18. Di quali la terza parte multiplicata nel triangolo .abg., cioé<lb></lb> |
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7043 in .90., fanno .540. per l’ area di tutta la piramide .dabg. dela quale misura, se se ne trae l’ area<lb></lb> |
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7044 dela piramida .dezi., cioé .20. che pervengono del dutto del tertio del catetto .dt. nel trian-<lb></lb> |
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7045 golo .ezi., cioé .di.2. in .10., rimarranno .520., commo di sopra per la corta piramide .eg. Si-<lb></lb> |
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7046 milmente se dimostra con le ditte demostrationi se la basa d’ alcuna corta piramide sirá qua-<lb></lb> |
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7047 drilatera over multilatera over circulare quel medesimo evenire. Nientedimeno acioché<lb></lb> |
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7048 questa opera sia piú perfecta alcuna corta piramide havente la basa e il capo circulare pro-<lb></lb> |
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7049 porremo cosí. |
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7050 </p> |
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7051 <p class="main"> |
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7052 L‘ é una piramide corta dela quale la basa é ’l circulo .abcd. e il suo capo sia il<lb></lb> |
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7053 cerchio .efgh. e la sua altezza sia la linea .ik., dela quale altezza li termini sonno e<lb></lb> |
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7054 centri de’ ditti cerchi. E menise li diametri loro .bd. e .fh. e sienno e ditti circuli in-<lb></lb> |
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7055 fra loro equedistanti. Onde si multiplicará el terzo del .ik. nela summa dele su-<lb></lb> |
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7056 perficie de’ cerchi .abcd. e .efgb. e dela superficie che è in mezzo dela proportione di cia-<lb></lb> |
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7057 scuno cerchio. Verbi gratia: sia il cerchio .omn., cioé cosí .efh. al .mno. cosí il cerchio .mno.<lb></lb> |
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7058 al cerchio .abc. Onde, multiplicando la mitá di ciascun diametro de’ ditti cerchi per .3 1/7., ha-<lb></lb> |
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7059 remo l’ area de’ ditti cerchi la quale, multiplicata per lo terzo dela sua alteza, cioé nel terzo del<lb></lb> |
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7060 .ik., haremo l’ area dela piramide corta .ec. che voglio si mostri con numeri. Sia il mezzo dia-<lb></lb> |
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7061 metro .bk.4. e il mezzo diametro .if. sia uno, unde il mezzo diametro .lm. sia .2. imperoché<lb></lb> |
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7062 gli é cosí .4. a .2. cosí .2. a uno e giongnamo li quadrati di questi .3. mezzi diametri, sonno .21.,<lb></lb> |
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7063 cioé agiongni .16.4.1. Li quali, multiplicati per .3 1/7., fanno .66. e questo, multiplicato nel terzo<lb></lb> |
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7064 dela sua altezza che sia .5., cioé il terzo del .lk., vienne per l’ area di tutta la piramide .ac.330. E<lb></lb> |
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7065 volendo sostituire tutta la piramide .qabcd., intendiamo il triangolo .qbd., segare la pira-<lb></lb> |
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7066 mide .qabcd. in .2. parti iguali nela quale superficie è il catetto .ik. el quale menato infi-<lb></lb> |
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7067 <lb></lb> |
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7068 <pb></pb></p> |
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7069 <p class="folio"> |
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|
7070 folio 48r |
|
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7071 </p> |
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7072 <p class="main"> |
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7073 <lb></lb> |
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7074 </p> |
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7075 <p class="runhead"> |
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7076 Distinctio sexta. Capitulum tertium. 48 |
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7077 </p> |
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7078 <p class="main"> |
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|
7079 <lb></lb> |
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7080 no al .q. sia la linea .qk. catetto del triangolo .qbd. nel quale, menato la linea .fr. equedistan-<lb></lb> |
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7081 te al .ik., sirá .fr. equale ala linea .ik., perché equedistante è la linea .fi. ala linea .kc.bk. e sia<lb></lb> |
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7082 .rk. iguale al .fi. e il triangolo .qif. e .frb. sonno simili. Onde se traremo .rk., cioé .if. del .kb., ri-<lb></lb> |
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7083 marranno .br.3. e, perché e gli é cosí .br. al .rf. cosí .fi. al .iq. Onde multiplicando .rf. per .fi. e<lb></lb> |
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7084 dividendo per .br., vienne .5. per lo catetto .qi. Onde tutta .qf. è .20. che è l’ altezza dela pira-<lb></lb> |
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7085 mide .qabcd. |
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7086 </p> |
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7087 <p class="main"> |
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7088 Se inn una spera si piglia un ponto dal quale .4. rette linee si menino infra loro<lb></lb> |
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7089 iguali e vadino ala superficie dela spera e quelle linee non sienno in una superficie<lb></lb> |
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7090 piana, quel ponto sia il centro dela spera. Verbi gratia: sia la spera .ab. e in quella<lb></lb> |
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7091 sia il ponto .z. dal quale sienno menate .4. linee infra loro iguali .zb.zg.zd.ze. e non<lb></lb> |
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7092 sienno li ponti .b.g.d.e. in una superficie piana. Dico il ponto .z. essere centro dela ditta spera e<lb></lb> |
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7093 questo evidentemente appare e peró nonn’ á bisogno de dimonstratione.<lb></lb> |
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7094 Quando sirá menato dal ponto del capo d’ ogni piramide colonnale al centro dela<lb></lb> |
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7095 basa sua perpendiculare sopra la sua basa, alora le linee rette che sonno menate<lb></lb> |
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7096 dal ponto del capo suo al cerchio contenente la superficie dela sua basa sonno in-<lb></lb> |
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7097 fra loro iguali. E la multiplicatone d’ una di quelle linee che sonno menate dal ca-<lb></lb> |
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7098 po loro al cerchio contenente la sua basa nela mitá del cerchio contenente la ditta basa é l’ a-<lb></lb> |
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7099 rea dela superficie de ditta piramide colonnale. Verbi gratia: sia la piramide colonnale .abgd.<lb></lb> |
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7100 dela quale la sua somitá sia .a. e la sua basa sia il circulo .bgd. del quale sia il centro .e. E la li-<lb></lb> |
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7101 nea .ae. ortogonalmente sia ritta sopra il piano del cerchio .bgd. E dal ponto .a. ala linea cir-<lb></lb> |
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7102 cunferentiale contenente il circulo .bgd. dela basa dela data piramide di colonna se meni<lb></lb> |
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7103 molte linee .ab.ag.ad. Dico certamente le rette .ab.ag.ad. infra lloro essere iguali. La prova: me-<lb></lb> |
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7104 nise dal centro .e. le rette .eb.eg.ed. che sonno tutte iguali infra loro. E, perché .ae. è perpen-<lb></lb> |
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7105 diculare sopra il piano del circulo .bgd. fienno gli angoli .aeb.aeg.aed. retti. Onde li trian-<lb></lb> |
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7106 goli sonno ortogonij .aeb.aeg.aed. e hano le base iguali che sonno .eb.eg.ed. e il lato .ae. è<lb></lb> |
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7107 commune. Onde li lati subtendenti agli angoli retti, che sonno .ab.ag.ad., sonno infra loro<lb></lb> |
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7108 iguali. E per questo è manifesto che tutte le rette linee che si possono menare dal .a. ala linea<lb></lb> |
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7109 circunferente .bgd. essere iguali ala linea .ab. |
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|
7110 </p> |
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7111 <p class="main"> |
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7112 Ancora dico che, multiplicato .ab. nela mitá dela linea circunferente .bgd., fará<lb></lb> |
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7113 l’ area dela superficie dela piramide, cioé l’ area di fuora dela superficie .abgd., la<lb></lb> |
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7114 quale superficie é dal circulo dela basa .bgd. infino ala sua summitá. E, se non fos-<lb></lb> |
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7115 se cosí, alora sia la multiplicatione dela linea .ab. dela mitá del circulo .bgd. ma-<lb></lb> |
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7116 giore o minore quella che facia l’ area dela superfie. Dico che quella quantitá che se mul-<lb></lb> |
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7117 tiplica per .ab., a fare l’ area dela superficie, sia minore over magiore dela mitá dela linea cir-<lb></lb> |
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7118 cunferente .bgd. E sia la quantitá .iz. e il doppio del .iz. é piú che ’l cerchio .bgd. Adonca fa-<lb></lb> |
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7119 ró sopra il cerchio .bgd. una figura rettilinea havente e lati e gli angoli iguali contenente<lb></lb> |
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7120 quello. E fienno li lati insiemi agionti meno che lo doppio del .iz. che sia la figura .lkt. E me-<lb></lb> |
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7121 neró la linea .ab., la quale è perpendiculare sopra la linea .bk., in questo modo. Meneró la li-<lb></lb> |
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7122 nea .et. Fienno li quadrati dele linee .eb. et .bt. iguali al quadrato dela linea .et. e commune a<lb></lb> |
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7123 tutti sia il quadrato dela perpendiculare .ae. Siranno li quadrati dele linee .eb. e .ba. igua-<lb></lb> |
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7124 li al quadrato dela linea .et. E communamente s’ agionga il quadrato dela perpendiculare .ae.<lb></lb> |
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7125 Fienno li quadrati dele linee .ae.eb.bt., cioé li quadrati dele linee .ab. e .bt., iguali al qua-<lb></lb> |
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7126 drato .at., onde l’ angolo .abt. è retto. Perpendiculare è adonca la linea .ab. sopra la linea .tk.<lb></lb> |
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7127 Similmente si mostra la linea .ag. essere perpendiculare sopra .kl. e .ad. sopra la linea .tl. E,<lb></lb> |
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7128 perché le rette .ab.ag.ad. sonno infra loro iguali, virrá dela multiplicatione d’ una di quel-<lb></lb> |
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7129 le, commo del .ab., nela mitá de’ lati del triangolo .tkl., l’ embado over area dela superficie de-<lb></lb> |
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7130 la piramide .atkl. magiore dela superficie dela piramide .abgd. conciosiacosaché la con-<lb></lb> |
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7131 tenga quella, cioé quello ch’ é infra ’l cerchio .bgd. e il ponto .a. E la mitá de’ lati del triangolo<lb></lb> |
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7132 .tkl. é minore che la quantitá .iz. Adonca giá fo la multiplicatione dela linea .ab. quello che<lb></lb> |
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7133 è meno dela linea .iz. é magiore dela superficie dela piramida di colonna che è impossibi-<lb></lb> |
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7134 le. Adonca nonn’ é possibile che la multiplicatione dela linea .ab. nela linea che sia magio-<lb></lb> |
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7135 re dela mitá del cerchio .bgd. sia l’ embado over continentia dela superficie .abgd. Anco-<lb></lb> |
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7136 ra porró la linea .iz. minore dela mitá dela circunferentia del circulo .bgd. e, se possibile è che del<lb></lb> |
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7137 dutto .ab. in .iz. ne pervenga l’ area dela superficie dela piramide .abgd. A multiplicare adonca dela<lb></lb> |
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7138 .1/2. dela circunferentia del circulo .bgd. fará la superficie d’ una minore piramide dela piramide .abgd.<lb></lb> |
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7139 <lb></lb> |
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7140 <pb></pb></p> |
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7141 <p class="folio"> |
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7142 folio 48v |
|
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7143 </p> |
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7144 <p class="main"> |
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7145 <lb></lb> |
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7146 </p> |
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7147 <p class="runhead"> |
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7148 Distinctio sexta. Capitulum tertium. |
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7149 </p> |
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7150 <p class="main"> |
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7151 <lb></lb> |
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7152 che sia la piramide .acfh. dela quale la somitá sia .e. la basa sua sia il circulo .fcb. E descrive-<lb></lb> |
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7153 ró nel cerchio .fch. la figura rettilinea .cfh. e meneró, dal centro .e. sopra la linea .cf. el catetto<lb></lb> |
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7154 .el. che dividerá la linea .cf. in .2. parti iguali e comporró la retta .ac. al .af.ah. E, per quello che<lb></lb> |
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7155 s’ é detto, si mostrará la linea .al. essere perpendiculare sopra la linea .fc. e fienno iguali le perpendi-<lb></lb> |
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7156 culari cadenti sopra le linee .ch.fh. ala perpendiculare .al. Sia la retta .al. magiore che .ab., impero-<lb></lb> |
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7157 ché magiore è .el. che .eb. e la mitá de’ lati dele figure .cfh. è magiore dela mitá dela linea<lb></lb> |
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7158 circunferentiale del cerchio .bgd. e la mitá del cerchio .bgd. è magiore che .iz. E, a multiplicare<lb></lb> |
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7159 .al. nela mitá deli lati dele figure di rette linee .cfh., fanno l’ area dela piramide .acfh., dela<lb></lb> |
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7160 quale la basa è il triangolo .cfh. e dela multiplicatione del .ab., che è minore del .al., nel .iz., che<lb></lb> |
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7161 è minore de’ lati .cfh., ne proviene l’ area dela piramide magiore che la piramida .acfh., che è in-<lb></lb> |
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7162 conveniente. Adonca, a multiplicare .ab. nela mitá dela circunferentia del cerchio .bgd., fa-<lb></lb> |
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7163 ranno l’ area dela superficie dela piramide .abgd., ch’ era da dimostrare, la quale superficie é dal<lb></lb> |
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7164 summo dela piramide, cioé dal ponto .a., e il cerchio .bgd. Onde, se poniammo la perpendicula-<lb></lb> |
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7165 re .ae. essere .24. e il mezzo diametro .eb.7., sirá la linea .ab.25. la quale, multiplicata nela mitá de-<lb></lb> |
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7166 la circuferentia del circulo .bgd., che è .22., faranno .550. per l’ area dela superficie dela piramide .abgd.<lb></lb> |
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7167 Se sia una piramide di colonna corta dela quale la basa sia uno cerchio e il capo sia<lb></lb> |
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7168 il cerchio e sia equedistante la basa al capo e tu voglia la superficie di quella piramide<lb></lb> |
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7169 corta, dico che te bisogna sapere la linea che si mena dala circunferentia del cerchio<lb></lb> |
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7170 del capo dela basa ala circonferentia dela basa dela ditta piramide. E quella nela mi-<lb></lb> |
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7171 tá dela circunferentia de’ .2. cerchi, cioé del capo e dela basa, multiplica. E la mulplicatione si-<lb></lb> |
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7172 rá la superficie dela ditta piramide corta. Commo sia la piramide corta dela quale la basa sia il<lb></lb> |
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7173 cerchio .abg. e il suo capo sia il cerchio .dez. li quali cerchi sienno infra loro equedistanti e da’ cen-<lb></lb> |
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7174 tri loro si meni la retta .it. perpendiculare ad amendoi li cerchi e l’ estremitá di ditti cerchi si<lb></lb> |
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7175 congionga per le linee .be. e .gz. Dico che la multiplicatione del’ una de linee .be.gz. nela mitá de-<lb></lb> |
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7176 la circunferentia de’ cerchi .abg. e .dez. ne pervenga l’ area dela superficie dela corta piramide<lb></lb> |
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7177 cioé l’ area dela superficie che è infra ’ditti .2. cerchi. Commo sia il diametro .bg.14. e il diametro<lb></lb> |
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7178 .ez. sia gli .2/5., cioé .5 3/5. Per la linea .be.15. e .it. sia .14 2/5. E comporró la retta .mi. e, per li gli ponti .ez., me-<lb></lb> |
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7179 neró le linee .ec.zf. equedistanti ala retta .it. E fienno le rette .ct. et .tf. iguali commo sonno .ei. e<lb></lb> |
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7180 .iz. l’ altre .bc. e .gf. infra loro iguali. Onde li triangoli .ecb. e .zfg. sonno simili infra loro e igua-<lb></lb> |
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7181 li. E gli angoli al .b. e al .g. sonno infra loro iguali. Onde il triangolo .mbg. è equicurio haven-<lb></lb> |
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7182 te lati .bm. e .mg. iguali. E, perché la retta .ez. è equedistanti ala retta .bg., sia il triangolo .mez. si-<lb></lb> |
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7183 mile al triangolo .mbg. e equicurio havente gli angoli che sonno al .e. e .z. iguali. Onde la ret-<lb></lb> |
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7184 ta .mi. sia catetto sopra .ez. conciosiacosa sia in sul mezzo .ez., fo adonca l’ angolo .tib. retto.<lb></lb> |
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7185 Onde è manifesto la linea .mit. essere continuata. Adonca .me. è perpendiculare dela pira-<lb></lb> |
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7186 mide .mbag. e passa per lo centro del cerchio .edz. E, perché e la retta .mb. e .mg. sonno igua-<lb></lb> |
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7187 li, se da quelli togli .me. e .mz., rimarranno .eb. e .zg. infra loro iguali. Adonca .gz. è .15. E, per-<lb></lb> |
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7188 ché .ez. al .bg. è como .2. a .5. E per questo .me. al .mb. è commo .2. a .5. Onde .me. del .eb. è gli .2/3.,<lb></lb> |
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7189 cioé .10.; onde .mb. sia .25. del quale il quadrato, che è .625., se se ne togli el quadrato .tb., che è<lb></lb> |
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7190 .49. rimarrano .576. per lo quadrato del catetto .mt. e l’ arco .mag. sia .22., che vengono del .tb. in .3 1/7.<lb></lb> |
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7191 E l’ arco .edz. sia .8 14/5., cioé li .2/5. del’ arco .bag., li quali archi agionti fanno .30 4/5., li quali multiplicati<lb></lb> |
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7192 nela linea .eb., cioé in .15., fanno .462. per l’ area dela superficie che è infra ‘circuli .abg. e .dez. E s-<lb></lb> |
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7193 se del’ area dela superficie dela piramide .mabg., che viene del .mb. nel’ arco .bag., cioé del .25.<lb></lb> |
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7194 in .22., si togli l’ area dela superficie dela piramide .mez., che viene del .me. nel’ arco .dz., cioé de .10.<lb></lb> |
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7195 in .8 4/5., rimarranno similmente .462. per l’ area contenta infra ’l circulo .edz. e il cerchio .bag. e<lb></lb> |
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7196 questo volia mostrare. |
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7197 </p> |
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7198 <p class="main"> |
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7199 Quando uno cerchio è diviso in .4. parti iguali dali doi diametri li quali fanno<lb></lb> |
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7200 sul centro .4. angoli retti. Dico che, se ’l si divide l’ arco d’ un di quelli .4. quarti in par-<lb></lb> |
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7201 ti iguali quante voi e de ciascuna di quelle parti si meni nel cerchio la linea eque-<lb></lb> |
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7202 distante alo diametro. E dal ponto del diamentro in sula circunferentia si pona<lb></lb> |
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7203 il regolo et passi per lo primo ponto dela prima parte e menisi in infinito, infino si congion-<lb></lb> |
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7204 ga colla linea del diametro menata infinitamente. Commo sia il cerchio .abcd., diviso in .4. par-<lb></lb> |
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7205 ti iguali dal diametro .ab. e .cd., che fanno .4. retti angoli al centro .e. e dividase una parte,<lb></lb> |
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7206 cioé il quarto, che è l’ archo .da., in .4. parti iguali che fienno .df.fh.hk. e .ka. E menise le li-<lb></lb> |
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7207 nee da’ ditti ponti .fhk. equedistanti alo diametro .ab. e fienno .fg. e .hi. e .kl. Dico che, menan-<lb></lb> |
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7208 do la linea .df. in infinito infino a tanto si congionga con lo diametro .ab., menato in continuo<lb></lb> |
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7209 <lb></lb> |
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7210 <pb></pb></p> |
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7211 <p class="folio"> |
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7212 folio 49r |
|
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7213 </p> |
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7214 <p class="main"> |
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7215 <lb></lb> |
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7216 </p> |
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7217 <p class="runhead"> |
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7218 Distinctio sexta. Capitulum quartum. 49 |
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7219 </p> |
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7220 <p class="main"> |
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7221 <lb></lb> |
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7222 con la linea del diametro menata infinitamente. Comme sia il cerchio .abcd., diviso in .4. par-<lb></lb> |
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7223 ti iguali dal diametro .ab. e .ed., che fanno .4. retti angoli al centro .e. e dividase una parte, cioé<lb></lb> |
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7224 il quarto, che è l’ arco .da., in .4. parti iguali che fienno .df.fh.hk. e .ka. E menise le linee da-<lb></lb> |
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7225 i deti ponti .f.h.k. equedistanti alo diametro .ab. e fienno .fg. e .hi. e .kl. Dico che, menando<lb></lb> |
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7226 la linea .df. in infinito infino a tanto si congionga con lo diametro .ab., menato in continuo<lb></lb> |
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7227 e diritto, el quale ponto del congiongnimento sia il ponto .m. Dico che la linea che è dal pon-<lb></lb> |
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7228 to dove si congiongono, cioé dal ponto .m. infino al centro del detto cerchio, cioé la linea<lb></lb> |
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7229 .me., essere iguale al’ agiongnimento di tutte le linee .fg.hi.kl. e ala mitá del diametro .ea. E<lb></lb> |
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7230 questo te sia manifesto. |
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7231 </p> |
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7232 <p class="main"> |
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7233 De modo inveniendi superficiem sphericam eiusque capacitatem et portionum suarum.<lb></lb> |
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7234 Capitulum quartum. |
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7235 </p> |
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7236 <p class="main"> |
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7237 Se la superficie d’ una spera voi havere multiplicarai l’ area del magiore cerchio<lb></lb> |
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7238 per .4. e haremo l’ area dela superficie d’ una spera. E il diametro del magiore cerchio ca-<lb></lb> |
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7239 dente nela spera è il diametro overo l’ axe dela spera. Onde, dicendo e gli é una<lb></lb> |
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7240 spera dela quale il diametro è .7., adimandasse quanto è la superficie dela spera. Di-<lb></lb> |
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7241 co che l’ area del cerchio che il suo diametro sia .7. truovi. E questo harai per gli modi dati.<lb></lb> |
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7242 Overo multiplicando la mitá del diametro in sé e la multiplicatione in .3 1/7. e haremo .38 1/2. e .38 1/2.<lb></lb> |
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7243 è l’ area del detto circulo, la quale multiplica per .4., fanno .154. e .154. è l’ area dela superficie<lb></lb> |
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7244 dela spera. Overamente el quadrato del diametro del detto cerchio multiplica in .3 1/7. e<lb></lb> |
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7245 haremo quello medesimo, cioé .49. multiplicati in .3 1/7., fanno .154. Overo il diametro mul-<lb></lb> |
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7246 tiplicato per la circonferentia, cioé .7. in .22. e haremo quel medesimo cioé .154. E, se vuoi l’ a-<lb></lb> |
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7247 rea dela mitá dela superficie dela spera, quando harai l’ area dela superficie dela spera, in .2. la dividi<lb></lb> |
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7248 e harai l’ area dela predetta mezza spera. |
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|
7249 </p> |
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|
7250 <p class="main"> |
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7251 E, se l’ area corporale d’ una spera vuoli, l’ area dela superficie di detta spera nel sex-<lb></lb> |
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|
7252 to del diametro di detta spera multiplica e la multiplicatione sia l’ area corpora-<lb></lb> |
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|
7253 le predetta. Comme nelo exemplo dato multiplicarai .154. per lo sexto di .7., cioé<lb></lb> |
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|
7254 per .1 1/6., fanno .179 2/3. per l’ area corporale di deta spera. Imperoché gli é provato<lb></lb> |
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|
7255 dagli antichi savi. (Como de sotto di lei piú amplamente nel tratato particulare de’ corpi regulari<lb></lb> |
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7256 parlaremo) che la multiplicatione dela terza parte dela superficie dela spera nella mitá di tutto il diametro sará<lb></lb> |
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7257 l’ area di tutta la spera, cioé l’ area corporale di tutta la spera che ancora te ’l mostreró. Sia la spera .ab.<lb></lb> |
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7258 e la mitá del diametro suo sia la linea .ag. e il centro suo sia il ponto .g. Dico adunque che la<lb></lb> |
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7259 multiplicatione del .ag. nel terzo della superficie dela spera .ab. è iguale ala misura del corpo dela spera .ab.<lb></lb> |
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7260 Della quale questa è la dimostratione. Se non è la multiplicatione del .ag. nel terzo dela superficie<lb></lb> |
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7261 dela spera .ab. iguale al corpo dela spera, sirá iguale a uno corpo magiore che la spera .ab. o-<lb></lb> |
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7262 vero minore. Sia adunque prima iguale a una magiore spera che la spera .ab. e sia la spera<lb></lb> |
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7263 .de. che sia con la spera .ab. in su uno medesimo centro. Possibile adonca è che nela spera .de. sia<lb></lb> |
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7264 uno corpo di piú lati, dele quali quelli lati che li diciammo base non sienno contingenti ala superficie<lb></lb> |
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7265 dela spera .ab. Onde sirá ciascuna dele perpendiculari, overo le linee cadenti dal centro .g. so-<lb></lb> |
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7266 pra le superficie di quelle magiore che ’l mezzo diametro .ag. Se adonca si continuan gli angoli di quel<lb></lb> |
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7267 corpo che caggiano nela spera .de., col centro della spera faranno piramide, dele quali el lo-<lb></lb> |
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7268 ro capo fienno li centri dela spera e le loro base fienno base di corpi. E la misura di tutte le det-<lb></lb> |
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7269 te piramidi viene dela multiplicatione dela sua perpendiculare nel terzo dela sua basa. E, per-<lb></lb> |
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7270 ché la linea .ag. è mezzo il diametro della spera .ab., e la è minore di ciascuna del’ altre perpen-<lb></lb> |
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7271 diculari. E sirá quello la multiplicatione dela linea .ag. nel terzo di caduna basa minore, che<lb></lb> |
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7272 la misura dela piramide dela quale la basa è quella. Adunque la multiplicatione dela linea<lb></lb> |
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7273 .ag. nel terzo dela superficie di quel corpo è magiore della superficie .ab. La multiplicatione del .ag. nel ter-<lb></lb> |
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7274 zo di quella superficie è minore di quel corpo. E giá fo posto la multiplicatione dela linea .ag. nel ter-<lb></lb> |
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7275 zo dela superficie dela spera .ab. essere iguale al .de. spera. Adunque è di bisogno che la spera .de. sia<lb></lb> |
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7276 minore molto del corpo che è infra quella: che è impossibile. Non adonca la multiplicatione<lb></lb> |
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7277 dela linea .ag. nel terzo dela superficie dela spera .ab. è magiore dela spera .ab. E an-<lb></lb> |
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7278 cora dico che nonn’ é minore della spera .ab. Comme è la spera .zh. che sia sopra il cen-<lb></lb> |
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7279 tro .g. E possibile ancora è che sia nella spera .ba. uno corpo di piú base de’ quali le base non sien-<lb></lb> |
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7280 no contingenti la superficie dela spera .ch. Adonca sirá ciascuna dele perpendiculari cadenti sopra le superficie<lb></lb> |
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7281 di quelli corpi moventisi dal centro dela spera .ab. meno dela mitá del diametro dela spera .ab.,<lb></lb> |
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7282 che è la linea .ag. Sirá adonca la multiplicatione del terzo di ciascuna superficie loro magiore che<lb></lb> |
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7283 la misura dela piramide, dela quale la basa è quella superficie e dela quale é il centro .g. La multiplicatione<lb></lb> |
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7284 <lb></lb> |
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7285 <pb></pb></p> |
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7286 <p class="folio"> |
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7287 folio 49v |
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7288 </p> |
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7289 <p class="main"> |
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7290 <lb></lb> |
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7291 </p> |
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7292 <p class="runhead"> |
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7293 Distinctio sexta. Capitulum quartum. |
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7294 </p> |
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7295 <p class="main"> |
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7296 <lb></lb> |
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7297 adunque dela linea .ag. nel terzo di ciascuna superficie .ab. è magiore del corpo di piú base.<lb></lb> |
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7298 Giá posto fo equale al’ embado, overo capacitá spere .zh. é molto magiore del corpo detto e<lb></lb> |
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7299 quella è infra quello e questo è impossibile. Non adunque la multiplicatione dela linea .ag.,<lb></lb> |
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7300 che è la mitá del diametro della spera .ab., nel terzo dela superficie sua è magiore del corpo<lb></lb> |
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7301 suo: quella adunque è iguale di quello corpo e quello é quello vogliamo la dischiaratione.<lb></lb> |
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7302 E, quando questo dichiarato sie e vogliamo havere la mitá dela spera, e multiplicaremo l’ a-<lb></lb> |
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7303 rea dela superficie sua nel sexto delo suo diametro. Overo la mitá del diametro suo multi-<lb></lb> |
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7304 plicaremo nel terzo dela sua superficie. Verbi gratia: sia il diametro dela data spera .10., la<lb></lb> |
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7305 quale, per la mitá sua, fanno .50. li quali, in .3 1/7. multiplicati, fanno .157 1/7., che sonno l’ area dela<lb></lb> |
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7306 superficie dela mezza spera. La quale, se multiplicaremo per lo sexto di quello, vienne .261 19/21., cioé<lb></lb> |
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7307 multiplicando .157 1/7. via .1 2/3., che è il .1/6. di .10., fanno .261 19/21., comme disse per l’ area dela mitá<lb></lb> |
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7308 dela spera. |
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7309 </p> |
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7310 <p class="main"> |
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7311 E, se sia de bisogno a noi misurare una portione de spera che sia magiore o mino-<lb></lb> |
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7312 re dela mezza spera, comme sonno le fonte ritonde e gli vasi i quali hano li fondi<lb></lb> |
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7313 tondi, l’ altezza del detto corpo che è una linea che si extende dal centro del cer-<lb></lb> |
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7314 chio dela bocca di quella tale parte e va infino al ponto del polo del detto cerchio colla mi-<lb></lb> |
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7315 tá del diametro dela spera, proportionare curiamo. E quella parte dela superficie dela mitá<lb></lb> |
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7316 dela spera e ancora dela misura sua togliamo. E haremo il desiderio. Verbi gratia: sia la linea<lb></lb> |
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7317 .ab. el diametro dela bocca d’ una fonte ritonda e .g. sia il suo centro e il ponto .d. sia il polo del<lb></lb> |
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7318 detto cerchio. Onde la linea .dg. sta ortogonalmente sopra la superficie del cerchio del qua-<lb></lb> |
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7319 le il diametro è .ab. Dove il quadrato dela mitá del diametro .ab. divideremo per .gd. e hare-<lb></lb> |
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7320 mo quello che resta di tutto il diametro dela spera sopra la linea .gd. Verbi gratia: sia la linea<lb></lb> |
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7321 .ab., cioé il diametro, radici di .160. Onde .gb., che è la mitá del .ab., sia radice di .40., del quale<lb></lb> |
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7322 il quadrato sia .40. El quale, se lo dividiamo per .gd., che lo poniamo .4., vienne .10. per lo avan-<lb></lb> |
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7323 zo del diametro sopra la linea .dg. El quale diametro, in .2. parti iguali diviso sopra il ponto .z.,<lb></lb> |
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7324 sia il detto .z. centro del cerchio grande cadente nela spera, el quale cerchio sia il cerchio .aebd.<lb></lb> |
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7325 Proportioneró adonca la linea .gd. con lo mezzo il diametro dela spera, che è .zd., cioé .4. con<lb></lb> |
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7326 .7. sia .gd. del .zd. gli .4/7. Onde torremo li .4/7. del’ area dela superficie dela mitá dela spera, cioé di<lb></lb> |
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7327 .308., che viene dela superficie del .dz. in .de. multiplicato in .3 1/7., viene .176. per l’ area dela super-<lb></lb> |
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7328 ficie dela portione dela spera, dela quale la basa è il cerchio del quale il diametro è .ab. e il suo<lb></lb> |
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7329 polo è .d. e l’ arco cadente in quella portione è l’ arco .adb. fatto dal cerchio grande cadente<lb></lb> |
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7330 nela spera. E la sua misura corporea, cioé la misura corporea dela detta portione, s’ ará se ’l ter-<lb></lb> |
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7331 zo del’ area dela sua superficie si multiplica in .7., cioé nelo mezzo diametro .zd., e di quella se<lb></lb> |
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7332 ne traga la misura corporea dela piramide di colonna, dela quale la sommitá è uno ponto .z.<lb></lb> |
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7333 e la sua basa è uno cerchio del quale il diametro è la linea .ab. e la sua altezza è la linea .zg., che<lb></lb> |
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7334 è .3. E la sua misura, cioé la misura dela detta piramide, é .125 5/7. Rimangano adonca, tratti di<lb></lb> |
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7335 .410 2/3., che è la multiplicatione del mezzo diametro in el terzo di .176., rimangono .285. meno<lb></lb> |
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7336 .1/21. per l’ area corporale dela portione detta. |
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7337 </p> |
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7338 <p class="main"> |
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7339 E, se l’ area del’ avanzo dela spera vuoli, cioé l’ area d’ una portione magiore dela mi-<lb></lb> |
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7340 tá dela spera dela quale sia la basa, cioé la bocca, quel medesimo cerchio del qua-<lb></lb> |
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7341 le il diametro è .ab. e la sua alteza è la linea .eg., che si chiama saetta, che la poniamo<lb></lb> |
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7342 essere .10. E vogliamo la misura dela detta portione. Dico che operiamo in questa<lb></lb> |
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7343 comme nella minore portione dela mitá dela spera facemmo. Cioé el quadrato dela linea<lb></lb> |
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7344 .gb., che è .40., dividiamo per la saetta .ge. e haremo .4. per la linea .gd., che è la saetta del’ altra<lb></lb> |
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7345 portione dela spera. Dove tutto il diametro dela spera è .ed. che è .14., nel quale, se multipli-<lb></lb> |
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7346 caremo la saetta .eg. e quello multiplicaremo per .3 1/7., overo se del’ area dela superficie dela mitá dela spera che è <lb></lb> |
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7347 .308.<lb></lb> |
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7348 torremo li .10/7., cioé la parte che á la saetta .eg. al mezzo il diametro<lb></lb> |
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7349 .cz., haremo .440. per l’ area dela superficie di questa parte magiore dela mitá dela spera. La<lb></lb> |
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7350 quale area, se nel sexto del diametro dela spera la multiplicarai, overo se ’l terzo di quella area<lb></lb> |
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7351 nela mitá del diametro multiplicarai, haremo .1026 2/3., al quali agionto la piramide di colon-<lb></lb> |
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7352 na sopra detta, che è .125 5/7., cioé la piramide .abz., fanno .1152 5/21. per la grandezza di quella magior<lb></lb> |
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7353 portione. Poterebesi fare la demostratione, dove se demostrarebbe queste cose dette dela par-<lb></lb> |
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7354 te dele spere essere chiare, ma perché el tempo poco non lo patisci per l’ altre cose che seguano,<lb></lb> |
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7355 le lascieró. Adunque a questa distinctione faremo fine e seguendo del’ altra diremo<lb></lb> |
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7356 Deo gratias.<lb></lb> |
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7357 <lb></lb> |
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7358 <pb></pb></p> |
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7359 <p class="folio"> |
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7360 folio 50r |
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7361 </p> |
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7362 <p class="main"> |
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7363 <lb></lb> |
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7364 </p> |
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7365 <p class="runhead"> |
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7366 Distinctio septima. Capitulum primum. Et .secundum. 50 |
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7367 </p> |
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7368 <p class="main"> |
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7369 <lb></lb> |
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7370 Distinctio .7a. de instrumentis quibus mediantibus solo aspectu rerum longitudines la-<lb></lb> |
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7371 titudines et altitudines habentur. capitulum primum.<lb></lb> |
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7372 <lb></lb> |
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7373 Alcuni per misurare con l’ ochio compongano uno strumento decto gnomone e di<lb></lb> |
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7374 questi forono alcuni antichi. El quale strumento è uno quadrato, cioé una figura qua-<lb></lb> |
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7375 drata, fatta d’ alcuno metallo, overo alcuno legno sodo e duro, di quantitá con-<lb></lb> |
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7376 vonevole, de magiore puoi, imperoché quanto è mangiore tanto meglio. E quel-<lb></lb> |
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7377 lo constituto quadrato con gli angoli retti e tu lo dividi in .60. parti per uno bracio, cioé che,<lb></lb> |
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7378 se gli é uno bracio per ogni verso, quello si divida in .60. parti, commo qui da lato appare, e quel-<lb></lb> |
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7379 le parti igualmente le constituisci e chiamarai ciascuna parte ponto. Adunque in .60. ponti<lb></lb> |
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7380 lo dividerai, e questo fatto, piglia una regola, al modo di quella del’ astrolabio, overamente<lb></lb> |
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7381 togli uno cannone con piccolo foro per lo qual passi lo tuo visuale, e con quello, comme ti<lb></lb> |
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7382 mostraró, trouverai quel vuoi. Alcuni hano .2. virgole insiemi commese in modo che l’ u-<lb></lb> |
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7383 na si puó andare in alto e in basso e con quelle facilmente vengono allo efetto loro. Al-<lb></lb> |
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7384 cuni con lo quadrante ordinano di sapere la longhezza e altezza dele cose vedute. Alcuni con<lb></lb> |
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7385 lo strolabio fanno medesimo. Alcuni con l’ ombra del sole hano quello che desidera-<lb></lb> |
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7386 no. Alcuni fanno quel medesimo con lo spechio, comme distesamente di sotto vederai.<lb></lb> |
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7387 Non mi pare necessario dividere questa distinctione in alcune parti, ma una sola e quella<lb></lb> |
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7388 vene e diligentemente mostraremo secondo l’ aiuto haró da chi puó. Dobiamo in questa<lb></lb> |
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7389 parte mostrare il modo a misurare una longhezza o altezza, ala quale non si possi andare, so-<lb></lb> |
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|
7390 lamente con l’ ochio. E peró è opportuno stare attento ale cose future lasciando quello che fa-<lb></lb> |
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7391 rebbe superfluo, cioé non dimostrando quello che dá impedimento al’ ochio, ma ragionamo<lb></lb> |
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|
7392 che l’ ochio sia chiaro e chiaramente comprenda. Dico adonca. |
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|
7393 </p> |
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|
7394 <p class="main"> |
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7395 De diversis casibus exemplaribus circa aspectuum dimensiones. Capitulum secundum<lb></lb> |
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7396 Sia un piano dato .ab. el quale sia de bisogno misurare; conciosiacosaché sia il pie’<lb></lb> |
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7397 tuo in sul ponto .a. Questo voglio per lo strumento gnomo. In questo modo pon-<lb></lb> |
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7398 gasi il detto gnomo in sul ponto .a., in modo che la basa .pq. sia uno con lo piano<lb></lb> |
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|
7399 .ab. E porrai l’ ochio tuo in sul ponto .s., cioé nel’ angolo .s. e guarda per la regola, cioé<lb></lb> |
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7400 per lo foro del detto strumento, e nota, overo fa notare, in che parte dela linea, overo lato .rq.<lb></lb> |
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7401 quella regola passa: che sia il ponto .c. E questo te bisogna perfettamente havere, imperoché ogni<lb></lb> |
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7402 poco de errore ne generaebbe gran quantitá. Adonca, bene compreso questo ponto .c., e tu<lb></lb> |
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|
7403 considera che parte e gli é dal .rc. al .rq. che, se la regola occupa un ponto, é .1/60., se .2., é 1/30. e, se .3., è 1/20.<lb></lb> |
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7404 e cosí de singulis. Ora, al presente, diciamo .rc. occupare un ponto del .rq., adonca è .1/60. Dico<lb></lb> |
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7405 adonca el .pq., che è iguale al. rq., essere .1/60. di tutta .ab. Dove, se ’l .qp. è uno bracio, .ab. sia .60.<lb></lb> |
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7406 bracia. E, se .qp. fosse .2.bracia., sarebbe .ab.120.bracia. se ’l .pq. fosse .2/3. di bracio, sarebbe<lb></lb> |
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7407 .ab.40. braccia. E cosí de singulis. Aliter, questo medesimo si puó fare con .2. virgole in<lb></lb> |
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7408 questo modo. Rizzise una virgola in sul ponto .a. e sia la virgola .ac., la quale sia rita in mo-<lb></lb> |
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7409 do che ’l ponto .a., cioé che l’ angolo .a. sia retto. E notato il ponto .c. e da quello si ponga l’ ochio.<lb></lb> |
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7410 E, notato lo raggio visuale in su che ponto dela virgola .mn. passa, la quale virgola sia dirit-<lb></lb> |
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7411 ta in sul piano .ab. in sul ponto .n. in modo che l’ angolo .n. sia retto. E adonca, saputo in che<lb></lb> |
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7412 ponto dela linea .mn. lo raggio visuale pasa, lo quale ponto sia .o., e questo bene notato e tu<lb></lb> |
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7413 considera che parte è .on. del .ca. Comme sia .ca.3.bracia e la linea .on. sia .2.bracia.9/10., che son-<lb></lb> |
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7414 no .2 9/10. di .3. gli .29/30. Dove considera quanto manca allo intero .1/30. E questo è la parte che l’ .an. é<lb></lb> |
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7415 del .ab. Comme sia .an.4.bracia adonca .ab. sia .30. cotanti che .an., che sia .120.bracia. An-<lb></lb> |
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7416 cora puoi sapere quanto è meno .on. che .ca. E comme ó detto sia .1/10. di bracio. Dove diremo: per<lb></lb> |
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7417 4.bracia. si scema .11/10. di bracia, per quante bracia si scemerá .3.bracia? Multiplica .3. per .4. e parti in .1/10., vienne<lb></lb> |
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7418 .120.bracia. E .120.bracia. sia la linea .ab. comme dicemmo. 2a.<lb></lb> |
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7419 <lb></lb> |
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7420 Se fosse in un piano e volesse misurare quanto è dal’ ochio tuo, overo da’ pie’ tuoi in-<lb></lb> |
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7421 fino ala somitá d’ una altezza per la linea uscente dal ponto dove voli misurare infi-<lb></lb> |
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7422 no al’ ochio tuo. Come sia l’ ochio tuo nel ponto .a. nel piano e volesse sapere quan-<lb></lb> |
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7423 to è dal ponto .a. infino al ponto .b. che è in sula somitá d’ una altezza, comme sonno e<lb></lb> |
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7424 campanili, monti o simile cose. Per lo modo passato lo puoi fare. Exempli gratia. Poni il ponto<lb></lb> |
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7425 .p. delo strumento gnomico in sul ponto .a. e dal ponto .p. si guardi, essendo la regola del det-<lb></lb> |
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7426 to strumento in sulo lato .pq., anzi uno col detto lato, e guarda per lo foro dela detta regola in<lb></lb> |
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7427 modo che vegga il ponto .b. E, quando l’ ái bene notato, e tu fa stare lo detto strumento senza<lb></lb> |
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7428 muoverlo di nulla. Et etiamdio guarda che non fosse mosso. E, questo fatto, dal ponto .f., con la<lb></lb> |
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7429 regola, guarda il medesimo ponto .b. e nota in che parte delo lato del .rq. la regola passa. E qual be-<lb></lb> |
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7430 <lb></lb> |
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7431 <pb></pb></p> |
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7432 <p class="folio"> |
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7433 folio 50v |
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7434 </p> |
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7435 <p class="main"> |
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7436 <lb></lb> |
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7437 </p> |
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7438 <p class="runhead"> |
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7439 Distinctio septima. Capitulum secudum. |
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7440 </p> |
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7441 <p class="main"> |
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7442 <lb></lb> |
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7443 ne notato, che pongo sia il ponto .d. E tu considera .rd. che parte sia del .sp. e tal parte sia .aq.<lb></lb> |
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7444 del .ab., adonca sia .rd. il terzo d’ un ponto. Conciosiacosaché noi habiamo detto che .rq. sia<lb></lb> |
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7445 .60. ponti, adonca .1/3. de ponto é .1/180. del .sp., per la qual cosa .aq. sia .1/180. del .ab. Onde, quando .aq.<lb></lb> |
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7446 fosse .1o. bracio, la linea visuale che si parte dal ponto .a. e va in verso .b., cioé la linea .ab., è .180.bracia.<lb></lb> |
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7447 E questo era da mostrare. 3a<lb></lb> |
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7448 E similmente è da operare quando tu fosse in sul’ altezza .b. e volesse sapere quan-<lb></lb> |
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7449 to fosse dal ponto .b. infino al ponto .a. Alora porrai lo ponto .p. col ponto .b. e l’ o-<lb></lb> |
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7450 chio tuo poni al ponto .p. e manda la regola secondo lo lato .pq. e, per lo foro de-<lb></lb> |
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7451 la regola, fa divedere il ponto .a. E perfettamente aconcio lo strumento in modo<lb></lb> |
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7452 che non si possa muovere né mutare e tu poni l’ ochio in sul ponto .s. e poni l’ uno lato, overo<lb></lb> |
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7453 capo, dela regola in sul ponto .s. e fa di vedere per lo suo foro il ponto .a., lo quale veduto, ferma<lb></lb> |
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7454 la regola e sappi in che parte del lato .rq. e lla passa, che pongo sia il ponto .d. E truova che par-<lb></lb> |
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7455 te é .dr. del .sp., che pongo sia .1/180., cioé che .dr. sia .1/180. del .sp. Adonca .sp. sia .1/180. del .ab., che,<lb></lb> |
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7456 essendo noto .qp., haremo nota tutta .ab. comme vogliamo. 4a<lb></lb> |
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7457 Se fosse in un piano, cioé in sun un prato e volesse misurare una altezza dela quale il<lb></lb> |
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7458 pie’ si vede con l’ ochio. Comme volendo misurare l’ altezza dela torre .bc. la quale<lb></lb> |
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7459 altezza è .bc. e fosse in sul ponto .a., dal quale ponto si vedesse il ponto .c. e volesse sa-<lb></lb> |
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7460 pere quanto sia .bc. Prima te bisognia sapere la distantia dal .a. al .c., la quale, se al-<lb></lb> |
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7461 tramente non potesse, misurala secondo la prima di questo trattato, cioé collo strumento: o-<lb></lb> |
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7462 vero con le .2. virgole, che pongo che dal .a. al .c. sia .240.bracia. Porrai lo tuo strumento in mo-<lb></lb> |
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7463 do che ’l lato .pq. sia in sul detto piano uno con lo piano .ac. E, questo fatto, e tu poni l’ ochio tuo<lb></lb> |
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7464 al ponto .p. e guarda per la regola, acioché vegga il ponto e noti in che parte delo lato .rq. la<lb></lb> |
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7465 regola passa, cioé quanto é dal .d. al .q., che pongo sia .40. ponti, li quali proportionerai collo<lb></lb> |
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7466 lato .aq., che è .60. ponti. Onde .40. sonno di .60. gli .2/3. Dico che .bc. è gli .2/3. del .ac. E detto é che<lb></lb> |
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7467 .ac. è .240.bracia. Dove gli .2/3. sonno .160. Adonca .bc. è .160.bracia. commo volavamo. Overo mul-<lb></lb> |
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7468 tiplicarai .40. per .240.bracia. e partirai in .60., che ancora ne viene .160.bracia. per l’ altezza .bc. Cioé<lb></lb> |
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7469 multiplicarai .dq. per .ac. e partirai per .aq. e cosí di tutte poi fare. 4a<lb></lb> |
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7470 Ancora questo medesimo si puó fare col quadrante. Commo nel detto exemplo<lb></lb> |
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7471 diciamo che havesse a misurare il .bc., cioé la sua alteza. Dico che pigli el quadran-<lb></lb> |
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7472 te e, per lo verso del’ arco, guarda il ponto .b. E sempre poni l’ ochio per gli buchi<lb></lb> |
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|
7473 del quadrante. E, se lo filo con lo piombo che passa per lo dorso del quadrante é in<lb></lb> |
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7474 mezzo .fe. ale .2. ombre, cioé in sul ponto .n., alora l’ altezza è iguale ala distantia del pie’ tuo, o-<lb></lb> |
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7475 vero all’ ochio tuo ala radici dela torre. E, quando il filo cadese in sula linea .no., alora l’ altezza è<lb></lb> |
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7476 minore dela detta distantia e, quando cade nela linea .mn., alora l’ alteza è magiore. Onde,<lb></lb> |
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7477 adonca, tanto indrieto overo innanzi te farai che il filo caggia in sul mezzo dele dette ombre, cioé<lb></lb> |
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7478 in sul ponto .n., cioé a .45. gradi. E fermati e misura dal tuo pie’ ala radici dela torre. Overo di<lb></lb> |
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7479 quella alteza e, a quella somma, agiongni la tua alteza e harai l’ alteza dela detta torre e que-<lb></lb> |
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7480 sto è assai buon modo. 4a<lb></lb> |
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7481 Ancora questo medesimo si puó fare con l’ ombra del sole. Exempli gratia: rizza<lb></lb> |
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7482 una misura conosciuta equedistante ala torre e quando il sole è in modo che l’ om-<lb></lb> |
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7483 bra dela detta torre si possa bene comprendere, e tu piglia la detta ombra. E simil-<lb></lb> |
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7484 mente piglia l’ ombra dela tua conosciuta misura, e proportionerai l’ ombra dela<lb></lb> |
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7485 tua misura al’ ombra dela torre. E quella parte e veramente sia la tua misura dela torre. Ver-<lb></lb> |
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7486 bi gratia: sia la misura conosciuta .4.bracia., cioé .ef. e l’ ombra dela torre sia .ed. e sia .120. bra-<lb></lb> |
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7487 cia. E la torre sia .bc., sí comme vedi segnato apresso. E l’ ombra dela tua misura sia .fg. e sia .3.<lb></lb> |
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7488 bracia. Dove proportionerai .3.bracia. con .120.bracia.: sonno .1/40. e 1/40 è la tua misura .ef. dela torre.<lb></lb> |
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7489 Adunque la torre è .160.bracia. E questo habia a mente. 4a<lb></lb> |
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7490 Questo medesimo ancora si puó fare col spechio. Cioé dico, chi volesse misurare<lb></lb> |
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7491 una altezza comme è uno albero, el quale è situato in sul piano .ac. Dico che pon-<lb></lb> |
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7492 ga lo spechio in sul piano e da quello ti parti in modo che l’ ochio tuo vegga la ci-<lb></lb> |
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7493 ma del’ albero. E quando ái veduto la cima, cioé il ponto .b., e tu considera che parti è la<lb></lb> |
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7494 tua altezza, cioé .fg. del .ge., tal parte é .cb., cioé l’ altezza del’ albero, del .ce. Comme sia la tua altezza<lb></lb> |
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7495 .3.bracia. e .ge. sia .2.bracia.1/2. e .ce. sia .12.bracia. Dove, a volere l’ altezza del’ albero, .cioé .bc., multipli-<lb></lb> |
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7496 carai .12. per .3., fanno .36. E questo parti in .2 1/2., vienne .14 2/5. E .14.bracia.2/5. sia l’ altezza .bc. e<lb></lb> |
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7497 cosí habia a mente sempre. 5a<lb></lb> |
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7498 <lb></lb> |
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7499 <pb></pb></p> |
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7500 <p class="folio"> |
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|
7501 folio 51r |
|
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7502 </p> |
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7503 <p class="main"> |
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7504 <lb></lb> |
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|
7505 </p> |
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7506 <p class="runhead"> |
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7507 Distinctio septima. Capitulum secundum. 51 |
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7508 </p> |
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7509 <p class="main"> |
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7510 <lb></lb> |
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7511 S e tu fosse in sul ponto .b. d’ altezza del monte o altra cosa e vedese nel piano il pon-<lb></lb> |
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7512 to .a. e volesse sapere quanto é dala linea che si parte dal ponto .a. e sia equedistan-<lb></lb> |
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7513 te al piano del monte, overo dela cosa alta, cioé al piano .be. e dal ponto .b. perpen-<lb></lb> |
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7514 diculare al detto piano del .a., cioé quanto sia dal .b. al .c., che è il congiongnimento<lb></lb> |
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|
7515 del piano .ac. con la radice del’ altezza .bc. Porrai lo tuo strumento in sul piano del .bc. e, quan-<lb></lb> |
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7516 do quello è bene situato e aconcio, che non sia per alcuna cagione movente, e guarda per la<lb></lb> |
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7517 regola, cioé per li fori dela regola, havendo postola al canto .r. E guarda con quella il ponto .a.<lb></lb> |
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7518 e quello, senza niuno impedimento veduto, e tu considera in che parte del lato .sp., overo .pq.,<lb></lb> |
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7519 la detta regola passa, che a questa proposta pongo passi per lo ponto .d., lo quale sia nel lato .sp.<lb></lb> |
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7520 E sia .sd.36. ponti e .dp. sia .24. ponti. E il .rs. sia .60. ponti, cioé lo iguale del .qp.; sia .rd. cir-<lb></lb> |
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7521 ca .70. ponti, cioé piccola cosa meno di .70. ponti, cioé la radici di .4896., la quale é di poca dif-<lb></lb> |
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7522 ferentia a .4900. Queste cose note, e tu per la .3a. di questo sappi quanto é dal .r. al .a., che pon-<lb></lb> |
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7523 go sia .150.bracia. E dipoi adonca dirai, se .36. ponti, cioé .ds., danno .70. ponti, cioé .rd., overo .dr.,<lb></lb> |
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7524 adimando che daranno .150.bracia. Dove con minori numeri dirai, se .18. danno .35., cioé, se<lb></lb> |
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7525 .ds. é .18. e .rd. é .35. e .ra. é .150., che sirá .rc. Multiplicarai .18. via .150., fanno .2700. E parti in<lb></lb> |
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7526 .35., vienne .77 1/7. E .77.bracia.1/7. sia l’ altezza .rc., dela quale altezza traremo l’ altezza del .rq., cioé .1o.<lb></lb> |
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7527 braccio, rimane .76 1/7. E .76.bracia. 1/7. é dal .b. al .c., la quale cosa era da mostrare. 6a<lb></lb> |
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7528 E, volendo sapere quanto è dal .a. al .c., convienti operare comme nela passata, cioé<lb></lb> |
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7529 sapere la longhezza del .r. al .a., per .3am. precedenten, che pongo ancora sienno .150.bracia.<lb></lb> |
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7530 E sappi quanto è dal .r. al .s. e quanto é dal .s. al .d., che pongo sia .60. ponti. E dal<lb></lb> |
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|
7531 .r. al .d. sienno .70. ponti. E dirai, se .70. potumisa danno .60. de piano, cioé, se .7. danno .6.,<lb></lb> |
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|
7532 che daranno .150. Multiplicarai .150. via .6., fanno .900. e partirai in .7., vienne .128 4/7. E .128.<lb></lb> |
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7533 bracia .4/7. sia dal .a. al .c. E questo sempre ti stia a mente. 7a<lb></lb> |
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7534 S e l’ ochio tuo è nel piano e tu voglia misurare una altezza in su uno monte, comme<lb></lb> |
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7535 sonno roche o castella. Comme sia l’ ochio al ponto .d. nel piano del .dc. e in sul<lb></lb> |
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7536 monte .cb. sia la torre .ba., la quale vogliamo misurare. Prima è di bisogno sapere<lb></lb> |
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7537 quanto è dal .d. al .b. E questo, per la seconda di questo, chiaramente l’ arai, che pongo<lb></lb> |
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7538 che dal ponto .d. al .b. sia .320.bracia. E dipoi, ponendo l’ ochio al ponto .d., debbi trovare la lon-<lb></lb> |
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7539 gezza .ad., per .2am. huius, la quale pongo sia .400.bracia. E, tutte queste cose notate, e tu poni l’ ochio<lb></lb> |
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7540 tuo al ponto .d. e poni la regola su per lo lato .pq. e fa di vedere dal ponto .d. il ponto .b. e, quan-<lb></lb> |
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7541 do lo vedi senza alcuno impedimento, e t’ aponi l’ ochio ancora al ponto .d. e mena la regola in<lb></lb> |
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7542 modo che, guardando per la regola, vegga il ponto .a. E quello veduto senza alcuno impedi-<lb></lb> |
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7543 mento, e tu nota in che parte del lato .rp. la regola passa, che dico sia il ponto .f. E questo bene<lb></lb> |
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7544 notato, tu poi, per piú agilitá, levare lo strumento e con questo argomento entrare. Noi<lb></lb> |
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7545 dicemmo che .db. era .320.bracia. e .da. era .400.bracia. Dove, se è possibile, con interi nume-<lb></lb> |
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7546 ri minori in detta proportione havere, togli. Adunque che haremo nei minori numeri, cosí<lb></lb> |
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7547 .4. a .5. comme .320. a .400. Adonca, quando .db. fosse .4.bracia, .da. sarebbe .5.bracia. Per la<lb></lb> |
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7548 qual cosa, io torró delo lato del strumento .qp.30. ponti, cioé il mezzo che sia il ponto .h. dela<lb></lb> |
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7549 linea .pf., ne piglieró tanti quanti é di bisogno sienno in detta proportione a .30. comme .5. a .4. |
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7550 </p> |
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7551 <p class="main"> |
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7552 Dove sia .37 1/2. Dico che dela linea .pf. ne piglieró .37. ponti .1/2., che sia .pu. Dove dipoi menerai<lb></lb> |
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7553 una linea dal .u. al .h., che sienno .20. ponti. E questo perfettamente inteso e compreso, e tu di-<lb></lb> |
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7554 rai, se .37 1/2. danno .20., che daranno .400. Dove multiplicarai .400. via .20., fanno .800. e parti-<lb></lb> |
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7555 ralo in .37 1/2., vienne .213 1/3. E .213.bracia.1/3. sia alta la detta torre, cioé dal .b. al .a. Potresti anco-<lb></lb> |
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7556 ra argomentare in questo modo. Detto é che .ph. è .30. ponti e .hu. è .20. ponti. Onde dirai, se<lb></lb> |
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7557 .30. mi danno .20., che mi dará .320., cioé .cb. Multiplicarai .320. per .20. e partirai in .30. e<lb></lb> |
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7558 vienne .213.bracia.1/3. E tanto è l’ altezza dela torre, comme di sopra trovammo. E nota che l’ al-<lb></lb> |
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7559 tezza .ba. é equedistante ala linea .hu. E questo era da mostrare. 8a<lb></lb> |
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7560 S e fosse in su un monte o altra altezza e vedesse .2. ponti in un piano, comme .2. ca-<lb></lb> |
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7561 se o .2. fonti o .2. albori o quel che si sia. Diciamo .2. case, le quali sonno in su’ ponti<lb></lb> |
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7562 .c. e .d. nel piano .db. E pongo sia in sul’ altezza .ab. nel ponto .a. E vogli sapere<lb></lb> |
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7563 quanto è dal ponto .d. al ponto .c. Piglierai lo strumento atto a questo e con quel-<lb></lb> |
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7564 lo, per la .2a. di questo, sappi la longhezza .ac., la quale pongo sia .120.bracia. E dipoi sapere ti bisognia<lb></lb> |
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7565 quanto è dal .a. al .d. per lo detto modo, cioé per la seconda di questo. Dove pongo sia .140.bracia.<lb></lb> |
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7566 E questo bene conosciuto, e tu ferma l’ ochio in sul ponto .p. ponendo el lato .pq. in sula linea.<lb></lb> |
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7567 .ca. E, volendo conoscere se hai bene adattato lo strumento, poni la regola in sul lato .pq. e, per<lb></lb> |
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7568 gli buchi di detta regola, guarda il ponto .c. E, quando bene lo strumento è adattato, e tu guarda<lb></lb> |
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7569 <lb></lb> |
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7570 <pb></pb></p> |
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7571 <p class="folio"> |
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|
7572 folio 51v |
|
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|
7573 </p> |
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7574 <p class="main"> |
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|
7575 <lb></lb> |
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|
7576 </p> |
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7577 <p class="runhead"> |
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7578 Distinctio septima. Capitulum secundum. |
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|
7579 </p> |
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|
7580 <p class="main"> |
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|
7581 <lb></lb> |
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|
7582 per lo ponto .p. in modo che per gli buchi dela regola tu vegga il ponto .d. E considera in che<lb></lb> |
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7583 ponto del lato .rq. passa la regola, che pongo sia il ponto .g. E quello bene notato, e tu leva lo<lb></lb> |
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7584 strumento e andrai per lo lato .pq.24. ponti, ai quali per la linea .pg. togli tanti ponti che si-<lb></lb> |
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7585 enno nella proportione a .24. comme .140. a .120., che fienno .28. ponti, che pongo sia il pon-<lb></lb> |
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7586 to .o. E questo perfettamente inteso, e tu mena una linea dal ponto .o. agli .24. ponti, che sia<lb></lb> |
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7587 il ponto .f. e perfettamente misura la linea .fo., che pongo truovi sia .12. ponti. Dove questo com-<lb></lb> |
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7588 preso, e tu in questo modo argomenta. E dirai, .se.24. ponti mi danno .12. ponti, che mi da-<lb></lb> |
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|
7589 ranno.120.bracia. Cioé ne’ numeri minori: se .2. danno .1o., che daranno .120.bracia. Multi-<lb></lb> |
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7590 plica .120. per .1o. e parti per .2., vienne .60. e .60.bracia. fienno dal .c. al .d. Overo ancora di-<lb></lb> |
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7591 rai, se la linea .po., che è .28. ponti, mi danno .fo., che è .12. ponti, che mi daranno .140. ponti. Multiplica<lb></lb> |
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7592 .12. per .140., overo per minori numeri, multiplica .3. per .140. e parti per .7., che è quanto a mul-<lb></lb> |
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7593 tiplicare .12. per .140. e partire per .28., vienne sempre .60. E .60.bracia. é dal ponto .c. al ponto<lb></lb> |
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7594 .d., commo volavamo sapere. 9<lb></lb> |
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7595 Se fosse in su n’ una altezza, comme è monte o altra alta cosa e volesse misurare .1a.<lb></lb> |
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7596 altezza posta in piano. Comme essendo in sul’ altezza .ab. e vedesse una altezza in<lb></lb> |
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7597 piano. Comme fosse torre overo albero, del quale si vedesse il pie’, la quale altezza<lb></lb> |
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7598 è .dc., cioé il capo è .d., il pie’ è .c. Convienti prima, per la seconda di questo, misurare dal<lb></lb> |
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7599 ponto .a. infino al ponto .c., cioé infino al pie’ dela torre overo altezza che vuoi misurare, la<lb></lb> |
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7600 quale altezza, comme ó detto, è .dc. Dove pongo che dal .a. al .c. sia .360.bracia. E dipoi ti bi-<lb></lb> |
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7601 sogna che misuri quanto é dal ponto .a. al ponto .d., pure observando il modo de la seconda di que-<lb></lb> |
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7602 sto. Onde dal ponto .a. al ponto .d. pongo sia .320.bracia. E questo fatto, e tu poni lo strumen-<lb></lb> |
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7603 to in modo che ’l lato .pq. sia in sula linea .ac. E questo bene operato, e tu guarda per lo pon-<lb></lb> |
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7604 to .p. lo ponto .d., guardando per gli fori dela regola. E, quando ti pare bene diligentemente<lb></lb> |
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7605 vedere lo ponto .d., e tu nota in che parte del lato .rq. la regola passa, che sia il ponto .g. Dove<lb></lb> |
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|
7606 questo bene segnato, e tu misura del lato .pq.36. ponti e dela linea .pg. ne misura .32. E in cia-<lb></lb> |
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|
7607 scuna linea poni un segno, che sia per me signato .pu.36. ponti e .po.32. ponti e dipoi, dal .o.<lb></lb> |
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7608 al .u., mena una linea retta la quale misura, pongo sia .16. ponti e sia la llinea .ou., la quale è certa-<lb></lb> |
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7609 mente equestistante ala altezza che disideri misurare. Adonca dirai: se .pu., ch’ é .36. ponti, mi dan-<lb></lb> |
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7610 no .ou., cioé d’ altezza .16. ponti, che in minore numero, quanto .9. danno .4., che mi dará .360. |
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|
7611 </p> |
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|
7612 <p class="main"> |
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7613 Dove multiplicarai .360.bracia. per .4. e partirai in .9., vienne .160.bracia. Cioé dicendo se .9.<lb></lb> |
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|
7614 danno .4., che dará la linea .ac. Dará li .4/9. di detta linea, dove gli .4/9. del .ac. sonno .160. Adonca<lb></lb> |
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|
7615 .dc. è .160.bracia. Ancora dirai, se la linea .op. é .8., la linea .uo. sia .4. Onde la linea .ou. è la mitá<lb></lb> |
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7616 dela linea .po. Adonca .dc. è la mitá del .ac., che è .160., comme volavamo intendere. 10.<lb></lb> |
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7617 Se fosse in su n’ un monte overo altezza, la quale sia sopra .1o. monte, comme una tor-<lb></lb> |
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7618 re o albero. Comme sia il monte .ab., sopra il quale sia l’ ochio tuo e vegga il mon-<lb></lb> |
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7619 te .cd., sopra il quale sia la torre dela quale sia de bisogno di sapere l’ altezza. Dico<lb></lb> |
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7620 che per la seconda di questo misuri quanto é dal .a. al .c., che pongo sia .360.bracia. E di-<lb></lb> |
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7621 poi trova, per la medesima ragione di questa, linea .ad., che pongo sia .300.bracia. E questo<lb></lb> |
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7622 fatto, e tu poni lo tuo strumento in modo che ’l lato .pq. sia uno con la linea .pc., cioé .ac. E que-<lb></lb> |
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7623 sto bene adattato e aconcio, e tu poni l’ ochio al ponto .p. e guarda per gli fori dela regola, la<lb></lb> |
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7624 quale sia situata in sula linea, overo lato, .pq. e fa di vedere il ponto .c. E, quando questo è fatto,<lb></lb> |
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7625 e tu muovi la regola tenendo l’ ochio al ponto .p., in modo che tu vegga per gli fori il ponto<lb></lb> |
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7626 .d. E, quando perfettamente lo vedi, e tu fa notare lo passamento dela regola per gli ponti de-<lb></lb> |
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7627 lo lato .rq., el quale noto che pongo sia .g. E tu leva lo strumento e con queste .2. linee, overo mi-<lb></lb> |
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7628 sure, harai quello vuoli. Cioé piglierai dela linea .pg.30. ponti, che fienno .po. E del lato .pq.<lb></lb> |
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7629 ne piglia .36., che fienno .pu. E dal .o. al u. muovi la linea .ou., che pongo sia .6. ponti. Onde in<lb></lb> |
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7630 questo modo arguisci .po. è .30. ponti e danno .ou. che sonno .6. ponti, che daranno .pd. che<lb></lb> |
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7631 sonno .300. ponti. Dove, perché .6. ponti sonno di .30. ponti il .1/5. Ancora l’ altezza .cd., che è eque-<lb></lb> |
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7632 distante alla linea .ou. che sia il .1/5. del .ad. E il .1/5. del .ad. sonno .60.bracia. Overo in questo mo-<lb></lb> |
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7633 do arguirai .ac. è .360.bracia. e .pu. è .36. ponti e .ou. è .6. ponti. Adonca .ou. é il .1/6. del .pu. Cosí<lb></lb> |
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7634 .cd. sia il .1/6. del .ac. e il .1/6. del .ac. é .60.bracia, cioé il .1/6. di .360.bracia, comme era de bisogno tro-<lb></lb> |
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7635 vare, cioé l’ altezza .dc. 11<lb></lb> |
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7636 Se fosse a’ pie’ d’ una altezza, comme sonno torre, e volesse sapere quanto è dal capo<lb></lb> |
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7637 della detta torre infino al capo d’ una altra torre. Comme dicendo che fosse nel<lb></lb> |
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7638 piano .ac. a lato ala torre .ab. e volesse sapere quanto é dal ponto .b. al ponto .d.,<lb></lb> |
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7639 capo d’ un’ altra torre situata in sul piano .ac. Dico prima che, per la .4a. di questo, truovi quanto<lb></lb> |
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7640 <lb></lb> |
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7641 <pb></pb></p> |
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7642 <p class="folio"> |
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|
7643 folio 52r |
|
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7644 </p> |
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7645 <p class="main"> |
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7646 <lb></lb> |
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7647 </p> |
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7648 <p class="runhead"> |
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7649 Distinctio septima. Capitulum secundum. 52 |
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7650 </p> |
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7651 <p class="main"> |
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7652 <lb></lb> |
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7653 é .ab., che pongo sia .100.bracia. e dipoi, per la .2a. di questo, trova quanto è .ad., che pongo sia<lb></lb> |
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7654 .300.bracia. E questo bene fatto e constiuito, e tu poni lo strumento in modo che lo lato .sp.<lb></lb> |
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7655 sia uno con la linea .ba. E, quando perfettamente tutto questo è adatatto, e tu poni l’ ochio<lb></lb> |
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7656 tuo al ponto .p. e, per la regola, fa divedere il ponto .d. E, quando perfettamente lo vedi,<lb></lb> |
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7657 fa notare il ponto dove la regola passa per lo lato .rq., che sia il ponto .g. E questo bene con-<lb></lb> |
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7658 preso, leverai lo strumento e in questo modo adaterai: che pigliarai del lato .ps. lo lato .po.,<lb></lb> |
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7659 che sia .10. ponti. E del lato, overo linea, .pq. ne piglia .30. ponti che sia la linea .au. E dipoi, dal<lb></lb> |
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7660 ponto .o. al ponto .u., menerai la linea che sia la linea .ou., la quale pongo sia .15. ponti. E que-<lb></lb> |
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|
7661 sto inteso, in questo modo argomenta: se .po. è .10. ponti e .ou. é .15. ponti e la linea .ou. è eque-<lb></lb> |
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7662 distante ala linea .bd. e la linea .ab. è equedistante ala linea .po. e la linea .ab. è .100.bracia. On-<lb></lb> |
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|
7663 de cosí comme é .15. a .10. cosí sia .bd. al .ab. Adunque .bd. sia .150.bracia. Imperoché .15.<lb></lb> |
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|
7664 contiene .10. una volta e .1/2. E cosí .bd. contiene .1a 1/2. volta .ab. E peró .db. è .150.bracia. Ove-<lb></lb> |
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|
7665 ramente puoi dire cosí. La linea .pu. é .30. ponti e la linea .ou. è .15. ponti. Adunque .pu. è .2.<lb></lb> |
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7666 cotanti del .uo. Cosí ancora .ad. che è equedistante al .pu. sia .2. cotanti del .bd. che è equedi-<lb></lb> |
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|
7667 stante al .ou. Adunque .bd. sia la mitá del .ad., che sia .150.bracia., cioé la mitá di .300.bracia., che<lb></lb> |
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|
7668 è .ad. E cosí in simili sempre usare poi questi modi. 12<lb></lb> |
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7669 A ncora quando fosse infra .2. altezze, cioé infra .2. torri o .2. alberi overo .1o. albe-<lb></lb> |
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7670 ro e in una torre. E volesse sapere quanto é dal’ uno capo del’ una torre al’ altro<lb></lb> |
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|
7671 capo. In questo exemplo sia il ponto .a. infra .2. torri. In sul ponto .a. sia la tua per-<lb></lb> |
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7672 sona e sienno in sul piano .ead. doi torre: una .dc. e l’ altra .eh. Voglio sapere quan-<lb></lb> |
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|
7673 to é dal ponto .c. al ponto .h. Prima è de bisogno sapere, per la .2a. di questo, quanto è .ac.,<lb></lb> |
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|
7674 che pongo sienno .360.bracia. E dipoi è da sapere pure, per la seconoda, quanto è dal .a. al .h., che pon-<lb></lb> |
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|
7675 go vi sia .200.bracia. E questo bene havuto, e tu poni lo tuo strumento in modo che lo lato<lb></lb> |
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|
7676 .ps. sia uno con la linea .ac. Cioé che, guardando per gli fori dela regola, la quale sia in sul la-<lb></lb> |
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|
7677 to .as., si vegga il ponto. E quello bene constituito, e tu guarda per lo ponto .p. delo strumen-<lb></lb> |
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|
7678 to, adattando la regola senza muovere lo strumento. E, per gli fori di quella, guarda il pon-<lb></lb> |
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|
7679 to .h. E quando perfettamente lo vedi, e tu nota in che parte dela linea, overo lato .qr., la regola<lb></lb> |
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|
7680 passa. E quello nota, che pongo sia il ponto .g. E questo bene fatto, e tu leva lo strumento e<lb></lb> |
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|
7681 togli del lato .ps.36. ponti e dela linea .pg. ne togli .20. ponti, che fienno .pu.36. ponti e .po.<lb></lb> |
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|
7682 fienno .20. ponti. Dipoi mena la linea dal .u. al .o., che sia la linea .uo., la quale pongo sia .40. pon-<lb></lb> |
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|
7683 ti. La quale linea .ou. è equedistante ala linea .hc. Onde con questo argomento entra, cioé,<lb></lb> |
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|
7684 se .pu. che sonno .36. ponti mi danno .40. ponti, che per numeri minori e comme a dire, se .9. pon-<lb></lb> |
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|
7685 ti mi danno .10. ponti, che mi daranno .36.bracia. Multiplica .360. per .10.bracia. e partirai in<lb></lb> |
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7686 .9., vienne .400.bracia. E .400.bracia. sia dal .c. al .h. Overo dirai: se .po., che è .20. ponti, dan-<lb></lb> |
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7687 no .ou., che è .40. ponti, cioé se .1o. dá .2., che dará la linea .ah. che è .200.bracia. Multiplica .200.<lb></lb> |
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7688 per .2. e parti in .1o., vienne .400.bracia. e .400.bracia. é dal .c. al .h. comme dicemno. 13<lb></lb> |
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7689 Se la profunditá d’ un pozzo o altra cosa vuoi misurare. Comme sia il pozzo .acbd.<lb></lb> |
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7690 e la bocca sia larga, cioé il diametro .ad.3.bracia. e similmente il diametro .bc. del<lb></lb> |
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7691 fondo sia .3.bracia. Adimandasi quanto sia dal .a. al .c. Tieni questo modo, che por-<lb></lb> |
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7692 rai una virgola in sul diametro .ad.; in su quella poni lo tuo strumento. E, per lo pon-<lb></lb> |
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7693 to .r., aconciavi la regola e guarda lo ponto .b. E quello perfettamente havuto, e tu considera in<lb></lb> |
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7694 che parte delo lato .pq. passa la regola, che pongo sia il ponto .g. E sia dal .q. al .g. 20. ponti, cioé .1/3.<lb></lb> |
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7695 di bracia. E in questo modo arguirai: se .qg., che è .1/3., mi danno .qr., ch’ é uno bracio, che mi da-<lb></lb> |
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7696 rá .ad. che è .3.bracia. Multiplica .3.bracia. per .1o. e parti in .1/3., viene .9.bracia. E .9. braccia sia .rc., dela qua-<lb></lb> |
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7697 le somma trai .qr., rimane .ac.8.bracia. E questo volavamo sapere. |
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7698 </p> |
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7699 <p class="main"> |
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7700 Se ti fosse di bisogno di misurare uno canapo del quale non potessi vedere il pie’,<lb></lb> |
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7701 comme diciamo si fosse detto: misura el canapo .ab., del quale uno capo .a. non si<lb></lb> |
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7702 vede, ma il .b. si vede. Dico che col capo .b. faccia uno arco, comme ó fatto, che sia<lb></lb> |
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7703 l’ arco .dbe. E sia il ponto .b. nela mitá di detto arco, cioé che tanto sia .db. quanto .be.<lb></lb> |
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7704 E bene notato tutto, e tu misura quanto è .dc. che pongo sia .20.bracia. Poi misura .bc., che po-<lb></lb> |
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7705 go sia .4.bracia. E questo bene detto e fatto, e tu argomenta per la distintione di questo, cioé<lb></lb> |
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7706 che tanto fa .dc. in .ce. quanto .bc. in .co. e .ce. é quanto .cd. Adunque .dc. in .ce. fanno .400. |
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7707 </p> |
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7708 <p class="main"> |
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7709 E .400. debono fare .bc. in .ca. e il .bc. è .4. adunque .ca. sia .100. e tutto .bo. sia .104. E un .ba. é<lb></lb> |
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7710 mitá del .bo. Adunque .ba. sia .52.bracia. E questo era da mostrare. Benché io havessi a<lb></lb> |
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7711 dire di molte altre misure, nientedimeno tutte con queste misurerai. E peró piú dicendo sa-<lb></lb> |
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7712 rebbe superfluo. Adunque a questa distinctione faremo fine et cetera.<lb></lb> |
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7713 <lb></lb> |
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7714 <lb></lb> |
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7715 <pb></pb></p> |
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7716 <p class="folio"> |
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|
7717 folio 52v |
|
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7718 </p> |
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7719 <p class="main"> |
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7720 <lb></lb> |
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|
7721 </p> |
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7722 <p class="runhead"> |
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7723 Distinctio octava. Casus primus. |
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7724 </p> |
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7725 <p class="main"> |
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7726 <lb></lb> |
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7727 Distinctio octava de diversis casibus utilissimis in differenter positis non capitulis sed so-<lb></lb> |
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7728 lo numero casuum distincta.<lb></lb> |
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7729 <lb></lb> |
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7730 In questa ultima distintione s’á a scrivere certi casi trovati in sule distintione pas-<lb></lb> |
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7731 sate, gli quali alcuna sottilitá e pratica al fare ingienerano commo vederari. E, per-<lb></lb> |
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7732 ché infra loro hanno molta diversitá non voglio la presente distinctione in altre<lb></lb> |
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7733 parti dividere. Anzi una sola, onde é di bisogno stia attento a quello che si deba dire.<lb></lb> |
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7734 Poterei dire molti casi in sul misurare dele muraglie. Imperoché ’l tetto se misura<lb></lb> |
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7735 la superficie di fuori. E del muro grosso si misura l’ area corporale e del sottile la su-<lb></lb> |
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7736 perficie, de’ palchi la superficie, dele volte murate la loro superficie di sopra dele pie-<lb></lb> |
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7737 tre concie la loro superficie, de’ davanzali la loro longhezza, e simili cose, le quali em-<lb></lb> |
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7738 pirebbonno uno quaderno di fogli e nullo effecto ingienererebbono. E peró gli lascieró. E<lb></lb> |
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7739 diremo per dare principio. po. |
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7740 </p> |
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|
7741 <p class="main"> |
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7742 E gli é uno cassone longo .3.bracia, alto .1 1/2.bracio, largo .3/4. di bracio. Adimandasi<lb></lb> |
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7743 quanto grano tiene, che tiene il bracio quadro .9. stara di grano. Prima hai<lb></lb> |
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7744 a recare questo cassone a bracia quadri corporei. Per lo modo detto ne’ solidi il fa-<lb></lb> |
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|
7745 rai, cioé multiplicarai .3. via .1 1/2. e quello che fanno per .3/4., fanno .3 3/8. E .3.bracia.3/8. é<lb></lb> |
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|
7746 quadro il detto cassone. E noi diciamo che ’l bracio quadro tiene .9. stara. Dove .3.bracia.3/8.<lb></lb> |
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7747 terrano: .3 3/8. via .9. stara, fanno .30. stara .3/8. E .30. stara .3/8. terrá il detto cassone. E sappi che<lb></lb> |
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7748 da’ nostri antichi fo trovato che uno bracio quadro di corpo, cioé uno bracio longo, uno<lb></lb> |
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7749 bracio largo, un bracio alto, tiene .9. stara di grano, .5. barile di vino, .6. orcia da olio. E, se fosse<lb></lb> |
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7750 pietra, pesa .1600. libre E cosí habia a mente. 2<lb></lb> |
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7751 E gli é una botte che ciascuno fondo è alto .1o. bracio .1/3. e nel cocchiume é .1o. bracio .1/2.<lb></lb> |
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7752 E dal’ uno fondo al’ altro è .2.bracia. Adimando quanto vino tiene. Questa pro-<lb></lb> |
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7753 posta non è altro se non é di quadrare .2. piramide corte. E, comme dicemmo, l’ area<lb></lb> |
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7754 di ciascuna piramide corta viene del multiplicare l’ area dela superficie dela basa<lb></lb> |
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7755 e l’ area del capo dela piramide e quello che è in mezzana proportione fra quelle .2. superfi-<lb></lb> |
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7756 cie per l’ altezza. E di quello pigliare il .1/3. Onde adunque, segando la detta botte per lo cochiu-<lb></lb> |
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7757 me, haremo fatto .2. piramide, dele quali la basa loro é ’l cerchio il cui diametro è .1o. bracio<lb></lb> |
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7758 .1/2. E la sua altezza è uno cerchio lo cui diametro è .1o.bracio.1/3. Dove, per piú facilitá, multiplica .1 1/2. in sé,<lb></lb> |
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7759 fanno .2 1/4. E .1 1/3. in sé, fanno .1 7/9. Dove la superficie mezzana, infra .2 1/4. e .1 7/9., è .2. Adunque habia-<lb></lb> |
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7760 mo a multiplicare .2 1/4. e .2. e .1 7/9., cioé .6 1/36. per lo terzo di .1o., fanno .2 1/108. E di questo piglierai gli<lb></lb> |
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7761 .11/14., che sonno .1 125/216. e tanto è l’ area corporale del’ una piramide. Onde fra amendui fienno .3 17/108. E <lb></lb> |
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7762 diremo la <lb></lb> |
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7763 detta botte essere quadra .3 17/108.bracia. E noi dicemo che ’l bracio quadro tie-<lb></lb> |
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7764 ne .5. barili. Dove in questa botte sia .15 85/108. barili di vino. E cosí sempre in simiglianti è da oprare. 3<lb></lb> |
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7765 E gli é uno tino che ’l fondo è per lo mezzo .4.bracia. e la sua bocha .á. per diametro<lb></lb> |
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7766 .3.bracia. ed é alto .2.bracia. Adimando quanto è quadro, cioé quanto vino tiene<lb></lb> |
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7767 essendo pieno. Ancora questa figura è una piramide corta, dela quale il diame-<lb></lb> |
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7768 tro dela basa è .4.bracia. E dal capo è .3.bracia. Dove multiplicarai .4. in sé, fan-<lb></lb> |
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7769 no .16., e .3. in sé, fanno .9. E la superficie mezzana infra .16. e .9. é .12. Adonca hai a mutiplicare<lb></lb> |
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7770 .16.9.12., cioé .37.37., per lo terzo di .2., cioé per .2/3., fanno .24 2/3., del quale piglia gli .11/14., sonno .19 8/21. E<lb></lb> |
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7771 .19.bracia.8/21. é quadro il detto tino che terrebe, quando fosse pieno di vino, .96. barili .19/21. E, se fos-<lb></lb> |
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7772 se pieno d’uve di piano, terrebbe, overo n’ uscirebbe e .2/3. di vino. E, se di monte, il .1/2. 4<lb></lb> |
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7773 E gli é uno monte di grano che è in su n’ una aia piana che gira intorno .22.bracia.<lb></lb> |
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7774 e nel mezzo è alto .3.bracia. Adimandasi quanto grano v’é, che tiene el bracio qua-<lb></lb> |
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7775 dro .9. stara. Prima quadra, o vogliamo dire truova l’ area del cerchio. Dove<lb></lb> |
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7776 multiplicarai .22. in sé, fanno .484. e partirai in .12 4/7. viene .38 1/2. E questo multi-<lb></lb> |
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7777 ca contro al .1/3. dela altezza, cioé contro al .1/3. di .3., che è .1o., fanno .38 1/2. E .38.bracia.1/2. é quadro il<lb></lb> |
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7778 detto monte di granno. Dove multiplicarai .38 1/2. via .9. stara, fanno .346 1/2. stara. E .346. stara<lb></lb> |
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7779 .1/2. fanno il detto monte. E cosí in simili aopera. 5<lb></lb> |
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7780 Una torre è alta .40.bracia. e da pie’ li passa un fiume largo .30.bracia. Adiman-<lb></lb> |
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7781 dasi quanto sia dala cima dela torre infino al’ orlo del fiume. Questa è propo-<lb></lb> |
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7782 sta in questo modo: che si truova il lato del triangolo oposto al’ angolo retto, che<lb></lb> |
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7783 habiamo mostro che il quadrato di quello lato è quanto e .2. quadrati degli al-<lb></lb> |
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7784 tri .2. lati. Adonca in questa agiongnerai el quadrato della torre col quadrato del fiume, cioé .1600.<lb></lb> |
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7785 con .900., fanno .2500. E .2500. è il quadrato dela misura dala cima della torre infino al’ orlo del fiu-<lb></lb> |
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7786 <lb></lb> |
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7787 <pb></pb></p> |
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7788 <p class="folio"> |
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|
7789 folio 53r |
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7790 </p> |
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7791 <p class="main"> |
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7792 <lb></lb> |
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7793 </p> |
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7794 <p class="runhead"> |
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|
7795 Distinctio octava. 53 |
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7796 </p> |
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7797 <p class="main"> |
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7798 <lb></lb> |
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7799 me, adunque è .50.bracia. 6<lb></lb> |
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7800 Ma dicendo e gli é una torre alta .40.bracia. e da pie’ li passa un fiume che non so quan<lb></lb> |
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7801 to s’é largo, ma ben so che ponendo una fune dala cima dela torre infino al’orlo<lb></lb> |
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7802 del fiume è .50.bracia. Adimandasi quanto è largo il fiume. Multiplicarai .40.<lb></lb> |
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7803 bracia, cioé l’ altezza dela torre in sé, fanno .1600. E multiplicarai la longhezza dela<lb></lb> |
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7804 fune in sé, fanno .2500. Del quale trai .1600., rimane .900. per lo quadrato del fiume. Adon-<lb></lb> |
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7805 ca il fiume è .30. braccia. 7<lb></lb> |
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7806 Dicendo ancora e gli é una torre che non so quanto è alta e da pie’ li passa un fiume lar-<lb></lb> |
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7807 go .30.bracia. E pongo una scala dal’ orlo del fiume infino ala cima dela torre, la quale<lb></lb> |
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7808 è .50.bracia. Adimando quanto è alto la torre. Multiplicarai .50. in sé, fanno .2500., e poi multiplica<lb></lb> |
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7809 .30. in sé, fanno .900. Tra’ de .2500., rimane .1600. per lo quadrato dela torre. Adonca la torre è 40.bracia. 8<lb></lb> |
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7810 L’ é un papiglione ch’é alto .8.bracia. e il diametro è .12.bracia. Adimando quanto pan-<lb></lb> |
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7811 no v’é dentro, che è largo il panno .1o.bracio.1/4. Prima truova l’ area dela superfi-<lb></lb> |
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7812 cie d’ una piramide dela quale la basa sia uno circulo no che ’l diametro è .12.bracia e<lb></lb> |
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7813 la sua altezza è .8.bracia. Prima è di bisogno sapere quanto è dala sommitá de-<lb></lb> |
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7814 la detta piramide infino al’ orlo del circulo, cioé ala circonferentia. Dove multiplicarai .8. in sé,<lb></lb> |
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7815 fanno .64., e il mezzo diametro in sé, fanno .36., agiongni a .64., fanno .100., del quale quadrato è<lb></lb> |
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|
7816 la .R.10. per la longhezza dela circonferentia infino ala sommitá, la quale longhezza multiplica per la mi-<lb></lb> |
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7817 tá dela circonferentia del cerchio dela basa, cioé per .18 6/7., fanno .188 4/7.E .188.bracia. quadro e .4/7.<lb></lb> |
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7818 di panno é nel detto padiglione. E noi habiamo detto che ’l panno è largo bracia .1 1/4. Dove par-<lb></lb> |
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|
7819 tirai .188 4/7. in .1 1/4., vienne .150 6/7. E .150.bracia. di panno .6/7. v’é dentro. 9<lb></lb> |
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7820 E gli é un tondo il cui diametro è .7.bracia., adimando quanto sarebbe per faccia il quadrato che<lb></lb> |
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7821 magiore potesse essere, che entrasse in detto tondo. Comme vedi ciascuno canto di det-<lb></lb> |
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7822 to quadrato tocca il cerchio di detto tondo. Adonca il quadrato è per lo suo diametro quan-<lb></lb> |
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|
7823 to è il diametro del tondo. Onde, se multiplicarai il detto diametro per sé, fanno .49., del quale<lb></lb> |
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|
7824 la mitá è .24 1/2. e .R. di .24 1/2. é la faccia del quadro. E cosí farai le simili. 10<lb></lb> |
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7825 E gli é uno quadro che per faccia è .7.bracia., adimando quanto sia il diametro del<lb></lb> |
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7826 tondo el magiore che si puó fare dentro al detto quadro. Comme chiaro vedi, el tondo<lb></lb> |
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7827 è contingente al detto quadro. E, per quello che s’é detto, movendo la linea dalo ponto del<lb></lb> |
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7828 contato infino al contato dirimpetto, quella linea sará el diametro del detto tondo, el quale<lb></lb> |
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7829 diametro è equidistante al ciascuno de’ .2. lati dal lato del quadro. Adonca è iguali alla faccia del<lb></lb> |
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|
7830 quadro. E peró el diametro del tondo è .7.bracia. comme la faccia del quadrato. 11<lb></lb> |
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7831 E gli é uno scudo che per ciascuno lato è .10.bracia. Vuovi mettere dentro el magiore<lb></lb> |
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7832 tondo che posso. Adimando quanto sirá el diametro del tondo. Comme vedi e gli é<lb></lb> |
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7833 certa cosa che ciascuno lato del triangolo è contingente al cerchio. E, per quello<lb></lb> |
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7834 che s’é detto, movendo una linea dal ponto del contatto infino al centro del det-<lb></lb> |
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7835 to tondo, quella linea sia perpendiculare al detto lato. Adonca menerai una linea da ciascu-<lb></lb> |
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7836 no angolo del detto scudo infino al centro del detto tondo e haremo diviso el detto scudo<lb></lb> |
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7837 in .3. triangoli e quali infra loro sonno iguali e ciascuno á perpendiculare la mitá del diame-<lb></lb> |
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7838 tro del tondo. Adunque porró che la mitá del diametro sia .1a. cosa e quadreró ciascuno di .3.<lb></lb> |
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7839 scudi piccoli multiplicando .1a. cosa per la mitá dello lato dove cade la perpendiculare, cioé .1a.<lb></lb> |
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7840 cosa per la mitá di .10.3. volte, sonno .15.cose. per l’ area superficiale di tutto il triangolo. La qua-<lb></lb> |
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7841 le area è, secondo il modo è .R. di .1875. Adonca .15.cose. sonno iguali a .R. di .1875. Dove la cosa<lb></lb> |
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7842 vale radice de .8 1/3. Adonca la mitá del diametro del tondo è radice di .8 1/3. e tutto il diametro è<lb></lb> |
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7843 radice di .33 1/3. E cosí in simili opera. 12<lb></lb> |
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7844 L’ é uno tondo che ’l diametro suo è radice di .33 1/3. Adimando quanto sarebbe<lb></lb> |
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7845 lo scudo donde tale tondo si traesse. Si bene noterai, troverai, per le cose dette,<lb></lb> |
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7846 che ’l diametro del tondo è ala perpendiculare delo scudo equilatero comme .2.<lb></lb> |
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7847 a .3. Adonca el quadrato delo diametro del tondo allo quadrato dela perpendi-<lb></lb> |
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7848 culare è comme .4. a .9. Onde, multiplicando .33 1/3. per .9., fanno .300., partendo per .4., vienne<lb></lb> |
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7849 .75. per lo quadrato della perpendiculare. Adonca la perpendiculare è radici di .75. E, per le<lb></lb> |
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7850 cose dette, la perpendiculare è ala faccia del triangolo equilatero comme radice di .3. a radice<lb></lb> |
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7851 di .4. Onde il quadrato dela perpendiculare è al quadrato delo lato del triangolo equilate-<lb></lb> |
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7852 ro comme .3. a .4. Dove, multiplicando .75. per .4. e partendo per .3., haremo .100. per lo qua-<lb></lb> |
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7853 drato del lato del triangolo. Adunque lo lato del triangolo è .10. 13<lb></lb> |
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7854 <lb></lb> |
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7855 <pb></pb></p> |
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7856 <p class="folio"> |
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|
7857 folio 53v |
|
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7858 </p> |
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7859 <p class="main"> |
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7860 <lb></lb> |
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7861 </p> |
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7862 <p class="runhead"> |
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7863 Distinctio octava. |
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7864 </p> |
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7865 <p class="main"> |
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7866 <lb></lb> |
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7867 E gli é un tondo che ’l diametro è .7.bracia. vovi mettere dentro il magiore triango-<lb></lb> |
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7868 lo che è possibile. Adimando quanto sia per faccia. Comme chiaro è che dal cen-<lb></lb> |
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7869 tro del tondo a ciascuno angolo del triangolo è la mitá del diametro del tondo.<lb></lb> |
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7870 Adunque dirai: e gli é uno triangolo equilatero che dal centro a ciascuno angolo<lb></lb> |
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7871 è .3 1/2., cioé la mitá del diametro. E noi habiamo detto che quello che è dal’ angolo d’ alcuno tri-<lb></lb> |
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7872 angolo equilatero a ponto di mezzo è ala perpendiculare comme .2. a .3. Adunque multiplica-<lb></lb> |
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7873 rai .3 1/2. per .3. e partirai in .2. e haremo .5 1/4. E .5.bracia.1/4. sia la perpendiculare. La quale perpen-<lb></lb> |
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7874 diculare è ala faccia comme la radici di .3. ala radice di .4. Adunque multiplicarai .5 1/4. in sé<lb></lb> |
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7875 fanno .27 9/16. E questo multiplica per .4., fanno .110 1/4. E questo parti in .3., vienne .36 3/4. e tan-<lb></lb> |
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7876 to sia el quadrato delo lato del triangolo. Adunque il lato sia la radice di .36 3/4. E cosí dirai che<lb></lb> |
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7877 il lato del triangolo equilatero che entra nel tondo il cui diametro è .7.bracia. 14<lb></lb> |
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7878 E gli é uno tondo che ’l suo diametro è .4.bracia. Voglio, nel quarto di quel tondo,<lb></lb> |
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7879 fare una figura quadrata. Adimando quanto sia per faccia. Comme vedi nel di-<lb></lb> |
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7880 segnio, il canto dela figura che è opposto al canto fatto dali diametri. Adunque,<lb></lb> |
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7881 movendo el diametro della detta superficie, sia mezzo il diametro del cerchio. E<lb></lb> |
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7882 peró dirai: e gli é un quadrato del quale il diametro è .4., adimando quanto è il quadrato per<lb></lb> |
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7883 faccia. Multiplicarai .4. in sé, fanno .16., del quale piglia il .1/2., ch’é .8. e tanto è il quadrato dela faccia del quadro.<lb></lb> |
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7884 Adunque il quadro sia per facia la radici di .8. E cosí farai le simili. 15<lb></lb> |
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7885 E gli é un mezzo tondo del quale e il suo diametro è .12.bracia. Adimando quanto<lb></lb> |
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7886 sia la faccia d’ un quadro el magiore che dentro vi cape. Adimando quanto è<lb></lb> |
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7887 per faccia il detto quadro. E gli é vera cosa che, faciendo nel’ altro mezzo ton-<lb></lb> |
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7888 do una figura al detto modo, tu haresti in tutto il tondo una figura quadrilate-<lb></lb> |
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7889 ra che á gli angoli retti e é .2. tanto longa che larga. E ancora è assai chiaro che ’l diametro<lb></lb> |
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7890 dela detta figura è el diametro del tondo. E peró diremo: e gli é una figura quadrilatera che<lb></lb> |
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7891 è .2. tanti longa che larga e il suo diametro è .12.bracia. Adimando quanto è longa e quanto<lb></lb> |
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7892 è larga. Dí: io fo position sia larga .1a. cosa, sia longa .2. cose. Dove il suo diametro sia la .R.<lb></lb> |
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7893 di .5.cose. Adonca .R. di .5.cose. sonno iguali a .12. Onde la cosa varrá la .R. di .28 4/5. E diremo<lb></lb> |
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7894 fosse larga la .R. di .28 4/5. Adonca fo per faccia la .R. di .28 4/5. E cosí farai le simiglianti. 16<lb></lb> |
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7895 E gli é uno vivaio che è longo .8.bracia, largo .6.bracia, alto .4.bracia. Ed é pieno<lb></lb> |
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7896 d’ aqua; cadendovi dentro una pietra ch’ é quadrata ed é per ogni verso .3.bra-<lb></lb> |
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7897 cia, adimando quanta aqua n’ uscirá, che comme ó detto il brachio quadro tiene .5.<lb></lb> |
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7898 barili. Noi habbiamo detto che ’l vivaio è pieno. E peró, non per altro, si pone la mi-<lb></lb> |
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7899 sura del vivaio, se non per mostrare che la pietra facilmente nel vivaio entra sotto l’aqua. Adon-<lb></lb> |
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7900 ca troverai l’ area corporale di detta pietra: che multiplicarai .3., che è longa, via .3., che è alta, via<lb></lb> |
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7901 .3., che è grossa, fanno .27. E .27.bracia. solide è la detta pietra. E ogni bracio tiene .5. barili. Do-<lb></lb> |
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7902 ve .27.bracia. terranno .135. barili. Adonca n’ uscirá .135. barili e cosí fa le simili. 17<lb></lb> |
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7903 E gli é uno vivaio che è longo .8.bracia. e largo .6.bracia. e alto .6.bracia. Ed évi alta<lb></lb> |
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7904 l’ aqua .4.bracia; cadevi dentro .1a. pietra quadrata, che è per ogni verso .3.bracia.<lb></lb> |
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7905 Adimando quanto vi debbe cresciere l’ aqua. Quadrerai prima la pietra, multiplicando .3.<lb></lb> |
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7906 che è longa, via .3., che è larga, via .3., che è grossa, fanno .27. E .27.bracia. è quadra. Le quali .27.<lb></lb> |
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7907 partirai per quello che fanno a multiplicare la longhezza del vivaio, per la larghezza, cioé per .48., vienne .9/16. E .9/16.<lb></lb> |
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|
7908 di bracio alzerá l’ aqua nel vivaio. Adonca vene sia .4.bracia.9/16. Cosí farai le simili. 18<lb></lb> |
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7909 E gli é una pietra tonda che ’l diametro è .2.bracia. Adimando quanto pesa. Prima ái a tro-<lb></lb> |
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7910 vare l’ area corporale di detta pietra. Dove multiplica il diametro .3. volte in sé, fanno .8. e di que-<lb></lb> |
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7911 sto piglia .11/21., che sonno .4 4/21. E .4. bracia .4/21. é quadra la detta pietra e detto habia-<lb></lb> |
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7912 mo che ’l bracio quadro pesa .1600. libre. Adonca multiplicarai .1600. via .4 4/21. fanno .6704 16/21.<lb></lb> |
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7913 E .6704. e .16/21. sia il peso della detta pietra. 19<lb></lb> |
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7914 Se ti fosse proposto un tondo e tu ne voglia fare uno altro di .2. tanta possessione, fa-<lb></lb> |
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7915 rai un quadro el minore che puoi, il quale contenga in sé el detto tondo. E dipoi farai un al-<lb></lb> |
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7916 tro tondo, il quale contenga in sé el detto quadro. Dico che ’l tondo di fuora è .2. cotanti di pos-<lb></lb> |
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7917 sessione che ’l tondo dentro. Imperoché ’l quadrato del diametro di quel di fuora è .2. tanti che’ l<lb></lb> |
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7918 quadrato del tondo dentro, comme chiaro appare per la figura et cetera. 20<lb></lb> |
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7919 Se ti fosse detto di fare un quadro di .2. cotanta possessione che uno quadro dato,<lb></lb> |
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|
7920 comporrai il minore quadro. E di fuora a quello farai un minore cerchio puoi, il quale con-<lb></lb> |
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7921 tenga il detto quadro e, di fuora a quel tondo, comporrai uno quadro el minore si puó,<lb></lb> |
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7922 che contenga quel tondo. Dico che ’l quadro di fuora è .2. cotanti che ’l quadro dentro.<lb></lb> |
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|
7923 <lb></lb> |
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|
7924 <pb></pb></p> |
|
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|
7925 <p class="folio"> |
|
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|
7926 folio 54r |
|
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|
7927 </p> |
|
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|
7928 <p class="main"> |
|
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|
7929 <lb></lb> |
|
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|
7930 </p> |
|
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|
7931 <p class="runhead"> |
|
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|
7932 Distinctio octava. 54 |
|
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|
7933 </p> |
|
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|
7934 <p class="main"> |
|
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|
7935 <lb></lb> |
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7936 Imperoché ’l quadrato dela faccia di quel di fuora è .2. cotanti che ’l quadrato della faccia di<lb></lb> |
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7937 quel di dentro, comme chiaro appare nela figura. 21<lb></lb> |
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7938 E gli é una figura quadrata che per ogni faccia è .2.bracia; vovi mettere dentro .2.<lb></lb> |
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7939 tondi e magiori che io posso. Adimando quanto sia il loro diametro. E gli é co-<lb></lb> |
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7940 sa assai chiara che li detti tondi debbono essere collocati in sul diametro del qua-<lb></lb> |
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7941 dro in modo che infra loro sonno contingenti. Dico adonca, se si menerá per lo<lb></lb> |
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7942 ponto dela loro contingentia il diametro, certamente quel quadrato è diviso in .2. triangoli equi-<lb></lb> |
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7943 curij e in ciascuno triangolo è collocato uno de’ detti tondi. Ma perché io nonn’ ó posto il mo-<lb></lb> |
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7944 do a mettere in uno triangolo equicurio il tondo porremo questo caso. Cioé: 22<lb></lb> |
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7945 E gli é uno triangolo equicurio che per ciascuna dele .2. faccie iguali è .10.bracia. e<lb></lb> |
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7946 per l’ altra è .12.bracia. Vovi mettere il magiore tondo che vi cape. Adimando quan-<lb></lb> |
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7947 to sia il suo diametro. Certamente, se dal centro del detto tondo a caduno ango-<lb></lb> |
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7948 lo del triangolo menerai una linea, tu dividerai il tuo triangolo in .3. triangoli e<lb></lb> |
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7949 ciascuno lato del grande triangolo sia la basa d’ uno triangolo piccolo, ale quali base menan-<lb></lb> |
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7950 do una linea dal centro del detto tondo al ponto dove il tondo è contingente, sia quella li-<lb></lb> |
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7951 nea perpendiculare. E questo dichiarato, e tu dirai: io pongo che dal centro del tondo infi-<lb></lb> |
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7952 no al ponto dove il cerchio tocca la basa sia una cosa, cioé che la mitá del diametro sia una co-<lb></lb> |
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|
7953 sa. Dove troverai l’ area de’ .3. triangoli, la quale s’ ará multiplicando la mitá di ciascuna basa in<lb></lb> |
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7954 una cosa. Adunque harai in tutto l’ area de’ .3. triangoli piccoli é quanto l’ area del gran trian-<lb></lb> |
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|
7955 golo, la quale è .48. Adunque .16.cose. sonno iguali a .48. Dove la cosa vale .3. Adunque la mi-<lb></lb> |
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|
7956 tá del diametro è .3. e tutto è .6. e questo era bisogno mostrare. 23<lb></lb> |
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7957 Hora al proposito nostro diremo: e gli é uno triangolo equicurio che per ciascu-<lb></lb> |
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|
7958 na dele faccie iguali è .6.bracia. e per l’ altra è la radice di .72., cioé il diametro del<lb></lb> |
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|
7959 quadro. Adimando quanto sia il diametro del tondo che magiore vi cappia.<lb></lb> |
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7960 Comme ó detto, dividendo (per la linea che si muova dal centro del tondo e vada a<lb></lb> |
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|
7961 ciascuno angolo) il deto triangolo, si fará .3. triangoli, de’ quali l’ area s’á del multiplicare la mi-<lb></lb> |
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|
7962 tá del diametro del tondo in mitá dela basa sua. Onde io porró il mezzo diametro del ton-<lb></lb> |
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|
7963 do sia una cosa. Adunque il detto triangolo, che è risoluto in .3. triangoli, é quadro .6.cose. e<lb></lb> |
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|
7964 radici di .18.censi. E noi sappiamo che gli é quadro .18., cioé la mitá di .36., che è tutto il quadro.<lb></lb> |
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|
7965 Adunque .6.cose. e radice di .18.censi. sonno iguali a .18. Dove la cosa vale .6. meno radice di<lb></lb> |
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|
7966 .18. E tu ponesti il mezzo diametro essere una cosa. Adunque tutto il diametro del tondo è .12.<lb></lb> |
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|
7967 bracia meno radice di .72. e diremo che ’l diametro del tondo sia .12. meno radice di .72. e co-<lb></lb> |
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|
7968 sí sempre opera. 23<lb></lb> |
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|
7969 E sonno .2. tondi iguali, de’ quali il diametro di ciascuno è .4.bracia. Vogli collocare in<lb></lb> |
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|
7970 nel minore quadro possono, in modo che gli stieno inn ischiso. Adimando quan-<lb></lb> |
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|
7971 to sia per faccia il quadro. Comme vedi, io divideró il detto quadro per lo dia-<lb></lb> |
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|
7972 metro passante per lo ponto del loro toccamento, non segando niuno degli ton-<lb></lb> |
|
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|
7973 di e haró diviso il detto quadro in .2. triangoli iguali. E in ciascuno sia collocato uno de’ .2.<lb></lb> |
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|
7974 tondi. E questo inteso, io faró positione che ’l quadro sia per faccia una cosa. Adonca .1o. de’ .2.<lb></lb> |
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|
7975 triangoli sirá per le .2. faccie, per ciascuna, una cosa. E per l’ altro lato sia radice di .2.censi. El qua-<lb></lb> |
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7976 le triangolo è risoluto in .3. triangoli per le linee che escono dal centro del tondo e vanno a<lb></lb> |
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7977 ciascuno angolo. Onde quadra el detto triangolo multiplicando .2.bracia., cioé la mitá del<lb></lb> |
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7978 diametro, per la mitá di ciascuna basa. E haremo il triangolo, che è risoluto in .3. triangoli, es-<lb></lb> |
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7979 sere quadro .2.cose. e .R. di .2. censi. E detto è che uno di quelli triangoli è la mitá del quadro. Adonca<lb></lb> |
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7980 del quadro è la sua area .4.cose. e .R. di .8.censi. E detto è che ’l quadro è per faccia .1a.cosa. Dove sia<lb></lb> |
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7981 la sua area .1o.censo. Adonca .1o. censo è iguali a .4.cose. e .R. di .8.censi. Dove la cosa varrá .4. e<lb></lb> |
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7982 radice di .8. è .4. e radice di .8. sia per faccia il quadro e cosí fa sempre. 24<lb></lb> |
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7983 E gli é uno quadro che è per faccia .5.bracia.; metto nel canto una colonna che gira<lb></lb> |
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7984 .6.bracia.2/7. Cioé è il suo diametro .2.bracia. Adimandasi quanto sia il quadro per<lb></lb> |
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7985 faccia che rimane dentro ale colonne. Prima è da vedere di quanta grandez-<lb></lb> |
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7986 za è uno quadro dove entrasse un tondo il cui diametro sia .2.bracia. dove che sia .2.bracia.<lb></lb> |
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7987 per faccia. E di questo quadro trova il diametro che è radice di .8. e radici di .8. è quello ch’ é in-<lb></lb> |
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7988 fra gli .2. canti che sonno tra la circonferentia e del tondo e il canto del quadro con uno<lb></lb> |
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7989 diametro d’ uno de’ tondi. Adunque trarai .2. e radice di .8. del diametro del quadro, el<lb></lb> |
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7990 quale è radice di .50. Dove, tratto radice di .8. di radice di .50., rimangono radice .18.,<lb></lb> |
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7991 dela quale trai .2., rimangono radice di .18. men .2. E questo è il diametro del quadro<lb></lb> |
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7992 <lb></lb> |
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7993 <pb></pb></p> |
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7994 <p class="folio"> |
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7995 folio 54v |
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7996 </p> |
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7997 <p class="main"> |
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7998 <lb></lb> |
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7999 </p> |
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8000 <p class="runhead"> |
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8001 Distinctio octava. |
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8002 </p> |
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8003 <p class="main"> |
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8004 <lb></lb> |
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8005 che rimane. Adonca hai a trovare uno quadro che il suo diametro sia radice di .18. men .2. |
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8006 </p> |
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8007 <p class="main"> |
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8008 Dove, a trovarlo, multiplicarai radice di .18. men .2. in sé, fanno .22. men .R. di .288., la qual somma par-<lb></lb> |
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8009 ti in .2. viene .11. men radice di .72. e la radici di questa somma sia il lato del quadrato, cioé<lb></lb> |
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8010 presa la radice del .72., tratta de .11., e del rimanente preso la radice, comme volavamo. 52<lb></lb> |
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8011 E gli é una scala di non so che longhezza la quale è ritta a un muro che è iguale a-<lb></lb> |
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8012 la longhezza dela scala; discostola da pie’ .6.bracia. e abassossi la vetta dela scala da-<lb></lb> |
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8013 la sommitá del muro .2.bracia. Adimandasi quanto è la scala longa. Benché<lb></lb> |
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8014 alcum maestro che pone questi casi gli asolva per altri modi, noi per l’ algebra gli<lb></lb> |
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8015 asolveremo. Porremo adonca la longhezza dela scala sia una cosa. Sirá adonca dal .a. al .b.<lb></lb> |
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8016 Sia la radice d’ uno censo meno .36. Adonca dal .b. al .d. sia .2. e radice d’ uno censo men .36. E<lb></lb> |
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8017 noi dicemo che dal .b. al .d. era .1a. cosa. Adonca .1a. cosa è iguale a .2. e radice di .1o. censo men<lb></lb> |
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8018 .36. Onde raguaglia le parti e harai .1a. cosa meno .2. iguali a radice d’ uno censo meno .36. Do-<lb></lb> |
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8019 ve recha ciascuna quantitá a quadrato e harai .1o. censo meno .36. iguali a un censo meno .4.<lb></lb> |
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8020 cose e .4. piú. Dove raguaglia le parti, levando da ciascuna parte uno censo e .4. e dando a cia-<lb></lb> |
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8021 scuna parte .36. per numero. E harai che .4.cose. sonno iguali a .40. Dove la cosa vale .10. Adon-<lb></lb> |
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8022 ca la scala fu .10.bracia. E questo volavamo mostrare. 26<lb></lb> |
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8023 E gli é una scala alta .10.bracia., la qual è congionta al muro iguali ala scala. Disco-<lb></lb> |
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8024 stola dal pie’ tanto che, agionto a quello ch’ é lo capo dela scala discende dal mu-<lb></lb> |
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8025 ro, sia .8. Adimandasi quanto la scostai dal muro. Pure per l’ algebra l’ asolvere-<lb></lb> |
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8026 mo dicendo: io pongo che io la scostasi una cosa. Cioé che dal .b. al .c. sia .1a. cosa.<lb></lb> |
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8027 Adonca dal .a. al .d. sia .8. meno .1a. cosa. E dal .b. al .a. sia radice di .100. meno uno censo. Adon-<lb></lb> |
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8028 ca, dal .b. al .d., é .8. meno .1a. cosa e radice di .100. meno .1o. censo. E noi dicemo che gli era .10.<lb></lb> |
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8029 bracia. Adonca .10.bracia. sonno iguali a .8. men .1a. cosa e radice .d.100. men .1o. censo. Do-<lb></lb> |
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8030 ve, da ogni parte, leva .8. men .1a. cosa, haremo che .2. e .1a. cosa sonno iguali a radice di .100.<lb></lb> |
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8031 meno .1o. censo. Dove multiplicarai ciascuna parte in sé, haremo che .1o. censo e .4.cose. e .4. son-<lb></lb> |
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8032 no iguali a .100. men .1o. censo. Dove reguaglia le parti, haremo che .2.censi. e .4.cose. sonno<lb></lb> |
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8033 iguali a .96. Dove opera secondo la regola, harai che la cosa vale .6. Adonca dal .b. al .c. fo .6.<lb></lb> |
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8034 e dal .a. al .d. è .2. e cosí farai le simili. E nota che sempre quello che la discende dal capo è me-<lb></lb> |
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8035 no che quello che la si discosta dal pie’ del muro. 27<lb></lb> |
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8036 E gli é una scala che acosto al muro d’ iguali altezza. La quale scala è .10.bracia.<lb></lb> |
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8037 Discostola di sotto in modo che quello che la vetta venne in verso terra tratto di<lb></lb> |
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8038 quello si discostó dal pie’, rimangono .4. Adimando quanto si discostó da pie’ e quan-<lb></lb> |
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8039 to da capo. Dirai pure per la regola d’ algebra. Io pongo che dal .a. al .d. sia .1a. co-<lb></lb> |
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8040 sa. Adunque dal .a. al .b. sia .10.bracia. meno .1a. cosa. Onde è da sapere quanto è dal .a. al .c.<lb></lb> |
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8041 Mulitiplicarai .ab. in sé, cioé .10. men .1a. cosa, fará .100. men .20.cose. e piú .1o.censo. E multiplica<lb></lb> |
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8042 .bc. in sé, fanno .1o. censo e .8.cose. e .16. Agiongni al .1o. censo e .100. men .20.cose., fanno .2. censi<lb></lb> |
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8043 e .116. meno .12.cose. e la .R. di .2.censi. e .116. men .12.cose. è .ab. E noi dicemo era .10. Adonca<lb></lb> |
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8044 .10. è iguali ala radice di .2.censi. e .116. meno .12.cose. Dove multiplica ogni parte in sé e hare-<lb></lb> |
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8045 mo .100. essere iguali a .2.censi. e .116. meno .12. cose. Dove, da ogni parte, leva .100. e, a ogni par-<lb></lb> |
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8046 te, darai .12.cose. e haremo che .2.censi. e .16. sonno iguali a .12.cose. Dove, secondo la regola,<lb></lb> |
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8047 la cosa vale .2. Adonca dal .a. al .d. è .2. e dal .b. al .c. è .6. e cosí farai le simili. 28<lb></lb> |
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8048 E gli é una scala che è longa .10.bracia. ed é appogiata a uno muro di simile altezza;<lb></lb> |
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8049 discostola dal pie’ tanto che la vetta viene in verso terra il .1/3. di quello che io la disco-<lb></lb> |
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8050 sto. Adimando quanto la discostai. Dirai: io fo positione che dal .a. al .d. sia .1a.<lb></lb> |
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8051 .cosa.; sia dal .b. al .c.3.cose. E, se .ad. è .1a.cosa., .ab. sia .10. meno .1a.cosa. Onde diremo<lb></lb> |
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8052 .ab. è .10. men .1a.cosa. e .bc. è .3.cose. Dove, a sapere che è .ac., multiplicarai .10. meno .1a. cosa in<lb></lb> |
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8053 sé, fanno .100. e .1o. censo meno .20.cose. E dipoi multiplicarai .bc. in sé, fanno .9.censi.; agiongni<lb></lb> |
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8054 a .100. e .1o. censo meno .20.cose., fanno .100. e .10.censi. meno .20.cose. e questo è iguali al quadra-<lb></lb> |
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8055 to del .ac., cioé a .100. Dove reguaglia le parti, harai che .10.censi. sonno iguali a .20.cose. Dove la cosa<lb></lb> |
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8056 vale .2. Adonca dal .a. al .d. fo .2.bracia. e dal .b. al .c. forono .6.bracia. E questo era da mostrare. 29<lb></lb> |
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8057 L’ é una scala di non so che longhezza, acostata a un muro di simile altezza. Ólla di-<lb></lb> |
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8058 scostato di sotto tanto che, agionto a quello che la s’ abassa giú pel muro, fanno .8. e multiplicato<lb></lb> |
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8059 quello che io la discostai per quello che ’l capo venne in giú feciono .12. Adimando quanto era lon-<lb></lb> |
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8060 ga la scala. Prima ái a ffare di .8. due parti che, multiplicata l’ una nel’ altra, facciamo .12. |
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8061 </p> |
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8062 <p class="main"> |
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8063 Dove torrai la mitá di .8., che è .4., e multiplica in sé, fanno .16., trane .12., rimangano .4., de’ quali la<lb></lb> |
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8064 radice sonno .2., trala di .4., riman .2. Adunque l’ una parte é .2., l’ altra è .6. Ora diremo<lb></lb> |
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8065 <lb></lb> |
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8066 <pb></pb></p> |
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8067 <p class="folio"> |
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|
8068 folio 55r |
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8069 </p> |
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8070 <p class="main"> |
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8071 <lb></lb> |
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8072 </p> |
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8073 <p class="runhead"> |
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8074 Distinctio octava. 55 |
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8075 </p> |
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8076 <p class="main"> |
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8077 <lb></lb> |
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8078 e gli é una scala che è longa non so quanto, la quale è agionta, cioé achosto, a uno muro d’ iguali al-<lb></lb> |
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8079 tezza. Óla dischoto di sotto .6.bracia. e lo capo venne in giú .2.bracia. Adimando quanto è la<lb></lb> |
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8080 scala. Dove, per la .15a. di questo, troverai che la scala è .10.braza. 30<lb></lb> |
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8081 E gli é una scala di non so che altezza, congionta a un muro d’ iguali altezza. Ó discosto<lb></lb> |
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8082 il pie’ tanto che lo capo venne in giú tanto che, tratto di quello che ’l pie’ si discostó rima-<lb></lb> |
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8083 sono .4. e, multiplicato quello discostai di sotto per quello che la vetta andó in giú, fecino .12. Adi-<lb></lb> |
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8084 mando quanto è la scala longa. Prima dirai truova .2. numeri che l’ uno sia piú che l’ altro .4.<lb></lb> |
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8085 e multiplicato l’ uno per l’ altro faccino .12. Che, per gli modi dati multotiens, harai l’ uno essere .2. e l’ altro<lb></lb> |
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8086 .6. Onde dirai: quando si discosta .6.bracia., la vetta vien .2.bracia. in verso terra che, per la .25a.<lb></lb> |
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8087 di questo, harai la scala era .10.bracia. 31<lb></lb> |
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8088 E gli é uno albore alto .30.bracia. Talgliasi in tal parte che la vetta sta in terra apresso al pe-<lb></lb> |
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8089 dale a .10.bracia. e l’ altro capo è appiccato al’ albore. Adimandasi in che parte si rup-<lb></lb> |
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8090 pe lo detto albore. Di molti modi che pone maestro Gratia, questo mi pare piú bel-<lb></lb> |
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8091 lo e di sottile inventione. Cioé poniamo che l’ albero, che è .30.bracia., sia la saetta .ag., la quale passi per<lb></lb> |
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8092 lo centro .d., che sia in luogo dove l’ albore si rompe. E .bg. sia quello ch’ é discosto la vetta quando ela<lb></lb> |
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8093 é in terra, che sirá la mitá della corda del’ arco .bkc. E, per quello che s’ é detto de’ tondi, tanto fa a multiplica-<lb></lb> |
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8094 re .bg. in .gc. quanto .ag. in .gk. E noi dicemo che .bg. è la mitá del .bc. Adonca, a multiplicare .bg. in .gc.,<lb></lb> |
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8095 fanno .100., el quale partirai in .30., vienne .3 1/3. e tanto è .gk. Adonca tutto .ak. é .33.bracia.1/3. E, perché tan-<lb></lb> |
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8096 to é .ad. quanto .dk., sirá .ad. la mitá del .33 1/3., cioé .16 2/3. E .16.bracia.2/3. é dal .a. al .d. e tante ancora<lb></lb> |
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8097 fienno dal .b. al .d., cioé quella parte che cadde. E quello che rimase sia l’ avanzo infino in .30.bracia., che<lb></lb> |
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8098 sia .13.bracia.1/3. e .13.bracia.1/3. sia quello che rimase, cioé .dg. e questo era da mostrar. 32<lb></lb> |
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8099 .4. bracia di corda legono uno fastello de .100. verghe. Adimandasi quante verghe si-<lb></lb> |
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8100 mili fie legate da una corda di .10.bracia. Multiplicarai .4. in sé, fanno .16. e .10. in sé, fanno .100. |
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8101 </p> |
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8102 <p class="main"> |
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8103 E diremo: se .16. legano .100. verghe, quanto legheranno .100. Multiplica .100. via .100. verghe,<lb></lb> |
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8104 fanno .10000. verghe, patirale in .16., vienne .625. e .625. verghe legheranno .10.bracia. di cor-<lb></lb> |
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8105 da. E cosí farai le simiglianti. E la cagione è che tal parte debba essere le verghe ale verghe com-<lb></lb> |
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8106 me el quadrato dele .4.bracia. al quadrato dele .10.bracia. E dicendo e gli é .4.bracia. che le-<lb></lb> |
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8107 gono uno fastello di .100. verghe. Adimando quante bracia legheranno .625. vergole. Mul-<lb></lb> |
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8108 tiplicarai .625. via .16.bracia., fanno .10000. Dico .16., cioé el quadrato di .4. E questo .10000. parti-<lb></lb> |
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8109 rai in .100., vienne .100. e la .R. di .100. braccia fienno quelle che legheranno .625. verghe e cosí et cetera. 33.<lb></lb> |
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8110 F irenze gira intorno .5. miglia. Ed é grosso il muro .3.bracia.1/2. e il fosso è largo .14.<lb></lb> |
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8111 bracia. Adimando quanto girerá di fuori in sul’ orlo del fosso. Perché il diametro è<lb></lb> |
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8112 cresciuto .2. cotanti, imperoché da ogni parte è agionto .3.bracia.1/2. e .14.bracia.<lb></lb> |
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8113 dele mura, adonca il diametro crescie .35.bracia. Dove diremo .35.bracia. de diame-<lb></lb> |
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8114 tro danno .35. via .3.bracia.1/7., fanno .110.bracia. E .110.bracia. girerá piú di fora. Adunque gi-<lb></lb> |
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8115 rerá .5. miglia e .110.bracia. e cosí se usa in simili. 34<lb></lb> |
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8116 Firenze gira .7. miglia ed é tonda; Prato é tondo e gira .2. miglia. Adimando quante<lb></lb> |
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8117 volte entrará Prato in Firenze. Como è mostro per lo .11o. libro de Euclide, tu debbi<lb></lb> |
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8118 multiplicare il giro di Firenze in sé e partire per la multiplicatione del giro di Prato in sé. Cioé<lb></lb> |
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8119 per .4. vienne .12 1/4. e .12. volte .1/4. entrará Prato in Firenze. E cosí farai le simili. 35<lb></lb> |
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8120 Roma è tonda e gira .33.miglia. Constantinopoli è per triangolo e gira .42. miglia,<lb></lb> |
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8121 in questo modo, ch’ é per l’ una faccia .15. miglia, per l’ altra .13. e per l’ altra .14. miglia. Adiman-<lb></lb> |
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8122 do chi possiede piú terreno. Quadrerai Roma multiplicando il giro per sé e pigliando di par-<lb></lb> |
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8123 tire in .12 4/7., vienne .86 5/8. E .86. miglia e .5/8. é il terreno che possiede Roma. E dipoi fa-<lb></lb> |
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8124 rai el terreno che possiede Constantinopoli. Dove quadrarai Constantinopoli per lo modo che dicemmo<lb></lb> |
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8125 ne’ triangoli, che troverai che possiede .84. miglia. Dove Roma possiede magiore terreno e peró<lb></lb> |
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8126 partirai .86 5/2. in .84., vienne .1 1/32. e .1a. volte e .1/32. entrará Constantinopoli in Roma e cosí habi a mente.<lb></lb> |
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8127 36<lb></lb> |
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8128 E gli é un mantello che è alto .2.bracia. Vo’ sapere quanto panno v’ entrará, che è largo il<lb></lb> |
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8129 panno .1o. bracio .1/2. E gli é vera cosa che il mantelo è .2. tanti il suo diametro che la<lb></lb> |
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8130 sua altezza. Imperoché, ponendolo in terra, farebbe un tondo, adonca dirai: e gli é .1o.<lb></lb> |
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8131 tondo che il suo diametro è .4.bracia. Adimando quanto é quadro. Multiplicarai .4. in sé, fanno<lb></lb> |
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8132 .16., de’ quali piglia li .11/14., che sonno .12 4/7. E .12.bracia.4/7. quadre sonno le bracia del panno del detto man-<lb></lb> |
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8133 tello. E noi diciamo che ’l panno è largo .1o.bracio.1/2. Dove partirai .12 4/7. in .1 1/2., viene .8.bracia.<lb></lb> |
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8134 .8/21. E .8.bracia.8/21. sia bisognio di panno a ffare il detto mantello. 37<lb></lb> |
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8135 E gli é uno che vole fare uno vestire e dice al sarto: Io ó trovato uno panno ch’ é lar-<lb></lb> |
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8136 go .1o.bracio.2/3. E ‘l detto sarto gli dice: .9.bracia. di panno te ’l faranno e, tornando al fondecho<lb></lb> |
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8137 <lb></lb> |
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8138 <pb></pb></p> |
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8139 <p class="folio"> |
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8140 folio 55v |
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8141 </p> |
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8142 <p class="main"> |
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8143 <lb></lb> |
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8144 </p> |
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8145 <p class="runhead"> |
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8146 Distinctio octava. |
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8147 </p> |
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8148 <p class="main"> |
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8149 <lb></lb> |
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8150 truova quel panno essere venduto. Ma éve, di quel colore, panno che è largo .1o. bracio .1/2.<lb></lb> |
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8151 Adimandasi quanto panno debba torre. Bisogniasi multiplicare .9. via .1o. bracio .2/3., che è la<lb></lb> |
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8152 largheza del primo panno, fanno .15.bracia. E questo partirai in .1 1/2., viene .10.bracia. E .10.bracia. di pan-<lb></lb> |
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8153 no sia di bisognio a ffare la detta roba, overo vestire, del secondo panno e cosí usa et cetera. 38.<lb></lb> |
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8154 Uno toglie a cavare uno pozzo hadentro .10.bracia. e debene havere .10. denari. Álo cava-<lb></lb> |
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8155 to .6.bracia. e non n’ é di bisognio lo cavi piú. Adimando quanto debba havere. Benché<lb></lb> |
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8156 in diversi modi per alcuni si dica, io voglio tenga questo, cioé che gli é cosa assai ma-<lb></lb> |
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8157 nifesta che al .2o.bracio. egli .á. la fatica del .primo. E al terzo bracio egli á la fatica del .primo. e .2o.<lb></lb> |
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8158 E cosí al quarto bracio egli .á. la fatica del .primo. 2o. 3o.bracio. e cosí degli altri. Adonca, al diecimo, egli á<lb></lb> |
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8159 le fatiche degli altri. E peró, a volere dare absolutione al caso, dirai s’ abbia a ragiongnere insiemi<lb></lb> |
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8160 tutti e numeri che sonno da .1o. infino in .10., che, per gli modi passati, sonno .55. e .55. fatiche diremo<lb></lb> |
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8161 sia in .10.bracia. Ora è da vedere quante fatiche sonno in .6., dove hai a giongnere tutti e numeri che son-<lb></lb> |
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8162 no da .1o. infino in .6., che sonno .21. e .21. fatica sonno in .6.bracia. Onde dirai: se .55. fatiche vagliono.<lb></lb> |
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8163 10.denari., che varrano .21. fatica. Multiplicarai .21. via .10. e parti in .55., vienne .3 9/11. e .3 9/11. debba havere per <lb></lb> |
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8164 .6.<lb></lb> |
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8165 bracia e cosí farai le simiglianti. 39<lb></lb> |
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8166 Uno toglie a cavare un pozzo adentro .10.bracia. e debbane havere .10.denari. Állo cavato<lb></lb> |
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8167 tanto che n’ á avuto .4.denari. Adimando quanto lo cavó. Comme ó detto in .10.bracia. sonno.<lb></lb> |
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8168 .55. fatiche, onde dirai: se .55. fatiche hano .10.denari., quante fatiche fienno a .4.denari. Multiplica .4.<lb></lb> |
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8169 via .55. fatiche, fanno .220. fatiche e partirai in .10., vienne .22. fatiche .e. 22. fatiche sonno quelle che ne<lb></lb> |
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8170 debba havere .4.denari. Ora è da saper quanti numeri son quegli che agionti insieme fanno .22. Dove, di<lb></lb> |
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8171 molti oppenioni, togli questo: tu sai che da uno infino in .6. fanno .21. e per infino in .22. v’ é .1o. e il nume-<lb></lb> |
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8172 ro seguente e .6. é .7. Dove .1o. è .1/7. di .7. Adonca lo caverá .6.bracia.1/7. E cosí farai le simiglianti. 40<lb></lb> |
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8173 E gli é uno albero che è longo .50.bracia. Uno lo vole tagliare e, a ogni colpo che gli dá,<lb></lb> |
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8174 la vetta sciende in verso la terra, cioé piega in verso la terra, uno bracio. Adimando in quan-<lb></lb> |
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8175 ti colpi la vetta sia in terra. E gli é cosa assai manifesta che, se la via che fa la vetta<lb></lb> |
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8176 fosse visibile, comme la via delo razo quella sarebbe uno arco e, possendo la vetta taglia-<lb></lb> |
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8177 re la terra e ritornare al primo luogo, ella farebe uno tondo, del quale quello che si fa insino in ter-<lb></lb> |
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8178 ra è il .1/4. del tondo. Adonca dirai: e gli é uno tondo che il mezzo diametro è .50.bracia. Adiman-<lb></lb> |
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8179 dasi quanto gira. Radoppiarai .50., fanno .100., el quale per .3 1/7. multiplica, fanno .314 2/7. E .314.bracia.2/7. gire-<lb></lb> |
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8180 rebbe se lla vetta ritornasse in suo luogo, ma noi habiamo detto che gli é il .1/4. Adonca piglierai<lb></lb> |
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8181 il .1/4. di .314 2/7., che è .78 4/7. E .78.bracia.4/7. girerá il detto quarto di tondo. Adonca penerá la vetta a ve-<lb></lb> |
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8182 nire in terra in .78. colpi .4/7. E cosí farai le simili. 41<lb></lb> |
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8183 <lb></lb> |
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8184 E gli é uno albero che è alto .40.bracia.; lego la vetta con una fune che è longa .50.<lb></lb> |
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8185 bracia e dipoi lo fo tagliare e, quando credo che sia tagliato, mi scosto tanto quan-<lb></lb> |
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8186 to posso e tiro, credendo caggia. Ma e non cade e misuro la fune che io ó in mano.<lb></lb> |
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8187 E é .10.bracia. Adimando, pionbando uno piombino, in che parte cadrá, cioé quanto apresso al<lb></lb> |
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8188 pedale e quanto sia longo il filo del piombino. Questa é una gran favola e non altro vuol dire<lb></lb> |
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8189 che questo: e gli é uno scudo ch’ é per li due lati .40.bracia. e per l’ altro é .30.; adimandasi quanto è<lb></lb> |
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8190 la perpendiculare che cade in sula facia dele .30.bracia, la quale sia la radice di .1375. E radice di<lb></lb> |
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8191 .1375. sia lo filo del piombino e cosí fa sempre. 42<lb></lb> |
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8192 E gli é una macina che ’l suo diametro é .6.bracia. la quale é di .3. huomini e ciascuno ne<lb></lb> |
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8193 debba havere la terza parte. Adimando quanto sia il diametro di ciascuno. Debbi<lb></lb> |
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8194 multiplicare .6. in sé, fano .36. del quale piglia .2/3., sonno .24. e .R. di .24. debba rimanere del dia-<lb></lb> |
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8195 metro, quando l’ ará logro il .primo. E, per lo .2o., trai overo piglia, il .1/3. di .36., che è .12. e .R. di .12. sia il<lb></lb> |
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8196 diametro dela macina, quando il secondo l’ará logro. E cusí ái diviso la detta macina in .3. parti iguali. E<lb></lb> |
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8197 piú stesa di sotto subtiliter ne porró un’ altra per algebra ala ragion .77. 43<lb></lb> |
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8198 Se volesse quadrare uno corpo iregulare, cioé senza regola, si dá questo modo. Poniamo che<lb></lb> |
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8199 volessi quadrare uno bove, overo una pietra di strana statura. Metterala in uno viva-<lb></lb> |
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8200 io in modo che l’ aqua sia sopra di quella tale figura e poi ne la trai e quello che isciema è<lb></lb> |
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8201 l’ area corporale di quella tale figura. 44<lb></lb> |
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8202 E sonno .2. sacchi d’ iguali altezza; l’ uno tiene .6. stara. e l’ altro .24. stara. Vogli scuscirli<lb></lb> |
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8203 e farne uno di simile altezza, cioé di .2. farne uno che sia alto quanto era prima. Adi-<lb></lb> |
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8204 mando quanto terrá il sacco grande. Debi agiognere .24. e .6., fanno .30. e questi<lb></lb> |
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8205 serba. Dipoi multiplica .24. via .6., fanno .144., dela quale somma si piglia la radice e rad-<lb></lb> |
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8206 doppiasi. Overamente el .144. multiplica per .4., fanno .576. e di questo piglia la radice,<lb></lb> |
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8207 che è .24., el quale agiongni a .30., fanno .54. E .54. staia terrá dapoi el grande sacco.<lb></lb> |
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8208 <lb></lb> |
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8209 <pb></pb></p> |
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8210 <p class="folio"> |
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8211 folio 56r |
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8212 </p> |
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8213 <p class="main"> |
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8214 <lb></lb> |
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8215 </p> |
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8216 <p class="runhead"> |
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8217 Distinctio octava. 56 |
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8218 </p> |
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8219 <p class="main"> |
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8220 <lb></lb> |
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8221 Qui é a dire ragiongni la radice di .24. con la radice di .6., fanno radice di .54., comme si disse<lb></lb> |
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8222 nel trattato dele radici. Adunque terrá .54. stara. 45<lb></lb> |
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8223 E gli é uno triangolo ch’ é .12.bracia. per faccia; voglio farvi a ciascuna faccia uno<lb></lb> |
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8224 muro grosso due bracia. Adimando quanto sia il vano di dentro, cioé quanto gi-<lb></lb> |
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8225 rerá dentro per faccia. Prima truova el catetto del detto triangolo, che è ra-<lb></lb> |
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8226 dice di .108., dela quale radice trai la grosseza del muro .de., cioé .2.bracia., rimango-<lb></lb> |
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8227 no .ae. radice di .108. men .2. Dipoi ne trarai .fa. e truoverai quanto è la quantitá del .fa. in que-<lb></lb> |
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8228 sto modo. Tu sai che .fg. è .2.bracia. e ancora è equedistante al .cb. Adonca dirai: se .db., che sonno<lb></lb> |
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8229 .6.bracia, mi danno .da., che è radice di .108., che mi dará .fg.,, che è .2.bracia. Multiplica .2. via<lb></lb> |
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8230 radice di .108.,, fanno radice di .432. e parti in .36., vienne radice di .12., la quale trai dela radi-<lb></lb> |
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8231 ce di .108., rimane radice di .48. Adunque .fe. è la radice di .48. men .2. Ora, per sapere quan-<lb></lb> |
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8232 to é per faccia dirai: e gli é uno triangolo equilatero ed é il suo catetto radice di .48. men .2.; quan-<lb></lb> |
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8233 to è per faccia. Multiplica la radice di .48. men .2. in sé, fanno .52. meno la radice di .768. Ponvi<lb></lb> |
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8234 sul terzo, fanno .69 1/3. men radice di .1365 1/3. e la radice di questo sia per faccia. Cioé preso la ra-<lb></lb> |
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8235 dice di .1365 1/3. e tratta di .69 1/3. e di quel preso la radice. 46<lb></lb> |
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8236 E gli é una piramide, di che forma sia non fa alcuna cosa, la quale è alta .6.bracia.<lb></lb> |
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8237 Vorrela segare per lo mezzo, cioé che da una parte rimanesse una piramide intera<lb></lb> |
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8238 e l’ altra parte fosse una piramide corta. Debbi multiplicare l’ altezza in sé, fanno<lb></lb> |
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8239 .36. e ancora questo multiplica per .6., fanno .216., cioé a dire debi cubicare l’ altezza<lb></lb> |
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8240 che è .216. il suo cubo. E di quello piglia il mezzo, che è .108. e .R. cubica di .108. sia l’ altezza dela pira-<lb></lb> |
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8241 mide intera e de l’ altra sia .6.bracia. meno la .R. cubica di .108. E cosí usa sempre di che parte voi .47. |
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8242 </p> |
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8243 <p class="main"> |
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8244 E gli é uno scudo che per l’ una faccia è .15. e per l’ altra è .14. e l’ altra non so. Ma ben<lb></lb> |
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8245 so che gli é quadro .84.bracia. Adimando quanto è per l’ altra faccia. In questo<lb></lb> |
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8246 modo farai. Tu sai che a multiplicare la basa dove cade il catetto per la mitá del ca-<lb></lb> |
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8247 tetto fa l’ area del triangolo. Onde poniamo la basa sia la faccia del .14. Adonca,<lb></lb> |
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8248 a partire .84.bracia., cioé l’ area del detto triangolo, in .14., ne verrá la mitá del catetto, che è .6. |
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8249 </p> |
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8250 <p class="main"> |
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8251 Adunque tutto è .12. Ora, trovato il catetto, e tu truova quanto cade presso ala faccia dele .15.<lb></lb> |
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8252 bracia, cioé ala faccia nota. Dove multiplicarai .12. in sé e .15. in sé e haremo .144. e .225.; trai<lb></lb> |
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8253 .144. di .225., rimane .81., la cui radice é .9. per lo cadimento del catetto apresso al lato dele<lb></lb> |
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8254 .15.bracia., dove cadrá apresso a quella dele bracia non sapute a .5.bracia., cioé da .9. a .14. E per<lb></lb> |
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8255 sapere quanto è, multiplica .5. in sé e .12. in sé e haremo .25. e .144. che, e insiemi agionti, fan-<lb></lb> |
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8256 no .169., la cui radice è .13. per la facia non saputa. 48<lb></lb> |
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8257 E gli é una linea che è longa .8. bracia, la quale é .ab. e in su ciascuna extremitá pon-<lb></lb> |
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8258 go una linea la quale faccia angolo retto con la linea .ab.; e sia la linea .da. e .bc. E<lb></lb> |
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8259 sia .bc.6. e .da. sia .4. E dal ponto .a. meno la linea .ac. e dal ponto .b. meno la linea<lb></lb> |
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8260 .db., le quali s’ incrocichiano, overo si segono, dal ponto .n., dal quale meno la linea<lb></lb> |
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8261 .no., equedistante a ciascuna dele linee .ad. e .bc. Adimando quanto è .no. Qui è d’ arguire co-<lb></lb> |
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8262 sí: simile é il triangolo .ano. al triangolo .acb. Adunque tal parte é .ao. del .ab. quanto .no.<lb></lb> |
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8263 del .bc. Adunque tanto fa .no. in .ab. quanto .ao. in .bc. E questo dichiarato, e tu va al’ altra<lb></lb> |
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8264 parte e dirai: simile è il triangolo .bno. al triangolo .bda. Adunque tal parte è .bo. del .ba.<lb></lb> |
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8265 quanto .no. del .da. Adunque tanto fa a multiplicare .no. in .ab. quanto .ob. in .da. E di so-<lb></lb> |
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8266 pra truovamo che tanto faceva a multiplicare .no. in .ab. quanto .ao. in .bc. Onde tanto fa<lb></lb> |
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8267 .ob. in .da. quanto .ao. in .bc. Cioé tanto fa a multiplicare .ob. per .4. quanto .ao. per .6. e tut-<lb></lb> |
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8268 to .ab. é .8. Onde s’ á a ffare di .8.2. parti che, multiplicata l’ una per .4., faccia quanto l’ altra per .6. |
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8269 </p> |
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8270 <p class="main"> |
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8271 Che haremo l’ una .4 4/5. e l’ altra .3 1/5. Adonca .ao. è .3 1/5. e .ob. è .4 4/5. Ora per sapere quanto è .no.,<lb></lb> |
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8272 di molti modi detto é che gli é tal parte .ao. del .ab. quanto .no. del .bc. E peró tanto fa .ao.<lb></lb> |
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8273 in .bc. quanto .no. in .ab. E la multiplicatione del .ao. in .cb. fanno .19 1/5. Adunque a multipli-<lb></lb> |
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8274 care .no. in .ab. fará .19 1/5. e noi habiamo detto che .ab. è .8. Adonca .no. in .8. fanno .19 1/5. E peró<lb></lb> |
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8275 .no. sia .2 2/5. e tanto diremo sia la linea .no. E cosí opera in simili. 49<lb></lb> |
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8276 E gli é uno quadrangulo rettangulo del quale la longhezza è piú che la larghezza .6.bracia. e la<lb></lb> |
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8277 sua area, col diametro, è .100. Adimandasi quanto è la sua longhezza e quanto la larghezza.<lb></lb> |
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8278 El quale quadrilatero rettangulo é qui da lato designato. Poni el lato magiore una cosa<lb></lb> |
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8279 piú .3. Onde il minore lato sia una cosa meno .3. E nota che tu non ponga .1a. cosa e l’ altro .1a. co-<lb></lb> |
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8280 sa e .6., perché, per la confusione dele cose, censi e cubi, la questione non si potrebe asolvere. Ma per questa via si<lb></lb> |
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8281 levono via quelli nomi. E peró in ogni questione é da prociede secondo meno e piú. Multiplica adonca<lb></lb> |
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8282 .ab. in .bc., cioé .1a. cosa e .3. in una cosa meno .3., haremo .1o. censo meno .9. per l’ area del detto quadran-<lb></lb> |
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8283 <lb></lb> |
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8284 <pb></pb></p> |
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8285 <p class="folio"> |
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8286 folio 56v |
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8287 </p> |
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8288 <p class="main"> |
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8289 <lb></lb> |
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8290 </p> |
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8291 <p class="runhead"> |
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8292 Distinctio octava. |
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8293 </p> |
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8294 <p class="main"> |
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8295 <lb></lb> |
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8296 gulo. Dipoi trova el diametro .ac. dove multiplicherai .ab. in sé, cioé .1a. cosa men .3. in sé fanno<lb></lb> |
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8297 . 1o. censo e .9. men .6. cose. Dipoi multiplica .bc. in sé, cioé .1a. cosa e .3. in sé, fanno .1o. censo e .6. cose e .9.<lb></lb> |
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8298 che, con .1o. censo e .9. men .6. cose agionte, fanno .2. censi e .18., de’ quali la radici è il diametro.<lb></lb> |
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8299 Adunque .1o. censo men .9. e radice di .2. censi e .18. sonno iguali a .100. Raguaglia le parti,<lb></lb> |
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8300 cioé darai a iongni parte .1o. censo meno .9., harai radice di .2. censi e .18. iguali a .1o. censo e<lb></lb> |
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8301 .109. Dove multiplica ogni parte in sé harai .2. censi e .18. iguali a .11881. meno .218. censi e piú<lb></lb> |
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8302 . 1o. censo di censo. Raguaglia levando el superfluo e giongnendo el diminuito e haremo .11863.<lb></lb> |
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8303 e uno censo di censo iguali a .220. censi. Dimezza li censi, sonno .110., multiplica in sé, fa .12100.; tra-<lb></lb> |
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8304 ne .11863., rimane .237. e la radice di .237. non si truova. Adunque dirai il censo valere .110.<lb></lb> |
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8305 men radice di .237. e la cosa varrá la radice di quello. Cioé preso la radice di .237. e tratta di<lb></lb> |
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8306 .110. e del rimanente piglia la radice. Adunque fu largo presa la radice di .237. e tratta di .110.<lb></lb> |
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8307 e di quel preso la radice men .3. E, per longhezza, sia preso la radice di .237. e tratta di .110. e di<lb></lb> |
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8308 quel preso la radice piú .3. e cosí farai l’ altre. La prova che la superficie con lo diametro fa-<lb></lb> |
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8309 cia .100. è bella e farala cosí. Multiplica ciascun di doi lati propinqui in sé: cioé il largo<lb></lb> |
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8310 che fará .119. men radici .237. e men .6. siate l’ incrociamento di quella radice. E il longo fa-<lb></lb> |
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8311 rá .119. piú .R.237. piú .6. volte ditto incrociamento che, gionti ditti quadrati, fanno .R.v.238. men<lb></lb> |
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8312 .R.948. per la linea diagonale. Poi trova la superficie multiplicando el largo per lo longo, fa-<lb></lb> |
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8313 rá .101. men .R.237. che, gionta al diametro, deve esser .100. Ma serebe travaglio asai acozzar dit-<lb></lb> |
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8314 te .R., perché l’ una è .R.v. e l’ altra puro reciso. Ma proverai che fanno .100. in questo modo.<lb></lb> |
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8315 Cava .101. men .R.237. di .100., resta .R.237. men .1. E questo sirá equale al ditto diametro che si pro-<lb></lb> |
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8316 va per la comune scientia: cioé le linee che fanno quadrati equali sonno equali, unde l’ uno e<lb></lb> |
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8317 l’ altro in sé fará .238. men .R.948. Donca chi havesse gionto possiando el diametro ala superficie<lb></lb> |
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8318 harebe fatto .100. et cetera dignum nota. 50<lb></lb> |
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8319 E gli é uno triangolo .abc. del quale il lato .ab. è .7.bracia. e il lato .ac. è .9.bracia. e il<lb></lb> |
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8320 lato .bc è .8., nel quale voglio collocare il magiore tondo che si puó. Adimandasi<lb></lb> |
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8321 quanto sia il suo diametro. Prima si debba quadrare il detto triangolo per lo mo-<lb></lb> |
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8322 do giá dato e harai sia quadro la radice di .720. E dipoi dobbiamo, dal centro del<lb></lb> |
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8323 detto cerchio al tocamento de’ lati del triangolo con la circonferentia del cerchio, produrre<lb></lb> |
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8324 .3. linee, cioé .of.oh.om. E, da quello centro agli angoli .a.b.c., debbi produrre .3. linee, cioé<lb></lb> |
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8325 .ao.bo.co. E queste linee dividono tutto il triangolo in .3. triangoli, cioé .aob. e .boc. e .aoc.<lb></lb> |
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8326 E hora è di bisogno quadrare questi .3. triangoli per questa via, perché quelle .3. linee che son-<lb></lb> |
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8327 no dal centro al toccamento de’ lati sonno perpendiculare sopra e lati de’ triangoli, comme per<lb></lb> |
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8328 la .17. del .3o. de Euclide è manifesto. Onde poniamo che ciascuna di queste linee sia una<lb></lb> |
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8329 cosa e, perché ciascuna è perpendiculare, ciascuna è semidiametro di questo circulo e le son-<lb></lb> |
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8330 no infra loro iguali. Adunque habiamo posto sia ciascuna una cosa. Adunque multiplica-<lb></lb> |
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8331 rai una cosa per la mitá di ciascuno lato del triangolo, cioé la mitá del .ab. per una cosa e la mi-<lb></lb> |
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8332 tá del .bc. per .1a. cosa e la mitá dela .ac. per .1a. cosa e in tutto haremo .12. cose per lo quadrato<lb></lb> |
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8333 di questo triangolo. Ma noi habiamo detto el detto triangolo essere quadro radice di .720. |
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8334 </p> |
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8335 <p class="main"> |
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8336 Onde .12. cose sonno iguali .720. cose, dove la cosa vale la radici di .5. E noi dicemmo che l’.1/2.<lb></lb> |
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8337 diametro era una cosa. Adunque il .1/2. diametro fo la radice di .5. e tutto fo radice di .20. E co-<lb></lb> |
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8338 sí habiamo fatto che la radici di .20. sia il diametro del magiore tondo che intra nel detto<lb></lb> |
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8339 triangolo. 51. |
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8340 </p> |
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8341 <p class="main"> |
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8342 E gli é uno triangolo .abc. del quale il lato .ab. sia .14. e il lato .ac. sia .13. e il lato<lb></lb> |
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8343 .bc. sia .15., nel quale voglio constituire .2. circuli iguali e magiori che è possibile.<lb></lb> |
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8344 Cercho la quantitá de’ loro diametri. Prima dal ponto .a. produrró la perpendi-<lb></lb> |
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8345 culare ala linea .bc., che sia la linea .ax., che comme e gli é manifesto sia .11 1/5. Dipoi<lb></lb> |
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8346 dal centro di questi cerchi al tocamento dela loro circonferentia al triangolo predetto me-<lb></lb> |
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8347 na le linee, che fienno .pn.ps. e le linee .mo. e .ot. E dipoi da’ loro centri a’ loro canti mena<lb></lb> |
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8348 le linee, cioé .ob.ao.ap.pc. Dipoi quadreró tutto il triangolo .abc. multiplicando la perpen-<lb></lb> |
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8349 diculare .11 1/5. nela mitá di .15., fanno .84. e .84.bracia. é quadro el detto triangolo .abc. e que-<lb></lb> |
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8350 sto serba. Dipoi quadrerai tutte le figure che sonno nel detto triangolo. Cioé il triangolo .aob.<lb></lb> |
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8351 e il triangolo .apc. e il triangolo .otb. e .pnc. e il quadrangolo .optn. Ma quando questi cir-<lb></lb> |
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8352 coli fienno iguali e loro diametri sonno iguali. Onde poniamo che ’l mezzo diametro di ques-<lb></lb> |
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8353 ti circoli sia .1a.cosa. Hora incomenciamo a quadrare questi triangoli. E prima quadreró il<lb></lb> |
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8354 triangolo .aob. multiplicando el mezzo diametro .om., che è .1a. cosa, per la mitá del .ab., che è .14., fa .7.<lb></lb> |
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8355 cose. E dipoi quadra il triangolo .apc. multiplicando il mezzo diametro .ps., che è una cosa,<lb></lb> |
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8356 <lb></lb> |
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8357 <pb></pb></p> |
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8358 <p class="folio"> |
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8359 folio 57r |
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8360 </p> |
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8361 <p class="main"> |
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8362 <lb></lb> |
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8363 </p> |
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8364 <p class="runhead"> |
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8365 Distinctio octava 57 |
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8366 </p> |
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8367 <p class="main"> |
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8368 <lb></lb> |
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8369 per la mitá del .ac. che è .13., fanno .6.cose.1/2. Ora ci resta a quadrare el quadrangolo .opnt.<lb></lb> |
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8370 multiplicando .ot., ch’ é mezzo diametro, cioé una cosa, in .tn., che è .2.cose., fanno .2. censi. Anco-<lb></lb> |
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8371 ra è di bisogno quadrare li .2. triangoli .obt. e .pnc. Ma, perché .ot. e .pn. ciascuno è una cosa, do-<lb></lb> |
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8372 biamo multiplicare una cosa per la mitá dela basa, cioé .bt. e .nc. che, insiemi agionti, fanno<lb></lb> |
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8373 .15. meno .2.cose., fanno .7 1/2. cosa meno uno censo. Ancora è di bisogno quadrare el triango-<lb></lb> |
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8374 lo .aop., che cosí il quadreró, che multiplicaró .ax., ch’ é perpendiculare del triangolo .abc.<lb></lb> |
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8375 et è .11 1/5. meno el lato del quadrangolo, che è una cosa. Adonca la perpendiculare ditta è .11 1/5.<lb></lb> |
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8376 meno una cosa. Et, perché .fx. è una cosa, multiplicarai .11 1/5. meno una cosa per .2.cose., fanno<lb></lb> |
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8377 .22.cose.1/5. meno .2.censi., de’ quali la mitá è .11.cose.1/5. meno uno censo. E cosí hai quadrato tut-<lb></lb> |
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8378 te le figure che sonno nel ditto triangolo, che in tutto sonno .32. 1/5. e tanto è quadro il ditto tri-<lb></lb> |
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8379 angolo .abc., che è quadro, commo dicemno, .84.bracia. Onde dividerai .84.bracia. per .32 1/5., vien-<lb></lb> |
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8380 ne .2 14/23. e tanto val la cosa. Et noi ponemmo che mezzo il diametro del tondo era una cosa. Adon-<lb></lb> |
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8381 ca fu .2 14/23. e tutto fu .5 5/23. e cosí farai le simili. .52. |
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8382 </p> |
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8383 <p class="main"> |
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8384 E gli é uno triangolo .abc., del quale il lato .ab. è .10. e il lato .ac. è .10. e il lato .bc. è .12.,<lb></lb> |
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8385 cioé á .2. lati iguali e l’ altro lato non è iguale. Commo ó detto nel quale voglio met-<lb></lb> |
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8386 tere .3. tondi, e magiori che posso, e iguali. Adimando quanto sia il diametro di-<lb></lb> |
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8387 ciascuno de’ ditti .3. tondi. Prima noi dobiamo collocare questi .3. cerchi nel trian-<lb></lb> |
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8388 golo .abc., in modo che dal centro .m. al centro .o. si meni la linea .mo. e dal centro .o. al centro<lb></lb> |
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8389 .n. si meni la linea .on. e dal centro .n. al centro .m. si facia la linea .mn., in modo che queste .3.<lb></lb> |
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8390 linee facino uno triangolo e sonno equedistanti ali lati del triangolo grande .abc. e, per con-<lb></lb> |
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8391 sequente, sonno proportionali imperoché cosí é il lato .ab. alo lato .mo., a lui equedistante, co-<lb></lb> |
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8392 sí é il lato .bc. allo lato .mn., a quello equedistante, e similmente .ac. al .on. sará. Onde questi<lb></lb> |
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8393 .2. triangoli sonno proportionali secondo e lati suoi. Ora é di bisogno trovare la perpendicu-<lb></lb> |
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8394 lare .ak. e, per lo modo ó detto, la farai e harai essere .8. La quale, multiplicata per la mitá dela<lb></lb> |
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8395 basa sua, cioé per .6., fanno .48. e tanto è quadro el triangolo .abc. Dapoi che habiamo trova-<lb></lb> |
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8396 ta l’ area del ditto triangolo, é di bisogno quadrare tute quelle figure che sonno nel ditto<lb></lb> |
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8397 triangolo, che sonno .3. quadrangoli e .7. triangoli e haremo, se agiongneremo insiemi .48.,<lb></lb> |
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8398 cioé l’ area de tutto il triangolo. Poniamo adonca che ’l diametro di ciascuno cerchio sia .2. co-<lb></lb> |
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8399 se, onde mezzo il diametro sia una cosa. Onde incominciamo a quadrare le ditte figure scritte<lb></lb> |
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8400 dentro al triangolo. E prima quadraremo il quadrilatero .mnze., che è longo .2.cose.2/5., im-<lb></lb> |
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8401 peroché la facia .bc. é piú il .1/5. che l’ altre doi facie e in longhezza é una cosa, cioé mezzo il dia-<lb></lb> |
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8402 metro. Onde l’ area sua è la multiplicatione d’ una cosa in .2.cose.2/5., che fanno .2.censi.2/5. E da-<lb></lb> |
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8403 poi quadriamo gli altri .2. quadrilateri, cioé .mrto. e .nuof., che ciascuno è, per larghezza, una co-<lb></lb> |
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8404 sa e per longhezza, .2.cose., dove ciascuno sia quadro .2.censi. E fra amendoi sonno .4. censi. Ora<lb></lb> |
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8405 è di bisogno quadrare e triangoli. E prima e .2. triangoli, cioé .mzb. e .nec. e, perché le base<lb></lb> |
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8406 di questi .2. triangoli e la linea .bc. meno la linea .ze., cioé .12. meno .2.cose.2/5., e il catetto di ciascu-<lb></lb> |
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8407 no è una cosa. Onde multiplica una cosa contro ala mitá di .12. meno .2.cose.2/5., fanno .6. cose<lb></lb> |
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8408 meno uno censo .1/5. E dapoi quadra e .2. triangoli, cioé .un.c. e .afo. E, perché le base loro, insie-<lb></lb> |
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8409 mi agionte, sonno tutta la linea .ac., che è .10. meno .2.cose., e la loro perpendiculare è una co-<lb></lb> |
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8410 sa, onde multiplica una cosa per la mitá di .10. meno .2.cose., fanno .5.cose. meno uno censo e que-<lb></lb> |
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8411 sto serba. E dapoi quadra e .2. altri triangoli, cioé il triangolo .aot. e il triangolo .mrb. E,<lb></lb> |
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8412 perché questi .2. triangoli sonno sopra la basa .ab., adonca la basa loro è .10. meno .2. cose,<lb></lb> |
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8413 cioé meno la linea .er. Adonca multiplica una cosa per la mitá di .10. meno .2.cose., fanno .5.<lb></lb> |
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8414 cose meno uno censo. E questo serba. E dapoi quadrarai el triangolo .mon., che sai che gli é<lb></lb> |
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8415 per li .2. lati ciascuno .2.cose. e il terzo lato è .2. co .2/5., dove è di bisogno trovare la perpendicula-<lb></lb> |
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8416 re. Che farai in questo modo: multiplica .om. in sé, fanno .4.censi. e dapoi multiplica .mx. in sé,<lb></lb> |
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8417 fanno uno censo .11/25., e quali trai di .4.censi., rimangono .2.censi.14/25., de’ quali la radice è una cosa<lb></lb> |
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8418 .3/5. E tanto è la perpendiculare .ox. Adonca multiplicare bisogna .ox. per la mitá del .mn., cioé<lb></lb> |
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8419 una cosa .3/5. per una cosa .1/5., fanno uno censo .23/25. e tanto è quadro il ditto triangolo. Onde agion-<lb></lb> |
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8420 gni l’ area di queste figure insiemi, fanno .5.censi.3/25. e .16. cose. E tanto è l’ area del grande<lb></lb> |
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8421 triangolo. E noi dicemmo che ’l gran triangolo è quadro .48.bracia. Adonca .5.censi.3/25. e .16. cose<lb></lb> |
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8422 sonno iguali a .48., dove la cosa vale .1 7/8. e il diametro del tondo sia .3 3/4. e cosí farai le simili. 53.<lb></lb> |
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8423 E gli é uno cerchio del quale il diametro è .12., nel quale voglio collocare .3. cerchi i-<lb></lb> |
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8424 guali e magiori che posso. Adimando quanto è il diametro del’ uno di loro. Tu de-<lb></lb> |
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8425 bi producere .3. linee dal centro d’ uno cerchio al centro del’ altro, cioé la linea .dc.<lb></lb> |
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8426 E la linea .de. e la linea .ce., che fanno uno triangolo equilatero, cioé el triangolo<lb></lb> |
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8427 <lb></lb> |
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8428 <pb></pb></p> |
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8429 <p class="folio"> |
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8430 folio 57v |
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8431 </p> |
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8432 <p class="main"> |
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8433 <lb></lb> |
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8434 </p> |
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8435 <p class="runhead"> |
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8436 Distinctio octava |
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8437 </p> |
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8438 <p class="main"> |
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8439 <lb></lb> |
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8440 .dce. Ora poniamo che il lato di questo triangolo sia .2.cose. E, perché ciascuno lato di que-<lb></lb> |
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8441 sto triangolo è fatto di .2. mezzi diametri, onde questo triangolo è .2.cose. per ciascuno la-<lb></lb> |
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8442 to e tanto è il diametro del tondo. E dapoi meneremo la linea .ab. del diametro del gran cer-<lb></lb> |
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8443 chio, che passa per lo ponto .d. e per lo centro .r., centro del gran cerchio. Ed é la ditta linea, cioé<lb></lb> |
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8444 diametro, .12. commo habiamo ditto. Ora è di bisogno trovare quanta sia la linea .ar., che è<lb></lb> |
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8445 mezzo il diametro del gran cerchio, che è fatto del .ad., mezzo diametro del cerchio picolo, e<lb></lb> |
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8446 dela linea .dr., che è .2. parti dela linea .df., la quale linea .df. è radice di .3.censi., commo è ma-<lb></lb> |
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8447 nifesto per Euclide. Onde .dr. è radice d’ uno censo .1/3. Onde la linea .ar. è fatta del .ad., che è<lb></lb> |
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8448 una cosa, e dela linea .dr., che è radice d’ uno censo .1/3. E cosí diremo che tutta .ar., che è mezzo dia-<lb></lb> |
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8449 metro del gran cerchio, è una cosa e radice di .1 1/3. censo. E noi habiamo posto che ’l diametro<lb></lb> |
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8450 era .12. e il .1/2. diametro è .6. Adonca una cosa e radice di uno .1/3. censo é iguali a .6., che partirai<lb></lb> |
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8451 .6. in radice di .1 1/3. e uno, che ne viene radice di .432. meno .18. e tanto vale la cosa. E noi ponen-<lb></lb> |
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8452 mo che il mezzo diametro era una cosa e tutto il diametro era .2. cotanti, che sia radice di .1728.<lb></lb> |
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8453 meno .36. E cosí farai li simili. .54. |
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8454 </p> |
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8455 <p class="main"> |
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8456 S ia uno cerchio .ahp. del quale il diametro è .40., nel quale voglio collocare qua-<lb></lb> |
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8457 tro cerchi magiori chi posso. Adimando quanto sia il diametro di ciascuno. Tu<lb></lb> |
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8458 vedi commo questi .4. cerchi sonno congionti per uno quadrato segnato .qkxm.,<lb></lb> |
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8459 del quale gli angoli sonno terminati ne’ centri di questi .4. circuli, del quale il dia-<lb></lb> |
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8460 metro con lo diametro d’ uno de’ piccoli tondi è iguali al diametro del gran tondo, cioé il dia-<lb></lb> |
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8461 metro .ap., che è .40., é iguali al diametro del quadrato .mxqk., el quale diametro è .mk., e al<lb></lb> |
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8462 diametro d’ uno de’ tondi piccoli. E, questo inteso, e tu dirai: io pongo che ’l diametro d’ uno de’<lb></lb> |
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8463 ditti tondi sia una cosa. Adonca il quadrato preditto é per facia una cosa. Onde il suo dia-<lb></lb> |
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8464 metro sia radice di .2.censi., cioé el diametro .mk. al quale, agionto el diametro d’ uno de’ tondi<lb></lb> |
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8465 piccoli, cioé .ak., che è mezzo diametro, e .mp., che è anco mezzo diametro, haremo radici di .2.<lb></lb> |
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8466 censi e una cosa. E questo è la quantita del diametro del tondo, cioé .40.bracia. Adonca ra-<lb></lb> |
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8467 dice di .2.censi. e una cosa è iguale a .40. Che, a sapere quello vale la cosa, parti .40. in radice de .2.<lb></lb> |
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8468 piú uno, vienne radice .3200. meno .40. E radice di .3200. men .40. val la cosa e tanto adonca sia<lb></lb> |
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8469 il diametro d’ uno de’ ditti tondi e cosí in simili aopera. .55. |
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8470 </p> |
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8471 <p class="main"> |
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8472 Adimandasi el diametro d’ uno de’ .5. magiori cerchi collocati in un cerchio del qua-<lb></lb> |
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8473 le il diametro è .12. Commo sia el circulo .anc., del quale il diametro .ac. è .12. e il cen-<lb></lb> |
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8474 tro de ditto tondo è il ponto .d. E commo vedi nela presente figura, e ditti tondi, con-<lb></lb> |
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8475 tingenti l’ uno l’ altro hano tale natura che, menando da’ centri loro le linee passan-<lb></lb> |
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8476 ti sopra le contingentioni, farebono uno pentagono. Dove dimostraremo che, essendo il diame-<lb></lb> |
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8477 tro .ac. noto, che tutto .ah. sia noto. E acioché con brevitá parliamo, manifestaremo una au-<lb></lb> |
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8478 toritá di Ptholomeo, lo quale dice che, quando el diametro del tondo fosse .4., el lato del pen-<lb></lb> |
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8479 tagono sarebbe la radice di questa linea, cioé di .10. meno radice di .20. Adonca, quando il dia-<lb></lb> |
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8480 metro del cerchio fosse .4., el lato del pentagono sia radici di .20. tratta di .10. e di quel preso la<lb></lb> |
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8481 radici. Onde, per havere piú noticia, togli el quadrato del diametro e il quadrato delo lato<lb></lb> |
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8482 del pentagono e haremo .16. e .10. men radice di .20. Dico che la proportione del quadrato del<lb></lb> |
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8483 diametro al quadrato del lato del pentagono è commo .16. a .10. men radice di .20. Onde io<lb></lb> |
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8484 porró che il lato del pentagono, cioé .zo., che è quanto el diametro d’ uno de’ minori tondi, esse-<lb></lb> |
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8485 re una cosa, dove il quadrato è uno censo. E dirai: se .16. val .10. men radice di 20., che varrá,<lb></lb> |
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8486 cioé che è quello che vale uno censo. Multiplica uno censo via .16. e parti in .10. men radice<lb></lb> |
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8487 di .20., vienne .2.censi. piú radice di .4/5. di censo di censo. E questo è per lo quadrato dela linea<lb></lb> |
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8488 .zp. E, perché la linea .ac. è il diametro del gran cerchio, el quale agiongni al diametro .zp. il<lb></lb> |
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8489 diametro d’ uno de’ piccoli tondi. Adonca .2.censi. e radice di .4/5. di censo di censo sonno quan-<lb></lb> |
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8490 to il quadrato di .12. meno una cosa, che è .144. e uno censo men .24. cose. Dove raguagliarai<lb></lb> |
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8491 le parti e harai che uno censo e radice di .4/5. di censo di censo e .24.cose. sonno iguali a .144. |
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8492 </p> |
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8493 <p class="main"> |
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8494 Onde riduci a uno censo, dove partirai tutto in uno censo e radice de .4/5. di censo di censo e<lb></lb> |
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8495 haremo .120.cose. e radice di .11520.cose. e per numero .720. meno radice di .414720. Dividi<lb></lb> |
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8496 le cose, haremo .60. meno radice di .2880. e il loro quadrato, al numero, e haremo .7200. meno<lb></lb> |
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8497 radice di .414720. e meno radice di .41472000. e preso la radice di tutto e trattone .60. men<lb></lb> |
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8498 radice di .2880. E, acioché con piú brievitá diciamo, noi possiamo ragiongnere radice di<lb></lb> |
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8499 .414720. meno con meno radice di .41472000. e tutto questo è radice di .50181120. E peró<lb></lb> |
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8500 haremo che la cosa vale la radice di .50181120. tratta di .720. e del rimanente preso la radice<lb></lb> |
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8501 e agionta ala radice di .2880. e tratta di .60. e tanto adonca sia il diametro de ciascuno de’ .5.<lb></lb> |
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8502 <lb></lb> |
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|
8503 <pb></pb></p> |
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8504 <p class="folio"> |
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|
8505 folio 58r |
|
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8506 </p> |
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8507 <p class="main"> |
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8508 <lb></lb> |
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8509 </p> |
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8510 <p class="runhead"> |
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8511 Distinctio octava. 58 |
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8512 </p> |
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8513 <p class="main"> |
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8514 <lb></lb> |
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8515 ditti cerchi. E cosí opera sempre.<lb></lb> |
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8516 <lb></lb> |
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8517 Possiamo ancora in un cerchio collocare .6. cerchi, imperoché, commo dice Eucli-<lb></lb> |
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8518 de nela .15a. del quarto, infra ’l proposto cerchio, cioé dentro al proposto cerchio,<lb></lb> |
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8519 uno exagono equilatero e equiangolo collocare. E per questo è manifesto che il la-<lb></lb> |
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8520 to del exagono è il .1/2. diametro del cerchio al quale lo exagono si scrive. E peró pos-<lb></lb> |
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8521 siamo in uno cerchio fare .6. triangoli iguali e, in ciascuno triangolo, possiamo scriverne uno,<lb></lb> |
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8522 cioé mettervi uno circulo. E cosí adonca, in ciascuno cerchio, possiamo scrivere .6. cerchi senza<lb></lb> |
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8523 tropo fatica.. E, perché quelli cerchi sonno insiemi contingenti, fanno, menate le linee da’ centri loro,<lb></lb> |
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8524 uno exagono. Onde, nel mezzo di loro, rimane uno spacio che vi si pó collocare un altro cer-<lb></lb> |
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8525 chio iguale a uno di quelli .6. cerchi. Onde sia l’ exagono .a.b.c.d.e.f. e dal’ angolo di ciascuno lato<lb></lb> |
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8526 infino al centro si meni le linee che con lati delo exagono faranno .6. triangoli iguali e equilate-<lb></lb> |
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8527 ri, che sonno congionti nel ponto .o. e i loro lati sonno igual commo manifesta Euclides.<lb></lb> |
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8528 Onde ancora negli angoli di questo exagono si ponga il pie’ dele sexte inmobile e l’ altro pie’<lb></lb> |
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8529 si extenda infino ala mitá delo lato di quello exagono per ciascuna parte e risultane uno<lb></lb> |
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8530 cerchio del quale il diametro è iguale alo lato del exagono. E, cosí facendo in ciascuno an-<lb></lb> |
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8531 golo del ditto exagono e a questo modo, haremo .6. cerchi iguali descritti nel cerchio del qua-<lb></lb> |
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8532 le è il centro .o. centro del exagono. Adonca la linea .xo. é una volta e .1/2. il lato delo exagono,<lb></lb> |
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8533 cioé una volta .1/2. il diametro d’ uno de’ cerculi piccoli. E peró dala circunferentia d’ uno di quel-<lb></lb> |
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8534 li cerchi infino al centro .o. rimane uno spacio nel quale spacio, computando tutti e .6. e detti<lb></lb> |
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8535 cerculi, si constituirá un altro cerculo iguale a ciascuno de’ .6. cerculi. E peró si pó costituire<lb></lb> |
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8536 .7. cerchi e ’l diametro di ciascuno di questi .7. cerchi è la terza parte del diametro del gran cerchio<lb></lb> |
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8537 che contiene e detti cerchi. E peró, quando il diametro .xt. fosse .30., il diametro de ciascuno di cer-<lb></lb> |
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8538 chi piccoli sarebbe .10., commo si manifesta per la linea .xt., diametro che passa per gli centri<lb></lb> |
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|
8539 del ditto cerchio. .75. |
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8540 </p> |
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|
8541 <p class="main"> |
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8542 E, volendo mettere uno ottagono in uno cerchio del quale il diametro fosse .2. bra-<lb></lb> |
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8543 cia. Commo sia il cerchio .agbd. del quale il diametro .ab. sia .2. Adimando quan-<lb></lb> |
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8544 to sia il lato del ottagono .ac. Dico prima doversi trovare il lato del quadrato che<lb></lb> |
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8545 si pó fare in ditto circulo. Dove multiplicherai il diametro del cerchio in sé, fanno<lb></lb> |
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8546 .4., del quale la mitá è .2., e radici di .2. sia il quadrato per facia, cioé il quadrato .hcef. E, questo<lb></lb> |
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8547 fatto, e tu dividi la linea, cioé lo lato del quadrato qual voi, per .2. parti iguali, commo sia il la-<lb></lb> |
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|
8548 to .hc. diviso in .2. parti iguali nel ponto .k. E faciasi la linea .ka., menata infino in nel ponto .i.<lb></lb> |
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8549 E sia semidiametro del cerchio, la quale linea .ai. sia uno bracio. Adonca .ka. sia uno bracio<lb></lb> |
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8550 meno radici di .1/2., imperoché .ki. è radici di .1/2. bracio, imperoch’ é la mitá dela facia del qua-<lb></lb> |
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8551 drato. Adonca sapiamo che .ak. è uno bracio meno radici de .1/2. e .kc. è radice di .1/2. Onde, a vo-<lb></lb> |
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8552 lere la linea .ac., che è la facia delo ottagono, multiplicarai .ak. in sé e .kc. in sé e haremo, per la<lb></lb> |
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8553 loro agiontione, il quadrato .ac. Imperoché l’ angolo fatto dala linea .ai. in sul ponto .k. ala<lb></lb> |
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8554 facia del quadrato .hc. qual voi è retto, imperoché lo lato .hc. è diviso in .2. parti iguali. E la<lb></lb> |
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8555 linea .ai. passa al centro. Onde multiplicarai uno meno radici di .1/2. in sé, fanno .1 1/2. meno radi-<lb></lb> |
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8556 ci di .2. E dapoi multiplicarai .kc. in sé, cioé radici di .1/2. in sé, fanno .1/2., agiongni a .1 1/2. men radici di .2.<lb></lb> |
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8557 fanno .2. meno radici di .2. per lo lato .ac., lato del ottagono constituto e fatto nel cerchio il<lb></lb> |
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8558 cui diametro è .2.bracia. |
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8559 </p> |
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8560 <p class="main"> |
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8561 E, volendo sapere quante bracia é quadro il ditto ottagono, di molte vie e modi, pi-<lb></lb> |
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8562 glia questo el quale da Lionardo P. è mostro: cioé di multiplicare el diametro del<lb></lb> |
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8563 cerchio per la facia del quadrato. Commo in questo passato caso multiplicarai<lb></lb> |
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8564 il diametro, che è .2.bracia., via la facia del quadrato, che è radici di .2.bracia., fanno radici<lb></lb> |
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8565 di .8. e la radici de .8. è quadro il ditto ottagono. E, acioché chiaro appaia in questo caso passato, di-<lb></lb> |
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8566 remo si voglia recare a bracia quadre l’ ottagono preditto, el quale è diviso in uno quadrato ret-<lb></lb> |
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8567 tangulo in .8. triangoli ortogonij. Del quadrato se ha l’ area multiplcando lo lato in sé, cioé mul-<lb></lb> |
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8568 tiplicando radici de .2. in radici di .2., che fanno .2. e .2.bracia. é quadro el quadrato. E dapoi qua-<lb></lb> |
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8569 dra ciascuno de’ ditti triangoli, de’ quali l’ area loro s’ á del multiplicare la saetta nela mitá dela<lb></lb> |
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8570 mitá dela facia del quadrato. E la saetta è, secondo che habiamo trovato, uno men radici di<lb></lb> |
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8571 .1/2. e la facia è radici di .2., cioé la facia delo quadrato è la mitá dela mitá, cioé il .1/4., dela ditta fa-<lb></lb> |
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8572 cia, è radici di .1/8. Adonca habiamo a multiplicare uno men radici di .1/2. via radici di .1/8., fanno<lb></lb> |
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8573 radici .1/8. meno .1/4. E tanto è uno de’ ditti triangoli. Dove .8. fienno .8. via radici di .1/8. meno .1/4.,<lb></lb> |
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8574 fanno radici del .8. men .2. E questa è l’ area degli .8. triangoli, e quali, agionti a .2., che è l’ area<lb></lb> |
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8575 del quadrato, fanno radici del .8. E radici di .8. é quadro el ditto ottagono commo volavamo<lb></lb> |
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8576 <lb></lb> |
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8577 <pb></pb></p> |
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8578 <p class="folio"> |
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|
8579 folio 58v |
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8580 </p> |
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8581 <p class="main"> |
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8582 <lb></lb> |
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8583 </p> |
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8584 <p class="runhead"> |
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8585 Distinctio octava. |
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8586 </p> |
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8587 <p class="main"> |
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8588 <lb></lb> |
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8589 Ancora secondo la dimostratione. Comme vedi, il ditto ottagono è diviso in uno quadra-<lb></lb> |
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8590 to e .8. triangoli ortogonij e l’ area del quadrato s’ á del multiplicare una dele facie in sé, cioé<lb></lb> |
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8591 diciamo del multiplicare .hf. in sé. E tu sai certo .ku. essere iguali al .hf. Adonca l’ area del dit-<lb></lb> |
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8592 to quadrato s’ á del multiplicare .hf. in .ku., cioé del .hc. in .ku. E l’ area de’ ditti triangoli s’ anno<lb></lb> |
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8593 del multiplicare .ak., cioé dela saetta, nella mitá dela mitá dela facia del quadrato, cioé il trian-<lb></lb> |
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8594 golo .akh. si quadra del multiplicare .ak. nela mitá del .hk. e il triangolo .akc. si quadra<lb></lb> |
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8595 del multiplicare .ak. nela mitá del .kc. e il .kc. é quanto .kh. Adonca e .2. triangoli ditti si<lb></lb> |
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8596 quadrano del multiplicare .ak. in .hk. E similmente e .2. triangoli .cdm. e .emd. si quadrano<lb></lb> |
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8597 del multiplcare .md. nela mitá dela facia .ce. e la mitá dela facia .ce. é quanto .kc. Adonca,<lb></lb> |
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8598 a quadrare il ditto quadro e gli .4. preditti triangoli, si multiplica .au. nela facia .hc., cioé facia<lb></lb> |
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8599 del quadrato, e gli altri .4. triangoli nel medesimo modo s’ ánno del multiplicare la loro saet-<lb></lb> |
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8600 ta nela mitá dela mitá dela facia e la loro saetta è iguale al .ub. Adonca l’ area loro s’ á de mul-<lb></lb> |
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8601 tiplicare .ub. nel .fe., cioé .ub. nel .hc. E peró, adonca l’ area de tutto l’ ottagono s’ á de multipl-<lb></lb> |
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8602 care il diametro .ab. nela facia del quadrato, la qual cosa era da mostrare. 58<lb></lb> |
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8603 Ancora proporró: e gli é uno ottagono la cui facia è .6.bracia. Adimando quanto è<lb></lb> |
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8604 il diametro del tondo dove tale ottagono s’ iscrive. Dico che di molti modi que-<lb></lb> |
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8605 sto mi pare il migliore. Commo diciamo e gli é uno ottagono .abcdefgh, del qua-<lb></lb> |
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8606 le il lato suo è .6.bracia. Adimandase quanto è il diametro .ae. Dico che in questo<lb></lb> |
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8607 modo facia, che dirai: ditto é che, quando il diametro è .2.bracia., che l’ ottagono è per facia la<lb></lb> |
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8608 radici di .2.bracia. meno la radici di .2. E noi diciamo che l’ ottagono è .6.bracia. Dove, con pro-<lb></lb> |
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8609 portione, andremo inperoché ciascuna figura constituta nel cerchio, di simile sito, è proportiona-<lb></lb> |
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8610 le e simile. Como se in .2. cerchi in ciascuno si fa uno triangolo di simile sito, cioé .2. lati iguali e l’ an-<lb></lb> |
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8611 golo del’ uno sia iguale al’ angolo del’ altro, over sienno .2. quadrati, over qual figura voi, infra<lb></lb> |
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8612 loro sonno simili e proportionali. Adonca dirai: se .2.bracia. fanno la .R. di .1. e meno radici<lb></lb> |
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8613 di .2., che fu quello che feci .6.bracia. Multiplicarai .6. via .2.bracia., fanno .12.bracia., e quali parti<lb></lb> |
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8614 in radici di questo, cioé in radici di .2. tratto di .2. e di quel preso la radici. Dove arrecarai cia-<lb></lb> |
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8615 scuna parte a quadrato multiplicando .12. in sé, fanno .144. e multiplica preso la radici di .2. e<lb></lb> |
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8616 tratta di .2. e di quel preso la radici in sé, fanno .2. meno la radici di .2. Onde partirai .144. in<lb></lb> |
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8617 .2. meno radici di .2., dove multiplicarai .2. meno radici di .2. per lo suo binomio, cioé per .2. e<lb></lb> |
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8618 radici di .2., fanno .2. e questo è il partitore. Dapoi multiplica .144. via .2. e radici di .2., fanno<lb></lb> |
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8619 .288. e radici di .41472., e quali parti in .2., vienne .144. e radici di .10368. E questo è il quadra-<lb></lb> |
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8620 to del diametro. Adonca il diametro del ditto tondo è la radici di questo, cioé di .144. e radici di .10368., cioé preso la <lb></lb> |
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8621 radici di <lb></lb> |
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8622 .10368. e posta sopra .144. e di quel preso la radici. .59. |
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8623 </p> |
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8624 <p class="main"> |
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8625 E gli é un exagono .abgdez. el quale voglio quadrare. Dico che meni la linea .ag.,<lb></lb> |
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8626 la quale linea è la facia del triangolo che cade nel cerchio dove si constituisse<lb></lb> |
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8627 tale exagono, la quale si seghi in .2. parti iguali dal diametro .bc., sopra il ponto<lb></lb> |
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8628 .k. Sará il triangolo .iab. la sexta parte delo exagono .abgdez. e la corda .ag. é<lb></lb> |
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8629 il lato del triangolo equilatero cadente nel cerchio agd. Conciosiacosaché l’ arco .ag. sia la<lb></lb> |
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8630 terza parte dela circunferentia del circulo e dividasi .kg. in doi parti iguali sopra il ponto<lb></lb> |
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8631 .l. sia .ak. doppio al .kl. Multiplicato .ak. in .bi., fanno el doppio del’ area del triangolo .abi.<lb></lb> |
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8632 Ma il diametro .be. è .2. cotanti del .bi. Onde, multiplicato .ak. in .be., fanno .4. cotanti de-<lb></lb> |
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8633 l’ area del triangolo .abi. e il .kl. é il quarto del .ag. Adonca sia la mitá del .ak. E peró, mul-<lb></lb> |
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8634 tiplicato .kl. nelo diametro .be., fará .2. cotanti del’ area del triangolo .abi. Onde, multiplica-<lb></lb> |
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8635 to .al. nel .be., fará .6. cotanti del’ area del triangolo .abi. E .6. cotanti del’ area del triangolo .abi.<lb></lb> |
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8636 e .6. cotanti del’ area del triangolo .abi. é quanto l’ area del’ exagono .abgdez. Adonca si con-<lb></lb> |
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8637 chiude che, a multiplicare il diametro per gli .3/4. dela corda del’ angolo exagono, cioé per gli<lb></lb> |
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8638 .3/4. dela facia del triangolo, over del multiplicare tutta la corda ditta negli .3/4. del diametro. E<lb></lb> |
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8639 questo è chiaro che sara la sua area. .60. |
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8640 </p> |
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8641 <p class="main"> |
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8642 Se in uno cerchio s’ iscrive uno triangolo, de’ quali e .3. angoli tochino la circunferen-<lb></lb> |
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8643 tia del cerchio, è possibile, per la noticia de’ lati del ditto triangolo, trovare il dia-<lb></lb> |
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8644 metro del ditto tondo. E, acioché questo chiaro appaia, sia el cerchio nel quale<lb></lb> |
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8645 voglio fare il triangolo .abg., de’ quali e .3. angoli, tocando el cerchio ne’ ponti .a.b.g.<lb></lb> |
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8646 E dal ponto .a. si meni lo diametro .ad. segante il lato del triangolo .bg. nel ponto .e. Dico<lb></lb> |
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8647 che, per la noticia de’ lati del triangolo .abg., esser possibile di trovare la quantitá delo diame-<lb></lb> |
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8648 tro .ad. Sienno prima e .2. lati del triangolo .ab. e .ag. infra loro iguali e compisi la retta .bd.<lb></lb> |
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8649 e .dg. E sia ciascuno de’ triangoli .abd. e .agd. ortogonio, imperoché ciascuno é nel mezzo cerchio<lb></lb> |
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8650 <lb></lb> |
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8651 <pb></pb></p> |
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8652 <p class="folio"> |
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8653 folio 59r |
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8654 </p> |
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8655 <p class="main"> |
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8656 <lb></lb> |
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8657 </p> |
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8658 <p class="runhead"> |
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8659 Distinctio octava 59 |
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8660 </p> |
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8661 <p class="main"> |
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8662 <lb></lb> |
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8663 e, perché lo lato .ab. è iguale alo lato .ag., sará il lato .bd. iguali alo lato .dg. Onde la periferia<lb></lb> |
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8664 .bd .ala periferia .dg. è iguale. Sopra le iguali periferie si fanno iguali angoli, adonca l’ ango-<lb></lb> |
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8665 lo .dag. è iguale al’ angolo .bad. e la basa .be. è iguale ala basa .eg. Adonca il diametro .ad.<lb></lb> |
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8666 sega la retta .bg. in .2. parti iguali e ad angoli retti lo sega, commo Euclide nel terzo libro mo-<lb></lb> |
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8667 stra. Ortogonii sonno e triangoli e iguali .aeb. e .aeg. e simili, perché gli angoli del’ uno agli<lb></lb> |
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8668 angoli del’ altro sonno iguali e perché il triangolo .abd. á l’ angolo .bad. commune col triangolo<lb></lb> |
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8669 .abe. e l’ angolo .aeb. al’ angolo .abd. é iguale perché ciascuno di loro è retto. L’ altro adonca, che<lb></lb> |
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8670 è .adb., al’ altro, che è .abe., iguale. Adonca sonno e triangoli .abd. e .aeb. iguali. Similmente si<lb></lb> |
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8671 mostra el triangolo .agd. essere equangolo col triangolo .aeg. e’ .4. triangoli adonca .aeb. e .aeg.<lb></lb> |
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8672 .abd.agd. sonno infra loro simili. E gli triangoli simili intorno agli angoli iguali hano<lb></lb> |
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8673 e lati proportionali. Unde é cosí .da., che è sotto al’ angolo retto che è .abd., al .ab., contenen-<lb></lb> |
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8674 te quello, cosí .ab., over .ag., che sonno opposti agli angoli retti alla retta .ae. Unde la multi-<lb></lb> |
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8675 plicatione del .ad. nel catetto .ae. è iguale a ciascuno de’ quadrati dela linea .ag., over .ab., o vo-<lb></lb> |
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8676 gliamo dire ala multiplicatione del .ab. in .ag. Unde, multiplicando .ab. in .ag., over piglian-<lb></lb> |
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8677 do il quadrato dela linea .ab. o quello del .ag. e quella quantitá divideremo per lo catetto<lb></lb> |
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8678 .ae., ne perverrá la quantitá del diametro .ad. E il catetto .ae. é nota, quando e lati del trian-<lb></lb> |
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8679 golo sonno noti. Onde e il diametro .ad. sia noto. E, acioché habia piú chiaramente, sia cia-<lb></lb> |
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8680 scuno lato .ab. e .ag.10.bracia. e .bg. sia .12. Onde il catetto .ae. sará .8. Onde, multiplicato .ab.<lb></lb> |
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8681 in .ag., over el quadrato dela linea .ab., over .ag., che è .100., el quale diviso per .ae., cioé per .8., vien-<lb></lb> |
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8682 ne .12 1/2., per lo diametro .ad. Over altramente .ad. e .bg. infra loro si segano nel circulo .abgd.<lb></lb> |
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8683 Sará la multiplicatione del .ae. in .ed. commo la multiplicatione del .be. in .eg. Onde, multi-<lb></lb> |
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8684 plicando. be. in .eg. e divideremo per .ae., cioé .36. per .8., vienne .4 1/2. per la linea .ed. onde tutta<lb></lb> |
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8685 .ad., che è diametro, sia .12 1/2. commo dicemmo.<lb></lb> |
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8686 <lb></lb> |
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8687 E, quando e fossino note le .2. facie .ab. e .ag. e l’ altra facia non fosse nota, ma il diame-<lb></lb> |
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8688 tro .ad. fosse noto. Onde multiplicarai .ab. in .ag., che fanno .100. e divideralo per<lb></lb> |
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8689 .12 1/2., che è il diametro, vienne .8. per lo catetto .ae., del quale, il quadrato tratto del qua-<lb></lb> |
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8690 drato .ab., rimangono .36. per lo quadrato dela linea .be. Onde. be. è .6. e tutta .bg. è .12. |
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8691 </p> |
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8692 <p class="main"> |
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8693 Ma non sienno iguali li lati del triangolo .abg. Ma sia lo minore .ab., commo in que-<lb></lb> |
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8694 sta figura appare, e menise nel triangolo .abg. el catetto .az. e, perché nel segamen-<lb></lb> |
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8695 to .bd.ga. sonno .2. angoli, de’ quali uno è .bga. e l’ altro .bda. e fienno infra loro i-<lb></lb> |
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8696 guali, imperoché ciascuno è retto e l’ angolo .azg. al’ angolo .abd. iguali, perché<lb></lb> |
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8697 sonno retti. E l’ altro, che è .zab., al’ altro, .bad. Et equiangoli sonno e triangoli .azg. e .abd. Si-<lb></lb> |
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8698 milmente si dimostra il triangolo .azb. essere simile al triangolo .agd. Sonno certamente ne-<lb></lb> |
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8699 la settione contenta dala retta .ga. e dal’ arco .abgd. gli angoli che sonno .abg. e .adg. Onde quelli<lb></lb> |
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8700 angoli sonno infra loro iguali e gli angoli .azb. e .agd. E, perché simili sonno e triangoli .abd.<lb></lb> |
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8701 e .azg., sia cosí .de.al.ab., cosí .ga.al.az. Onde, multiplicando .ab. in .ag. e dividendo per .az.,<lb></lb> |
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8702 ne perverrá il diametro .ad. Exemplo con numeri: sia .ab.13. e .ag.15. e .bg.14. Voglio intorno al det-<lb></lb> |
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8703 to triangolo fare il minore cerchio posso. Adimando quanto sia lo suo diametro. Trovato dove cade<lb></lb> |
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8704 il catetto in sula facia del .bg. che sia .z. e sia .bz.5. e .zg. sia .9., il catetto .az. sia .12., adonca multiplicarai<lb></lb> |
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8705 .ab. in .ag., cioé .13. in .15., fanno .195. che, divisi per .12., viene .16 1/4., cioé divisi per lo catetto .az., el quale .16.<lb></lb> |
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8706 .1/4. è il diametro del detto tondo. |
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8707 </p> |
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8708 <p class="main"> |
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8709 Sia noto il diametro .ad. e .sia noto ciascuna dele rette .ab. e .ag. e la retta .bg. che<lb></lb> |
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8710 è la corda del’ archo .bdg., overo .bag. non sia manifesta. Perché simili sonno e trian-<lb></lb> |
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8711 goli .adg. e .azg. e intorno a simili angoli sonno e lati in proportione e sia cosí<lb></lb> |
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8712 .ad. al .dg., cosí .ab. al .bc. Onde la multiplicatione del .ab. in .dg. è iguale ala mul-<lb></lb> |
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8713 tiplicatione del .ad. in .bz. Ancora, perché simili sonno e triangoli .abd. e .azg., sia cosí .ad. al<lb></lb> |
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8714 .db., cosí .ag. al .gz. Onde la multiplicatione del .ad. in .zg. è iguali ala multiplicatione del .db.<lb></lb> |
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8715 in .ag. Ma la multiplicatione del .ab. in .gd. fo iguale ala multiplicatione del .ad. in .bz.<lb></lb> |
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8716 Adonca la multiplicatione del .ab. in .gd., con la multiplicatione del .ag. in .bd., è iguale ale<lb></lb> |
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8717 .2. multiplicationi del .ad. in .bz. e del .ad. in .zg., le quali .2. multiplicationi sonno iguali a quel<lb></lb> |
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8718 ch’ é fatto del .ad. in .bg. Adonca la multiplicatione del .ab. in .gd., con quello ch’ é fatto del<lb></lb> |
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8719 .ag. in .db. è iguale a quel ch’ é fatto del .ad. in. bg. Adonca, se la multiplicatione del .ab. in .dg.<lb></lb> |
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8720 congiongnerai con quello ch’ é fatto del .ag. in .bd. e la summa dividerai per .ad. ne perverrá<lb></lb> |
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8721 nota la corda .bg., commo dicemmo, che adonca del quadrato del diametro, ch’ é .264 1/16., si tolga<lb></lb> |
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8722 el quadrato del .ab. e .ag., cioé .95 1/4. e .39 1/16., dele quali le radici sonno le corde .bd. e .dg. Adonca<lb></lb> |
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8723 .bd. é .9 3/4. e .dg. é .6 1/4. Onde, multiplicando .6 1/4. per .13., cioé .gd. per .ab., e .9 3/4. per .15., cioé .bd. per<lb></lb> |
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8724 <lb></lb> |
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8725 <pb></pb></p> |
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8726 <p class="folio"> |
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8727 folio 59v |
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8728 </p> |
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8729 <p class="main"> |
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8730 <lb></lb> |
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8731 </p> |
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8732 <p class="runhead"> |
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8733 Distinctio octava. |
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8734 </p> |
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8735 <p class="main"> |
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8736 <lb></lb> |
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8737 .ag., ‘remo in tutto .227 1/2., che dividi per .ad., cioé per .16 1/4., haremo .14. per la corda .bg. E que-<lb></lb> |
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8738 sto è molto utile nel trovare le corde del’ archo agregato di .2. archi de’ quali la corda sonno<lb></lb> |
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8739 note, cioé forono note le corde deli archi .ba. e .ag., cioé la retta .ab. e .ag. per la quale noticia<lb></lb> |
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8740 troviamo la corda .bg. del’ archo .abg., che è l’ agionto di .2. archi .ba. e .ag. E questo dimostró<lb></lb> |
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8741 Ptolomeo nela compositione dele taole deli archi e dele corde nel’ almgesto. .61. |
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8742 </p> |
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8743 <p class="main"> |
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8744 E gli é uno triangolo, commo vedi disegnato, .abc. e .bc. é uno piú che .ac. e .ab.<lb></lb> |
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8745 é uno bracia piú che .bc. e trovo che gli é quadro .84.bracia. Adimandasi quanto<lb></lb> |
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8746 é per ciascuna facia. Abi questa per una sottile e bella domanda. E questo modo<lb></lb> |
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8747 ti bisogna tenere. Che porrai che .bc. sia una cosa, dove .ac. sia una cosa meno uno<lb></lb> |
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8748 e .ab. sia una cosa piú uno. Ora, a volere trovare l’ area sua, tu hai prima a trovare il catetto .ad.,<lb></lb> |
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8749 dove prima è di bisogno sapere il ponto .d. quanto presso al ponto .b. e quanto è presso al<lb></lb> |
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8750 ponto .c., che cosí lo saperai: multiplica .ab. in sé, cioé una cosa e uno, fanno uno censo e .2. co-<lb></lb> |
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8751 se e uno. E poi multiplica .ac. in sé, fanno uno censo e uno meno .2. cose. Trallo del’ uno censo .2.<lb></lb> |
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8752 cose e uno, rimane .4. cose, le quali parti nel .bc., cioé in una cosa, vienne .4., agiongni a una cosa, fan-<lb></lb> |
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8753 no una cosa .e.4., cioé al .bc. tone la mitá, ch’ é .1/2.co. e .2. E tanto é dal .b. al .c. Overo trai .4. d’ una<lb></lb> |
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8754 cosa, rimane una cosa men .4., de’ quali la mitá è .1/2. cosa men .2. e questo è .dc. Ora, saputo qu-<lb></lb> |
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8755 anto é .bd. overo .cd., et tu voglia .ad., del quadrato del .ab. trai il quadrato .bd., overo del qu-<lb></lb> |
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8756 adrato del .ac. trai il quadrato .dc., che uno medesimo numero te rimarrá. Dove, tratto il qua-<lb></lb> |
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8757 drato .bd. del quadrato .ab. rimane .3/4. di censo meno .3., la cui radice è il catetto .ad. Adonque<lb></lb> |
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8758 .ad. è la radice di .3/4. di censo e .3. meno. E, a volerlo quadrare, multiplicarai la radice di .3/4. di cen-<lb></lb> |
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8759 so men .3. per la mitá del .bc., cioé per .1/2. cosa, fanno la radice di .3/16. di censo di censo meno .3/4. di cen-<lb></lb> |
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8760 so. E tanto é quadro el detto scudo. E noi dicemmo che gli era quadro .84.bracia. Adonca<lb></lb> |
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8761 la radice di .3/4. di censo di censo meno .3/4. è iguali a .84. Dove areca ciascuna parte a quadrato e<lb></lb> |
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8762 harai che .3/16. di censo di censo meno .3/4. censo sonno iguali a .7056., dove raguaglia le parti dando<lb></lb> |
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8763 .3/4. di censo ciascuna e arai .3/16. censo di censo iguali a .3/4. di censo e .7056., dove areca a .1.cen. di censo e arai<lb></lb> |
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8764 uno censo di censo iguali a .4.censi. e .37632. Dimeza e .censi., sono .2., multiplica in sé, fanno .4., po-<lb></lb> |
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8765 ni sopra .37632., fanno .37636., la cui radice è .194. e diremo la cosa valere la radice di .37632., ci-<lb></lb> |
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8766 oé .194., posta sopra .2. e di quel preso la radici, cioé la radice di .196., che è .14. Adonca la cosa<lb></lb> |
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8767 vale .14. e tanto é .bc. e .ab. sia .15. e .ac. sia .13. e cosí fa sempre. .62. |
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8768 </p> |
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8769 <p class="main"> |
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8770 E sono .2. torri, l’ una distante al’ altra .150. e l’ una è alta, cioé .ab., 100.bracia. e .cd è .70.<lb></lb> |
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8771 bracia e il piano .bd. è .150. In sule quali torri, cioé in ciascuna extremitá, é uno uccelo<lb></lb> |
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8772 e a uno tratto si movano e a uno modo volano e uno tratto giongano a una fon-<lb></lb> |
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8773 te situata infra ’l .b. e .d. Adimandasi quella fonte quanto é apresso al .b. e quan-<lb></lb> |
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8774 to è apresso al .d., cioé. a dire trova uno ponto, infra ’l .b. e .d., che sia da quello alla somitá .a.<lb></lb> |
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8775 quanto da quello alla somitá .c. El quale ponto diciamo sia el ponto .e. E diremo che dal .e. al<lb></lb> |
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8776 .b. sia una cosa. Seguita dal .e. al .d. essere .150. meno una cosa. Ora, per sapere quanto è .ae., multi-<lb></lb> |
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8777 plica .ab. in sé, cioé .100., fanno .10000. e multiplica .be. in sé, cioé una cosa, fanno uno censo che,<lb></lb> |
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8778 con .10000., fanno .10000. e uno censo. E la radice di questo è .ae. Ora é da sapere quanto è<lb></lb> |
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8779 dal .e. al .c., dove multiplicherai .ed. in sé, cioé .150. meno una cosa in sé, fanno uno censo e .22500.<lb></lb> |
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8780 meno .300.cose. E di poi multiplica .dc. in sé, cioé .70. in sé, fanno .4900. Agiongni a uno censo<lb></lb> |
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8781 e .22500. meno .300.cose., fanno uno censo e .27400. meno .300.cose. E la radice di questo é la<lb></lb> |
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8782 linea .ec., la quale è iguali ala linea .ea. Adonca la radice di uno censo e .27400. é iguali ala ra-<lb></lb> |
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8783 dice di uno censo e .10000. Unde multiplica ogni parte in sé e arai che uno censo e .27400.<lb></lb> |
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8784 meno .300.cose. è iguali a uno censo e .10000. Dove raguaglia li parti levando da ogni parte<lb></lb> |
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8785 uno censo e .10000. e dando a ogni parte .300.cose. E aremo che .17400. sonno iguali a .300.<lb></lb> |
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8786 cose, dove la cosa vale .58. E noi facemmo posicione che dal .b. al .e. fosse una cosa. Adonca fo .58.<lb></lb> |
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8787 bracia, cioé dala torre dele .100.bracia. alla fonte. E dala torre del .70.bracia. alla fonte fo l’ avan-<lb></lb> |
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8788 zo infino in .150. che fo .92. bracia. E cosí aopra sempre. Potresti ancora multiplicare cias-<lb></lb> |
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8789 cuna torre in sé e trare il quadrato dela minore del quadrato dela magiore e quello avanza<lb></lb> |
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8790 partire nel dopio dela distancia dal’ una torre al’ altra e quello ne viene poni sopra la<lb></lb> |
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8791 mitá dela distantia che è dal’ una al’ altra e quella somma è quello che la fonte è presso ala mi-<lb></lb> |
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8792 nore torre. Puó intervenire che la fonte è fora del piano .bd., o dala parte del .b.,<lb></lb> |
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8793 overo dala parte del .d. et cetera. |
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8794 </p> |
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8795 <p class="main"> |
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8796 Sonno doi arbori over torri equedistanti in un piano levate, l’ una .ab., alta .30., l’ altra<lb></lb> |
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8797 .ec., alta .40.; distante l’ una dal’ altra .20. Io tiro una corda ale loro cime, longa<lb></lb> |
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8798 .25. Dimando, lasciando un piombino dala cima dela piú alta, quanto si fermará distante<lb></lb> |
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8799 <lb></lb> |
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8800 <pb></pb></p> |
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8801 <p class="folio"> |
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8802 folio 60r |
|
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8803 </p> |
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8804 <p class="main"> |
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8805 <lb></lb> |
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8806 </p> |
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8807 <p class="runhead"> |
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8808 Distinctito octava 60r |
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8809 </p> |
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8810 <p class="main"> |
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8811 <lb></lb> |
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8812 dal’ una al’ altra e quanto saran del piombino a terra perpendiculariter. Fa cosí. Poni che dal<lb></lb> |
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8813 piombino al ponto .f. sia .1.co., che .gi. ven a essere .20. men .1.co. Ora, se tu ben guardi, tu ái doi triangoli<lb></lb> |
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8814 ortogonii mediante la equedistante .gf., tirata al al ponto dela gravezza del piombo. La quale è<lb></lb> |
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8815 quanto bc., cioé .20., per la .34a. del primo de Euclide di quali triangoli, l’ uno éne .agi., l’ altro .efi.<lb></lb> |
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8816 E l’ angolo .g. e l’ angolo .f. son retti per la .29a. del primo. Ora, se tu imagini una equedistante ala basa<lb></lb> |
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8817 .bc., tirata dal ponto .a. fin al lato .ec., sará ancor lei equale e equedistante al .gf. e causará el triangolo<lb></lb> |
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8818 ortogonio .aek., che l’ angolo .k. è retto, che, per non confonderte, in linee .ak. non tiro. E lo ponto .i. sia<lb></lb> |
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8819 pure nel medesimo luogo, distante dal .k.l. cosí commo dal .f. Peró, commo de sotto apresso ave-<lb></lb> |
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8820 rai, se dirá: se .20., cioé .ak., de basa, mi dá .25., cioé .ae., de potunissa, che mi dará .1.co. de basa, cioé<lb></lb> |
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8821 .if. Adonca aveno che gli é cosí .ai.alo.gi. commo .ei. alo .fi. e cosí .ag. alo .gi. commo .ef. alo .fi. E peró<lb></lb> |
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8822 dirai, per modo sotietatis, se .20. me dá .25., che me dará .1.co. Multiplica e parti, arai che .1.co. te dará .1 1/4.co.<lb></lb> |
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8823 e tanto sia .ei. L’avanzo fin .25., che è .25. men .1 1/4.co., sará .ai. Ora quadra .ei., fará .1 9/16.ce. e di questo chava<lb></lb> |
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8824 el quadrato .fi., cioé .1.ce., resta .9/16.ce., la cui .R. sia .ef., cioé .3/4.co. E poi multiplica ancora .gi. in sé, cioé .20. men<lb></lb> |
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8825 .1.co., fa .400. men .40.co. piú .1.ce. e questo cava del quadrato .ai., cioé de .625. piú .1.9/16.ce. men .62 1/2.co., re-<lb></lb> |
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8826 sta .225. piú .9/16.ce. men .22 1/2.co. La .R. de questo fo .ag., cioé .15. men .3/4.co. Donca, se giogni .10. sopra .ag., fa-<lb></lb> |
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8827 rá .25. piú .3/4.co. e sirá equale alo .ef., cioé a .3/4.co., peroché, siando ex ypotesi .ec.10. piú alta che .ab. e da-<lb></lb> |
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8828 l’ una e dal’ altra se remove quantitá equale, cioé da .ab. se cava .gb. e dalo .ec. si leva .fc., che sonno<lb></lb> |
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8829 equali che ’l mostra la equedistante .gf. Adonca li remanenti sonno .ag. e lo .ef. inequali e peró, gionto<lb></lb> |
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8830 .10. sopra .ag., cioé sopra .15. men .3/4.co., fa .25. men .3/4.co. equali a .3/4.co., che è .ef. Aguaglia e sequi la equa-<lb></lb> |
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8831 tione, arai la cosa valere .16 2/3. E tanto sia distante el piombino dala torre .ec. piú alta che sia .fi.<lb></lb> |
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8832 E la distantia dal piombino ala torre bassa sia .3 1/3., cioé .gi. fatta et cetera. Ora farai de .25. doi tal parti<lb></lb> |
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8833 che tanto sia l’ una a .3 1/3. quanto l’ altra a .16 2/3. O voi dire che tal parte sia l’ una del’ altra qualche .3 1/3. de<lb></lb> |
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8834 .16 2/3., overo che, multiplicata l’ una per .3 1/3., facia quanto l’ altra multiplicata per .16 2/3., che tanto fará a una via <lb></lb> |
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8835 quanto<lb></lb> |
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8836 al’ altra. E averai .ai.4 1/6. e lo .ie.20 5/6. E lo .id.27 1/2., perché .ag. sia .2 1/2. et cetera. 64<lb></lb> |
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8837 Sono .2. albori, o voi dir torri, l’ una alta .10.ab., l’ altra alta .15.dc., distanti a livello<lb></lb> |
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8838 .12.bc. Vengo e tiro doi corde, l’ una .db., l’ altra .ac., cioé dalla cima del’ una al pie’ del’ al-<lb></lb> |
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8839 tra. Dimando in che parte se intersegaranno e quanto lonzi da terra e a che parte de ciascuna<lb></lb> |
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8840 corda. Fa chosí. Poni che .ei. sia .1.co., poi dí: se .dc., ch’ é .15., mi dá de basa .12., che<lb></lb> |
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8841 me dará .1.co. Opra. Te dará .4/5.co. e tanto sara .ib. Poi voltate al triangolo .abc. e dirai: se .ab. ch’ é<lb></lb> |
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8842 .10. me dá de basa .12., che me dará .1.co. Che te virrá a dare .6/5.co. E tanto sirá .ic., peroché .ei. perpendicu-<lb></lb> |
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8843 lare fue al’ un triangolo e l’altro, movendose dal ponto .e., sotto le potunisse .ac. e .db. Ora cava .4/5.co.,<lb></lb> |
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8844 che prima avesti per .bi., de .12., cioé .de.bc., resta .12. men .4/5.co. e tanto viene essere .ic. Et tu ái trovato che lo .ic., <lb></lb> |
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8845 per<lb></lb> |
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8846 ragion del triangolo .abc., essere .6/5.co., donca arai .12. men .6/5.co. equale a .6/5.co. Aguaglia e sequi. Arai la cosa va-<lb></lb> |
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8847 ler .6. E tanto sia .ei. Ora, per sapere quanto sia .eb., multiplica .bi. in sé, che é .4 4/5., fa .576/25. e multiplica in sé .6. <lb></lb> |
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8848 cioé <lb></lb> |
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8849 .ei. fa<lb></lb> |
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8850 .36. che, gionti insiemi, fanno .59 1/25. e la .R. di questo sia .eb. L’ avanzo fin .d., cioé .ed., sia .R.369. men .R.59 1/25.<lb></lb> |
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8851 e lo .ec. sia .R.87 21/25. e lo resto fin .a. sia .R.244. men .R.87 21/25. facta et cetera. 65<lb></lb> |
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8852 E gli é un alboro de nave che tende in cono uniformiter, ma schapezzo in ponta, alto<lb></lb> |
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8853 .25. El diametro dela basa de sotto è .3., cioé .cg., el diametro dela basa de sopra è<lb></lb> |
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8854 .1., cioé .ab. Vogliolo segare in doi parti equali atraverso, che l’ una metto tochi al<lb></lb> |
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8855 patrone e l’ altra al nochieri. Dimando quanto distante dale base se doverá segare.<lb></lb> |
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8856 Fa chosí. Perché le spetie de piramide troncata, fornesila tutta fin in ponta aguzza in questo<lb></lb> |
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8857 modo. Imagina doi perperpendiculari equidistanti dale extremitá del diametro dela basa de so-<lb></lb> |
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8858 pra verso la cima al diametro dela basa de sotto e sieno l’ una .ad. l’ altra .bf. che ciascuna sia .25., cioé<lb></lb> |
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8859 quanto l’ altezza. Ora tu ái che .df. similmente è .1. commo .ab. Donca .cd. sia .1. e anco .fg.1., secon-<lb></lb> |
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8860 do el tema. Ora tu ái da’ lati doi triangoli ortogonij, se ben guardi, cioé .acd. e l’ altro .bfg., che<lb></lb> |
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8861 li angoli .d. éne retto e cosí l’ angolo .f. causati dal caso dele perpendiculari. Or prendine uno<lb></lb> |
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8862 qual voli, che non fa caso, e sia .acd., del qual li doi lati .ad. e .cd. sonno noti. Ora per for-<lb></lb> |
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8863 nire la piramide al suo vertice supremo, che sia el ponto .h., prendi la .1/2. de .cg., cioé dela ba-<lb></lb> |
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8864 sa del pede, che è .1 1/2. e sia .ce., nella cui mitá imagina cadere l’ assi .he. perpendiculariter, qual<lb></lb> |
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8865 similmente fará un triangolo ortogonio .hce. simile al primo .acd., per la .2a. del .6o. de Euclide:<lb></lb> |
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8866 peroché la linea .ad., equedistantemente al terzo lato .he., taglia li doi lati .hc. el lato .ce. nelli pon-<lb></lb> |
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8867 ti .a. e .d. che li sega in proportione, per la giá adutta conclusione. Donca, per trovare la quan-<lb></lb> |
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8868 titá .he. dirai: per la regola del .3., se .1. de basa, cioé .cd., me dá de catetto .25., cioé .ad., che me<lb></lb> |
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8869 dará de catetto .1 1/2., cioé la basa .ce. Multiplica .25. via .1 1/2. e parti in .1. Arai che te dará .37 1/2. per tutto l’ a-<lb></lb> |
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8870 xis .he. Ora quadra la basa .cg., con la ragione del cerchio, arai de sua quadratura .7 1/14., qual<lb></lb> |
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8871 multiplica via la longhezza tutta del’ alboro, cioé via .he., che è .37 1/2., fará .265 5/26. e tanto sia l’ aria corpo-<lb></lb> |
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8872 <lb></lb> |
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8873 <pb></pb></p> |
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8874 <p class="folio"> |
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8875 folio 60v |
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8876 </p> |
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8877 <p class="main"> |
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8878 <lb></lb> |
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8879 </p> |
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8880 <p class="runhead"> |
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8881 Distinctio octava |
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8882 </p> |
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8883 <p class="main"> |
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8884 <lb></lb> |
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8885 rea de tutto el chelindro rotondo. Del qual prendi el .1/3. per lo suo cono, ne ven .88 11/28. per tut-<lb></lb> |
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8886 ta la soliditá del’ alboro, qual serba da parte. E poi quadra tutta la piramide piccola agion-<lb></lb> |
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8887 ta al ponto .h., dela quale hai noto tutti ’soi lati. Cioé altezza e basa .ab. ch’ é 1. E l’ axis .hk. che<lb></lb> |
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8888 è .12 1/2. Cioé el rimanente de .37 1/2. trattone .25., che fo posto longo l’ alboro. Multiplica in sé .ab.,<lb></lb> |
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8889 fa .1., prendene li .11/14., ne ven .11/14. E tanto sia quadra la basa .ab. dela suprema piramidella, qual<lb></lb> |
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8890 multiplica via la sua altezza .hk., cioé via 12 1/2., fa .9 23/28. per lo chelindro; prendine el .1/3. harai .3 23/84.<lb></lb> |
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8891 per la piramide .hab. La qual cava dela piramide tutta che sopra serbasti, cioé de .88 11/28., re-<lb></lb> |
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8892 sta .85 5/42. per tutta la soliditá del’ alboro proposto. E questo parti in .2., ne ven .42 47/84. E tanta<lb></lb> |
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8893 soliditá del ditto alboro tochará per uno, cioé al patrone e nochiero. Hora e da vedere in<lb></lb> |
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8894 che parte se deverá segare, commo dici el tema. La qual cosa te conven fare per positione al-<lb></lb> |
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8895 gebratica in questo modo. Poni che ’l diametro dela basa dove se debia segare sia .1a. cosa e<lb></lb> |
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8896 di questo trova l’ axis fin ala ponta .h., ala rata de tutta la gran piramide, dicendo: se .ab.,<lb></lb> |
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8897 che è .1., diametro me dá .12 1/2., cioé .hk. de altezza, che me dará .1.co. Opera multiplicando .1.co.<lb></lb> |
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8898 via .12 1/2. e parti in .1., ne ven .12 1/2.co. E tanto sia l’ axis dela piramide la cui basa fosse .1.co. E po’<lb></lb> |
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8899 devi ancora dire: se .3., cioé .cg. basa del grande, me dá .37 1/2., che me dará .1.co. Operando te<lb></lb> |
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8900 darebe el medesimo che di sopra, cioé .12 1/2.co. Perché ditte piramidi sonno fra loro proportio-<lb></lb> |
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8901 nali, perché stanno sotto medesima altezza e medesima basa .cg., avenga che l’ una naschi ma-<lb></lb> |
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8902 gior del’ altra et cetera. Hora quadra la basa posta. Cioé multiplica .1.co. in sé, fa .1.cen., prendine li<lb></lb> |
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8903 .11/14., ne ven .11/14. cen., qui multiplica via l’ altezza, cioé via .12 1/2.co., fa .3 23/84.cu. E tanta sia la soliditá dal<lb></lb> |
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8904 taglio in cima. E giá tu sai che al’ uno ne tocha .42 47/84. de soliditá fra le doi base .ab. e .cg. Don-<lb></lb> |
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8905 ca, per venire ala equatione del capitolo, giongni .3 23/84., che è la piramidella .hab. dela cima,<lb></lb> |
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8906 sopra .42 47/84., fará .45 25/42. per tutta la soliditá dela piramide, la cui basa hai posto essere .1.co.<lb></lb> |
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8907 Doncha harai .3 23/84.cu. essere equali a .45 23/42., sequi, la cosa varra .R.cu.13 51/55. E tanto conven che<lb></lb> |
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8908 sia el diametro .lm. dela basa dove si deve segare. Ora, per sapere quanto distante dal capo<lb></lb> |
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8909 grosso se debia segare, prima trova l’ axis da questo diametro trovato fin ala cima. Poiché<lb></lb> |
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8910 tu sai la valuta dela cosa essere .R.cu.13 51/55. e giá tu sai comme sopra trovasti che l’ axis .hn. de<lb></lb> |
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8911 questa basa .fo.12 1/2.co. Donca multiplica .12 1/2. via la valuta de .1.co., cioé via .R.cu.13 51/55., fará.<lb></lb> |
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8912 .R.cu.27201 31/44. E tanto distante dala ponta se segará l’ axis interiore del qual cava .12 1/2., che<lb></lb> |
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8913 sai de certo essere l’ axis .hk., che de ponto te forní la gran piramide de tutto l’ alboro, restará<lb></lb> |
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8914 aponto .R.cu .27201 31/44. men .12 1/2. E tanto distante sia la segatura dala basa suprema .ab., cioé<lb></lb> |
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8915 .kn. E dala basa inferiore .cg. sia la distantia del rimanente de .25. trattone .R.cu.27201 31/44.<lb></lb> |
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8916 men .12 1/2. Overamente el rimanente de .37 1/2., che è tutto l’ axis grande .he., cavatone .R.cu.<lb></lb> |
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8917 .27201 31/44. per l’ axis piccolo, che l’ uno e l’ altro tanto fa, cioé. 37 1/2. men .R.cu.27201 31/44. E tanto lon-<lb></lb> |
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8918 tano dal capo grosso del’ axis interiore se segará. E sia .ne. Mo é da vedere quanto distan-<lb></lb> |
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8919 te dala parte defore da ciascun capo se doverá metter la sega, che cosí lo troverai tutto per la<lb></lb> |
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8920 penultima del primo de Euclide, perché tu hai un triangolo ortogonio .hln., del qual l’ an-<lb></lb> |
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8921 golo .n. éne recto. E l’ un di lati che lo contiene sia mezzo diametro dela basa .lm., cioé .ln.e<lb></lb> |
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8922 l’ altra l’ axis .hn. Peró multiplica .hn. in sé, cioé .R.cu.27201 31/44. E multiplica .ln. in sé, cioé la<lb></lb> |
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8923 .1/2. de .R.cu.13 51/55. E queste doi quadrature giongnerai insiemi faranno el quadrato dela ypo-<lb></lb> |
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8924 tomissa .hl., dela cui .R. caverai la quantitá .ha. che ancora, per ditta penultima, la troverai,<lb></lb> |
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8925 perché tu hai pur un altro triangolo ortogonio .hak., del qual l’ angolo .k. éne retto. E l’ axis<lb></lb> |
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8926 .hk. sia aponto in mezzo el diametro .ab., in ponto .k. Peró multiplica .ak. in sé, .che sai che è<lb></lb> |
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8927 .1/2., perché tutto .ab. fo posta .1., fa .1/4. E multiplica .hk. in sé, che trovasti essere .12 1/2., fa .156 1/4. che,<lb></lb> |
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8928 gionte queste doi quadrature insiemi, fanno .156 1/2. per lo quadrato dela ypotumissa .ha. La<lb></lb> |
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8929 cui .R., tratta dela gran ypotumissa .hl., remarratte la quantitá nota .al. per la distantia dala<lb></lb> |
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8930 segatura al capo sutile, la cui basa fo posta .ab. per diametro. E cosí poi, havuto la notitia de-<lb></lb> |
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8931 lo .al., quella caverai dela quantità .ac., qual ancora éne ypotumissa de un altro triangolo or-<lb></lb> |
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8932 togonio .acd., che sia .R.626., perché .ac., uno di lati continenti el rettangolo, è .25., cioé la lon-<lb></lb> |
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8933 ghezza del’ alboro e l’ altro .cd., che è .1., li cui quadrati, gionti insiemi, fanno .626. per lo quadra-<lb></lb> |
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8934 to dela ypotumissa .ac., la cui .R. sia essa .ac., cioé .R.626. Siché dirai che la segatura sirà distan-<lb></lb> |
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8935 te dal capo grosso .R.626. men la quantitá .al. et cetera. La qual a te lascio trovare al modo ditto, per non<lb></lb> |
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8936 mi stender piú in operatione et cetera. E questa sia molto utile e bella ragione e de gran stima<lb></lb> |
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8937 in l’ arte geometrica. Dí che: nota che, stu ponesse ditto alboro sopra un taglio de coltello, in<lb></lb> |
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8938 modo che ’l taglio fosse nel ponto .l., overo .m., staria in equilibra, cioé in bilancia, che da niun<lb></lb> |
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8939 capco trabuchari. Perché tanto peso hará dal taglio al capo piccolo quato dal taglio al ca-<lb></lb> |
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8940 <lb></lb> |
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8941 <pb></pb></p> |
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8942 <p class="folio"> |
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8943 folio 61r |
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8944 </p> |
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8945 <p class="main"> |
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8946 <lb></lb> |
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8947 </p> |
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8948 <p class="runhead"> |
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8949 Distinctio octava 61 |
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8950 </p> |
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8951 <p class="main"> |
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8952 <lb></lb> |
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8953 po grosso, peroché la mesura e ’l peso sempre sonno fra loro in tutte le cose proportionate et cetera.<lb></lb> |
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8954 E, in tal modo, reggerate in tutte sorte de piramidi scapezze tonde e laterate et cetera. 66<lb></lb> |
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8955 E gli é un prato triangulare .abc., del quale el lato .ac. è .130. e ’l lato .ab.150. e ’l lato<lb></lb> |
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8956 .cb.140. e in sun l’ angolo .a. é una torre .ad. alta .125. e in sun l’ angolo .b. un’ altra .bf.,<lb></lb> |
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8957 alta .135. E in sun l’ angolo .c. un’ altra .ce. alta .125. Io fermo un ponto in ditto pra-<lb></lb> |
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8958 to, equidistante dale cime di dette .3. torri. Dimando quanto sia dal ditto ponto .a.<lb></lb> |
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8959 ciascuna dele cime de ditte .3. torri e quanto da ditto ponto al pie’ di ciascuna de ditte .3. tor-<lb></lb> |
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8960 ri. Fa cosí. Prima ferma un ponto a tuo modo, che sia equidistante dale cime dele doi tor-<lb></lb> |
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8961 ri, quali voli, che non fa caso. E poi sequirai per la terza torre. Or pigliamo le .2., cioé .bf. e<lb></lb> |
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8962 quella .ce., che sai che .bf. è alta .135. e .ce. alta .145. e la distanza dal’ una a l’ altra, cioé .cb. é<lb></lb> |
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8963 .140. E in su questa linea .cb., de necessitá, sirá fermato el ponto equidistante dale lor cime, ació<lb></lb> |
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8964 possamo sequire la nostra operatione, overo aguaglimento, che aliter serebe difficile avenga<lb></lb> |
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8965 che ’l potesse essere ancora fore de ditta linea .cb. E seria pure equidistante, ma non serebe al<lb></lb> |
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8966 proposito del nostro aguaglimento. E, per questo trovare, farai commo quella dela fonte<lb></lb> |
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8967 fra doi torri e doi ucelli in cima ciascuna, volando pari a un tempo, zonzano ala fonte et cetera. E pe-<lb></lb> |
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8968 ró poni che ’l ditto ponto sia distante dal pede dela torre .ce.1.co. E sia .k. Sirá adonca .kb.140.<lb></lb> |
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8969 men .1. cosa. Hora quadra .ce. ch’ é. che .145., fará .21025. e quadra .ck., che è .1. cosa, fa .1. censo, zonzi questi<lb></lb> |
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8970 doi quadrati insiemi, fanno .21025. piú .1. censo, qual salva. Poi quadra .bf., ch’ é .135., fará .18225. e<lb></lb> |
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8971 quadra .kb. che è .140. men .1. cosa, fará .19600. men .280.co. piú .1. censo. E questo zonzi con .18225., fa-<lb></lb> |
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8972 rá .37825. men .280.cose. piú .1. censo. E le .R. di queste .2. summe., cioé .R.21025. piú .1. censo che serbasti<lb></lb> |
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8973 e la .R.37825. men .280.co. piú .1. censo. E l’ una prima sirá la linea .ek. e la seconda sirá la linea .kf., che<lb></lb> |
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8974 vengono essere .2. ypotumisse et cetera. Adonca multiplica li extremi in sé, harai li lor quadrati equa-<lb></lb> |
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8975 li, cioé che .21025. piú .1. censo sirá equale a .37825. men .28.cose. piú .1. censo. Aguaglia le parti e sequi, harai<lb></lb> |
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8976 la cosa valere .60. E tanto sirá .ck. L’ avanzo fin .140. sirá .kb., cioé .80. E tanto sirá .kf. quan-<lb></lb> |
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8977 to .ke., che l’ una e l’ altra sirá .R.24625. Ora, poiché abiam trovato in la linea .cb. el ponto .k. e-<lb></lb> |
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8978 quidistante dale cime, bisogna trovare el catetto del prato triangulare. Cioé .abc. cadente in<lb></lb> |
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8979 sula basa .cb., che sia .al. E sirá longo .120. E poi, nel ponto .k., leva una pararella equedistan-<lb></lb> |
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8980 te a ciascuna torre .bf. e .ce., che sia .ko., sopra la quale, de necessitá, convien sia el ponto eque-<lb></lb> |
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8981 distante da tutte .3. le cime. E peró sapi: poiché .kf. e .ke. sonno trovate equali, similmente, su<lb></lb> |
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8982 per la linea .ko., fermato un ponto dove si voglia, commo sia el ponto .s., ancora siran .cs. e<lb></lb> |
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8983 .sf. Avenga che non siran equali in la medesima quantitá, che prima era .kf. e .ke. E peró qui<lb></lb> |
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8984 fa la toa positione. E poni che lo .ks. sia .1.co. Ora mena la linea .rs. equedistante alo .lk., che<lb></lb> |
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8985 sirá ancora lei .10. Cioé la distantia dal catetto alo ponto .k., perché el ponto .l. cade distan-<lb></lb> |
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8986 te dal’ angolo .c.50. e dal’ angolo .b.90. Adonca sirá .rl.1.co., comme .ks., perché dele superficie de’<lb></lb> |
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8987 lati equedistanti, li lati oppositi e li angoli opositi, sempre sonno equali, per corelario .34e. primi<lb></lb> |
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8988 Euclidis. Donca sirá .ra., cioé l’ avanzo del catetto .120. men .1.co. Or, tutte queste cose notate,<lb></lb> |
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8989 tu vedi che noi habiamo doi triangoli ortogonij che l’ uno è .kbs., l’ altro .rsa. E l’ angolo .k.<lb></lb> |
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8990 del primo è retto e l’ angolo .r. del secondo anche è retto. Unde la linea .sf. ven a essere potumissa d’ un<lb></lb> |
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8991 triangolo ortogonio .bsf., del qual l’ angolo .b. è retto, perché la potumissa .bs. del triangolet-<lb></lb> |
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8992 to .ksb., movendose dal ponto .b. al ponto .s., fa squadro. Adunque li doi quadrati deli .2. lati<lb></lb> |
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8993 .bf. e .bs. fanno el quadrato dela potumissa .sf., per la penultima del primo de Euclide. Ma,<lb></lb> |
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8994 perché el quadrato de .bs. vale li .2. quadrati delo .ks. e .kb., per eandem penultimam, sequi-<lb></lb> |
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8995 ta la linea .sf. valere .3. quadrati dele .3. linee .ks. e .kb. e .bf. E peró quadra ciascuna dele dit-<lb></lb> |
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8996 te; harai per quella .bf.18225. e per quella .kb.6400. e per quella .ks.1.cen., che, gionti que-<lb></lb> |
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8997 sti .3. quadrati insiemi, fanno .1.cen. piú .24625., quali serba. Ora, per lo simile, tu hai dal’ altra<lb></lb> |
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8998 parte un triangolo .sad. pur ortogonio, del qual l’ angolo .a. e retto e la linea .sd. ven a esse-<lb></lb> |
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8999 sere potumissa. E li doi quadrati dele .2. linee .ad. e .as., gionti insiemi, fanno el quadrato de-<lb></lb> |
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9000 la linea .ds., per la ditta penultima. E cosí la linea .as., perché l’é potumissa del triangolo .ars.<lb></lb> |
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9001 del qual l’ angolo .r. è retto. Donca la linea .sd. ven a valere .3. quadrati de .3. linee .rs. e .ra. e<lb></lb> |
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9002 .ad. Peró quadra ciascuna de ditte linee, cioé .ad., che è .125., fa .15625. e .rs., che è .10., fa .100.<lb></lb> |
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9003 e .ra. che è .120. men .1.co. fará .14400. men .240.co. piú .1. censo. E giogni questi .3. quadrati insiemi, faranno in<lb></lb> |
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9004 tutto .30125. men .240.co. piú .1. censo e la .R. di questa summa convene essere equale .ala .R. dila sum-<lb></lb> |
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9005 ma deli altri .3. quadrati che sopra serbasti, cioé .a.R.1.ce. piú .24625., perché le doi .li.sd. e .sf.<lb></lb> |
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9006 hano essere equali per lo nostro ponere, perché si movano dal ponto .s. del’ altra che è .es., non<lb></lb> |
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9007 <lb></lb> |
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9008 <pb></pb></p> |
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9009 <p class="folio"> |
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9010 folio 61v |
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9011 </p> |
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9012 <p class="main"> |
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9013 <lb></lb> |
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9014 </p> |
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9015 <p class="runhead"> |
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9016 Distinctio octava |
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9017 </p> |
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9018 <p class="main"> |
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9019 <lb></lb> |
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9020 si fa chaso, perché, commo dicemo sempre, essa convene de necessitá essere equale ala linea .sf., quando<lb></lb> |
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9021 el ponto sia super la linea .ko. E peró basta aguagliare le .2. linee .sd. e .sf. e l’ altra seque de necessitá. Ora<lb></lb> |
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9022 multiplica li extremi in sé, arai .30125. men .240.co.piú .1.ce. equali a .1.ce. piú .24625. Aguaglia le parti e se-<lb></lb> |
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9023 qui el capitulo, arai in ultimo .5500. equali a .240.co. Parti .5500. per .240., ne ven .22 11/12. e tan-<lb></lb> |
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9024 to varrá la cosa e tanto sará .ks. e tanto .rl. e lo resto, fin .120., sia .97 1/12., ch’ é .ra. Ora poi trovare<lb></lb> |
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9025 tutte le linee. Per forza dele quadrature arai .sb.R.6925 25/144. e la linea .cs.R.4125 25/144. e .lo.as. sará<lb></lb> |
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9026 .R.9525 25/144. e cosí .ái. la distantia del ponto .s. dal pede de ciascuna dele .3. torri. Ora, per sapere<lb></lb> |
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9027 la distanza dele cime, prima, per quella .bf. quadra .sb.; fa .6925 22/144. e quadra .bf., fa .18225. qual zon-<lb></lb> |
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9028 zi con .6925 25/144., fa .25150 25/144. e la .R. de .25150 25/144. sia .sf. e altretanto sia .ds. e altretanto sia .es., com-<lb></lb> |
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9029 mo porrai trovare quadrando .as. e .ad. e zonzere insemi la .R. di quel zonto, fará .sf. e quadra .cs.<lb></lb> |
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9030 e .ce., zonzi insemi, fará la .R. di quella summa la linea .es. facta et cetera. 67<lb></lb> |
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9031 E gli é el triangolo .abc. che, per ciascuno lato, è .28. Meno una retta .df. distante da-<lb></lb> |
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9032 l’ angolo .a.10. e passa per lo centro del ditto triangolo, che è el ponto .e., e sega .bc.<lb></lb> |
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9033 in ponto .f. Dimandase che sia longa tutta .df. e quanto distante dagli angoli .bc.<lb></lb> |
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9034 chaderá, cioé che sia .bf e che .fc. Fa chosí exui .2e.6i. e poi per la chosa prima<lb></lb> |
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9035 trova el catetto .dl. ala basa .bc. in questo modo. Tu sai che ’l catetto .ag. è .R.588. del gran trian-<lb></lb> |
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9036 golo, peró dirai: se .28. de potumissa che è .ac., mi dá .R.588. per lo catetto .ag. che mi dará la po-<lb></lb> |
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9037 tunissa .de.18., cioé .dc. Opera; te dará .R.243. per lo catetto .dl. e caderá distante al .c.9. e lo .gl.<lb></lb> |
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9038 resta .5. fin .14., che è el chaso .g. Ora tira una equidistante a .gl., che sia .eh., qual sirá pur .5., per<lb></lb> |
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9039 la .33a.pi.e arai un triangoletto .edh. ortogonio e simile al triangolo fdl. Peró dirai: se .R.56 1/3.,<lb></lb> |
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9040 cioé .dh. catetto, over perpendiculare, mi dá de potumissa .R.81 1/3., cioé .ed., che mi dará .R.243.,<lb></lb> |
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9041 cioé .dl., che sonno sotto la medesima potomissa del triangolo .dfl. Opera per viam radicum. Multiplica<lb></lb> |
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9042 .R.81 1/3. via.R.243., fa .R.19764., qual parti per .R.56 1/3., ne ven .R.350 142/169. per la linea .df. quantum est <lb></lb> |
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9043 quesitum <lb></lb> |
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9044 et cetera. |
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9045 </p> |
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9046 <p class="main"> |
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9047 Potevi pigliarla per altra via, cioé per viam proportionum, cosí arguendo per .2am.6i.ed. cioé<lb></lb> |
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9048 .R.81 1/3. sia li .13/27. delo .df., perché .eh. sia li .13/27. delo .fl. e lo .dh. sia li .13/27. delo .dl. E peró prendi el<lb></lb> |
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9049 .1/13. delo .de., cioé .R.81 1/3., ne virrá .R. 244/507., qual multiplica per .27., fará .R.350 142/169. per tutto .df. E, <lb></lb> |
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9050 perché<lb></lb> |
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9051 .ed. sia li .13/27. delo .df., doncha .df. sia li .14/27. de tutto .df., che è lo resto fin .27., peroché .27/27. fanno un<lb></lb> |
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9052 tutto. Doncha prendi .1/13. delo .ed. che sia .R.244/507., qual multiplica per .14., fará .R.94 106/507. per lo .efe. Cosí, de<lb></lb> |
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9053 mano in mano, regite al resto le sue prove per ordine le poi parte fare. E se per la chosa la voli, ponte commo<lb></lb> |
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9054 ti pare, con le predicte evidenze e viratte. Or ponte che .df. sia .1.co. Poi dirai: se .R.243. mi dá .1.co.,<lb></lb> |
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9055 che mi dará .R.56 1/3. Opera, viratte che sirá equale a .R 8 1/3. e sequirai etc. Overo dirai: se .R.56 1/3.<lb></lb> |
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9056 mi dá .R.81 1/3., che mi dará .R.243. Opera a tuo modo, varrá la chosa .R.350 142/149. ut supra. Ma apo-<lb></lb> |
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9057 nendote .al.o .fe., la cosa te virrá a valere .R.94 166/507. Et cetera. 68<lb></lb> |
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9058 E gli é el triangolo .abc., ch’ é .bc. .14.ac.15.ab.13. Jo meno una linea dal’ angolo .b. ret-<lb></lb> |
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9059 ta al ponto .g., distante dal’ angolo .a.3. e meno l’ altra, dal’ angolo .a.2., che si segano in<lb></lb> |
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9060 ponto .h. Dimando che sia la retta .hc. e quanto cada distante da ciascuno angolo .b. e del<lb></lb> |
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9061 .c. Fa chosí. Prima trova el catetto .gf. del triangolo .bgc. dicendo: se .ac., ch’ é .15., po-<lb></lb> |
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9062 tomissa mi dá .9. de basa, cioé .dc., che mi dará .12. Arai che te dará .7 1/5. per .ef. E poi, se .ac. mi dá .ad.<lb></lb> |
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9063 catetto, che mi dará .gc. Multiplica .12.via.12., fa .144., parti in .15., ne ven. 9 3/5. per lo catetto .gf. E poi, per l’ altro<lb></lb> |
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9064 verso, dirai: se .13. de potomissa mi dá .12. de catetto, cioé .ad., che ditto catetto è come a doi potomisse.<lb></lb> |
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9065 .ac. e .ab., che è .13., che mi dará .11., cioé .mb. Multiplica .11. via .12., fa .132., parti in .ba., ne ven .10 2/13. per lo .mk.<lb></lb> |
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9066 E lo .bk. sará .4 3/13. e lo .kd.10/13. e arai .m.c.R.80 33/169. e lo .bg.R 138 2/5. e lo .bn.14 70/12. e lo .co.19 13/17. Ora,<lb></lb> |
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9067 per lo tema, poni che .de. sia .1.co., cioé dal caso del catetto .ad. al caso dela linea, che tu sai per l’ ordi-<lb></lb> |
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9068 nario che .bd. sia .5. e lo .dc.9, perché tutta .bc. è .14. Donca .be. sia .5. piú .1.co. e lo .bf. è .6 4/5., noto. per<lb></lb> |
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9069 la forza de tutte le linea protratte ut patet. Donca dirai per la regola del .3.: se .bf. ch’é .6 4/5., basa mi dá<lb></lb> |
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9070 .9 3/5., che è .gf., di catetto, che mi dará .5. piú .1.co. Multiplica .9 3/5. via .5. piú .1.co., fará .48. piú .9 3/5.co. A par-<lb></lb> |
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9071 tire in .6 4/5., cioé ne virá .7 1/17. piú .1 7/17.co., qual serba che sia .he. per la ragione del triangolo<lb></lb> |
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9072 .boc., peró che .he. e lo .gf. sonno sotto una potomissa .bo. alla basa .bc., che sonno in proportione<lb></lb> |
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9073 per esser loro equidistanti per la .2a. del .6o. Poi volta verso per ragione del triangolo .bnc. e dirai:<lb></lb> |
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9074 se .9 10/13., che è .ck., de basa mi dá .10 2/13. de catetto, che è .mk., che mi dará .9. men .1.co., cioé .cd., che<lb></lb> |
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9075 sonno base sotto la medesima potumissa .nc. del triangolo .bnc. Multiplica .10 2/13. via .9. men .1.co., fará .91 5/13.<lb></lb> |
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9076 men .10 2/13.co. a partire in .9 10/13., che ne ven .1 61/127. men .1 5/127.co. Sirá equale a quello che di sopra serba-<lb></lb> |
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9077 sti, cioé .a.7 1/17. piú .1 7/17.co., che l’ uno e l’ altro verso te dá el medesimo catetto .he. Aguaglia per la e sequi<lb></lb> |
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9078 el capitulo. Arai la cosa valere .59/63. e tanto .fu.de. e tutto .be. fo .5 59/67. L’ avanzo fin .14. fo .ec., cioé<lb></lb> |
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9079 .8 4/3. Poi, a saper .he., ci son piú versi, commo vedi mediante la forza dele linee note che in propor-<lb></lb> |
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9080 tione procedano equidistante. Dirai: se .64. de basa mi dá .9 5/3. de catetto, cioé .gf., che mi dará .5 59/67.,<lb></lb> |
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9081 <lb></lb> |
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9082 <pb></pb></p> |
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9083 <p class="folio"> |
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9084 folio 62r |
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9085 </p> |
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9086 <p class="main"> |
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9087 <lb></lb> |
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9088 </p> |
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9089 <p class="runhead"> |
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9090 Distinctio octava 62 |
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9091 </p> |
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9092 <p class="main"> |
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9093 <lb></lb> |
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9094 cioé .be. Opera, ti dará .8 8/21. per la linea .he. Anchora potevi dire per l’ altro verso, cioé .ck. mi<lb></lb> |
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9095 dá .10 2/13., che mi dará .ce. Opera, te dará quello medesimo, cioé .8 8/21. piú o men non staria bene.<lb></lb> |
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9096 E gli é un triangolo equilatero ch’ é per ciascuna faccia bracia .28. Vi metto dentro un altro trian-<lb></lb> |
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9097 golo piccolo equilatero .rds. e la linea .ad. è bracia .17. e .sb. è bracia .11. e .sc. è<lb></lb> |
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9098 bracia .17. e la linea.cr. è bracia .11. e la linea .ra.bracia.17. Dimandase quanto è grande per faccia el ditto<lb></lb> |
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9099 triangolo interior equilatero. Fa cosí. Trova lo catetto del ditto gran triangolo e fallo ca-<lb></lb> |
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9100 dere dal ponto .a. in sul ponto .e., che sirá .R.588. e serba. Or movi una linea dal ponto .d. al ponto .o. che<lb></lb> |
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9101 sia equedistante ala linea .cb. e farai de sopra un piccol triangoluzzo .ado. equilatero e sia per faccia bracia<lb></lb> |
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9102 .11., perché di sopra dicemmo che la linea .ab. era bracia .11. Or trova el catetto del ditto triangoluc-<lb></lb> |
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9103 cio e fallo cadere dal ponto .a. in sul ponto .n. e, cosí facto, trovarai del ditto catetto é .R.90 3/4., la<lb></lb> |
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9104 quale .R.90 3/4. cava de .R.588., che di sopra serbasti, restará .R.216 3/4. E tanto viene a essere la linea .ne.<lb></lb> |
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9105 Or movi una linea dal ponto .d. fin al ponto .f. che sia equedistanti e equale ala linea .ne. E questa linea<lb></lb> |
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9106 .df. viene a essere .R.216 3/4., cioé commo la linea .ne. Ora guarda quanto è dal ponto .s. al ponto .f. E per<lb></lb> |
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9107 questo fa cosí. Tu sai che dal ponto .n. al ponto .d. è bracia .5 /12., cioé la mitá de .11. E tanto viene a essere dal pon-<lb></lb> |
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9108 to .e. al ponto .f., perché le linee sonno paralelle. Adonca dal ponto .e. al ponto .f. è bracia .5 1/2. E dal ponto .e. al<lb></lb> |
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9109 ponto .s. conviene che sia bracia .3., peroché dal ponto .s. al ponto .b. sonno bracia .11., commo di sopra fo detto. E dal<lb></lb> |
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9110 ponto .e. al ponto .b. é la .1/2. dela faccia del gran triangolo, cioé bracia .14., siché, tratto l’ un del’ altro, resta<lb></lb> |
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9111 bracia .3., commo è ditto. Lo qual, cavato de bracia .5 1/2., resta bracia .2 1/2. e tanto vene a essere dal ponto .s. al ponto <lb></lb> |
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9112 .f.,<lb></lb> |
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9113 cioé bracia .2 1/2. E questo multiplica in sé e giongni con la multiplicatione dela linea .df. in .sé. E cosí agionto harai .223. |
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9114 </p> |
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9115 <p class="main"> |
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9116 E di questo prendi la .R. che è .R. 223. e tanto fo per faccia el ditto triangolo e sia facta et cetera. Nota<lb></lb> |
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9117 che ’l centro del gran triangolo è lo medesimo centro del piccolo, commo per te porrai provare. .70. |
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9118 </p> |
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9119 <p class="main"> |
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9120 U gli é il quadrato .abcd. quale è .4. per faccia. Mettovi dentro el magior triangolo che vi ca-<lb></lb> |
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9121 pa equilatero. Dimando quanto sirá per faccia aponto. Fa cosí. Poni che .eb. sia .1. co. Don-<lb></lb> |
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9122 ca .ae. sirá .4. men .1.co. e altretanto sirá .af., perché .fc. ancor lui sirá .1.co., peroché el trian-<lb></lb> |
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9123 golo conviene che habia un suo angolo in cantone, perché non è possibile che giacia in su-<lb></lb> |
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9124 la faccia del quadrato, perché la faccia del triangolo è magiore che quella del quadro. E peró pongo che<lb></lb> |
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9125 l’ un deli angoli s’ apoggi nel ponto .e. e dal ditto ponto al’ angolo .b. pongo sia .1. co. Ora, se ben guar-<lb></lb> |
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9126 di, tu hai un triangolo ortogonio .ebd. del quale l’ angolo .b. éne recto e ’l lato .ed. sia ypotomissa che,<lb></lb> |
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9127 per la penultima del primo de Euclide, é quanto li doi quadrati deli doi lati che contengono ditto an-<lb></lb> |
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9128 golo recto. E peró multiplica .1.co. in sé, fa .1. ce. E multiplica .4. in sé, fa .16. Giongni insiemi, fanno .16. piú .1.<lb></lb> |
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9129 censo. E .R.v.16. piú .1.ce. qual serba per lo lato .eb. E poi fa per lo lato .fe. qual troverai cosí. Tu ve-<lb></lb> |
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9130 di ancora che tu hai un altro triangolo ortogonio .afe. che l’ angolo .a. éne recto, .ef. potumissa.<lb></lb> |
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9131 Multiplica .4. men .1.co. in sé, fará .16. men .8.co. piú .1. ce. E multiplica .af. ancora in sé, fará pur .16. men .8.co. piú .1. ce <lb></lb> |
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9132 quali<lb></lb> |
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9133 giongni insiemi, fanno .32. men .16.co. piú .2.ce. per lo quadrato dela potumissa .ef., la cui .Rv. sirá equale a<lb></lb> |
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9134 quell’ altra che sopra salvasti, perché li lati del triangolo sonno equali ex ypotesi. Multiplica ciascuna parte in sé per<lb></lb> |
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9135 levare le .R., harai .32. men .16.co. piú .2. ce. equali a .16. piú .1. ce. Aguaglia, harai .16. piú .1. ce. equali a .16.<lb></lb> |
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9136 co. Smezza le cose, multiplica in sé, fa .64., cavane el numero, resta .48. e la mitá dele cose men ditta .R.<lb></lb> |
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9137 val la cosa, cioé .8. men .R.48. e tanto sia .eb. Lo .ae. sirá .R.48. men .4. E ’l lato del triangolo sirá .128.<lb></lb> |
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9138 men .R.12288. e sia facta. Aliter prima al ditto quadro circunscrivi un cerchio e tira li .2. diametri .ab.<lb></lb> |
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9139 e .cd. Poi, dal ponto .a., stendi .1/2. el diametro. Finirá in ponto .e. E, dal’ altro canto, in ponto .f., dico el ma-<lb></lb> |
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9140 gior triangolo nel cerchio esser .bef. e li so lati taglian li lati del quadro in ponti .g.h. quali fan il numero triangoli .71. |
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9141 </p> |
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9142 <p class="main"> |
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9143 E gli é el triangol .adg. equilatero che per ciascuna facia è .6. Dimando che sia per facia el ma-<lb></lb> |
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9144 gior quadro ch’ entro vi capa. Fa cosí. Prima tira el catetto .gh., qual di ponto caderá nel<lb></lb> |
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9145 mezzo del lato .ad. in ponto .h. per esser el triangulo equilatero, siché .ah. e anche .hd.<lb></lb> |
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9146 ciascuna sirá .3. Or poni el quadro sia .1. co. per facia, la qual facia sia .bc., perché<lb></lb> |
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9147 de necessitá el magior quadro fermará sua facia sopra uno lato del triangulo. E a questo tira<lb></lb> |
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9148 la equidistante .ef., finendo el quadrato .bcef. per la propositione penultima del primo. Ora tu ái .ab. equale<lb></lb> |
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9149 alo .cd., peroché delo .ah., remosso .bh., ch’ é mezza facia del quadro, per la ragion ditta e dalo .hd.,<lb></lb> |
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9150 remosso .hc., pur mezza facia del quadro, resta .cd. iquale alo .ab., per conceptionem si .ab. equalibus et cetera. E<lb></lb> |
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9151 arai .ab.3. men .1/2.co. e cosí .cd. E cosí ái formato el triangulo ortogonio .abe. del quale li doi lati con-<lb></lb> |
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9152 tinenti l’ angol retto .b. son noti e l’ un sia .eb., lato del quadro che è .1.co., l’ altro .ab.3. men .1/2.co. Quadra<lb></lb> |
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9153 l’ uno e l’ altro, arai el quadrato .ab.9. men .3. co. piú .1/4.ce. e per lo quadrato .eb.1. ce., che, gionti insie-<lb></lb> |
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9154 mi, fanno .9. men .3.co. e piú .1 1/4.ce. per lo quadrato dela potumissa .ae., per la penultima del pri-<lb></lb> |
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9155 mo, la cui .R. sia equale a .6. men la quantitá .eg., cioé a .6. men .1. co., peroché .eg. è iguale .abc. e, per con-<lb></lb> |
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9156 sequente, a .ef. e .cf. e anche a .fg., per la .2a. del .6o. Donca multiplica li extremi dela equatione in sé, per leva-<lb></lb> |
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9157 re le .R., harai .9. men .3.co. piú .1 1/4.ce. equale a .36. men .12.co. piú .1. ce. Aguaglia e seque ’l capitulo reducen-<lb></lb> |
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9158 <lb></lb> |
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9159 <pb></pb></p> |
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9160 <p class="folio"> |
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9161 folio 62v |
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9162 </p> |
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9163 <p class="main"> |
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9164 <lb></lb> |
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9165 </p> |
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9166 <p class="runhead"> |
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9167 Distinctio octava. |
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9168 </p> |
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9169 <p class="main"> |
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9170 <lb></lb> |
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9171 do prima tutto .a.1.ce. Harai in ultimo .36.co. piú .1.ce. equali a .108. La cosa varrá .R.432. men .18. e tan-<lb></lb> |
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9172 to sia per faccia el quadro. Facta. E tu per te farai le simili et cetera. .72. |
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9173 </p> |
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9174 <p class="main"> |
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9175 E gli é un triangolo equicurio .abc.ab.10.ac.10.bc.12. Metovi dentro un pentagono<lb></lb> |
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9176 equilatero magior vi possa. Dimando li lati. Prima trova el catetto .ab., sirá .8. Poi<lb></lb> |
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9177 forma il pentagono .adefg. e harai negli angoli .b.c. doi triangoletti .dbe. e l’ altro<lb></lb> |
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9178 .gfe. di quali trova lor catetti .di.gl., che li troverai in questo modo. Perché el triangolo .dbi.<lb></lb> |
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9179 è simili al triangolo .abh., che l’ uno e l’ altro è rectangolo e tal proportione éne .ah. alo .ab., cioé .4/5., quel<lb></lb> |
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9180 .di. alo .db., che anche .di. sia li .4/5. delo .db. Questo inteso, poni che ’l pentagono sia per faccia .1.co., si-<lb></lb> |
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9181 rá adonca .be.6. men .1/2.co., perché .ah. casca nel mezzo delo lato .ef. e lo .db. sirá .10. men .1.co., di qua-<lb></lb> |
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9182 li piglia li .4/5., ne vien .8. men .4/5.co. e tanto sia il catetto .di. E poi, perché .bh. éne li .3/5. delo .ab., peró .bi. si-<lb></lb> |
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9183 rá li .3/5. delo .db. Donca piglia li .3/5. delo .db., ne viene .6. men .3/5.co. per lo .bi. Donca quadra .bi. e anche .di<lb></lb> |
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9184 e giongni insiemi, harai .100. men .20. co. piú .1.ce., over dal’ altro lato te reggi, cioé .ie. e lo .de., che<lb></lb> |
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9185 tanto fará. E harai che la cosa vará et cetera harai .1.ce. piú .36 4/7.co. equali .a. 182 6/7. et cetera .73. |
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9186 </p> |
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9187 <p class="main"> |
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9188 Uno triangolo equilatero che per ciascuna faccia è tanto quanto che è dal centro ali angoli<lb></lb> |
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9189 piú la .R. di quanto é dal centro ali angoli. Dimandase quanto sia per faccia e quanto sia dal<lb></lb> |
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9190 centro ali angoli e che sia quadro. Fa cosí. Poni che dal centro ali angoli sia .1.ce.<lb></lb> |
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9191 Donca sirá per faccia .1.ce. piú .1.co. Ora trova il suo catetto: multiplica .1.ce. piú .1.co., che è un<lb></lb> |
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9192 di lati, in sé, fará .1.ce.ce. piú .2.cu. piú .1.ce., qual salva. Poi multiplica la .1/2. de una facia in sé, cioé .1/2.ce. piú .1/2.co. <lb></lb> |
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9193 fa-<lb></lb> |
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9194 rá .1/4.ce.ce. piú .1/2.cu. piú .1/4.ce. E questo cava de .1.ce.ce. piú .2.cu. piú .1.ce., che fo la potumissa, del catet-<lb></lb> |
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9195 to, restará .3/4.ce.ce. piú .1 1/2.cu. piú .3/4.ce. e la .R.3/4.ce.ce. piú .1 1/2.cu. piú .3/4.ce. sirá el catetto suo. Ora, per<lb></lb> |
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9196 venire ala questione, prendi li .2/3. del decto catetto conmo .R. Reca .2/3. a .R., fa .4/9., piglia li .4/9. del .R.3/4.ce.ce. piú <lb></lb> |
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9197 .1 <lb></lb> |
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9198 1/2.cu.<lb></lb> |
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9199 piú .3/4.ce., ne vien .R.1/3.ce.ce. piú .2/3.cu. piú .1/3.ce. E questa .R. universale sirá equale a .1.ce. che ponesti esser<lb></lb> |
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9200 el centro distanti dali angoli, perché in ogni triangolo equilatero el centro è distante dali angoli quan-<lb></lb> |
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9201 to che è li .2/3. del suo catetto, che vene a esser el semidiametro del tondo che ’l circundasse, commo se<lb></lb> |
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9202 manifesta per corelarium .8e.13mi. Euclidis. Donca sequi l’ aguaglimento, leva le .R. multiplicando ciascuno<lb></lb> |
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9203 extremo in sé e dirai: .R.1/3.ce.ce. piú .2/3.cu. piú .1/3.ce. in sé fa .1/3.ce.ce. piú .2/3.cu. piú .1/3.ce. Poi, per l’ altra, <lb></lb> |
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9204 multiplica .1.ce.<lb></lb> |
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9205 in sé, fa .1.ce.ce. Sirá equale a .1/3.ce.ce. piú .2/3.cu. piú .1/3.ce. Leva el superfluo e schisa doi volte le dignitá<lb></lb> |
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9206 e arecarai a .1.ce. la equatione. Harai .1.co. piú .1/2. equale a .1.ce., perché li censi de censi vengono censi<lb></lb> |
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9207 e li cubi vengon cose e li censi vengon numero et cetera. Smezza le cose, multiplica in sé, giongnici el numero,<lb></lb> |
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9208 fa .3/4. piú .1/2. e la .R.3/4. piú .1/2. per lo dimezzamento val la cosa e ’l censo val la sua quadratura, che è .R.3/4. piú<lb></lb> |
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9209 .1. E tanto dirai che sia dal centro a’ canti: cioé .R.3/4. piú .1. e per faccia sirá la summa del censo e dela<lb></lb> |
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9210 cosa insiemi, che fa .R.3. piú .1 1/2. e tanto sia per faccia. Facta. Poi, per sapere quanto sia quadro, tro-<lb></lb> |
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9211 verai el catetto, che sirá la .1/2. del censo gionta al censo per la ragion ditta e quello multiplicarai via<lb></lb> |
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9212 la mitá de un di lati. E sia fatta et cetera. |
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9213 </p> |
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9214 <p class="main"> |
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9215 Un triangolo equilatero che dal centro a’ canti é tanto quanto che è per faccia men la .R. de<lb></lb> |
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9216 quello che è per faccia. Dimando quanto sia per faccia e quanto dal centro a’ canti. Poni che per<lb></lb> |
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9217 faccia sia .1.ce. Donca, dal centro a’ canti, sirá .1.ce. men .1.co. Ora trova el catetto che è .R.<lb></lb> |
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9218 .3/4.ce.ce. Di questo piglia el .1/3., che ne vien .R.1/12.ce.ce., perché el centro sia nel terzo del ca-<lb></lb> |
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9219 tetto. Ora tu hai un triangoletto rectangolo che la potumissa sia la linea dal centro a’ canti, cioé<lb></lb> |
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9220 .1.ce. men .1.co. L’ altre doi continenti el rectangolo, l’ un sia el .1/3. del catetto, cioé .R.1/12.ce.ce., l’ altro la .1/2. dela<lb></lb> |
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9221 basa, cioé d’ una dele face .1/2.ce. Ora quadra li doi lati continenti el rectangolo e summa insiemi, ará l’ .1/5.<lb></lb> |
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9222 ce.ce. e .R.1/3.ce.ce. sirá equale ala potumissa, cioé a .1.ce. men .1.co. Leva .R. e aguaglia, schisa la equati-<lb></lb> |
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9223 one e arecarala a .1.ce., harai .1 1/2. piú .1.ce. equale a .3.co. Sequi el capitolo, harai la cosa valere .1 1/2.<lb></lb> |
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9224 men .R.3/4., over .1 1/2. piú .R.3/4. Proverala meglio tu, ma un di questi doi sirá et cetera. Donca el censo varrá .3. men<lb></lb> |
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9225 .R.6 3/4. e tanto sia per faccia. E dal centro a’ canti sia .1 1/2. men .R.3. et cetera. .74. |
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9226 </p> |
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9227 <p class="main"> |
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9228 E gli é il triangolo .abc. del quale il lato .ab. éne .13.ac.15.bc.14. Io voglio slongarn uno<lb></lb> |
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9229 de’ ditti lati, cioé quello che piú mi dará longhezza, non sminuendo la sua superficie. Di-<lb></lb> |
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9230 mando qual di loro virrá piú longo e quanto al piú si possa, non movendo niun degli altri .2. |
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9231 </p> |
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9232 <p class="main"> |
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9233 Questo è gentil caso e fasse cosí per solverla. Perché non te dici piú d’ un lato che de-<lb></lb> |
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9234 l’ altro, bisogna che tu, da te, provi qual sia quel lato che piú si possi slongare. E, per questo, te conviene trovare<lb></lb> |
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9235 li catetti che si movano dal’ angolo opposito a ciascun lato e cade in sun ciascun lato qual sia piú<lb></lb> |
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9236 longo e quanto sia la distantia de lor caso ali angoli dela basa dove cade. E poi te conviene trovare<lb></lb> |
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9237 la differentia dal caso ala mitá de ditta basa, peroché la linea che, dal medesimo angolo donde si move<lb></lb> |
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9238 el catetto, si parte e vene a ponto ala .1/2. dela decta basa quella é la .1/2. di quello che si porrá decta basa slonga-<lb></lb> |
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9239 re, commo intenderai. Verbi gratia: tirando el catetto dal’ angolo .a. ala basa .bc., che è .14., trovarai che<lb></lb> |
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9240 sirá .12., per li modi dati dinanze, e sia .ad. Dove, dal ponto .d. al’ angolo .b., éne .5. e al’ angolo .c.9. Di-<lb></lb> |
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9241 <lb></lb> |
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9242 <pb></pb></p> |
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9243 <p class="folio"> |
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9244 folio 63r |
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9245 </p> |
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9246 <p class="main"> |
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9247 <lb></lb> |
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9248 </p> |
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9249 <p class="runhead"> |
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9250 Distinctio octava. 63 |
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9251 </p> |
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9252 <p class="main"> |
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9253 <lb></lb> |
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9254 co ora che prenda la .1/2. dela basa .bc. nel ponto .e., che .be. sirá .7. Altretanto sia .ec., onde tu hai, se ben<lb></lb> |
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9255 guardi, un triangoletto .ade. ortogonio, che l’ angolo del cadimento del catetto éne retto, cioé<lb></lb> |
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9256 .d. E la potumissa sia la linea .ae., cioé quella che vien al ponto medio dela basa, qual dico essere<lb></lb> |
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9257 la mitá di quello che al piú el lato .bc. slongar si possi, non minuendo l’ area del gran triangolo .abc.<lb></lb> |
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9258 proposto, la cui notitia se haverá per la penultima del primo de Euclide, giognendo la potentia<lb></lb> |
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9259 del catetto .ad. con la potentia del lato .de., cioé dela differentia dal caso ala mitá dela basa.<lb></lb> |
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9260 La qual differentia sia .2., perché .be., sando .7., trattone .bd., che è .5., resta .2. per lo .de. Donca qua-<lb></lb> |
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9261 dra .ad., fa .144., quadra .de., fa .4.; giongni insiemi, fanno .148. per la potentia dela potumissa .ae.<lb></lb> |
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9262 La cui .R. sirá sua longhezza, .cioé .R.148. Ora questa si vol doppiar, fará .R.592. e tanto si po-<lb></lb> |
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9263 trá slongar il lato .bc. al piú. Qual salva da parte. E poi farai per lo lato .ac. che è .15., tenendo si-<lb></lb> |
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9264 mil modo. Cioé troverai el catetto del’ angolo .b. ala basa .ac. essere .11 1/5. e caderá distante al’ ango-<lb></lb> |
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9265 lo .a.6 3/5. E la differentia dal caso ala mitá dela basa: cioé .a.7 1/2. e .9/10. La potumissa del triangoletto<lb></lb> |
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9266 ortogonio sia .R.126 1/4., la qua doppiata, per la cagion ditta, fará .R.505. per el piú che slon-<lb></lb> |
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9267 gar si possa il lato .ac. de .15. Siché tu vedi che quel de .14. doventa piú longo che quel de .15. E<lb></lb> |
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9268 questo anche salva e, per le simil vie, vederai del lato .ab., cioé de .13. Trovarai el catetto che da-<lb></lb> |
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9269 l’ angolo .c. si move e cade in sula basa .ab. essere .12 12/13. e la distantia del caso al’ angolo .a. sia .7 18/13.<lb></lb> |
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9270 e dal’ angolo .b.5 5/13. e la differentia dal caso ala mitá de .13., cioé a .6 1/2., sia .1 3/26. e la potumissa del tri-<lb></lb> |
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9271 angoletto ortogonio sia .R.168 1/4., la qual doppiata, per la causa sopra ditta, fará .R.673. per lo<lb></lb> |
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9272 piú che slongar si possi il lato .ab., cioé .13. Ora tu vedi che il lato de .13. è quello che pó doven-<lb></lb> |
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9273 tare el magiore e non minuirá superficie al triangolo .abc. proposto, cioé che tanto sirá el trian-<lb></lb> |
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9274 golo deli lati .14.15. e .R.673. quanto de .13.14.15. e cosí de .15.13. e .R.592. e ancora de .13.14.<lb></lb> |
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9275 e .R.505., che l’ uno e l’ altro possederá .84. Facta. E per questo modo porrai sequire in tutti altri<lb></lb> |
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9276 simili. El perché è da notare sí commo si dá verso a slongare, cosí ancora si porrá scortare quan-<lb></lb> |
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9277 do el caso dicesse in questa forma. E gli é el triangolo .abc. proposto, del quale il lato .ab. éne .13. e<lb></lb> |
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9278 il lato .ac. éne .15. e il lato .bc. éne .R.592. e questo voglio scortare al piú si possa, non minuendo<lb></lb> |
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9279 .ab. e anche .ac. e né mancando superficie a tutto el triangolo. Dimando se si pó e possiandose<lb></lb> |
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9280 quanto si scorterá. Opera per la via di sopra trovando el catetto dal’ angolo .a. ala basa .bc. e no-<lb></lb> |
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9281 ta suo caso nel ponto .d. E poi dividi la basa .bc. in doi parti equali, cioé .R.592., ne vien .R.148. |
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9282 </p> |
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9283 <p class="main"> |
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9284 E trova la differentia dal caso a ditta mitá, cioé .de. la quale, a modo ditto, forma un triangoletto<lb></lb> |
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9285 ortogonio .ade. del qual li doi lati continenti l’ angolo retto l’ uno è .ad. l’ altro .de.; cioé il catetto<lb></lb> |
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9286 e la differentia trovarai e la potumissa .ae., per la penultima del primo, e sirá .7., qual doppia, fará .14. |
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9287 </p> |
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9288 <p class="main"> |
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9289 E fin tanto si porrá scortare ditto lato .bc. al piú. E sie in ceteris et cetera. .75. |
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9290 </p> |
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9291 <p class="main"> |
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9292 E gli é il triangolo .13.14.15. che in sul lato del .14. faccio un semicirculo magior che<lb></lb> |
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9293 vi capa. Dimando che sia suo diametro. Recordate del notando che pone la .35a.<lb></lb> |
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9294 del .3o., che tutte le linee che si partano da un ponto fuor del cerchio e vengono contingen-<lb></lb> |
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9295 ti sonno equali. E anche recordate che tutte le linee che si partano dal centro al luogo del<lb></lb> |
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9296 contatto cagiono a squadro. E, per consequente, poni che dal centro al contatto sia .1.co. e questo è semi-<lb></lb> |
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9297 diametro. Onde tu hai resoluto il triangolo .abc. in doi triangoli partiali che l’ uno éne .abd., l’ altro<lb></lb> |
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9298 .adc. e il catetto di ciascuno è .1.co. Per quel ch’ é ditto quadra ciascuno, harai per quel delo .abd.6 1/2.co.<lb></lb> |
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9299 e per quello .adc.7 1/2.co. Giongni insiemi, fanno .14.co. e questo è equale ala superficie de tutto el triangolo .abc.,<lb></lb> |
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9300 cioé .84. Sequi, harai la cosa valere .6. e tanto sia semidiametro. Fatta. Per te cerca il resto.<lb></lb> |
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9301 Ma chi absolute dicesse: e gli é il triangolo .abc., non facendo mentione d’ alcun lato, e domandasse<lb></lb> |
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9302 el diametro del magiore semicirculo che entro vi capa. Operando geometrice, a uno apre de sexte.<lb></lb> |
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9303 Farai cosí. Prima vedi qual sia el magiore suo angolo, che pongo sia l’ angolo .a. E questo dividi in doi parti<lb></lb> |
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9304 equali con la linea .ad., per la .9a. del secundo e il ponto .d. sia centro del cerchio e harai resoluto el gran triango-<lb></lb> |
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9305 lo in .2. triangoli partiali che l’ uno sia .adc. e l’ altro .dba. E d’ uno de questi qual voli trova el catetto che<lb></lb> |
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9306 si parte dal ditto ponto .d. in sulo lato opposito .ab., over .ac., e secondo la quantitá de ditto catetto apri<lb></lb> |
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9307 le sexte, fermando el pede in ponto .d. e, circinando, harai sempre el magiore semicirculo che<lb></lb> |
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9308 entro vi capa et cetera.76. |
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9309 </p> |
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9310 <p class="main"> |
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9311 E gli é el triangolo .13.14.15. Fovi dentro un mezzo tondo sula faccia del .14. magior<lb></lb> |
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9312 vi capi e tochi con la circunferentia li altri doi lati. Dimando che sia suo diametro del<lb></lb> |
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9313 tondo e quanto distante da ciascun angolo el cerchio segará la basa di .14. Fa cosí.<lb></lb> |
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9314 Prima trova el diametro del tondo. E poi arguesci che dal centro che sia in sul lato<lb></lb> |
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9315 .14. al’ angolo .a. si move una linea che divide tutto el triangolo in doi triangoli di quali l’ uno á di ba-<lb></lb> |
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9316 sa el lato .13. e l’ altro el lato .15., siando catetti sopra ditti lati el mezzo diametro, cioé la linea che si<lb></lb> |
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9317 parte dal centro e vene ali contatti, che sempre cade a squadro, per la .35a. del terzo. Donca tu sai che la<lb></lb> |
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9318 <lb></lb> |
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9319 <pb></pb></p> |
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9320 <p class="folio"> |
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9321 folio 63v |
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9322 </p> |
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9323 <p class="main"> |
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9324 <lb></lb> |
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9325 </p> |
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9326 <p class="runhead"> |
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9327 Distinctio octava. |
|
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9328 </p> |
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9329 <p class="main"> |
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9330 <lb></lb> |
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9331 linea del triangolo, che á basa .13., éne .39., che nasci dela multiplicatione de mezzo catetto, ch’ é .3., in tutta<lb></lb> |
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9332 la basa, che è .13., quel serba. E sai ancora che nel triangolo .abd. vi cade un’ altra linea ala basa .bd. El<lb></lb> |
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9333 resto sequi commo altre volte festi. |
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9334 </p> |
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9335 <p class="main"> |
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9336 E gli é una rota .c. che è di .3. compagni il cui diametro .ab. é pie’ .7. La vogliano partire in<lb></lb> |
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9337 questo modo che ciascuno aroti e, arotando, ciascuno consumi la sua parte, c’ altramente le cose<lb></lb> |
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9338 dannose non si debono spezzare secondo le leggi. Dimando quanto diametro doverá con-<lb></lb> |
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9339 sumare ciascuno di costoro da per sé. Fa cosí. Prima quadra tutta la rota conmo cerchio.<lb></lb> |
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9340 Harai che sirá .38 1/2., del qual prendi el .1/3., che è .12 5/6. E tanto del campo dela rota deve consumare<lb></lb> |
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9341 ognuno. E, quando el primo e secondo aran consumato la lor parte restará in ultimo un cerchietto, circa cen-<lb></lb> |
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9342 trum, che sirá .fg., del qual trova suo diametro .gf. in questo modo. Poni che ’l fosse .1.co., quadra, fa<lb></lb> |
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9343 .1.ce. Prendine li .11/14., son .11/14.ce. e questo sia equale a .12 5/6. Parti .12 5/6. per .11/14., ne ven .16 1/3. e .R.16 <lb></lb> |
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9344 1/3. sia el diametro<lb></lb> |
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9345 .fg. del piccolo cerchio ultimo circa centrum, che sirá tutta la parte del terzo compagno. Ora, per trova-<lb></lb> |
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9346 re quella del secondo, doppia .12 5/6., fa .25 2/3. per lo cerchio .de., che sono li .2/3. de tutta la rota, del qual<lb></lb> |
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9347 similiter trova suo diametro ponendo che ’l sia .1.co. Quadrala, fa .1.ce., prendine li .11/14., che son .11/14.ce.,<lb></lb> |
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9348 commo festi di sopra e siran equali a .25 2/3., partilo neli censi, virrá .32 2/3. e .R. di tanto sirá el diametro<lb></lb> |
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9349 .de., del qual cava .R.16 1/3., che è el diametro del terzo, resta .R.32 2/3. men .R.16 1/3. e tanto de diame-<lb></lb> |
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9350 tro, fra de sotto e di sopra, deve consumare el secondo. Qual dividi per mezzo, ne vien .R.8 1/6. men .R.4 1/12.<lb></lb> |
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9351 e tanto sirá .df. e tanto ancora .ge. suo opposito. Ora, per sapere quello del primo, acozza quel del se-<lb></lb> |
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9352 condo e terzo, cioé .R.32 2/3. men .R.16 1/3., con .R.16 1/3. fa .32 2/3. per lo diametro .de., qual cava delo .ab.<lb></lb> |
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9353 diametro grande, resta .7. men .R.32 2/3. e tanto deve consumare el primo del diametro, fra sotto e so-<lb></lb> |
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9354 pra; prendine la mitá, ne ven .3 11/2. men .R.8 1/6. e tanto sirá .ad. e altretanto sia .ab. parte opposita. Siché tu<lb></lb> |
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9355 hai ciascuno da l’ un lato e l’ altro ffacta. Et cetera. E cosí sequiristi se fossero quanti si vogliava, cavando, de ma-<lb></lb> |
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9356 no i mano, l’ uno del’ altro e per forza poi se conclude la parte del primo, la cui superficie sempre ce sirá no-<lb></lb> |
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9357 ta. Questa ó voluto ponere piú aperta che di sopra, in questo al numero .42., non posi quel-<lb></lb> |
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9358 la per piú tuo documento. .78. |
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9359 </p> |
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9360 <p class="main"> |
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9361 Uno si trova un palio longo e largo una quantitá e vollo longo e largo a tal proportione ch’ é<lb></lb> |
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9362 radoppiata la sua longhezza, cioé smezzatela e fattone larghezza e la larghezza e la larghezza fat-<lb></lb> |
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9363 tone longhezza havesse la prima proportione. Se dimanda quanto sia longo e largo. Di-<lb></lb> |
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9364 co che a simili, non ti strengendo piú a longo che a largo, tu poi a libito darli o longo o il largo, a<lb></lb> |
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9365 tuo modo, purché uno di questi li dia fermi, l’ altro investigarai. Or poniamo che ’l tema dica<lb></lb> |
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9366 che ’l voglia longo bracia .2.; la larghezza non ti pó dar se non piú larga del devere e sapere quanto si ne<lb></lb> |
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9367 debbia scemare et cetera. Or poni che voglia essere largo, ala ditta longhezza data, .1.co. de bra-<lb></lb> |
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|
9368 cio. Smezata la longhezza, che è .2., venga poi .1.co. essere in quella proportione .a.1. che è .2. a .1.co. Ora pie-<lb></lb> |
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9369 ga ditto panno, el qual è .abcd., ad angoli retti e longo quanto la linea .ab., che è .2., e largo .bc. che é<lb></lb> |
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|
9370 .1.co. Se tu lo doppi, virrá conmo in la .2a. disponne, appare longo .bc., cioé .1.co. e largo .ce., cioé .1.bracio., per-<lb></lb> |
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9371 ché nel doppiarlo el longo si vien a smezzare e doventa largo. Ora é da vedere se questo secondo lon-<lb></lb> |
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9372 go é in proportion al secondo largo commo era prima el primo longo al primo largo, cioé se gli é cosí .bc. al<lb></lb> |
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|
9373 .ce. commo .ab. al .bc. in la prima disponne. E per trovarlo dirai: se .2., che era longo prima mi dá .1.co. di largo,<lb></lb> |
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9374 che me dará, ala medesima ragione, .1.co., che è longo al presente. Opra, te dará .1/2.ce. per largo secondo<lb></lb> |
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|
9375 e giá tu vedi che l’ é .1. Donca harai .1/2.ce. equale a .1. La cosa vará .R.2. e tanto bisognará che ’l detto pa-<lb></lb> |
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9376 lio .abcd. sia largo, dobiando essere longo .2. Siché, se prima ti fosse proposto piú largo, dirai che bisogna<lb></lb> |
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|
9377 tagliarlo in su .R.2.bracia. Aliter et facilis. geometrice. Stu hai il longo fermo, ut puta .2., di questo forma-<lb></lb> |
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|
9378 te un quadro equilatero. Sirá .2. per facia, commo in la .5a. disponne, .abcd., del qual trova la diagonale .bc.,<lb></lb> |
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|
9379 che sia .R.8. Di questa sempre prendi la .1/2., ne ven .R.2. e tanto dirai che vorrá essere largo, siando longo<lb></lb> |
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9380 .2. e mai falla. E, se volesse longo .10., forma ditto quadro e tolli la sua diagonale che sirá .R.200. |
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|
9381 </p> |
|
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|
9382 <p class="main"> |
|
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|
9383 Pigliane la .1/2., ne ven .R.50. e tanto vorrá essere largo. E, se ti desse fermo el largo, alora dop-<lb></lb> |
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9384 pialo e harai una diagonale de un quadrato, trovarai il lato e tanto converrá essere longo,<lb></lb> |
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9385 quod dignissimum in arte. .79. |
|
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|
9386 </p> |
|
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|
9387 <p class="main"> |
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|
9388 Cava el quadrato .adg. del quadrato .abc. magior gnomice. Fa cosí. In sul lato .ac.<lb></lb> |
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|
9389 del grande fa el semicirculo .aoc. e dal ponto .a. tira .ao. equale alo .dg., cioé a un lato<lb></lb> |
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|
9390 del quadrato piccolo, e tira .oc. Manifesto é che l’ angolo .o. é retto, per la .30a. del .3o. e per la pe-<lb></lb> |
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|
9391 nultima del secundo, el quadrato .ac. val li .2. quadrati .ao. e .oc. Donca prendi dela linea ac., lato del qua-<lb></lb> |
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9392 drato grande, la quantitá .cf. equale alo .oc., dal qual ponto tira la equadistante alo .ab., che sia .fh. e altretanto<lb></lb> |
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9393 tagliarai per l’ altro verso del quadrato grande, nel ponto .k., e sia el gnomone .fhk. remosso equale<lb></lb> |
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9394 al ditto quadrato piccolo per la {coescia}, perché el quadrato .fchk. éne equale al quadrato .oc., qual<lb></lb> |
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|
9395 tratto del quadrato .ac., restará per forza la quantitá gnomonica equale al quadrato piccolo et cetera.<lb></lb> |
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9396 <lb></lb> |
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9397 <pb></pb></p> |
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9398 <p class="folio"> |
|
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9399 folio 64r |
|
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9400 </p> |
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9401 <p class="main"> |
|
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9402 <lb></lb> |
|
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|
9403 </p> |
|
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9404 <p class="runhead"> |
|
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|
9405 Distinctio octava. 64 |
|
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|
9406 </p> |
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9407 <p class="main"> |
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9408 <lb></lb> |
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9409 D’ un gnomone far quadrato. Prima reca un supplemento a quadrato, per la ultima del secondo e<lb></lb> |
|
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|
9410 . 9a. del .6o. E cosí farai del resto e haverai doi quadrati li quali poi giongni insiemi trovando un la-<lb></lb> |
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9411 to tetragonico che possa tutti doi quelli lati et cetera. 80. |
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|
9412 </p> |
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|
9413 <p class="main"> |
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9414 E son .2. cerchi, l’ uno magior del’ altro e stanno un dentro l’ altro in su uno medesimo<lb></lb> |
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9415 centro. Ora e gli é uno homo in sul gran cerchio che camina e circunda el ditto cer-<lb></lb> |
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9416 chio, in .12. die é tornato donde se mosse. E in sul piccol cerchio é un altro homo che<lb></lb> |
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9417 camina e circundalo in .24. hore, tornando dove si mosse. E questi .2. homi sonno rem-<lb></lb> |
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9418 petto, a corda, un con l’ altro e movansi a camino a un tratto, caminando tutti da un medesimo<lb></lb> |
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9419 lato. Domando in quanti giorni siranno de rimpetto l’ uno con l’ altro a corda. Fa cosí. Convien-<lb></lb> |
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9420 te considerare che costoro, quando si trovaran essere rimpetto a corda in tal parte del cerchio si<lb></lb> |
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9421 trovará quel del grande che quel del piccolo. Verbi gratia: se quel del gran cerchio, quando<lb></lb> |
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9422 si trovaran de rimpetto, poniamo ará caminato l’ octavo del cerchio, dico che l’ altro homo si mo-<lb></lb> |
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9423 vará ancor lui in sul’ octava parte del piccol cerchio. E anco te convien considerare che gli é<lb></lb> |
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9424 forza che passi un dí nanze che si scontrino, per la velocitá del piccolo e per la lentenzza del<lb></lb> |
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9425 grande. Peró farai positione in un dí .1.co. de dí. E poi dirai: se ’l primo camina in .12. dí un cer-<lb></lb> |
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9426 chio, che ne caminerá in un dí piú .1.co. de dí, che troverai che n’ averá caminato .1/12. de cerchio<lb></lb> |
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9427 piú .1/12.co. de cerchio e serba. Poi dirai: l’ altro che camina in .24. hore, che son un giorno, un cer-<lb></lb> |
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|
9428 chio, che arallo caminato in .1.co. de dí. Troverai che n’ ará caminato .1.co. Ecco adonca .1. co.<lb></lb> |
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9429 de cerchio è equale a .1/12. de cerchio piú .1/12.co. de cerchio, che sopra serbasti. Ora sequi l’ aguaglimen-<lb></lb> |
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9430 to. Harai da l’ un lato .11/12.co. equali a .1/12. de cerchio, peró parti .1/12. per .11/12., ne ven .1/11. e tanto val la cosa. <lb></lb> |
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|
9431 E<lb></lb> |
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9432 noi ponemmo un die .1.co., adonca é facta a un di e .1/11. de dí. E in tanti dí si scontraranno a cor-<lb></lb> |
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|
9433 da. E cosí fa le simili. .81. |
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|
9434 </p> |
|
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|
9435 <p class="main"> |
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9436 E son .2. cerchi, uno è grande e l’ altro piccolo e l’ uno dentro al’ altro e sonno in sul medesi-<lb></lb> |
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|
9437 mo centro. E in sul gran cerchio è un che camina e circundalo tutto in .12. di. E in sul pic-<lb></lb> |
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9438 colo cerchio é un che camina e circundalo tutto in un dí, cioé in .24. hore. Ora dit-<lb></lb> |
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9439 ti doi homi son a rimpetto a corda l’ uno con l’ altro e movansi tutti a una otta e camina-<lb></lb> |
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|
9440 no uno contra l’ altro su per ditti cerchi. Dimando in quanti dí seranno insiemi a un rimpetto a corda.<lb></lb> |
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9441 Fa cosí. Conviente considerare che, quando questi homi se scontrarannno a rimpetto, quella parte del gran cer-<lb></lb> |
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|
9442 chio che hará fatto el primo, l’ altro hará fatto el resto del suo piccolo cerchio. Verbi gratia: se ’l primo<lb></lb> |
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9443 hará fatto l’ octavo del gran cerchio, dico che l’ altro ará fatto li .7/8. del suo piccolo cerchio. Adonca<lb></lb> |
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|
9444 poniamo che questi se scontrino in una cosa de dí. E poi dirai: se ’l primo camina in .12. dí un cerchio, che<lb></lb> |
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|
9445 ne caminará in .1.co. de dí, che troverai che fará .1/12.co. de cerchio e serba. Poi dirai: per lo .2o. che ca-<lb></lb> |
|
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|
9446 mina un cerchio in un dí, che ne caminará in .1.co. dí, che troverai caminará .1.co. di cerchio. Adon-<lb></lb> |
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|
9447 ca acozza insiemi con .1/12.co., che sopra serbasti, fa .1 1/12.co. di cerchio. E doverebono fare un cerchio<lb></lb> |
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|
9448 solo, adonca .1 1/12.co. é iguale a un cerchio. Peró parti .1. per .1 1/12., ne ven .12/13. e tanto val la cosa. E in tanti gior-<lb></lb> |
|
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|
9449 <lb></lb> |
|
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|
9450 ni se scontrarano a un rimpetto a corda commo domandasti. .82. |
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|
9451 </p> |
|
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|
9452 <p class="main"> |
|
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9453 Sonno .2. nave in mare che si partano da un porto medesimo, l’ una va a sirocco e<lb></lb> |
|
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9454 l’ altra va a libeccio, aliter garbino. E, quando ciascuna á caminato cinquanta miglia, elle<lb></lb> |
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|
9455 si fermano. Adimando quante miglia le ditte navi sonno descoste una dal’ altra e<lb></lb> |
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9456 anche domando: se una nave volesse andare a trovar l’ altra nave, con che venti navi-<lb></lb> |
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|
9457 garanno, cioé per che vento l’ una e per che vento l’ altra, peroché l’ ánno a giostrare et cetera. Fa cosí. Con-<lb></lb> |
|
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|
9458 vienti cognoscere l’ ordine deli venti, quali sonno .8., commo li vedi qui designati. E ciascuno ven-<lb></lb> |
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|
9459 to á el suo vento opposito, siché questi venti vengono a fare un cerchio tondo con li spatij equali. Or<lb></lb> |
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|
9460 torniamo a nostro proposito. Queste doi navi vengano a fare un triangolo nel sopra ditto cerchio,<lb></lb> |
|
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9461 el qual ven a essere .50. miglia per ciascuno de’ doi lati e la porta del ditto triangolo, cioé el porto<lb></lb> |
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9462 donde si partano, vene a essere el centro del ditto cerchio e la basa del ditto triangolo viene a essere la<lb></lb> |
|
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|
9463 distantia de ditte doi navi e cosí vien a essere la faccia del quadro del sopra ditto cerchio, peroché,<lb></lb> |
|
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|
9464 da libeccio a sirocco, c’ é lo spatio de’ doi venti, che sonno el .1/4. de tutto el cerchio. Siché, a voler sape-<lb></lb> |
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9465 re quanto è lo spatio dal’ una nave al’ altra, multiplica le .50. miglia in se medesime, fa .2500. Ora la radop-<lb></lb> |
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9466 pia, fa .5000. e di questo prendi la .R. che è .R.5000. E tanto fo dal’ una nave al’ altra ed é fat-<lb></lb> |
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9467 ta. E, volendo sapere per che vento si trovaranno, fa cosí. Tu vedi che la sopra ditta basa del trian-<lb></lb> |
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9468 golo, che è la linea da libeccio a sirocco, commo nel cerchio qui si vede, va equedistante con la linea<lb></lb> |
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9469 de ponente e levante. E peró poi dire cosí, che le ditte doi navi se trovaranno per ponente e levante e<lb></lb> |
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9470 siran discoste, l’ una dal’ altra, .R.5000. commo é detto. E, se ’l caso havesse ditto: l’ una va a siroc-<lb></lb> |
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9471 co e l’ altra va a mezzodí, siando .50. miglia lontane dal porto, alora te convien trovare la distan-<lb></lb> |
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9472 tia .ab., cioé da mezzodí a sirocco, in questo modo, commo descrivo in questa di sotto. .83.<lb></lb> |
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9473 <lb></lb> |
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9474 <pb></pb></p> |
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9475 <p class="folio"> |
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9476 folio 64v |
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9477 </p> |
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9478 <p class="main"> |
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9479 <lb></lb> |
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9480 </p> |
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9481 <p class="runhead"> |
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9482 Distinctio octava. |
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9483 </p> |
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9484 <p class="main"> |
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9485 <lb></lb> |
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9486 Da mezzodí a sirocco è .1/8., che dicano li vulgari mezanino et cetera. Fa cosí. Prima trova lo diametro<lb></lb> |
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9487 .da., quale è .100. Ora tu sai che .eb. è .R.1250., cioé la .1/2. delo .fb., che è la faccia del gran quadra-<lb></lb> |
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9488 to e sai che ditto diametro taglia a squadro ditta faccia in ponto .e. e sai, commo per la .34a. del<lb></lb> |
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9489 terzo Euclide prova, che tanto fa .ea., saetta piccola, ducta in .eb., saetta grande, quanto .eb. dut-<lb></lb> |
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9490 ta in .ef., vel in sé. E peró, per trovare la quantitá .ea., dicendo famme de .100.2. parti che mul-<lb></lb> |
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9491 tiplicata una in l’ altra faccia .1250., cioé quanto .eb. in .ef. Opera per la cosa o commo voli.<lb></lb> |
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9492 Harai .ea. essere .50. men .R.1250. Ora tu hai un triangolo ortogonio, cioé .eab., del qual l’ an-<lb></lb> |
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9493 golo .e. éne retto e li .2. lati che lo contengono, cioé .ea.et.eb., sonno noti, che l’ uno è .R.1250. e<lb></lb> |
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9494 l’ altro .50. men .R.1250. Quadrali tutti .2. e giongnili insiemi, fanno .5000. men .R. 12500000. e la .R.<lb></lb> |
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9495 di questo legata sia .ab., che è la distantia de una nave al’ altra. Cioé in questo modo .Rv.5000.<lb></lb> |
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9496 men .R.12500000. Facta. La qual cosa molto serve a’ naviganti. Ma, se ’l caso dicesse che l’ una, mettia-<lb></lb> |
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9497 mo quella che va a mezzodí, havesse caminato .80. miglia e quella .50. e poi fermaro e vo-<lb></lb> |
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9498 lesse la distantia dal’ una al’ altra, commo se la fosse fermata in ponto .h., alora, secondo la quanti-<lb></lb> |
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9499 tá dela minor linea, cioé de .50., formarai el cerchio e in quello forma el magior quadro che<lb></lb> |
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9500 poi e dividi e taglia, commo di sopra, e harai noto tutto. E poi, per lo triangolo ortogonio .ehb.,<lb></lb> |
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9501 tu harai la potumissa .hb., cioé la distantia dal’ una al’ altra, peroché tu sai una volta .ea., che è<lb></lb> |
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9502 .50. men .R.1250., al qual giongni .ah., che è .30., sirá .eh. tutto .80. men .R.1250. e lo .eb. pure .R.1250. |
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9503 </p> |
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9504 <p class="main"> |
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9505 Quadra l’ uno e l’ altro e giongni insiemi, fará .7650. men .R.3200000. per lo quadrato dela potu-<lb></lb> |
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9506 missa .hb., cioé la distantia dal’ una al’ altra, che sirá cioé .Rv.7650. men .3200000., cioé presa<lb></lb> |
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9507 la .R. de .3200000. e quella tratta de .7650.R. di quel che restasse et cetera. E tu, per te, farai e for-<lb></lb> |
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9508 marai del’ altre et cetera con la evidentia sempre dela bossola e ‘venti ordinarii et cetera. .83. |
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9509 </p> |
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9510 <p class="main"> |
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9511 I o dividi un cerchio a sesta el quale á de diametro .10. e fessi una rosetta egual fo-<lb></lb> |
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9512 glie .6., commo a sesto se concria. Adimando la larghezza e superficie d’ una foglia. Fa<lb></lb> |
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9513 cosí: tolli la superficie de tutto el cerchio, che è .78 4/7., lo qual dividi in .6. egual parti,<lb></lb> |
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9514 che ne vien .13 2/21. E cotanto superficia la sexta parte, cioé linea .cb. e linea .cd. e cir-<lb></lb> |
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9515 culo .bd. E imperoché l’ é manifesto ch’ é .cb. egual .cd., vel .bd., linee per linee e pero dise .ab. è<lb></lb> |
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9516 .10., adonca lo triangalo .cbd. é per fazza .5., adonca állo de superficie .R.117 3/16. Adonca, par<lb></lb> |
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9517 che linea .bd. con lo suo circulo .bd. superficia .13 2/21. men .R.117 3/16., adonca adoppia questo, che<lb></lb> |
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9518 fa .26 4/21. men .R.468.3/4. e cotanto superficia .ec. Ed é fatta. .84. |
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9519 </p> |
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9520 <p class="main"> |
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9521 Le sonno doi sphere, una sopra l’ altra. Dela de sopra el diametro sia .ab., quale pon-<lb></lb> |
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9522 go .10. e sia tutta lucida. L’ altra de sotto, el cui diametro è .8.de. e sia tutta obscura<lb></lb> |
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9523 e densa, distante dala splendente .14., cioé .np. E la scura fa un’ ombra in terra il cui diame-<lb></lb> |
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9524 tro .gh., che sia .4. Dimando quanto sia la distantia dala obscura al’ ombra che fa in terra,<lb></lb> |
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9525 cioé quanto sia .qk. Questa è bella e serve a piú cose e fasse in questo modo. Te conviene immagina-<lb></lb> |
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9526 re una linea che passi per li centri de tutti tre questi tondi, cioé dele doi sphere e del’ ombra in terra, per-<lb></lb> |
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9527 ché anche lei se intende un cerchio. E giá tu sai che, in ciascuna spera, è un maximo cerchio, com-<lb></lb> |
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9528 mo meglio de sotto intenderai nel trattato particulare deli corpi regulari. E questa linea, che pas-<lb></lb> |
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9529 sa per ditti centri, sia .cfk., peroché ’l centro dela lucida pongo .c. e dela scura .f. e del’ ombra in terra .k.<lb></lb> |
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9530 Ora tira una equadistante a questa dal ponto .g., extremitá del diametro del’ ombra, fin al diametro<lb></lb> |
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9531 .ab. dela lucida, che lo tagli in ponto .l. E quello dela scura tagli in ponto .r. E questa linea, cosí protrat-<lb></lb> |
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9532 ta, sirá longa aponto quanto che .ck., perché tuti ditti diametri sonno fra loro equadistanti. E tanto an-<lb></lb> |
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9533 cora sirá .lc., quanto .rf. e quanto .gk., che ciascuno sirá .2., perché .gk. è semidiametro del’ ombra. Dapoi,<lb></lb> |
|
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9534 dal ponto .d., extremitá del diametro dela scura, tirarai un’ altra equedistante a questa, fin al diametro<lb></lb> |
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9535 dela lucida, che lo tagliará in ponto .m. e sirá questa equale .a.lr. parte del’ altra, che ciascuna sirá .23.<lb></lb> |
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9536 cioé la distantia dala lucida e scura, che è .14. E tanto piú quanto sonno li semidiametri loro sin al<lb></lb> |
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9537 centro interiore che ’l mezzo diametro dela lucida .cn. e .5. e ’l mezzo diametro dela scura .pf. éne<lb></lb> |
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9538 .4., che fra tutti doi fanno .9., qual gionto sopra .np., cioé sopra .np., cioé a .14., fa .23. per tuta .cf. E cosí sia ancora la pa-<lb></lb> |
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9539 ralella .md. Ora tu hai doi triangoli ortogonij simili, per la .2a. del sexto de Euclide che l’ u-<lb></lb> |
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9540 no sia .alg. l’ altro .amd. E di questo tutti li lati te sonno noti, per la penultima del secondo del ditto Eu-<lb></lb> |
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9541 clide. Cioé .am., che è .1., e .md., che è .23., che contengono l’ angolo retto .m. dela ypotomissa, non curo<lb></lb> |
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9542 al presente che sia .R.530. Ora te convien trovare la quantitá .lg. in questo modo. Tu sai .al. essere .3. |
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9543 </p> |
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9544 <p class="main"> |
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9545 E peró dirai, per la regola del .3., se .am., che è .1., me dá, de basa o de catetto, .md., che è .23., che me<lb></lb> |
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9546 dará, de basa over catetto, .al. che è .3. Multiplica .al. via .md., cioé .3. via .23., fa .69., qual parti<lb></lb> |
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9547 per .am., che è .1., ne ven quel medesimo. E tanto sia aponto .lg., cioé .69. E tanto ancora sia .ck., sua<lb></lb> |
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9548 paralella, commo fo detto. Del qual cava .27., cioé .23. per la quantitá .cf. e ancora .4. per lo<lb></lb> |
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9549 mezzo diametro .fq. dela scura, restará la parte .qk.42. E tanto dirai che sia distante la sphera<lb></lb> |
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9550 <lb></lb> |
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9551 <pb></pb></p> |
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9552 <p class="folio"> |
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9553 folio 65r |
|
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9554 </p> |
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9555 <p class="main"> |
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9556 <lb></lb> |
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9557 </p> |
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9558 <p class="runhead"> |
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9559 Distinctio quinta. Capitulum secundum. 65 |
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9560 </p> |
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9561 <p class="main"> |
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9562 <lb></lb> |
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9563 obscura dala pliana terra, dove reverbera sua umbra. E cosí farai le simili, che molto sonno<lb></lb> |
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9564 applicabili in astrologia et cetera. 8<lb></lb> |
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9565 E gli é una tavola longa bracia .12., larga bracia .2., piana in terra e io sto lontano da lei bracia .10. e,<lb></lb> |
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9566 dal’ ochio mio a terra, cioé la mia statura, é .bracia.2. Dimando quanto longa me se repre-<lb></lb> |
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9567 senta ditta tavola. Sapi che questa domanda è de perspectiva, ma perché questa<lb></lb> |
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9568 scientia è subaltternata a geometria e aritmetrica si la solveremo. Sia la tavola<lb></lb> |
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9569 .abcd. e ’l piano dela distantia da te ala tavola sia .eb. e la tua persona sia .fe. e la larghezza<lb></lb> |
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9570 dela tavola sia .ab. verso te e l’ ochio tuo sia el capo .f. Ora tu ái, con l’ ochio e con la per-<lb></lb> |
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9571 sona e con lo piano, fine al capo dela tavola .dc., causato un triangolo rettangolo che è<lb></lb> |
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9572 .fec., il cui catteto è la tua persona, cioé .fe. e la basa éne tutto el piano .ec. e la potumissa éne la li-<lb></lb> |
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9573 nea .fc. visuale e taglia la larghezza .ab. nel ponto .g. e causa un altro triangoletto similmente<lb></lb> |
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9574 rettangolo, zoé .gbc. e la tavola te s’ apresenta tanto longa quanto .egb., perché chi metesse un bastone ritto,<lb></lb> |
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9575 a filo dela quantitá delo .gb., ritto a te, e guardando tu per lo capo suo, che seria el ponto .g., non vederesti<lb></lb> |
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9576 niente dela tavola. Donca aponto te s’ apresenta quel tanto che è .gb. Ora, per saper la quantitá delo .gb., dirai cosí:<lb></lb> |
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9577 se l’ .ec., basa del triangolo grande, me dá de catetto .fe., che me dará la basa .bc. del triangoletto. Che te<lb></lb> |
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9578 virá a dare .gb. Zoé, per numero, dirai: se .22., ch’ é tutto el piano, me dá .2., che me dará .12. Opera e tro-<lb></lb> |
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9579 verai che te dará .1 1/11. e tanto sia .gb. e tanto te s’ apresenta longa ditta tavola, zioé .ba.1 1/11. et cetera. De-<lb></lb> |
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9580 la largezza non si fa caso peroché par quella medesima, perché sta in maiestá et cetera. 86.<lb></lb> |
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9581 E gli é un piano longo bracia .12. e in capo d’ esso é uno homo longo bracia .2. Vengo io che so<lb></lb> |
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9582 alto bracia .2. commo lui, me scosto dal’ altro bracia .10. e, dove prima io stava, vi msi un al-<lb></lb> |
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9583 tro hom pur alto bracia .2., cioé in sull’ altro capo del piano verso me. Dimando qual di lo-<lb></lb> |
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9584 ro me se mostrará magiore, el primo overo secundo, e quanto. Sapi che questa è<lb></lb> |
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9585 di perspectiva e fassi cosí. Sia el piano .eb.bracia.12., commo è ditto, e l’ omo sia .bc.bracia.2. e habia li piedi<lb></lb> |
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9586 fermi in ponto .b. e la distantia da me al ditto piano sia .ae.bracia.10. E la mia statura sia la linea .ad.,<lb></lb> |
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9587 bracia .2., cioé li piedi in ponto .a. e l’ ochio in ponto .d. E l’ altro homo sia la linea .eg.bracia.2., c’ abia li piedi in pon-<lb></lb> |
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9588 to .e. Or, questo stante, tu imagini un quadrilatero fatto dela tu’ persona e quella del’ altro piú<lb></lb> |
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9589 lontano, ch’ é .abcd. E si imagini una piramide, cioé un triangolo rettangolo .dcb., zoé facto da-<lb></lb> |
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9590 l’ ochio tuo .d. con la linea che va ali soi piedi .db., ch’ é potumissa e l’ altra, che è .dc., che va simil-<lb></lb> |
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9591 mente dal’ ochio tuo al suo, ch’ é tanto quanto tuto el piano .ab., cioé .22. Ora tu vedi che con l’ ochio<lb></lb> |
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9592 tuo tu tagli l’ omo .eg. nel ponto .f. e causi un altro triangoletto .dgf. il cui catetto ène la linea .gf.<lb></lb> |
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9593 e la basa sia .dg., che è tanto quanto la distantia e cioé bracia .10. Ora dico che quello homo che<lb></lb> |
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9594 sta in capo del piano, zoé .bc., te se mostra aponto quanto che è la linea .gf., cioé molto menore<lb></lb> |
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9595 che non fa l’ omo .eg. E, se voli sapere aponto per numero quanto è l’ uno e l’ altro, dirai cosí: se .dc., che è<lb></lb> |
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9596 bracia .22., me dá de catetto .cb., ch’ é bracia .2., che me dará .dg. che è bracia .10., cioé se .22. me dá .2., che me<lb></lb> |
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9597 dará .10. Multiplica e parti, che te dará .10/11. de bracia e tanto sirá el catetto .gf., zoé tanto te se representerá<lb></lb> |
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9598 l’ omo .bc. L’ altro che te sta piú presso, senz’ altro termino fra te e lui, cioé .ge., te se rapresenta quello me-<lb></lb> |
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9599 desimo che lui e zoé bracia .2. E la cagione che si mostri quello lontano manco sie che li piedi che te-<lb></lb> |
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9600 ne in ponto .b. se innalzano ala vista su per la potumissa .db., ma la testa .c. sta sempre para ala tua in<lb></lb> |
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9601 linea retta, siando tu longo quanto lui comme è ditto. Io et cetera. E sapi che questa va al contrario de<lb></lb> |
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9602 quella dela tavola che di sopra dicemmo in questo lato, perché la tavola giaci a piano in terra. Peró<lb></lb> |
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9603 ala vista vene alzare el capo lontano, zoé il capo .c., e l’ altro capo verso te, cioé el capo .b. vene a<lb></lb> |
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9604 stare fermo. E inperó vene a fare catetto de sotto, in lo proprio piano, ma el capo levato, cioé l’ o-<lb></lb> |
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9605 mo el capo alto sta fermo e quello che sta a pian, zoé li piedi se levano in alto a vedere su per la po-<lb></lb> |
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9606 tumissa, commo è ditto, e fanno catetto di sopra ala potumissa e la longhezza che è dal’ ochio tuo<lb></lb> |
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9607 al suo é simile al piano che è fra li toi piedi ali soi, cioé tanto una quanto l’ altra io diligenter adverte.<lb></lb> |
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9608 E cosí sequiresti quando te fosse detto sonno piú homini, uno dopo l’ altro; l’ uno è alto tanto<lb></lb> |
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9609 e l’ altro tanto et cetera, in linea retta dal tuo ochio, l’ uno é scosto dal’ altro tanto che commo per te porrai<lb></lb> |
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9610 formare sempre, imaginando la piramide e li triangoli visuali. Avenga che non fossero de pari<lb></lb> |
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9611 longhezza, che tanto fa, ma in questa te li ó messi tutti longhi l’ uno commo l’ altro, perché meglio li aprenda<lb></lb> |
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9612 per l’ altre. E cosí, quando la domanda dicesse: io e uno altro siamo in un piano, distante l’ u-<lb></lb> |
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9613 no dal’ altro bracia .10. e ciascuno di noi è alto bracia .2. Va el compagno, se slontana indrieto da me per<lb></lb> |
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9614 bracia .12. Dimando se me s’ apresenta magiore o minore de bracia .2. e quanto. Alora, per farla, farai commo<lb></lb> |
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9615 di sopra, perché dove lui stava prima, per imaginatione tu ci porai un altro simile di longhezza a lui e se-<lb></lb> |
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9616 quirai como di sopra e arai che parrá .10/11. de bracia. E cosí, quando dicesse: noi sian doi de par sta-<lb></lb> |
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9617 tura, ognuno bracia .2., distanti l’ uno dal’ altro bracia .12. Vengo e tirome indrieto bracia .10. e lui sta fer-<lb></lb> |
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9618 mo. Dimando quanto me se mostreran. Fa commo di sopra. Imagina nel luogo dove prima stavi<lb></lb> |
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9619 un altro a te simili e sequirai.<lb></lb> |
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9620 <lb></lb> |
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9621 <pb></pb></p> |
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9622 <p class="folio"> |
|
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9623 folio 65v |
|
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9624 </p> |
|
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9625 <p class="main"> |
|
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9626 <lb></lb> |
|
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9627 </p> |
|
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9628 <p class="runhead"> |
|
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9629 Distinctio quinta. Capitulum secundum. |
|
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9630 </p> |
|
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|
9631 <p class="main"> |
|
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9632 <lb></lb> |
|
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9633 E gli é uno longo bracia .5. in su ‘nn un piano con meco, distante da me bracia .22. E fra lui e me éne<lb></lb> |
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9634 una finestra quadra, overo balestrera, distante da me bracia .10. e distante da lui bracia .12. |
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|
9635 </p> |
|
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9636 <p class="main"> |
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9637 E, si vego de ponto colui, dala cima del capo fino ala pianta di piedi, dimando quan-<lb></lb> |
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9638 ti bracia sirá alta ditta finestra. Sapi questa essere gintil domanda e per farla fa cosí. Po-<lb></lb> |
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9639 niamo che ’l piano sia la linea .cb. e la fessura .ef. e colui, dela dal muro, sia la linea .ab. e la tua perso-<lb></lb> |
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9640 na sia la linea .dc., cioé l’ ochio tuo sia el ponto .d. E non fa caso sapere la tua altezza: avenga che in l’ ar-<lb></lb> |
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9641 te pictoria si metta communamente l’ omo longo bracia .3. E sia la distantia da te ala balestriera .dg. e da<lb></lb> |
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9642 lui ala finestra sia .gh. Ora, se tu ben consideri, tu causi doi triangoli rettanguli, fra te e lui con l’ ochio<lb></lb> |
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9643 tuo: l’ uno è .dhb., l’ altro éne .dha. L’ uno, cioé .dhb. á il catetto volto in giú, che è .hb. L’ altro áne el catet-<lb></lb> |
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9644 to volto in su, cioé .dha., il cui catetto éne .ha. e la linea .dh. è tanto quanto che el piano .cb.,<lb></lb> |
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9645 cioé .22. E taglia la finestra rettangolo nel ponto .g., siché co’ lui te s’ apresenta aponto tan-<lb></lb> |
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9646 to quanto ditta finestra. E, per aver ditta finestra, troverai il catetto soprano .ge. e ancho<lb></lb> |
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9647 di catetto sottano .gfe., giongnerali insiemi e tanto sia l’ alteza dela finestra. E, per trovar<lb></lb> |
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9648 ditti catetti, dirari cosí: se .dh., che è .22., me dá de catetto .ha., ch’ é .2., che me dará .dg. ch’ é .10. |
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9649 </p> |
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9650 <p class="main"> |
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9651 E troverai che te dará .10/11. e tanto sia .eg., qual salva. Poi, per l’ altro, dirai: se .dh., ch’ é .22., me dá de<lb></lb> |
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9652 catetto sottano .hb., che è .3., cioé quanto la mia statura, qual per la ragion ditta metto bracia .3., e, se piú<lb></lb> |
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9653 la volesse, piú la meteria, che me dará .dg. pure di catetto sottano. Multiplica e parti e troverai che te<lb></lb> |
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9654 dará .1 4/11. e tanto sia lo catetto .gf., quale giongni con .eg. che salvasti, cioé con .10/11., fa .2 3/11. per tutta l’ altez-<lb></lb> |
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9655 za dela finestra e tanto sia .ef. e anche tanto dirai che te s’ apresenta colui, cioé bracia .2 3/11. Fatta.<lb></lb> |
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9656 Vel aliter et brevius, perché l’ ochio tuo causa piramide e anche un triangolo fra doi linee equi-<lb></lb> |
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9657 distanti, che l’ una è la tua statura e l’ altra la statura del compagno, sonno pur simili fra loro. E, se<lb></lb> |
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9658 ditta finestra s’ abasasse in sul piano uniforme a squadro, cadaria fra .cb. in sul ponto .l., commo appa-<lb></lb> |
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|
9659 re in la .2a. figura. E alora diresti: se .22., cioé .cb. me dá de catetto .5., cioé .ba., che me dará .10., cioé .cl.<lb></lb> |
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9660 Opera e troverai che ten dará in un tracto .2 3/11., che tuta la quantitá dela finestra, senza farla in doi volte.<lb></lb> |
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9661 Ma la prima è piú demostrativa, questa piú maistrevoli et cetera.<lb></lb> |
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9662 E, se la domanda dicesse: e gli é uno alto bracia .5., longni da me bracia .22.; fra lui e me c’ é una<lb></lb> |
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9663 finestra che me lo mostra de ponto, da pie’ a capo, alto bracia .2 3/11. Dimando quanto sia lonta-<lb></lb> |
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9664 na ditta finestra da me e quanto da lui. Alora, per le medesime considerationi dirai: se .ab.<lb></lb> |
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9665 ch’ é .5. di catetto, me dá .22. de piano, ch’ é .cb., che me dará .2 3/11. Multiplica e parti. Troverai<lb></lb> |
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9666 che te dará .10.bracia. E tanto sia distante da te ditta finestra. E lo resto fin .22., ch’ é .12., sirá da lui, che<lb></lb> |
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9667 per l’ una poi aver l’ altra e per l’ altra l’ una e virate bene et cetera. E, se dicesse: e gli é uno lontano da me<lb></lb> |
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9668 bracia .22. e fra lui e me éne una finestra, distante da me bracia .10. e da lui distante bracia .12. La quali nestra<lb></lb> |
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9669 è alta bracia .2 3/11. e mostrame de ponto colui. Dimando quanto sirá longo quel tale. Alora dirai: se .10. de ba-<lb></lb> |
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9670 sa me dá de catetto .2 3/11., che me dará .22. de basa. Multiplica e parti. Troverá che te dará de catetto bracia<lb></lb> |
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9671 .5. e tanto sia longo quel tale. |
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|
9672 </p> |
|
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|
9673 <p class="main"> |
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9674 S e gli é .1o. piano longo una quantitá e in esso a filo ugualmente sonno .3. persone; la prima è lontana<lb></lb> |
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9675 dala .2a.6.bracia. e la .2a. dala .3a.12.bracia; ciascuna è alta .3.bracia. e io sto distante dal ditto pia-<lb></lb> |
|
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9676 no .6.bracia. e da me ala prima persona é .10 bracia. Dimando quanto me s’ arepresen-<lb></lb> |
|
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9677 ta ciascuna de ditte persone e quanto dissgrada una e quanto l’ altra. Sapi che questa<lb></lb> |
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9678 é gentil domanda in perspectiva e, per satisfarla, tien questa via .V.g. Poniano che ’l ditto piano sia la linea<lb></lb> |
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9679 .eb. e la tua persona sia nel ponto .o. e la linea ortogonale, da te al ditto quel piano, dicemo essere bracia .6.; sia<lb></lb> |
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9680 la linea .oa. contingente il piano .eb. nel ponto .a. e sia la prima persona a te prosimana nel decto piano in ponto<lb></lb> |
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9681 .d., qual dicemo esser distante da te bracia .10., che sia la linea .od. e questa viene a esser potomissa d’ un trian-<lb></lb> |
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9682 golo rettangolo .oad. che l’ un di lati continenti il rettangolo, cioé .oa., é ditto esser .6. Donque dela po-<lb></lb> |
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9683 sanza .od., quale è .100., trattone la posanza .oa., rimane .64. per la possanza .ad. Donca sia .ad.R.<lb></lb> |
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9684 di .64., cioé .8. Ora sia la .2a. pa. apresso questa pure in decto piano d’ uno paraggio nel ponto .c., qual di-<lb></lb> |
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9685 cemo essere distante dal ponto .d.bracia.6. Donca sia la potumissa .oc. del triangolo rettangolo .oac.R.<lb></lb> |
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9686 di .232. Poi apresso questa sia la .3a. nel ponto .b., che dicendo .cb. essere bracia .12. donca sia la potomissa .ob.<lb></lb> |
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9687 del triangolo rettangolo .oab.R. di .712., perché la basa tutta .ab. éne bracia .26. Ora, per trovare el<lb></lb> |
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9688 digradamento del’ uno e l’ altro, dirai cosí: se la potomissa .ob., che è .R. di .712., me dá .3. d’ altezza<lb></lb> |
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9689 che dicemo essere la statura degli omini, che me dará la potomissa .od., che è .10. Multiplica e parti e<lb></lb> |
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9690 trovarai che te dará .R. de .1o.47/178. E tanto te s’ apresenta colui che sta nel ponto .b., perché, se fosse nel primo<lb></lb> |
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9691 uoco a te proximano, cioé nel ponto .d., te s’ apresenteria pur .3., sí commo e gli é peroché, fra te e lui, non siri<lb></lb> |
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9692 termine alcuno. Donca, per essere lui slontanato dal decto ponto .d.bracia.18., che è tutto .db., me se rapre-<lb></lb> |
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9693 tenta molto manco che prima, cioé. R. 1o.47/178. Donca è digradato. El resto fine a .3., che è .3. men .R.1.<lb></lb> |
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9694 .47/178. Poi, per lo .2o. che sta nel ponto .e., dirai similmente: se .oe., che è .R. del .232., me dá d’ alteza bracia<lb></lb> |
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9695 <lb></lb> |
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9696 <pb></pb></p> |
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9697 <p class="folio"> |
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9698 folio 66r |
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9699 </p> |
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9700 <p class="main"> |
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9701 <lb></lb> |
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9702 </p> |
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9703 <p class="runhead"> |
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9704 Distinctio octava. 66 |
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9705 </p> |
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9706 <p class="main"> |
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9707 <lb></lb> |
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9708 .3., che me dará .od., che è .10. Multiplica e parti. Troverari che te dará .R.3 51/58. E tanto, per la ragion ditta, te<lb></lb> |
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9709 s’ apresentará quel che sta nel ponto .e. Donca, per essere lui dislontanato dal ponto .d.bracia.6., á disgrada-<lb></lb> |
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9710 to .3. men .R.3 51/56. E quel che sta in ponto .d. te s’ apresenta quel medesimo che gli é a ponto, cioé bracia .3. Ma,<lb></lb> |
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9711 quando nel ponto .a. vine fusse un altro dela medesima altezza, che virria a esser distante dal .d. bracia<lb></lb> |
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9712 .8., alora similmente quel che sta nel ponto .d. virria a degradare ancor lui. E quell che stesse nel pon-<lb></lb> |
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9713 to .a. staria saldo in sua quantitá, peroché gli é el termine a te piú proximano che sia. E, se tu volesse trova-<lb></lb> |
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9714 re il digradamento di quel che sta nel ponto .d., alor diresti commo negli altri hai facto, cioé: se .10.<lb></lb> |
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9715 me dá .3., che me dará .6. Che te veria a dare .1 4/5. Siché, per la mutation sua, dal ponto .a. fine al<lb></lb> |
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9716 ponto .d. che è .8., vene a esser digradato el resto da .1 4/5. fine a .3., che è .1 1/5. a ponto. Cosí te converia<lb></lb> |
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9717 poi similmente ritrovare gli altri del ponto .c. e ponto .b. dicendo: se .R.232. me dá .3., che me dará<lb></lb> |
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9718 .6. E tanto te se rapresenterá quello del ponto .c. E poi, per quello del ponto .b., dirai: se .R.712. me dá<lb></lb> |
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9719 .3., che me dará .6. E quello che ti dará, tanto ti parrá quello dil ponto .b. E cosí, se fossero infinite per-<lb></lb> |
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9720 sone sopra una medesima linea, o da man destra o da senestra, commo si voglia, o de equali o vero<lb></lb> |
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9721 de diverse distantie, o de equali overo diverse altezze over stature (commo torri, merli o simili co-<lb></lb> |
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9722 se che vano a filo) trovaresti di tutti lor potomisse per la via ditta con la notitia dela perpendicu-<lb></lb> |
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9723 lare da te al ditto piano. Avenga che l’ uno fosse alto .4.5.6..10.20. et cetera. Sempre dirai: se questa poto-<lb></lb> |
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9724 missa me dá d’ altezza (.0.4.0.5. che si voglia.), che me dará quella ch’ é piú proximana. E quello che<lb></lb> |
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9725 te dará tanto te rapresentará l’ alteza che sia in capo de ditta ypotomissa. E ’l digradamento puoi per<lb></lb> |
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9726 te facilmente trovarai. E cosí farai in tutte simili. Avenga che fossero o da man dextra o da sene-<lb></lb> |
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9727 stra de ditta perpendiculare: observarai similmente. E, per piú tua chiareza, qui da lato guarda la<lb></lb> |
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9728 figura et cetera. E, quando te fosse datto fermo l’ ultimo, quello del ponto .b., e per quello tu volesse sapere quan-<lb></lb> |
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9729 to te s’ apresenteran gli altri piú, alora faresti al contrario, perché ciascuno te virrá a crescere e dire-<lb></lb> |
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9730 sti: se .od., ch’ é .10., me dá .3., che me dará .R.712., che te virrá a dare .R.64 2/25. e tanto te se mostra-<lb></lb> |
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9731 rá quello del ponto .d. E cosí viresti per gli altri, a uno a uno. Per quello de .e. diresti: se .10. me dá .3., che<lb></lb> |
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9732 me dará .R.232., che te virá a dare .R.20 22/25. e tanto te mostrará quello del ponto .e. et cetera va per lo<lb></lb> |
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9733 contrario che di sopra festi .89. |
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9734 </p> |
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9735 <p class="main"> |
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9736 E gli é un quadro .abcd. per facia .10. Io lo voglio redure a .8. facce e consumarne meno<lb></lb> |
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9737 sia possibile. Dimando commo se habia a fare geometrice e ancora per arithmetica.<lb></lb> |
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9738 Dico che se de’ fare cosí per geometria. Tira el diametro in lo ditto quadro, che sia .bc., e divi-<lb></lb> |
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9739 dilo in mezzo in ponto .t. e super esso descrivi el cerhio torno al ditto quadro e poi l’ archo<lb></lb> |
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9740 .ab. dividi in mezzo in ponto .e. e cosí li altri archi: .ac. in .f.cd. in .g.db. in .h. E tira le liniee da ditti pon-<lb></lb> |
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9741 ti dividenti ditti archi per equali dal’ uno e l’ altro e harai un altro quadro .efgh. el qual taglia li ango-<lb></lb> |
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9742 li del primo quadro neli ponti .k.p.l.m.n.o.p.q.r. Queli sonno angoli del quesito octagono equilatero e an-<lb></lb> |
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9743 che equiangolo, commo vedi in figura, e questo è il magior che si possa fare del ditto quadro, perdendo man-<lb></lb> |
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9744 co si possa del quadro et cetera. Ancora potevi fare el tondo dentro al quadro e far similiter e haresti lo<lb></lb> |
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9745 octagon, ma si perde asai del .primo. quadro et cetera. Or prendila arithmetice. Poni che si levi del lato<lb></lb> |
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9746 .ab .1.co. per lo .ak., donca .1.co. ancora se ne leverá del lato .ac. per lo .am. Ora tu hai uno triangolo or-<lb></lb> |
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9747 togonio .akm. del quale l’ angolo .a. è retto, per la penultima del .po. Euclidis, el quadratto del lato .km.<lb></lb> |
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9748 fa li altri .2. Peró quadra .ak., fa .1.ce. e quadra .am. fa .1.ce., similiter queli giongni insemi fanno .2.ce. per lo qua-<lb></lb> |
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9749 drato delo .km. Donca la .R.2.ce. sia .km., lato delo octagono, quale serba. Ora prendi .kl., che sirá<lb></lb> |
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9750 .10. men .2.co., perché a ogni cantone perde .1.co., siché vien restar .kl.10. men .2.co. Multiplica li extremi in sé harai .2. <lb></lb> |
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9751 ce <lb></lb> |
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9752 equali a .100.<lb></lb> |
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9753 men .40.co. piú .4.ce. Aguaglia le parti levando li superflui e rendendo li debiti. Harai in ultima equatione, quan-<lb></lb> |
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9754 do la sia recata a .1.ce., .50. piú .1.ce. equali a .20.co. Smezza le cose, multiplica in sé, cavane el numero, resta .50.e.<lb></lb> |
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9755 .R. di questo cava del dimezzamento e harai la cosa valere .10. men .R.50. E tanto s’ é levato dela fa-<lb></lb> |
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9756 cia verso ciascun angolo del quadro e sirá .ak. e ancho .lb. e ancho .qd. e .rd. e .co. e .cm. et cetera. Ora<lb></lb> |
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9757 trova .km. multiplicando in sé .ak., ch’ é .10. men .R.50., fa .15. men .R.20000. e altretanto sia el quadrato<lb></lb> |
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9758 .am. che, gionti insemi, fanno .300. men .R.80000. E tanto sia el quadrato del lato opposito al’ ango-<lb></lb> |
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9759 lo retto .a., cioé de .k.m. Donca la sua .R. universale, overo legata, sia ditto lato .km., cioé .Ru.<lb></lb> |
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9760 .300. men .R.80000. e altretanto sia .kl. lato pure del quesito octagono. Donca harai quella .R. lega-<lb></lb> |
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9761 ta equal a .R.200. men .10., che sta bene, che tanto val l’ una commo l’ altra, peroché quella .R.<lb></lb> |
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9762 legata in sé fa .300. men .R.80000. e tanto ancora fa .R.200. men .10. in sé dutto, cioé .300. men<lb></lb> |
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9763 .R.80000., siché tanto sia l’ una commo l’ altra ratione quadrature quod est nota, dignissimum<lb></lb> |
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9764 in similibus ut patet quamvis denominationes sint diverse tamen quantitas est<lb></lb> |
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9765 eadem utrobique et cetera. 90. |
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9766 </p> |
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9767 <p class="main"> |
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9768 Una botte tien .10. barili; ne voglio ogni giorno trare la decima parte di quello che io vi<lb></lb> |
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9769 <lb></lb> |
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9770 <pb></pb></p> |
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9771 <p class="folio"> |
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9772 folio 66v |
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9773 </p> |
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9774 <p class="main"> |
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9775 <lb></lb> |
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9776 </p> |
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9777 <p class="runhead"> |
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9778 Distinctio octava |
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9779 </p> |
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9780 <p class="main"> |
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9781 <lb></lb> |
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9782 trovaró dentro. Dimando in quanti dí n’ aró tratto .5. barili. Fa cosí. Proponi che in questa botte sia<lb></lb> |
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9783 .10000000. E questa positione si fa per fugire la confusione deli rotti. Adonca, se ve siran diecimilion<lb></lb> |
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9784 de barili, e gli é forza che ’l primo dí io cavi un milion de barili. E peró sequita che in essa restaran<lb></lb> |
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9785 .9. milioni de barili. E l’ altro dí me conven trarne el .1/10. de nove milioni, che sonno novecento migliai<lb></lb> |
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9786 e tutto questo serba. El terzo giorno sirá in essa .8. milioni e centomigliai de barilie e di questo te conven ca-<lb></lb> |
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9787 cavar ne ‘l .1/10., che sonno octocento diecimigliai e similmente serba. Or vien el .4o. giorno, che non ve sirá piú che<lb></lb> |
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9788 .7. milioni. e .290000. de barili, de quali similiter ne cavi el .1/10., che sonno .729000. e similiter serba.<lb></lb> |
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9789 Or vien el .5o. dí, che in essa non si trovará piú che .6. milioni e .6561000., dela qual somma similiter ca-<lb></lb> |
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9790 varai el .1/10., che sonno .656100. e similiter serba. El .6o. giorno in essa non trovarai se non .5. milioni e<lb></lb> |
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9791 .5940900., deli quali cava el .1/10., che son .590490. e similiter serba .Ora recogli tutte queste serbanze insie-<lb></lb> |
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9792 mi, faranno .4685590., che, fin .5. milioni, che è la .1/2., mi manca .314410. E si volesse trare el .7o. dí el<lb></lb> |
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9793 .1/10., ch’ entro seria .5314410., deli qual serebbe el .1/10. barili .531441., siché seria troppo. E peró dico:<lb></lb> |
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9794 si se .1. dí me dá .531441. e io vorri’ .314410., che parte debio prendern de dí. Fa cosí. Multiplica .1. via .314410.<lb></lb> |
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9795 e parti in .531441., che nen ven questo rotto .431410/531441. de giorno. E, concludendo, dico che ditta botte sará<lb></lb> |
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9796 voita mezza in dí .6. e questo spezzato .314410/531441. de giorno. Facta. 91<lb></lb> |
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9797 Una botte á .4. canole ordinate, tutte in retta linea, secundum sub et supra e non equali-<lb></lb> |
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9798 ter iacentes vt pz in fila. Dala prima fin ala cima è el .1/3. del tenuto de tutta la botte; dala<lb></lb> |
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9799 . 2a. ala prima é il .1/4. de tutto el tenuto; dala .3a. ala .2a. é il .1/5. del tenuto e dala .4a. ala .3a.<lb></lb> |
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9800 el resto del tenuto. La prima voita la sua parte suprema in un dí, la .2a. la sua parte la voita in<lb></lb> |
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9801 .2., la .3a. voita la sua in .3. dí, cioé quello che è fra la .3a. e 2a. E la 4a. voita la sua in .4. dí. Dimando: apren-<lb></lb> |
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9802 dole tutte a un tratto, siando piena, in quanti giorni si voiterá. Prima te conven trovare un te-<lb></lb> |
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9803 nuto a tutta la botte, poiché ‘l thema non lo dá e, quando lo desse sequiristi quello commo faremo in<lb></lb> |
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9804 questo. Donca poni che la tenga una quantitá de barili a tuo modo, che non fa caso, ma per piú aconzo poni<lb></lb> |
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9805 un numero che habia .1/3, 1/4, 1/5, ació possi togliere le parti sane, che sia .60. Or dici che la parte suprema tien<lb></lb> |
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9806 el .1/3., donca tira barili .20. e la .2a. barili .15., per lo .1/4. dela .3a.12., per lo .1/5. la .4a. el resto fin .60., che son<lb></lb> |
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9807 .13. Ora te conven fare la ragione a parte a parte, perché facendola tutta insiemi non resciri,<lb></lb> |
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9808 peroché le ditte canole non lavoran tutte fin al fondo: perché, commo la parte soprana sia fini-<lb></lb> |
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9809 ta, quella canella non getta piú e cosí del’ altre successive. Siché la portion prima è aiutata da tutte<lb></lb> |
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9810 e la portion .2a. non è aiutata se non da .3. e la .3a. da .2. la .4a. solo da una vt pz intuenti. Donca fa-<lb></lb> |
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9811 rai prima per la parte soprana e vedi, lavorando ciascuna commo è ditto, in quanti dí la voitaranno, che<lb></lb> |
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9812 fai che la prima fa el dí .20. barili, la .2a.7 1/2., perché in .2. dí voita .15., la .3a. fa .4., che in .3. voita .12., la<lb></lb> |
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9813 . 4a. fa .3 1/4., che in .4. voita .13. Summa insiemi .207 1/2.4.3 1/4., fanno .34 3/4. e tanti barili voitaranno in un<lb></lb> |
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9814 dí tutte .4. E tu voli solo barili .20. per la prima parte. Peró dirai: se .34 3/4. sonno voitati da .1. giorno, da<lb></lb> |
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9815 quanti dí siram voitati .20. barili. Opera, harai .60/139. e in tanti esimi vedi tutte .4. haranno voitata<lb></lb> |
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9816 la prima parte, qual salva. Poi el simile farai per la .2a. parte e vedi le .3. canole che rimagano in quanti dí<lb></lb> |
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9817 voitaranno barili .15., che sai che la seconda voita el dí .7.1/2., la .3a.4., la .4a.3 1/4. Summa insemi, fanno<lb></lb> |
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9818 .14 3/4. e tanti barili ne voitaranno in un dí. E tu voi sapere de .15., peró dirai: se .14 3/4. sonno voitati<lb></lb> |
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9819 da un dí, da quanti saran voitati barili .15. Opera, harai che siran voitati in dí .1 1/59., qual salva. Poi farai per<lb></lb> |
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9820 la .3a. parte e vedi le .2. canelle rimanenti in quanti dí voitaraveno barili .12., che sai che la .3a. voita<lb></lb> |
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9821 .4. e la .4a.3 1/4. Summa, fanno .7 1/4. e tanti barili ne voitaranno in .1. dí la .3a. e .4a. insiemi. E tu voi sa-<lb></lb> |
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9822 pere de .12., peró dirai, ut supra, se .7 1/4. sono votati in .1. dí, in quanti dí siran voitati .12. Opera, siran voi-<lb></lb> |
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9823 tati in dí .1 10/29. Per la .4a. parte non bisogna che facia altro, peroché lei solo se voita da una canella<lb></lb> |
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9824 in .4. dí perché le precedenti non la possano aiutare .ut pz. Ora summa insiemi tutti questi dí che hai tro-<lb></lb> |
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9825 vato per la prima .2a. 3a. 4a. parte, cioé .80/139.1 1/59.1 10/29.4., fanno in summa dí .7 245235/2959139. e in tanti dí <lb></lb> |
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9826 dirai<lb></lb> |
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9827 che voitaranno la ditta botte. E tenga ció che si voglia, siando disposte in essa le zanole: ut sonat<lb></lb> |
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9828 in thenate et cetera. E cosí porresti imbratarla con .5. o .6. canolle et cetera e dirai la prima la voita in dí et cetera; la <lb></lb> |
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9829 prima <lb></lb> |
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9830 cannola-<lb></lb> |
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9831 tien el .1/3. del tenuto per .4. barili, la .2a. á l’ .1/4. men .12. barili .et cetera. E seria grande operare siché tu per te <lb></lb> |
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9832 formane.<lb></lb> |
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9833 .92. |
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9834 </p> |
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9835 <p class="main"> |
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9836 E gli é una fontana con doi pile l’ una sopra l’ altra, commo vedi. E l’ una e l’ altra ha .3.<lb></lb> |
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9837 canoni, cioé .3. che empi e .3. chi voita, di tal conditione che il primo di quelli di sopra, per.<lb></lb> |
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9838 se solo, empiria el pil di sotto in una ora. El .2o. lo empiria in .2. ore, el .3o. in .3. E deli<lb></lb> |
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9839 canoni del pil de sotto, el primo per sé, quando fosse pieno, lo voitarebe in .2. ore. E lo<lb></lb> |
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9840 . 2o. in .3. el .3o. in 4. Dimando, aprendo tutti a un tratto, cioé quelli che empiano e quelli che voi-<lb></lb> |
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9841 tano e lavorando ognuno a modo ditto, in quante ore quelli de sopra haveranno impito aponto el<lb></lb> |
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9842 pil de sotto. Fa cosí. Prima trova .1.no. che habia parti integre denominate da quelli numeri de ore, cioé<lb></lb> |
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9843 da .2.3. e .4., cioé che habian .1/2 1/3 1/4., che si trova multiplicando li denominatori uno in l’altro. E poi quel che<lb></lb> |
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9844 fa in l’ altro et cetera, commo nel trattato de’ rotti te insegnai. E dí: .2. via .3. fa .6. e poi .6. via .4. fa .24. e<lb></lb> |
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9845 <lb></lb> |
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9846 <pb></pb></p> |
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9847 <p class="folio"> |
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9848 folio 67r |
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9849 </p> |
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9850 <p class="main"> |
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9851 <lb></lb> |
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9852 </p> |
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9853 <p class="runhead"> |
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9854 Distinctio octava 67 |
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9855 </p> |
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9856 <p class="main"> |
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9857 <lb></lb> |
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9858 questo è quel numero che á .1/2 1/3 _. integraliter per l’ ordinario trovato. Ma al piú basso, .12. ancora<lb></lb> |
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9859 á le medesime parti e tanto te serve uno quanto l’ altro. Ma quanto piú basso possiamo operare<lb></lb> |
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9860 tanto è meglio e men briga. Or, facto questo, vedi in questo tempo de ore .12. quante volte li cannoni de<lb></lb> |
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9861 sopra (lavorando a modo ditto) inpirieno quella de sotto, che lo trovarai in questo modo. Tu fai che ’l<lb></lb> |
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9862 primo cannon per sé l’ enpi in un’ ora, donca lui l’ enpi .12. volte in .12. ore. E quello da .2. l’ enpirá .6. volte. E<lb></lb> |
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9863 quello da .3. l’ enpi .4. volte. Ora summa insiemi .12.6.4., fanno .22. Siché tu hai che fra tutti .3. ‘ditti cannoni<lb></lb> |
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9864 la se inpiria .22. fiade; quel salva da parte. E poi fa per quelli di sotto similmente. Cioé vedi in .12. ore quante<lb></lb> |
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9865 ne voitaria el cannon da .2., che ne voitaria .6. e quello da .3. ne voitaria .4. e quel da .4. ne voita-<lb></lb> |
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9866 ri .3.; queli suma insiemi, cioé .6.4.3., fanno .13. e tante volte serebe voitata da quelli de sotto nel tem-<lb></lb> |
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9867 po de ore .12. Ora cava queste .13. voite de .22. piene, resta .9. piene. E tu voli una sola piena. E<lb></lb> |
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9868 peró dirai, per la regola del .3., cosí. Se .9. me sonno inpite da .12. ore, da quante me sirá inpita una.<lb></lb> |
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9869 Multiplica e parti secondo la regola ditta. Harai che se inpirá in ore .1 1/3. Facta. La prova farai cosí commo de<lb></lb> |
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9870 sopra. Cioé, vedi in .1 1/3. quante n’ enpirá ciascuna a modo ditto che sai commo lavorano, pur per regola del<lb></lb> |
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9871 .3. dicendo: se un’ ora me dá .1a. piena per lo primo canonne, che me dará .1 1/3. d’ ora. Te dará .1 1/3. de pila pie-<lb></lb> |
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9872 na. E quel da .2. te dará .2/3. de pila. e quel da .3. te ne dará .4/9., che fra tutti .3. te vengono a dare .2 4/9.<lb></lb> |
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9873 E tante ne harai de piene in .1 1/3. d’ ora, qual salva. Poi vedi quante ne haverai de voite in ditto tem-<lb></lb> |
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9874 po per li cannon de sotto similmente dicendo: se in .2. ore el primo ne voita .1., donca in .1 1/3. ne voita .2/3.<lb></lb> |
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9875 El secondo ne voita .4/9., el .3o. ne voita .1/3. che, gionte insiemi .2/3 4/9 1/3., fanno .1 4/9. e tante ne votaranno ore .1 <lb></lb> |
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9876 1/3.<lb></lb> |
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9877 qual cava de .2 4/9., resta una sola aponto piena per quel che volemo. Siché aponto fie facta et cetera. Or<lb></lb> |
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9878 prendila per la cosa. Poni che in .1.co. d’ ore quelli de sopra ne inpissero una. Unde el primo, se in un ‘ora<lb></lb> |
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9879 n’ empí .1., in .1.co. d’ ore n’ enpirá .1.co. de pile. El .2o.1/2.co., el .3o.1/3.co., qual summe asiemi, cioé .1.co.1/2.co.<lb></lb> |
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9880 .1/3.co., fanno .1 5/6.co. e tante pile li .3. empiranno in .1.co. d’ ora. Qual salva da parte. E poi ved quelli de<lb></lb> |
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9881 sotto quante ne voitarebeno in .1.co. d’ ore, che troverai el primo d’ essi voitarne .1/2.co., el .2o.1/3.cose., el .3 1/4.co.,<lb></lb> |
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9882 quali summa insiemi, fanno .1 1/12.co. e tante ne voitarebeno in .1.co. d’ ore, qual cava de .1 5/6.co., resta .3/4.co.<lb></lb> |
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9883 de pile. E tu ne voresti .1a. sola., donca harai .3/4.co. equali a .1. Sequi el capitolo, harai la cosa valere<lb></lb> |
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9884 .1 1/3. e in tante ore quelli de sopra haveranno avanzato una pila piena, nonobstante che quelli de sot-<lb></lb> |
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9885 to gettino quello che lor danno et cetera. E cosí in simili te regerai. Ma in simili é d’ avertire, ació<lb></lb> |
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9886 tu non lavorasse alo impossibile, perché la domanda se porria ponere in modo che non si porebe mai in-<lb></lb> |
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9887 pirne alcuna parte dela de sotto. E questo averria quando quelli cannoni de sotto fosero piú velo-<lb></lb> |
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9888 ci al voitare che quelli de sopra a impire, che in tal caso mai li de sopra non porienno giongnere quelli<lb></lb> |
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9889 de sotto. E anche quando l’ uni e l’ altri lavorassero pari, anche non si giongnerienno. Io n’ ad-<lb></lb> |
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9890 verte. E commo habian detto de .3. che inpiano .e. 3 che voitano, cosí poi dire de .3. che empino<lb></lb> |
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9891 e .4. che votino, e .10. che empino e .6. che voitino et cetera, proportionando el tempo in modo che quelli de so-<lb></lb> |
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9892 pra possino giongnere quelli de sotto et cetera. E cosí, commo habian ditto de doi pile, cosí poi sequire<lb></lb> |
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9893 de .3. pile e de .4., travagliando, or piú or manco cannoni, tempo e pile, nondimeno al modo dato,<lb></lb> |
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9894 sequiristi e virram sempre bene et cetera. E per questa via ancora faresti quando dicesse solo deli can-<lb></lb> |
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9895 noni de sopra, cioé non voitandose quel di sotto in quant’ ore quelli per sé empirieno quel di sotto et é conver-<lb></lb> |
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9896 so, cioé siando pieno quel di sotto. E non inpiendo quelli de sopra in quanto tempo quelli voitarie-<lb></lb> |
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9897 no el pilo, commo se costuma dire de botti de .3. o .4. cannole et cetera, ponendo sempre .1. numero fermo<lb></lb> |
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9898 de ore, de giornoi et cetera e vedere in quel tempo quante volte se inpiria e quante volte se voitaria et cetera e tu<lb></lb> |
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9899 poi, per la regola del .3., sempre sequirai lo effecto dicendo: se tante sonno in tant’ ore impite o voi-<lb></lb> |
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9900 te et cetera in quant’ ore sira piena una. E cosí sempre operando farai bene et cetera. 93<lb></lb> |
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9901 E gli é una verga, o voi dir bastone, eguale qual è deritto e longo palmi .12. e pessa .libbre.12. |
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9902 </p> |
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9903 <p class="main"> |
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9904 Ora io lo messi in bilico in sul primo palmo e apiccai in sula sumitá de ditto primo palmo<lb></lb> |
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9905 una grossa petra con l’ oncino e dipoi tolsi un’ altra petra minore con l’ oncino di ferro<lb></lb> |
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9906 e apiccala in sula stremitá opposita del ditto bastone, che pesa ditta minor pietra<lb></lb> |
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9907 .libbre.2., cioé .libbre.2. con l’ oncino e trovo che stanno pari li contrapesi, cioé in equilibra. Dimando che<lb></lb> |
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9908 pesa la gran petra con l’ oncino. Fa cosí. Prima é da vedere quanto el sopradetto bastone sostien da<lb></lb> |
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9909 se stesso, senza l’ aiuto dela picola pietra. E, per questo fare, sappi che ogni bastone iguale, sia longo<lb></lb> |
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9910 o corto, a suo modo, se tu lo metti in bilico, cioé in equilibra, diciamo in su sexta parte dela<lb></lb> |
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9911 sua lunghezza, ellevará per se stesso el peso de doi bastoni. E, se tu lo metti in bilico in sula quarta<lb></lb> |
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9912 parte dela sua longhezza, levará per se stesso el peso d’ un bastone. E, per questo sapere, fa cosí. Noi dicia-<lb></lb> |
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9913 mo che ’l ditto bastone è in bilico in sul primo palmo e che in tutto è longo palmi .12., adonca el<lb></lb> |
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9914 ditto bastone è in bilico in sul duodecimo dela sua longhezza e peró prendi el mezzo di .12., che<lb></lb> |
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9915 sta sotto la verga, che è .6., del quale cava .1., ch’ é sopra alo detto .12., resta .5. e questo parti per lo<lb></lb> |
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9916 preditto .1., che sta sopra al .12., ne ven .5. E tanti bastoni sostirrá per se stessa la ditta verga, cioé<lb></lb> |
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9917 <lb></lb> |
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9918 <pb></pb></p> |
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9919 <p class="folio"> |
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9920 folio 67v |
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9921 </p> |
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9922 <p class="main"> |
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9923 <lb></lb> |
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9924 </p> |
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9925 <p class="runhead"> |
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9926 Distinctio octava. De Corporibus regularibus. |
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9927 </p> |
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9928 <p class="main"> |
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9929 <lb></lb> |
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9930 .5. E, se la ditta verga over bastone fosse in bilico dela quarta parte dela sua longhezza, quanto leva-<lb></lb> |
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9931 rá da se stesso. Fa commo di sopra t’ ó mostro. Prendi la .1/2. de .4., che è .2., e di questo tra’ .1., che è di so-<lb></lb> |
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9932 pra, resta .1., qual parti in .1., che era di sopra, ne ven .1. e .1. bastone di peso levará per se stesso ditto bastone.<lb></lb> |
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9933 Or torniamo al nostro proproposito. Noi trovamo che ’l ditto bastone da se stesso leva .5. bastoni. E co-<lb></lb> |
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9934 mo fo ditto el bastone pesa .$.12., donca li .5. pesamo .$.60. Adonca bisogna ora vedere quanto leva-<lb></lb> |
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9935 rá con l’ aiuto dele .$.2. Fa cosí. Sapi che l’ aiuto del bastone á questa natura: che, se ’l bastone è posto<lb></lb> |
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9936 in bilico in sula quarta parte del bastone, quella petra picola che tu metti per aiuto levará .3. picole petre o<lb></lb> |
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9937 voglian dire .3. aiuti e, se ’l bastone sirá messo in bilico in sula .5a. parte del bastone, l’ aiuto che li darai le-<lb></lb> |
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9938 vará .4. aiuti. E, se ’l bastone sirá in bilico in sul’ octava parte del bastone, l’ aiuto che li darai levará .7.<lb></lb> |
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9939 aiuti e sempre mai se ne cava .1. E se ’l bastone fosse in bilico diciamo in suli doi septimi del bastone,<lb></lb> |
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9940 dimando quanto l’ aiuto levará, cioé quella preta picola ditta. Fa cosí, commo di sopra festi siando el<lb></lb> |
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9941 bastone in bilico in sul’ octava parte, che cavasti .1. de .8., restó .7. e questo partisti in .1., che era sopra la ver-<lb></lb> |
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9942 ga, venne .7. Cosí hai ora a fare siando .2/7. cioé tra’ .2. de .7., resta .5. e questo parti in .2. che è sopra la<lb></lb> |
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9943 verga, ne ven .2.1/2. e .2 1/2. aiuti levará. Or torniamo al nostro proposito. Noi diciamo che ’l ditto basto-<lb></lb> |
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9944 ne è in bilico in sula dodecima parte. E peró tra’ .1. de .12., resta .11., parti in .1. che è sopra la verga, ne ven<lb></lb> |
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9945 .11. e .11. aiuti levará. E noi dicemmo che l’ aiuto era una picola preta che pesa .$.2. Adonca questi<lb></lb> |
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9946 .11. aiuti sonno .$.22. che, agionto ale .$.60. che levará el bastone per se stesso, fará .$.82. e .$.82. leva-<lb></lb> |
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9947 rá el ditto bastone con lo sopra ditto aiuto. Siché la sopra ditta gran preta pesava .$.82. con lo so-<lb></lb> |
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9948 pra ditto uncino. Fatta. Facsimiles et cetera. |
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9949 </p> |
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9950 <p class="main"> |
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9951 E gli é un bastone longo .12. palmi d’ una medesima grossessa e pesa .$.12. ed é tutto de-<lb></lb> |
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9952 ritto uniforme. Siché ciascuno palmo pesa .$.1. Ora io lo metto in equilibra in questa<lb></lb> |
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9953 forma: ch’ io lego una corda in sul segno del primo palmo e l’ altro capo dela corda ta-<lb></lb> |
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9954 cai al palgo e dipoi tolli un peso de petra con .1. uncino de ferro e apicolo ala ponta<lb></lb> |
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9955 del ditto bastonne, cioé ala vetta del primo palmo e trovo che gli sta paro in equilibra aponto commo sta<lb></lb> |
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9956 un peso in sula stadiera di mano. Che pesava quella preta con quello uncino de ferro. Fa cosí. Convien-<lb></lb> |
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9957 te considerare che ’l primo palmo se contrapesa con lo secondo e non vi resta se non .10. palmi de tracollo,<lb></lb> |
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9958 li quali .10. palmi pesano .$.10., peroché, pesando tutto el bastone .$.12., li .10. conven che pesi-<lb></lb> |
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9959 no .$.10. in proportione et cetera. Or ti convien cognoscere che ’l tracollo á questa natura: che leva de pe-<lb></lb> |
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9960 so la mitá dela sua multiplicatione. Adonca multiplica .$.10. per lo bastone tutto, fará .120.$., pigliane la .1/2., che<lb></lb> |
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9961 è .60. e parti per .1., ch’ é di lá dal bilico, viene .60. e tanto venne a pesare quella preta con quel ferro. Ed é<lb></lb> |
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9962 fatta, cioé che leverá .5. bastoni. E ancora poi dir cosí: che ’l bilico é in sul .12o. del bastone, commo dis-<lb></lb> |
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9963 si e levará .5. bastoni, cioé la sopraditta preta pesará .5. bastoni. E, se si ponesse in bilico in sul se-<lb></lb> |
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9964 sto del bastone, e levará doi bastoni. E, se ’l si mettesse in bilico in sul .4o. del bastone, levará un ba-<lb></lb> |
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9965 stone. E se ’l si mettesse in bilico in sul terzo del bastone, e levará mezzo bastone. E, se ’l si mettesse in<lb></lb> |
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9966 bilico in sul mezzo del bastone, non levará niente, perché pesará tanto di lá quanto di qua. E, se ’l si metesse<lb></lb> |
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9967 in bilico in sul .5o. del bastone, levará .1. bustone e mezzo e fie expedito ditto proposito et cetera.<lb></lb> |
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9968 <lb></lb> |
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9969 </p> |
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9970 <p class="head"> |
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9971 Modus fabricandi stateram ubilibet. |
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9972 </p> |
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9973 <p class="main"> |
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9974 <lb></lb> |
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9975 Fa el manico dela stadera longo corto a tuo modo, che sia uniforme de groseza, o ton-<lb></lb> |
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9976 do o quadro o intagliente. El romano d’ essa fallo ancora grande, picolo, commo a te pare e<lb></lb> |
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9977 l’ oncini con le catene d’ atacare el peso ancora falli grandi, picoli, a to modo. E gli ata-<lb></lb> |
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9978 catoi d’ essa, dal grosso e dal sotile, farali grandi, picoli, ancora o to modo e distanti da-<lb></lb> |
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9979 la testa e dal’ uno e l’ altro ancora a to modo, purché non sienno de rinpetto, che alora tanto seria l’ u-<lb></lb> |
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9980 no quanto l’ altro, ma con spatio. E sappi che, quanto piú discosti siranno ditti tacatoi dal’ uno al’ altro, magior<lb></lb> |
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9981 differenza faranno nel peso, cioé che quanto piú discosti, tanto magior peso levará la stadera da s’ uno lato che<lb></lb> |
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9982 dal’ altro. Or te bisogna provedere de far una libra de peso. Trova un saco o altra materia che pe-<lb></lb> |
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9983 si .1. libra, del campione paesano, secondo el paese che voli che la stadiera serva e quello poni in su l’ uncini longhi<lb></lb> |
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9984 de ditta stadiera e acosta el romano a quelli uncini di sopra, cioé ali atacatoi in qua e in lá, negotiando<lb></lb> |
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9985 tanto che stia in equilibra con lo peso e, commo l’ arai in bilico con esso, segna con una tacca el ditto ma-<lb></lb> |
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9986 nico e harai la tacca dela prima libra. E dapoi, per far la tacca dela .2a. libra, metti in suli ditti un-<lb></lb> |
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9987 cini grandi el peso de .2.$. e di novo poni el ditto romano in bilico con questo peso, quale trovato, sega l’ altra ta-<lb></lb> |
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9988 cca e sia la tacca dele .2. linee. E cosí farai per le .3.4.5. et cetera, finché decto manico s’ enpia de tacce. E vogliando<lb></lb> |
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9989 che l’ abia le mezze libre, overo once, giongnerai in suli l’ oncini longhi mezza libra, overo una<lb></lb> |
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9990 onci. E d’ altre parti de libra che tu volese che facesse e sia fatta bene. E, se piú presto<lb></lb> |
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9991 la vogli finire, senza giongnere peso all’ oncini, farai con le seste in questo modo, cioé, commo<lb></lb> |
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9992 tu hai la tacca de .2. libre, prendi con le seste el spacio dal’ una tacca al’ altra e con quello dividerai<lb></lb> |
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9993 ditto manico. E ciascuna de ditte tacche tirrá una libra di peso e segnarai le dicine e li centinara<lb></lb> |
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9994 et cetera. E, per havere le mezze libre, prendi con diligentia con lo sesto la mitá dal’ una tacca al’ altra e se-<lb></lb> |
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9995 <lb></lb> |
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9996 <pb></pb></p> |
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9997 <p class="folio"> |
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9998 folio 68r |
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9999 </p> |
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10000 <p class="main"> |
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10001 <lb></lb> |
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10002 </p> |
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10003 <p class="runhead"> |
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10004 Distintio octava 68 |
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10005 </p> |
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10006 <p class="main"> |
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10007 <lb></lb> |
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10008 gna con alquanta differentia dale prime tacche, che si cognosca le mezze dal’ intere et cetera, cioé l’ una fa<lb></lb> |
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10009 che tiri giú tutta e l’ altra non tanto. E, vogliendoli far l’ once, dividirai lo spacio dal’ una tacca al’ al-<lb></lb> |
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10010 tra dela libra in .12. parti equali, over lo spacio fra la libra e la mezza libra, dividi in .6., che ciascuna de<lb></lb> |
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10011 ditte parti sirá una oncia aponto, ma bisogna con gran diligentia se divida ditto manico. E questo ba-<lb></lb> |
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10012 sti quanto al lato piú legieri. Ora, per fare el lato grosso, cioé quello che porta piú peso e non grosso,<lb></lb> |
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10013 per havere magiori libre, excepto che ale volte in Vinegia con una medesima stadiera faranno un<lb></lb> |
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10014 lato per la sotile, che è .12. once e l’ altro per la grossa, ch’ é .18. once. Onde voltarai el ditto manico<lb></lb> |
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10015 e ataccarai l’ altro atacatoio e reggerate similmente commo di sopra. Ma, volendo far la libra de<lb></lb> |
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10016 .18. once, converebbete fare una pietra de .18. once del campione et cetera. 96.<lb></lb> |
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10017 E gli é un triangolo .abc. che lo lato .ab. è .13. e’l lato .ac.15. e’l lato .bc.14. e da un pon-<lb></lb> |
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10018 to intrinseco a lui, a ciascuno deli soi angoli, commo vedi in figura, meno .3. linee, cioé .db.dc.da.<lb></lb> |
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10019 La linea .db. è .9. e la linea .dc. è .10. Dimando quanto sia la linea .ad. Sappi che per aponerse<lb></lb> |
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10020 veresti in gran travagli de .R. e anche cubi, ma per forza de linee farai cosí. Prima tu ve-<lb></lb> |
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10021 di che tu hai doi triangoli certi, cioé el grande .abc., la cui basa ponemo .bc.15., e altro si é il trian<lb></lb> |
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10022 golo intrinseco a quello .dbc., sopra la medesima basa constituto. E peró, de ciascuno de questi, trova-<lb></lb> |
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10023 rai soi catetti sopra la medesima basa de .15. e vederai se tutti doi cascano in medesimo ponto de-<lb></lb> |
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10024 la basa o non. Se tutti doi cascassero in un luogo, la ragion seria facta perché .da. seria el resto del<lb></lb> |
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10025 catetto grande abattutone el catetto piccolo. Ma, se ditti catetti non caggiano in un luogo, alora<lb></lb> |
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10026 tirrai questa via. Trovarai la differenza che fa l’ un caso e l’ altro e trovarai che sirá .4/15., cioé el catetto pic-<lb></lb> |
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10027 colo caderá distante dal .c.8 2/15. e ’l catetto grande caderá distante pur dal .c.8 2/5. Cava l’ un del’ altro,<lb></lb> |
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10028 resta .4/15. e tanto sia fra l’ un caso e l’ altro. Ora dal ponto .d. tira la equedistante ala basa .bc. e harai<lb></lb> |
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10029 un triangolo rectangolo la cui potumissa sirá .ad. Oprando per la penultima del primo ha-<lb></lb> |
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10030 rai che sirá .Rv.159 81/225. men .R.16984 104/5625. .97. |
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10031 </p> |
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10032 <p class="main"> |
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10033 E gli é un alboro longo .R.61. fitto in terra che la perpendiculare dala cima, over vetta,<lb></lb> |
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10034 fine a terra è bracia .5. Vengo e alzo ditto alboro tanto che, per retta linea, dove stava prima la ci-<lb></lb> |
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10035 ma e dove sta ora, sonno bracia .4. Dimando quanto sia el pendicolo .2o. dala cima fin a ter-<lb></lb> |
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10036 ra et cetera. Fa cosí. Conviente considerare che questo vien essere un tondo el cui diametro é ’l doppio<lb></lb> |
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10037 de .R.61., cioé .R.244. e in esso è el quadrilatero che le doi menori facce ciascuna è bracia .4. e la basa de<lb></lb> |
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10038 ditto quadrangolo é doi volte .5., cioé .10. e la faccia di sopra, cioé la magior non so che sia. Diman-<lb></lb> |
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10039 dase che sia la .1/2. de ditta magior facia. Fa cosí. Multiplica .R.61. in sé, fa .61., cavane el quadrato de .5., resta<lb></lb> |
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10040 .36. e la .R.36. fo la linea .gn. El simile fo la linea .bp., adonca la linea .bt. fo .12. Ora poni che la linea .rc. fos-<lb></lb> |
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10041 se .1.co., multiplica in sé, fa .1.ce., cavalo del quadrato de .4., resta .16. men .1.ce. e .R.v.16. men .1.ce. fo la linea .rb.<lb></lb> |
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10042 Poi multiplica la linea .rb. via la linea .rt., cioé .12. men .Rv.16. men .1.ce. fará .Rv.2301. men .144.ce. men .16. men .1.<lb></lb> |
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10043 ce. per quantitá pura e serba. Poi multiplica .rc. via la linea .ra., che .rc. ponemo .1.co., peró la linea .ra. fo .10. piú .1.co.,<lb></lb> |
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10044 peró multiplicando .1.co.via.10. piú .1.co. fa .10.co. piú .1.ce. e questo é equale a quanto serbasti, cioé a .Rv.2304. men<lb></lb> |
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10045 .144.ce. men .16. men .1.ce. per quantitá pura. Aguaglia le parti daendo el debito e levando el mobile ch’ é el super-<lb></lb> |
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10046 fluo, cioé darai .16. men .1.ce. a ciascuna parte e harai .Rv.2304. men .144.ce. equale a .10.co. per .16. Multiplica<lb></lb> |
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10047 ciascuna extremo in sé harai .256. piú .100.ce. piú .320.co. equale a .2304. men .144.ce. Aguaglia e reca a .1.ce.<lb></lb> |
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10048 l’ ultimo equamento e smezza le cose, multiplica in sé, giogni el numero, la cosa varrá .R.8 3064/3721. piú .4 21/61. e <lb></lb> |
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10049 tanto fo <lb></lb> |
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10050 la<lb></lb> |
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10051 .1/2. de ditta gran faccia che vien a essere lo pendicolo del ditto alboro fin a terra. E la .R.8 3064/3721. men .40/61.<lb></lb> |
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10052 fo la linea .rc. e sopra vai a porre .5., per venire ala .1/2. dela faccia, e fará .R.8 3064/3721. piú .4 21/61. e fo la mi-<lb></lb> |
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10053 tá dela gran faccia. Facta. .98. |
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10054 </p> |
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10055 <p class="main"> |
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10056 Un albero alto .bracia.30., lego una fune ala vetta e tirola tanto che, si tirasse dritto, el pion-<lb></lb> |
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10057 bino cascaria apresso al pedale bracia .8. Or, per lo tirar ch’ io feci, non mossi piedi, anzi me ri-<lb></lb> |
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10058 colsi la fune in mano. Dimando quanta fune io recolsi in mano. Fa cosí. Tu vedi che<lb></lb> |
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10059 tu hai un triangolo ortogonio .abc. che ’l lato .ab. è .30. e lo .ad.50., che è potumissa, multiplicalo in sé, fa<lb></lb> |
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10060 .2500., cavane el quadrato delo .ab., che è .900., resta .1600. la cui .R. sia .bd., cioé .40. Ora, per esser .ab.<lb></lb> |
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10061 inclinato, e gli é pur .30. commo prima e causa un altro triangolo ortogonio con la perpendiculare .ac. e .bc.,<lb></lb> |
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10062 del quale .2. lati sonno noti, cioé .ab.bc., che è .8., multiplicalo in sé, fa .64., cavalo del quadrato de .ab., che è<lb></lb> |
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10063 .900., resta .836. la cui .R. sia el catetto .ac. E giá tu sai che lo .ed. è .32., perché .bd. è .40. Donca<lb></lb> |
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10064 quadra .32. e quadra .R.836. e giongni insiemi ditti quadrati, faranno .1860. la cui .R. sia .ad. del .2o.<lb></lb> |
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10065 triangolo inclinato la qual, tratta de .50., che è tutta la fune, resta .50. men .R.1860. per lo .de., cioé per la<lb></lb> |
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10066 parte recolta in mano per lo tirare et cetera. .99. |
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10067 </p> |
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10068 <p class="main"> |
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10069 Prendi questa per bella e degna. Un cerchio il cui diametro .ab. è .10. e la linea .ce. è<lb></lb> |
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10070 equidistante ala linea .fg. e la linea .fg. son lo quarto magior che la linea .ce. e lo .eg.<lb></lb> |
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10071 è .2. Dimando che sia .ce. Fa cosí. Recordate che le linee che se tagliano nel cer-<lb></lb> |
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10072 chio, per la .34a. del .3o. de Euclide, tanto fa .ad. in .db. quanto .cd. in .de. e tanto fa .ah. in .hb.<lb></lb> |
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10073 <lb></lb> |
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10074 <pb></pb></p> |
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10075 <p class="folio"> |
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10076 folio 68v |
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10077 </p> |
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10078 <p class="main"> |
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10079 <lb></lb> |
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10080 </p> |
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10081 <p class="runhead"> |
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10082 Distinctio octava. De Corporibus regularibus. |
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10083 </p> |
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10084 <p class="main"> |
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10085 <lb></lb> |
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10086 quanto .fh. in .hg. E peró poni che la linea .ce. sia .1.co., donca la mitá dutta in sé fa .1/4.ce., adonca .ad. per .db.<lb></lb> |
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10087 de’ fare .1/4. ce. E peró fa de .10.2. tal parti che, dutta l’ una in l’ altra, faccia .1/4.ce. e será la menor .5. men<lb></lb> |
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10088 .R.25. men .1/4.ce. e l’ altra, cioé .db. será .5. e .R.25. men .1/4.ce. Mo fa bisogno sapere quanto è .dh. in questo modo<lb></lb> |
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10089 Di quanto è una linea piú del’ altra, cioé lo quarto, adonca è la mitá de .1/8.co., lo qual adutto in sé, fa .1/64.ce., lo<lb></lb> |
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10090 qual batti de .eg. in sé, cioé de .4., resta .R.4. men .1/4.ce., cioé .dh. E peró se dirá che .ah. è .5. men .R.5. men .1/4.ce.<lb></lb> |
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10091 e piú .R.4. men .1/64.ce. E l’ altra .hb. fo adonca .5. e .R.25. men .R.4.1/64.ce., li quali multiplica l’ un per l’ altro, fa <lb></lb> |
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10092 .17/64.<lb></lb> |
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10093 ce. men .4. e .R.1/64.ce.ce. piú .400 men .5 9/16. de .ce. E questo è equale a .hg. in sé, cioé .5/8.co. in sé, fa .25/64.ce. <lb></lb> |
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10094 abatti ’censi<lb></lb> |
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10095 per .ce. e sirá .11/64.ce.ce. piú .400. men .9/16.ce. equal .1/8.ce. e .4. Mo du’ ciascuna parte in sé e será .1/64.ce.ce. 400. <lb></lb> |
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10096 men .5 9/16.ce.,<lb></lb> |
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10097 equal .16 1/3 1/64.ce.ce. Mo abatti ’censi de’ censi per censi de’ censi. E restará numero per numero anche .384., gual<lb></lb> |
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10098 .6 9/16.ce. Mo parti lo numero per li censi, che ne ven .58 18/39. e la .R. di questo fo la cosa, cioé la notitia dela<lb></lb> |
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10099 linea .ce. Et tu sie in ceteris et cetera. 100. |
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10100 </p> |
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10101 <p class="main"> |
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10102 E gli é una mezza spera .ab. il cui diametro è .12. Io vi voglio mettere il magior cu-<lb></lb> |
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10103 bo vi capa. Dimando quanto sia per faccia. Fa cosí. Prima farai la spera sana e in quel-<lb></lb> |
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10104 la loca un solido, doi sia longo co largo, che virrá essere .2. cubi uno sopra l’ altro. E<lb></lb> |
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10105 poni che sia largo .1.co. Donca longo sirá .2.co. Ora trova l’ axis over diametro di questo solido qual<lb></lb> |
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10106 sirá ancora diametro de ditta spera et cetera, haverai la valuta dela cosa. E tanto sirá per faccia ditto cubo et cetera.<lb></lb> |
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10107 <lb></lb> |
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10108 </p> |
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10109 <p class="head"> |
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10110 Particularis tractatus circa corpora regularia et ordinaria incipit. |
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10111 </p> |
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10112 <p class="main"> |
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10113 <lb></lb> |
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10114 E, benché di sopra, in questo nella distition .6a. al capitolo .4o., dela mesura dela spera suc-<lb></lb> |
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10115 cintamente fosse ditto abastanza, nondimeno me par qui, excelso Duca, particularmente<lb></lb> |
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10116 dire de alquanti corpi essentiali, in ditta spera locabili, deli quali, un angolo toccando, subi-<lb></lb> |
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10117 to tutti toccano. E principalmente lo so per la notitia deli .5. regulari di quali Euclide a-<lb></lb> |
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10118 pieno, nelli ultimi soi libri, scientificamente tratta. Di che me pare non inutile aponere de loro cer-<lb></lb> |
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10119 ti casi ació li pratici vulgari ancora essi qualche dolcezza di loro dimensioni sentino. Di questi fra’<lb></lb> |
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10120 phylosofi si fa gran discussioni. E maxime se ben el Thymeo del divin philosofo Platone (secondo lo Au-<lb></lb> |
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10121 relio doctor sancto Augustin) con diligentia s’ atende. Dove de universi natura diffusamente<lb></lb> |
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10122 parlando spesso a suo proposito li induci. Attribuendo lor forme separatamente ali .5. corpi sem-<lb></lb> |
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10123 plici, cioé Terra.Aqua.Aeri.Fuoco e Cielo. Sí commo apieno di sopra in questo, nella par-<lb></lb> |
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10124 te de Arithmetica, in la distinction prima, nel secondo suo trattato, al terzo articolo, di loro<lb></lb> |
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10125 parlammo. Questi son quelli, Magnanimo Duca, di quali le forme materiali (con assai adornez-<lb></lb> |
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10126 ze, nelle proprie mani di. U. D. S., nel sublime palazzo del Reverendissimo cardinale nostro,<lb></lb> |
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10127 protectore Monsegnor de San Piero in vincula, quando quella venne ala visitatione del summo pon-<lb></lb> |
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10128 tifice Innocentio Octavo, negli anni dela salute nostra 1489 del mese de aprile, che giá sonno .5.<lb></lb> |
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10129 anni elapsi. E, insiemi con quelli, vi foron molti altri da’ ditti regulari dependenti. Quali fabri-<lb></lb> |
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10130 cai per lo Reverendo monsegnor meser Pietro de Valetarij de Genoa, dignissimo vescovo de<lb></lb> |
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10131 Carpentras, al cui obsequio alora foi deputato in casa dela felicissima memoria del Reverendisimo Car-<lb></lb> |
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10132 dinale de Fois, nel palazzo ursino, in Campo de Fiore. Siché di questi le sequenti petitioni inqui-<lb></lb> |
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10133 riranno el lor modo operentivo abastanza pratica insegnaremo, commo apresso se intenderá.<lb></lb> |
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10134 E prima del primo regulare, ditto tetracedon, cioé .4. base triangolari, la cui figura Platone al fuoco<lb></lb> |
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10135 attribuí, contra la cui opinione AR, in quel de celo et mundo prese ardire et cetera. E diró cosí.<lb></lb> |
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10136 E gli é un .4. base triangolari equilatero .abcd. e ’l centro suo éne .e. e dal’ angolo .a. alo .e. éne .4.<lb></lb> |
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10137 ch’ é l’ axis. Dimando, del lato .ab., che è equale agli altri lati. Opra ut scis, harai .ab. esser .R.24. .3. |
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10138 </p> |
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10139 <p class="main"> |
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10140 E gli é un .4. base triangolari equilatero .abcd. e ciascun di soi lati é .R.24. e ’l suo axis<lb></lb> |
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10141 .4. Dimando quanto sirá quadrato. Trova prima el diametro d’ una dele base, che è .R.24.<lb></lb> |
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10142 per lato. Dividi per equali .R.24., ne ven. R.6., multiplica in sé, fa .6., cavalo de .24., resta .18. e<lb></lb> |
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10143 .R.18. sia diametro d’ una dele base, cioé .bg., diametro dela basa .bcd. Donca, multiplicando<lb></lb> |
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10144 .R.6., ch’ é la mitá d’ un lato dela basa, via .R.18., ch’ é lo diametro, fa .R.108. e tanto sia la superficie<lb></lb> |
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10145 dela basa, la qual se vole multiplicare con l’ axis, che è .R.16., fa .R.1728., el qual se vole partire per .3. E perché<lb></lb> |
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10146 .1728. è .R., reca .3. a .R., fa .9., parti .1728. per .9., ne ven .192. e .R.192. sirá quadrato. 4.<lb></lb> |
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10147 E gli é una figura corporea che á base .4. triangolare, de angoli e lati equali, che ciascun lato<lb></lb> |
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10148 è .R.24. e ’l suo axis è .4. Dimando quanto é da ciascuno angolo al centro. Tu hai la fi-<lb></lb> |
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10149 gura corporea de .4. base triangolari equilatera .abcd. e l’ axis suo é .ae. e ’l centro suo<lb></lb> |
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10150 è nell’ axis in ponto .f. E perché .ae. è .4. e .af. è sexquitertia alo .ae., adonca .af. é .3. Ala<lb></lb> |
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10151 prova, se e ss’ é ditto che un lato è .R.24., peró piglia la .1/2., ch’ é .R.6. e trallo de .24., resta .R.18. e .bh.<lb></lb> |
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10152 è l’ axis el qual cade in suli .2/3. del diametro .bh., ch’ é .R.18. e li .2/3. sonno .R.8. che, in sé multiplicato, fa .8. E ss’ é<lb></lb> |
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10153 ditto che .af. è .3., donca .fe. é .1., perché .ae. è .4., tranne .af. ch’ é .3., resta .1. per lo .fe. che, in sé multiplicato, fa .1.,<lb></lb> |
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10154 gionto con .8., fa .9. Adonca .bf.af.cf.df. é ciascun .R.9., cioé .3., commo volemo nel tema aponto facta. .5.<lb></lb> |
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10155 <lb></lb> |
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10156 <pb></pb></p> |
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10157 <p class="folio"> |
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10158 folio 69r |
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10159 </p> |
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10160 <p class="main"> |
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10161 <lb></lb> |
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10162 </p> |
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10163 <p class="runhead"> |
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10164 Distinctio octava. De Corporibus regularibus. 69 |
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10165 </p> |
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10166 <p class="main"> |
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10167 <lb></lb> |
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10168 E gli é un .4. base triangolari proposto del quale el suo axis è .4. Voglioci mettere dentro<lb></lb> |
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10169 el magior corpo sperico che ci capa. Dimando che sia suo diametro. E ss’é ditto che<lb></lb> |
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10170 del .4. base triangolari .abcd., che ’l suo axis è .4. e che il centro è nell’ axis in ponto .f. e<lb></lb> |
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10171 anche .af. è .3. e anche .fc. è .1., donca, ponendo un pie’ del sexto sul ponto .f. e con l’ altro<lb></lb> |
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10172 pie’ girare contingente, el contingerá tutte le altre facce, perché .af. è equale al .bf.cf. E cosí .fe. è equa-<lb></lb> |
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10173 le alo .fk.fl.fm. Onde per questo .fe. che è .1. ed é semidiametro del corpo sperico cadente in quella figura,<lb></lb> |
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10174 e tutto el diametro sia .2. braccia. Facta. .6. |
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10175 </p> |
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10176 <p class="main"> |
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10177 E gli é un cubo .abed.efgh. che, per ciascun lato, è .4.bracia. Dimando quanti braci sirá quadro. Multiplica<lb></lb> |
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10178 .4. in sé, fa .16. e poi .16. via .4., fa .64. é tanto e quadro. Facta. .7. |
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10179 </p> |
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10180 <p class="main"> |
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10181 E gli é un cubo ch’ é .4.bracia. per lato, cioé .abcd.efgh. Dimando dela sua superfcial linea diago-<lb></lb> |
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10182 nale. Opera, trovarai che sia .ad.R.32. .8. |
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10183 </p> |
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10184 <p class="main"> |
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10185 E gli é un cubo ch’ é per faccia .4. Dimando che sia la diagonale interiore passante per lo centro .k.,<lb></lb> |
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10186 cioé l’ axis. Opera e troverai che sirá la diagonale .ah.R.48. Facta. .9. |
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10187 </p> |
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10188 <p class="main"> |
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10189 E gli é un cubo ch’ é per faccia .4. Dimando che sia quadro la sua piramide e quanto sirá le<lb></lb> |
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10190 sue potumisse. Vedi prima quanto e gli é quadrato ditto cubo, che hai che è .64. Adonca<lb></lb> |
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10191 pigliane el .1/3. de .64., ch’ é .21 1/3., perché s’ é ditto ogni pirramide essere .1/3. de tutto el corpo,<lb></lb> |
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10192 essendo sopra la medesima basa, siché sia quadrata .21 1/3. E le sue potumisse trovarai<lb></lb> |
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10193 cosí, dividendo la linea .ad. in doi parti equali, ch’ é .R.32., che l’ una sirá .R.8. che multiplicata in sé, fa .8. Da poi<lb></lb> |
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10194 multiplica l’ altezza del cubo in sé, ch’ é .4., fa .16., giongni a .8., fa .24. e .R.24. sia ciascuna dele sue potu-<lb></lb> |
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10195 misse del ditto cono o voi dir pirramide. Facta. 10. |
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10196 </p> |
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10197 <p class="main"> |
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10198 E gli é un cubo ch’ é per faccia .4. Vi voglio metter dentro el magior triangolo corpo-<lb></lb> |
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10199 reo equilatero, cioé .4. base, che io possa. Dimando che sia per ciascun lato. Tu hai<lb></lb> |
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10200 el cubo dato .abcd.efgh. Tira una linea .ad. e .af. e .df.e. taglia via l’ angolo .b. Poi<lb></lb> |
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10201 linea .ag. e .dg., levando via l’ angolo .c., remarrá .adf. e .adg. che sonno doi lati del triangolo<lb></lb> |
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10202 domandato. Poi linea .fg. e .af. e .ag. e .df. e .dg., levando via l’ angolo .e. e l’ angolo .h. e remar-<lb></lb> |
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10203 rá .afg. e .dfg., doi altri lati dela figura .4. base triangolari. E, perché .ab. è .4. e .bd.4. e .ad. pó quan-<lb></lb> |
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10204 to tutte doi, perché ella è opposita al’ angolo recto contenuto da quelle, .ab. é .4., multiplicato in sé, fa .16. e<lb></lb> |
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10205 .bd. è .4., che fa pur .16., gionti insiemi fanno .32. e .R.32. è .ad., ch’ é un lato de ditta figura .4. base trian-<lb></lb> |
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10206 golari. Peró dirai del magiore triangolo corporeo che capa nel cubo .abcd.efgh. éne per<lb></lb> |
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10207 lato .R.32. Facta. .11. |
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10208 </p> |
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10209 <p class="main"> |
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10210 E gli é un corpo sperico che ’l diametro suo è .7. Dimando quanto sirá la sua superfice. Éc-<lb></lb> |
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10211 ci molti modi a saperlo. Primo é che tu multiplichi lo diametro suo, ch’ é .7., via la circunferentia,<lb></lb> |
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10212 ch’ é .22., fa .154. e tanto sia la sua superficie. E Archimede dici che ogni superficie de spera è .4.<lb></lb> |
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10213 tanto che la superficie del magiore cerchio di quella propria spera, commo in figura, di questo por-<lb></lb> |
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10214 remo sua demostratione: che è il suo diametro, che s’ é ditto che è .7., che la superficie sua è .38 1/2. che, multiplica-<lb></lb> |
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10215 to per .4., fa .154. aponto, commo di sopra. Siché dirai che la superficie dela spera, che ’l suo diame-<lb></lb> |
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10216 tro è .7. sie .154. Facta. .12. |
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10217 </p> |
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10218 <p class="main"> |
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10219 E gli é un corpo sperico che ’l suo diametro è .7. Dimando che sia quadrato tutto el corpo.<lb></lb> |
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10220 Tu hai che la superficie è .154. e ’l suo diametro è .7. Multiplica la superficie sua via la .1/2. del diametro,<lb></lb> |
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10221 over la .1/2. dela superficie via tutto el diametro, che ognuna fa .539. Del quale piglia el .1/3., che è<lb></lb> |
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10222 .179 2/3. Tanto sia quadro ditto corpo et cetera. Posse ancora fare per altra via, perché tu die sapere<lb></lb> |
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10223 che ogni cubo contene in sé un corpo sperico che éne li .11/21. del ditto cubo. Cioé, se ’l cubo fosse .7. per<lb></lb> |
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10224 faccia, possede in tutto .343., de questo se ne vol pigliare li .11/21., che ne ven .179 2/3., commo prima, over de .343. se<lb></lb> |
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10225 ne vol bugliare li .10/21., che remarrá .179 2/3. E tanto sirá el tenuto dela spera. La ragion perché se ne<lb></lb> |
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10226 getti li .10/21. si é che, stu hai un dado .7. per faccia e tu ne voglia fare una pallotta, tu lo vien a scan-<lb></lb> |
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10227 tonare per modo che gli é provato, che quello che se ne getta éne li .10/11. de tutto quello che prima era el cu-<lb></lb> |
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10228 bo e quello che remane vene a essere li .11/21. de tutto el ditto dado et cetera. Siché basti. .13. |
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10229 </p> |
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10230 <p class="main"> |
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10231 E gli é una palla el cui diametro è .7. Dimando che sia el lato del .4. base triangolari equi-<lb></lb> |
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10232 latero ch’ entro aponto vi capiste. Dividi prima el diametro suo, ch’ é .7., in .3. parti equa-<lb></lb> |
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10233 li, cioé prendi el .1/3., e quel multiplica via li .2/3., ch’ é .4 2/3., che ciascuna sia .2 1/3. Multiplica via l’ altra, cioé<lb></lb> |
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10234 .2 1/3. via .4 2/3., fa .10 8/9. e questo adoppia como .R., fa .R.43 5/9. dela quale fa .4. parti equali, ne ven<lb></lb> |
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10235 .R.2 32/72., trallo de .R.43 5/9., resta .R.24 1/2. che è diametro dela basa del corpo triangolare, al qual<lb></lb> |
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10236 giognici .1/3. de .24 1/2., la possanza del diametro, ch’ é .8 1/6., fa .32 2/3. e la .R. de .32 2/3. sirá per lato. Facta.<lb></lb> |
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10237 14. |
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10238 </p> |
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10239 <p class="main"> |
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10240 E gli é una palla el cui diametro è .7. Dimando che sia per facia el magior cubo che en-<lb></lb> |
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10241 tro vi capisse. Habian ditto nelli cubi che ’l diametro che se parte dali angoli e pas-<lb></lb> |
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10242 sa per lo centro pó quanto tre volte el lato suo. Adonca la possanza del suo diametro, divisa<lb></lb> |
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10243 per .3., e quello che ne ven sirá la possanza del suo lato. E, perché il cubo continge con li soi an-<lb></lb> |
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10244 goli la spera, donca é ’l diametro suo equale a quello dela spera, donca l’ uno e l’ altro sia .7., la cui possanza è<lb></lb> |
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10245 <lb></lb> |
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10246 <pb></pb></p> |
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10247 <p class="folio"> |
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10248 folio 69v |
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10249 </p> |
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10250 <p class="main"> |
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10251 <lb></lb> |
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10252 </p> |
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10253 <p class="runhead"> |
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10254 Distinctio octava. De Corporibus regularibus. |
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10255 </p> |
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10256 <p class="main"> |
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10257 <lb></lb> |
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10258 .49., qual dividi in .3. parti, ciascuna sia .16 1/3. e .R.16 1/3. sirá per lato.<lb></lb> |
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10259 E gli é un corpo sperico il cui diametro è .6. Voglioci mettere dentro un corpo de .8. ba-<lb></lb> |
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10260 se, .4. triangolari e .4. exagone, equilatere l’ una e l’ altra. Dimando che sia per lato. Tu<lb></lb> |
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10261 hai, per la .5a. di corpi .4. base triangolari, el quale è .abcd., che dal centro a ciascun suo<lb></lb> |
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10262 angolo è .3. E, perché ciascun suo angolo continge la circunferentia dela spera, sirá .3. semi-<lb></lb> |
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10263 diametro e tutto el diametro dela spera è .6., che contiene quel tal corpo .4. base triangolari. E l’ axis è<lb></lb> |
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10264 .4., che éne .ac. e ciascun suo lato è .R.24. E, perché questo corpo áne .4. base, volendolo redure a .8.,<lb></lb> |
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10265 è necessario tagliare li soi .4. angoli. E che remanga le facce equali e peró fa de’ lati soi .3. parti equali,<lb></lb> |
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10266 cioé dividi .ab. ch’ é .R.24., in .3., ne vene .R.2 2/3. e .R.2 2/3. sirá ciascuna parte, cioé .al.lm.mb. Ora fa<lb></lb> |
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10267 .3. parti del’ axis .ac., ch’ é .4., che sia .ag.gh.he., che sia ciascuna .1 1/3. Dividi .be., che è .2/3. del diametro,<lb></lb> |
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10268 che è .R.8., in tre ponti: .1. in ponto .i. et .k., che sirá ciascuno .R.8/9. e .ie., ch’ é .2/3., é equale .ad.mh. e .2/3. de .R.8.<lb></lb> |
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10269 ónno .R.3 5/9. Donca .mh. è .R.3 5/9. E tu hai che .ef. è diametro ed ái che .ef. è .1., trallo de .1 1/3., resta .fh.<lb></lb> |
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10270 .1/3. Tira la linea .mf., la qual pó quanto .fh. e .mh., perché .mh. è .R.3 5/9. che, multiplicato in sé, fa .3 5/9. e .fh. è .1/3., <lb></lb> |
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10271 che<lb></lb> |
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10272 quadrato, fa .1/9. Giognilo con .3 5/9. fa .3 2/3., donca .mf. è .R.3 2/3. Peró dirai cosí: se .R.3 2/3., che è semidiametro, <lb></lb> |
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10273 me <lb></lb> |
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10274 da .R.2 2/3.,<lb></lb> |
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10275 che è un lato, che me dará .R.9., ch’ é semidiametro dela spera dove habian a collocarn el corpo .8. base. Multiplica<lb></lb> |
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10276 e parti secondo .R. Trovarai che te dará .R.6 8/11. e tanto sirá per lato l’ otto base che .4. sienno<lb></lb> |
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10277 exagone e .4. triangole. Facta. .16. |
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10278 </p> |
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10279 <p class="main"> |
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10280 E gli é una spera il cui diametro è .6. e contiene un corpo .8. base, cioé .4. exagone e .4.<lb></lb> |
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10281 triangole equilatere e ciascuno suo lato éne .R.6 6/11. Domando quanto sirá quadrato. Tu<lb></lb> |
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10282 dei prima trovare el lato del triangolo donde nasce questo corpo, che ’l suo lato éne .1/3. del la-<lb></lb> |
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10283 to del triangolo. Donca el lato del triangolo donde è cavato ditto corpo è tre volte<lb></lb> |
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10284 .R.6 6/11., peró multiplica .3. via .R.6 6/11., fa .R.58 10/11. e tanto sia el lato del triangolo. E l’ axis éne sexquiter-<lb></lb> |
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10285 tia al suo lato, siché parti .58 10/11. per .3., ne ven commo numero .19 17/33. Cavalo de .58 10/11., resta .39 3/11. e <lb></lb> |
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10286 .R.39 <lb></lb> |
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10287 3/11.<lb></lb> |
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10288 éne el diametro dela sua basa sexquitertia a .58 10/11., che sia .R.44 2/11. El qual, multiplicato con la mitá de-<lb></lb> |
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10289 la basa, che è .R.14 8/11., fa .R.632 60/121., il qual multiplica con l’ axis, che è .R.39 3/11., fa .R.19839 85/121. Parti per <lb></lb> |
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10290 la<lb></lb> |
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10291 .R.9., ne ven .R.2204 448/1089. E tanto è quadro el triangolo. Del qual se de’ cavare .4. ponte: che ciascun è triango-<lb></lb> |
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10292 lo corporeo equilatero e ciascuno lato è .1/3. dela .R.58 10/11., ch’ é .R.6 6/11., del quale trova el diametro, cioé<lb></lb> |
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10293 tranne el .1/4., che è .1 7/11. de .6 6/11., resta .4 10/11. e .R.4 10/11. è el diametro. El qual, multiplicato con la .1/2. de <lb></lb> |
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10294 .R.6 <lb></lb> |
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10295 6/11.R.8 4/121.<lb></lb> |
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10296 el qual multiplica con l’ axis che é sexquitertia .a.6 6/11., ch’ é .R.4 4/11., fa .R. 35 871/14641., el qual parti per .R.9., ne <lb></lb> |
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10297 vene <lb></lb> |
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10298 .R.<lb></lb> |
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10299 .3 13111/14641. e tanto è quadrata una dele ditte ponte, che sonno .4., reca a .R., fa .16., multiplica .16. via .3 <lb></lb> |
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10300 13111/14641. fa<lb></lb> |
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10301 .R.62 4802/14641. e tante sonno quadre tutte le ponte, trallo de .R.2204 448/1089., restará .R.2204 448/1089. men .R. <lb></lb> |
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10302 62 <lb></lb> |
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10303 4802/14641.<lb></lb> |
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10304 e tanto è quadro el ditto corpo .8. base, cioé .4. exagone e .4. triangole, che è contenuto dala<lb></lb> |
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10305 spera il cui diametro è .6. Facta et cetera. .17. |
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10306 </p> |
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10307 <p class="main"> |
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10308 E gli é una spera il cui diametro è .6. Voglioci mettern dentro un corpo de .14. base: cioé .6.<lb></lb> |
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10309 quadrate e .8. triangolari de equali lati. Dimando che sia ciascuno de’ ditti lati. Questa<lb></lb> |
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10310 tal forma si cava del cubo, perché á .6. base e .8. cantoni che, tagliando li soi .8. canti, fa<lb></lb> |
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10311 .14. base. Cioé cosí tu hai el cubo .abcd.efgh., piú volte ’nanze posto; dividi ciascun lato per equali:<lb></lb> |
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10312 .ab. in ponto .i.cd. in ponto .l.bd. in ponto .k.ac. in ponto .m.ag. in ponto .n.gh. in ponto .o.hb. in ponto<lb></lb> |
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10313 .p.hf. in ponto .q.fd.in ponto .r.fe. in ponto .s.ec. in ponto .t.eg. in ponto .v. Tira una linea da .t. al .p., la qual pó quan-<lb></lb> |
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10314 to le doi linee .nt. e .np., perché .n. fane angolo retto opposto ala linea .tp. e l’ angolo .p. e l’ angolo .t. toca-<lb></lb> |
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10315 no la circunferentia dela spera. E cosí fanno li altri .mk.il.oq.ns.vr. Adonca .tp. è diametro dela<lb></lb> |
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10316 spera, ch’ é .6. e la possanza sua è .36., che è quanto pó le doi linee .np. e .nt., donca é ciascuna .R.18. Se .np. è<lb></lb> |
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10317 .R.18., la qual pó per le doi linee .ni. e .ip., donca è ciascuna .R.9. e la .R.9. è .3., peró .ip. é .3., che è un lato<lb></lb> |
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10318 del corpo .14. base sopra ditto e .3. anche sirá ciascuno degli altri lati, quando el diametro dela<lb></lb> |
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10319 spera si é .6. et cetera. Le linee tutte non se tirano, commo altre volte s’ é ditto, per la loro confusione in-<lb></lb> |
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10320 trinseca neli corpi, ma tu, per ymaginatione, bisogna supplesca et cetera. .18. |
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10321 </p> |
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10322 <p class="main"> |
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10323 E gli é una spera che contiene un corpo .14. base: cioé .8. triangolari. e .6. quadrate, che ciascun suo la-<lb></lb> |
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10324 to è .3. Dimando quanto sirá quadro. Tu hai di sopra che tal figura se cava del cubo ed ái<lb></lb> |
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10325 che è per lato .R.18. che, in sé multiplicato, fa .18. Poi multiplica in sé .18., fa .324. per quadrarlo. E que-<lb></lb> |
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10326 sto multiplica via l’ altezza del cubo, ch’ é .R.18., fa .5832. e .R.5832. é quadro el ditto<lb></lb> |
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10327 cubo. Dela quadratura hai a cavare .8. triangoli solidi: cioé le .8. ponti del cubo. Quadra prima la<lb></lb> |
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10328 basa de un de questi triangoli, ch’ é .3. per lato, dimezza .3., ch’ é un lato, ne ven .1 1/2. Multiplica in sé .2 1/4. trallo de-<lb></lb> |
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10329 la passanza de .3., cioé de .9., restará .6 3/4. e tanto sia el diametro dela basa, cioé .R.6 3/4. Multiplicalo via<lb></lb> |
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10330 .R.2 1/4., fa .R.15 3/16. e questo se die multiplicare via l’ axis che si vol trovare cosí. Tu hai che lo diametro è<lb></lb> |
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10331 R.6., dividilo in .3., ne ven .R.27/16., el quale radoppia, fa .R.3., il suo quadrato trallo de .9., resta .R.6. il qual<lb></lb> |
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10332 multiplica via .R.15 3/16., fa .R.91 1/8., partilo per .3., ne ven .R.10 1/8., multiplicalo per .R.64., fa R.648. e .R.648. sonno <lb></lb> |
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10333 qua-<lb></lb> |
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10334 <lb></lb> |
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10335 <pb></pb></p> |
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10336 <p class="folio"> |
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10337 folio 70r |
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10338 </p> |
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10339 <p class="main"> |
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10340 <lb></lb> |
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10341 </p> |
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10342 <p class="runhead"> |
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10343 De Corporibus regularibus. 70 |
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10344 </p> |
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10345 <p class="main"> |
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10346 <lb></lb> |
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10347 drati li .8. triangoli solidi, cioé le .8. ponti del cubo che, tratta questa quadratura dela .R.5832., el ri-<lb></lb> |
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10348 manente sia la possessione corporea del ditto .14. base cosí constituto, cioé sia quadro ditto corpo<lb></lb> |
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10349 R.5932. men .R.648. El qual corpo è per ciascun lato basale .3.bracia. contenuto dala spera che ’l diametro<lb></lb> |
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10350 suo sia .6. E sappi che gli é bisogno dire contenuto dala spera in simil domande, cioé vol dire che,<lb></lb> |
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10351 tocando un angolo, tocan tutti, altramente serebe difficultá expedire le questioni et cetera. .19. |
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10352 </p> |
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10353 <p class="main"> |
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10354 E e gli é una botte che li soi fondi per diametro ciascuno è .2.bracia. e al cocone .2 1/4. e tra li<lb></lb> |
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10355 doi fondi e ’l cochiume è .2 2/9. e longa bracia .2. Dimando quanto será quadra. Questa è de spe-<lb></lb> |
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10356 tie de piramide tagliata, peró fa cosí. Multiplica el fondo in sé, che fa .4., poi multiplica .2 2/9. in sé, fa<lb></lb> |
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10357 .4 76/81., ch’ é infra el cochiume e ‘l fondo; giongni insiemi, fa .8 76/81. Poi multiplica .2. via .2 2/9., fa .4 4/9.,<lb></lb> |
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10358 giongnilo con .8 76/81., fa .13 31/81., parti in .3., ne ven .4 112/243., cioé .R.4 112/243. che, in se multiplicato, fa .4 <lb></lb> |
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10359 112/243. e questo tieni<lb></lb> |
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10360 a mente.Tu hai che, multiplicato in sé, .2 2/9. fa .4 76/81. Ora multiplica .2 1/4. in sé, fa .5 1/16. giongni fa .10. 1/1296. <lb></lb> |
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10361 e <lb></lb> |
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10362 multiplica .2 2/9.<lb></lb> |
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10363 va.2 1/4., fa .5., giogni. insiemi, fa .15 1/1296., parti per .3., ne ven .R.5. 1/3888., quadralo, fa .5 1/3888., giongnilo con <lb></lb> |
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10364 quello de sopra<lb></lb> |
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10365 ch’ é .4 112/243., fará 9 1792/3888., el qual multiplica per .11. e parti per .14., cioé toglie li .11/14., ne verrá .7 <lb></lb> |
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10366 23600/54432. e tanto sia quadra-<lb></lb> |
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10367 ta la ditta botte. Questo modo se pó tenere quando tutte le mesure sonno equestistanti una al’ altra e sta-<lb></lb> |
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10368 rá bene. Ma, quando non fossero equidistanti, tieni quest’ altro modo che vale a tutti: cioé mettamo che li<lb></lb> |
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10369 fondi dela botte sienno de diametro .8.bracia. e al cochiume sienno .10. e .2.bracia. apresso ali fondi sia .9. e<lb></lb> |
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10370 sia la botte longa .10. El primo fondo sia el suo diametro .af. e l’ altro diametro apresso sia .bg. e quello<lb></lb> |
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10371 del cochiume sia .ch. e ’l terzo sia .di. e ’l fondo derieto sia .ek. Ora è da multiplicare prima quella del cochiu-<lb></lb> |
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10372 me .ch., ch’ é .10., in sé, .fa. 100. Poi multiplica .bg., ch’ é .9., in sé, fa .81.; giongi insiemi fa .181. Ora multiplica .ch. con <lb></lb> |
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10373 .bg.,<lb></lb> |
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10374 fa .90., giongnilo con .181., fa .271., el qual parti per .3., ne ven .90 1/3. e di questo togli li .11/14. che sonno .70 41/42., <lb></lb> |
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10375 el <lb></lb> |
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10376 qual<lb></lb> |
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10377 multiplica per .6., che è dala linea .bg. ala linea .di., fa .428 31/42. e questo serba. Tu hai multiplicato .9., che fa .81., ora <lb></lb> |
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10378 multiplica<lb></lb> |
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10379 el fondo .af., ch’ é .8., fa .64., giongni insiemi fa .145. Multiplica .8. via .9., fa .72., giongni insiemi, fa .217., partilo per<lb></lb> |
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10380 3, ne ven .72 1/3. il qual multiplica per .11. e parti per .14., cioé pigliane li .11/14., ne ven .56 35/42. el qual multiplica <lb></lb> |
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10381 per .4. <lb></lb> |
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10382 perché dala<lb></lb> |
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10383 linea .af. ala linea .bg. è .2. E dala linea .di. ala linea .ek. è .2., siché fa .4. via .56 35/42., fa .227 1/3.; gion-<lb></lb> |
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10384 gnilo con .428 31/42., fará .656 1/14. Tanto sia quadrata la ditta botte, cioé bracia .656 1/14. et cetera. Facta. <lb></lb> |
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10385 20<lb></lb> |
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10386 E perché ale volte acade a mesurare corpi inregolari quelli non se possano mesurare<lb></lb> |
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10387 per linee, commo statue de marmo o de metallo, cioé figure de animali regoli o irregolli dico che<lb></lb> |
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10388 tenga questo modo a quadrarli apresso quello che di sopra sucintamente dissi in la domanda .43. |
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10389 </p> |
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10390 <p class="main"> |
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10391 Verbi gratia. Metiamo che tu voglia sapere quanto è quadrata una statua de homo nuda,<lb></lb> |
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10392 che sia .3.bracia. de longhezza proportionata. Dico che tu faccia un vaso de legno, longo bracia .3 1/4. e largo<lb></lb> |
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10393 .1 1/2. e alto .1., el qual sia quadro, cioé con angoli retti e sia ben stagno, siché l’aqua non d’ esca ponto. Poi lo metti<lb></lb> |
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10394 ben piano a livello e metti dentro tant’ aqua che agionga a un terzo al’ orlo di sopra. Poi fa un segno<lb></lb> |
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10395 dove agiongni l’ aqua e mettivi dentro la statua che tu voi mesurare e vedi quanto è cresciuta l’ aqua e<lb></lb> |
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10396 fa un altro segno a sommo l’ aqua ritto a quello di prima. Poi mesura dal primo segno. al .2o. e vedi quanto e gli é. Met-<lb></lb> |
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10397 tiamo che sia .1/4. Ora multiplica la longhezza del vaso, ch’ é .3 1/4., con la larghezza, ch’ é .1 1/2., fará .4 7/6., el qual <lb></lb> |
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10398 multiplica<lb></lb> |
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10399 per .1/4. che creve l’ aqua, fa .1 7/32. e tanto è quadrata ditta statua. E cosí observa a mesurare tali corpi se fosse<lb></lb> |
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10400 ben un par de buoi con un carro de fieno et cetera. .21. |
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10401 </p> |
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10402 <p class="main"> |
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10403 E gli é un corpo sperico il cui axis è .R.48. e contiene in sé un corpo de .12. base pentago-<lb></lb> |
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10404 nali equilatero. Dimando de’ suoi lati. Tu die sapere che il lato del cubo descripto in .1a.<lb></lb> |
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10405 medesima spera, diviso secondo la proportione avente el mezzo e ’doi extremi, che la magior<lb></lb> |
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10406 sue parte éne il lato del corpo de .12. base pentagonali in la medesima spera descripto, com-<lb></lb> |
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10407 mo per la .18a. del .14o. libro de Euclide si prova. E anco hai, per la .13a. del .13o., che la possanza del diametro<lb></lb> |
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10408 dela spera è tripla ala possanza del lato di quel cubo da quella contenuto. Adonca dividi .48. per .3., ne ven<lb></lb> |
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10409 .16., perché .48. è la possanza del diametro dela spera, siché questo .16. éne la possanza de lato del cubo.<lb></lb> |
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10410 Donca il lato è .4. Peró dividi .4. secondo la proportione avente el mezzo: cioé fa de .4. doi parti che tal<lb></lb> |
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10411 parte sia la prima dela .2a., qual che la .2a. de tutto el numero, cioé de .4. E troverai che l’ una, cioé la menor parte,<lb></lb> |
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10412 sia .6. men .R.20. e la magiore sia .R.20. men .2. e .R.20. men .2. dico ch’ é el lato del pentagono corpo-<lb></lb> |
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10413 reo dimandato. Facta. 22. |
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10414 </p> |
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10415 <p class="main"> |
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10416 E gli é una spera che contene un .12. base pentagone che ciascun suo lato éne .4. Di-<lb></lb> |
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10417 mando che sia diametro de ditta spera. Tu hai per la precedente che ’l diametro dela<lb></lb> |
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10418 spera, ch’ é .R. 48., el lato del suo .12. base éne .R.20. men .2. Multiplica .R.20. men .2. in sé, fa .24.<lb></lb> |
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10419 men .R.320. Ora dirai: .se.24. men .R.320. me dá .48., che me dará .4. che è lo lato noto. Recalo a<lb></lb> |
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10420 .R. fa .16., poi multiplica .16. via .48. fa .768. el qual parti per .24. men .R.320. Opra per via de binomi trovando el re-<lb></lb> |
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10421 siduo e partirai, che ne virrá .72. piú .R.2880., cioé sia el diametro dela spera presa la .R. de .2880. e<lb></lb> |
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10422 posta sopra .72. e la .R. di quella summa sia diametro domandato, cioé .R.72. piú .R. 2880. E,<lb></lb> |
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10423 per via de proportion, riescanvi sempre questi corpi et cetera. .23.<lb></lb> |
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10424 <lb></lb> |
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10425 <pb></pb></p> |
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10426 <p class="folio"> |
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10427 folio 70v |
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10428 </p> |
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10429 <p class="main"> |
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10430 <lb></lb> |
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10431 </p> |
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10432 <p class="runhead"> |
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10433 Distinctio quarta. Capitulum secundum. |
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10434 </p> |
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10435 <p class="main"> |
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10436 <lb></lb> |
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10437 E gli é una spera che contiene un .12. base pentagonali equilatero il cui lato éne bracia .1. Di-<lb></lb> |
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10438 mando che sia la superficie de tutto el .12. base pentagonali. Tu hai che nel corpo<lb></lb> |
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10439 de .12. base ogni basa éne pentagona ed esse ditto che uno di lati di questo pentago-<lb></lb> |
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10440 no è .4.bracia. e tu voi la superficie de questi .12. pentagoni. Trova prima la superficie de uno,<lb></lb> |
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10441 per li modi che de’ pentagoni piani te mostrai ala terza distinzione nel .6º. capitolo. Cioé tu sai, per la .9a. del<lb></lb> |
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10442 14o., che, multiplicando li .3/4. del diametro del cerchio, dove sia descripto tal pentagono, via li .5/6. dela<lb></lb> |
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10443 sua corda pentagonica, quel producto sia la superficie de tutto el pentagono. E noi, per piú facilitá, dire-<lb></lb> |
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10444 mo che se debbia multiplicare li .5/8. del diametro del ditto tondo via tutta la corda ditta e quello che<lb></lb> |
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10445 fará sirá la superficie del pentagono. Over torremo li .5/8. dela corda e multiplicarli via tutto el diametro<lb></lb> |
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10446 del ditto tondo, che tanto vale, perche è manco briga a togliere li rotti de una sola quantitá che torli<lb></lb> |
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10447 d’ ambedoi Ior et cetera. Donca, prima te conviene trovare la corda de ditto pentagono e anco trovare<lb></lb> |
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10448 el suo cerchio, per li modi e ve’ che nel suo luogo ó ditto ma, per men briga, habi sempre familia-<lb></lb> |
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10449 re apresso te un pentagono con tutte sue indigentie: cioé cerchio, lato, corda, exagono, decagono,<lb></lb> |
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10450 tondo intrinseco ed extrinseco et cetera. E, mediante quello, porrai retrovare de qualunche altro che te fosse<lb></lb> |
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10451 proposto, per via de proportioni, che sempre riescano. E cosí de triangoli, quadrati, octagoni, tondi et cetera.<lb></lb> |
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10452 E mai fallano. Donca habia apresso de te quello che mette Ptolomeo in el Almegesto che dici: quan-<lb></lb> |
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10453 do il lato del pentagono fosse .10. men .R.20. che ’l diametro del cerchio dove fosse descrinto se-<lb></lb> |
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10454 rebbe .R.16. Donca piglia li .5/8. de .16., ne ven .6 1/4. Peró dí: se .10. men .R.20. me dá .6 1/4., che me dará .4. |
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10455 </p> |
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10456 <p class="main"> |
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10457 Reca .4. a .R., fa .16., poi multiplica .6 1/4. via .16., fa .100. e questo parti per .10. men .R.20. Trova el residuo<lb></lb> |
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10458 commo t’ ó mostro al suo luogo, nella arithmetica, e partirai, ne virrá aponto .12 1/2. piú .R.31 1/4. e tanto son-<lb></lb> |
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10459 no li .5/8. dela possanza del diametro del tondo che contenesse aponto una dele base del ditto .12.<lb></lb> |
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10460 base pentaganiche. Poi trova la corda del ditto pentagono. Tu fai che ’l suo lato è .4. e questo è<lb></lb> |
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10461 una parte de ditta corda, peró poni che l’ altra sia .1.co. Multiplica .1.co. via .4. piú .1.co., fa .4.co. piú .1.ce.<lb></lb> |
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10462 E multiplica .4. in sé, fa .16. E questo equale a .4.co. piú .1.ce. Parti e sequita, harai che la cosa varrá .R.20. men<lb></lb> |
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10463 .2. E questa é la menor parte, giognici .4., ch’ é la magior parte, fa .R.20. piú .2. e tanto sia tutta la dit-<lb></lb> |
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10464 ta corda, perché dici la .11a. del .13o. che, se la corda sia divisa secondo la proportione avente el mez-<lb></lb> |
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10465 zo, sempre la sua magiore parte sirá lato del pentagono. E peró te dissi lasú che .4. fo una par-<lb></lb> |
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10466 te de ditta corda. Poi, per trovare l’ altra, festi positione. Ma stu n’ avesse una nota, haresti sequito<lb></lb> |
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10467 per la regola del .3. e seri venuta ala prima et cetera. Ora multiplica .R.20. piú .2. in sé, fa .24. piú .R.320. il qual<lb></lb> |
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10468 multiplica via .12 1/2. piú .R.31 1/4., fa .400. piú .R.50000. piú .R.18000. e tanto è la superficie del’ una basa. E tu voi<lb></lb> |
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10469 de .12., donca multiplica .12. via .12., fa .144., poi fa .144. via .R.400. piú .R.50000. piú .R. 18000., fará<lb></lb> |
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10470 .R.57600. piú .R.720000. piú .R.216000. e tanto dirai che sia la superficie tutta del ditto corpo .12. base penta<lb></lb> |
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10471 gone che per lato suo sia .4. equilatero. Facta et cetera, cioé .R. dela summa, che fa la .R. de .1036800000.<lb></lb> |
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10472 e la .R. de .373248000. posta sopra .57600. et cetera. |
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10473 </p> |
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10474 <p class="main"> |
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10475 E gli é un .12. base pentagonali che il lato de ciascuna basa é .4. Dimando quanto sirá qua-<lb></lb> |
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10476 dro tutto el corpo. Prima trova el diametro dela spera dove tal corpo sia descrip-<lb></lb> |
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10477 to, cioé cosí. Tu hai, per la passata, che la linea che sotende al’ angolo pentagonico è .2<lb></lb> |
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10478 piú .R.20. e si hai che .2. piú .R.20. éne il lato del cubo descripto pure in tale spera, perché<lb></lb> |
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10479 sempre tanto è il lato del cubo quanto che la corda pentagonale in medesima spera locati, commo per<lb></lb> |
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10480 la .18a. del .14o. si manifesta. E fo ditto che la possanza del diametro era tripla al lato del cu-<lb></lb> |
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10481 bo, cioé ala possanza del lato del cubo locato in ditta spera. Adonca multiplica il lato del cubo, ch’ é .2.<lb></lb> |
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10482 piú .R.20., in sé, fa .24. piú .R.320. e tanto è la possanza del lato cubo. Ora triplala multiplicando per .3., fa<lb></lb> |
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10483 72. piú .R.2880. e tanto sia la possanza del diametro dela spera. Ora trova el diametro del circulo dove<lb></lb> |
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10484 è descripto una dele .12. base pentagonali al modo giá ditto: che il lato del pentagono era .10. men<lb></lb> |
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10485 .R.20., dava de diametro .4., che dará il lato ch’ é .4. Multiplica e parti per binomio, troverai che te dará .R. 204 4/5.<lb></lb> |
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10486 piú .32., siché dirai che ’l diametro del cerchio ambiente una dele base éne .32. piú .R. 204 4/5. Ora pi-<lb></lb> |
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10487 gliane la .1/2. de questi doi diametri, cioé .72. piú .R.2880., che quello dela spera è .32. piú .R.204 4/5. che è<lb></lb> |
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10488 quello del cerchio basale pentagono la .1/2. de .R.72. piú .R.2880. éne .R.18. piú .R.180. E la .1/2. de<lb></lb> |
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10489 .R.32. piú .R.204 4/5. éne .R.8. piú .R.12 4/5. Ora cava la possanza de .R.8. piú .R.12 4/5. dela possanza de<lb></lb> |
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10490 .R.18. piú .R.180. resta .18. piú .R.180. men .8. piú .R.12 4/5. Del qual piglia el .1/3., che ne virrá .2. piú .R.2 2/9. men <lb></lb> |
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10491 .8/9. <lb></lb> |
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10492 piú<lb></lb> |
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10493 .R.64/405. con lo qual hai a multiplicare la superficie del pentagono che il lato suo é .4. che hai, per la precedente, che l’ <lb></lb> |
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10494 é in<lb></lb> |
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10495 tutto ditta superficie .57600. piú .R.1036800000. piú .R.373248000. Donca multiplica .2. via .57600., fará<lb></lb> |
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10496 .115200. Poi reca .2. a .R., fa .4., multiplica .4. via .103680000., fa .R. 4147200000. e .4. via<lb></lb> |
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10497 .373248000. fa .R.1492992000. che, gionte insiemi queste doi .R., cioé .R.4147200000. e<lb></lb> |
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10498 .R.1492992000. fanno una .R. de .16588800000. Ora reca a .R.57600., fa .3317760000., multiplica con<lb></lb> |
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10499 2 .2/9., fa .R.7372800000. e multiplica .2 2/9. via .1036800000., fa .R.2304000000., che é .48000. E multiplica<lb></lb> |
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10500 <lb></lb> |
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10501 <pb></pb></p> |
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10502 <p class="folio"> |
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10503 folio 71r |
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10504 </p> |
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10505 <p class="main"> |
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10506 <lb></lb> |
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10507 </p> |
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10508 <p class="runhead"> |
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10509 Distinctio Octava. De Corporibus regularibus. 71 |
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10510 </p> |
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10511 <p class="main"> |
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10512 <lb></lb> |
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10513 .2 2/9. via .37348000. fa .R.829440000. che è .28800., giogni con .115200. e .48000. e .28800.<lb></lb> |
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10514 fa .192000. piú .R.16588800000. e .R. de .7372800000. da una parte. Ora, per l’ altra parte, ch’ é<lb></lb> |
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10515 el men, multiplica .8/9. via .57600., fa .51200., reca .8/9. a .R., fa .64/81., multiplica .64/81. via .1036800000. fa <lb></lb> |
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10516 .R.819200000. |
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10517 </p> |
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10518 <p class="main"> |
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10519 E multiplica .64.81. via .37348000., fa .R.29412000. E multiplica .64/81. via .R.3317760000. che<lb></lb> |
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10520 è .57600., recato a .R. fará radici .524288000. che, gionto con radici .292377600., fará una<lb></lb> |
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10521 .R. de .3787908080. E multiplica .64/81. via .1036800000., fa .R.163840000. che è .12800. Poi mul-<lb></lb> |
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10522 tiplica .64/405. con .37348000., fa .R.58982400., ch’ é .7680. Giogni insiemi prima .51200. e .12800.<lb></lb> |
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10523 e .7680. fa .71680. e .R.2624226954 8946/16651. e. R.294912000. Adonca dirai che il corpo de .12.<lb></lb> |
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10524 base pentagonali che per faccia dela basa sia .4. sirá quadro .192000. piú .R.16588800000. e .R. de<lb></lb> |
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10525 .7372800000. meno. 71680. e .R.2624226954 8946/16651. e .R. de .294912000., cioé la .R. dela sum-<lb></lb> |
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10526 ma che fa la .R. de .16588800000. e .R.7372800000. posta sopra .192000. meno. la .R. dela sum-<lb></lb> |
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10527 ma che fa la .R. de .2624236954 8946/16651. e la .R. de .294912000. posta sopra de .71680. Facta et cetera .25. |
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10528 </p> |
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10529 <p class="main"> |
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10530 E gli é un corpo de .20. base triangolari equilatero contenuto da una spera che il suo diametro<lb></lb> |
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10531 é .12.bracia. Dimando del lato de una dele base. Fa cosí. Fa una linea che sia .ab. dela<lb></lb> |
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10532 quantitá del diametro ch’ é .12. e dividila per equali in ponto .d. e descrivi el semicirculo de-<lb></lb> |
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10533 la quantitá delo .ad. che sia .aed. e, sopra .a., mena la perpendiculare .fa. dela quantitá de<lb></lb> |
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10534 .ab. E dal ponto .f. tira .fd. che segará il semicirculo in ponto .e. E dal ponto .e. mena la perpendicula-<lb></lb> |
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10535 re sopra .ab., che la segará in ponto .c. E habiamo doi triangoli simili .afd.ced., perché l’ ango-<lb></lb> |
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10536 lo .a. del triangolo .afd. è retto e l’ angolo .c. del triangolo .ecd. è retto. E l’ angolo .d. del’ uno é an-<lb></lb> |
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10537 golo del’ altro e li lati e le base sonno proportionali. Adonca, de necessitá, l’ angolo .f. é equale al’ angolo .e.<lb></lb> |
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10538 Conciosiacosaché ciascuno éne opposto ale base contenute da doi angoli equali. E, per l’ ultima del .13o.,<lb></lb> |
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10539 si prova che la linea .fd. divide el semicirculo .aeb. in ponto .e. che, preso la linea .ae., é il lato del .20.<lb></lb> |
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10540 base triangolari descripto nella medesima spera. Tu sai che .af. éne equale alo .ab., ch’ é .12., e<lb></lb> |
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10541 .ad. è .6., ch’ é la mitá de .ab. E perché .fd. del triangolo .afd. é opposta al’ angolo .a., ch’ é retto, per la<lb></lb> |
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10542 penultima del primo, pó quanto .af. e .ad. e la possanza de .af. è .144. e la possanza de .ad. éne .36. che,<lb></lb> |
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10543 gionte insiemi, fan .180. e .R.180. éne .fd., ch’ é .5. tanto dela posanza de .ad., che è .36. E tal pro-<lb></lb> |
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10544 portione é da .fd. alo .ad. che è da .ed. alo .cd. e .ed. é quanto .ad., che è .6., perché éne semidiametro, che la<lb></lb> |
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10545 sua posanza è .36., ch’ é .5. tanto dela posanza delo .cd. ch’ é .7 1/5. e la .R.7 .1/5. é .cd. e la posanza delo .ce.<lb></lb> |
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10546 è .28 4/5., ch’ é lo resto fine a .36. Siché .ce. é .R.28.4/5. E tu voi .ae. che pó quanto .ac. e .ce., peró multiplica .ac., cioé<lb></lb> |
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10547 cosí multiplica .6. men .R.7 1/5. via .6. men .R.7 1/5. fa .43 1/5., men .R.1036 4/5., giongni con la posanza delo .ce., ch’ é <lb></lb> |
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10548 .R.<lb></lb> |
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10549 .28 4/5., cioé tratta la .R.1036 4/5. de .43 1/5. e lo rimanente posta sopra .28 4/5. e la .R. di quella summa sirá<lb></lb> |
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10550 aponto el lato del .20. base triangulari, che l’ asis dela spera é .12. Fatta. 26a<lb></lb> |
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10551 E gli é un corpo de .20. base triangulari equlilatero che il lato de ciascuna basa è .4.bracia.<lb></lb> |
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10552 Dimando che sia diametro dela spera che aponto lo contenesse. Fa .1. linea che sirá<lb></lb> |
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10553 .ab. e dividila per equali in ponto .d. e sopra .d. centro descrivi el semicirculo che sia .aeb. e<lb></lb> |
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10554 sopra .a. mena la perpendiculare .af. dela quantitá che è .abl. Dapoi tira .fd., che seghi la cir-<lb></lb> |
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10555 conferentia .aeb. in ponto .e., poi linea .ae., che sia .4.bracia. che, per la precedente, é il lato dele base de .20. base tri-<lb></lb> |
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10556 angolare descripto in quella spera. Dapoi linea .eb. Dico che .ae. e .eb., gionto insiemi in directo, compon-<lb></lb> |
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10557 gano .1. linea divisa in ponto .a. secondo la proportione havente el mezzo e la magior parte é .be. e .ae.4.,<lb></lb> |
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10558 ch’ é la menore, sia e sia el lato dela basa del .20. base. E, per la penultima del primo, se prova che la pos-<lb></lb> |
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10559 sanza del’ axe d’ uno triangolo oposta al’ angolo recto é quanto la possanza dele .2. linee che contengo-<lb></lb> |
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10560 no l’ angolo retto gionto insiemi. E, perché s’ á a dividere secondo la proportione havente el mezzo e<lb></lb> |
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10561 la menor parte è .4., trovarai la magiore commo sai: trovarai che l’ altra parte sirá .R.20. piú .2. e tan-<lb></lb> |
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10562 to sirá .eb. e l’ altra parte è .4. Or multiplica .2. piú .R.20 via .2. per .R.20., fa .24. piú .R.320., ch’ é la posanza<lb></lb> |
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10563 delo .eb. Multiplica .ae., ch’ é .4., in sé, fa .16., giongni con .24. piú .R.320., fa .40. piú .R.320. Tanto é la posan-<lb></lb> |
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10564 za de .ab., ch’ é il diametro dela spera. Adonca el diametro dela spera che contene il corpo de .20. base triangulari<lb></lb> |
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10565 equilatero è .R. dela summa che fa la .R.320. posta sopra a .40. che è lo proposito. Facta et cetera. .27. |
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10566 </p> |
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10567 <p class="main"> |
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10568 E gli é un corpo de .20. base triangoli equilatero che per ciascuno lato è .4.bracia. Doman-<lb></lb> |
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10569 do dela sua superficie. Tu sai che la sua basa è triangulare equilatera .4. per ogni<lb></lb> |
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10570 verso. Trova el catetto de una basa che è .R.12. E giá tu sai che multiplicare el catetto ne-<lb></lb> |
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10571 la .1/2. dela basa ne ven la superficie de .1. triangolo ch’ é una dela base .20. E tu voi la super-<lb></lb> |
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10572 ficie de .20. base, donca piglia la .1/2. de .20., ch’ é .10. base, che ciascuna è .4., che fa .40. Recalo a .R.<lb></lb> |
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10573 perché l’ái a multiplicare con .R.1600. e questo multiplica con lo catteto de una basa ch’ é .R.<lb></lb> |
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10574 .12., fa .R.19200. e tanto è la superficie de tutte le .20. base triangolari che per fa-<lb></lb> |
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10575 cia sian .4. |
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10576 </p> |
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10577 <p class="main"> |
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10578 E gli é un .20. base triangulari la cui superficie è bracia .200. Dimando che sia ciascuno<lb></lb> |
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10579 <lb></lb> |
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10580 <pb></pb></p> |
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10581 <p class="folio"> |
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|
10582 folio 71v |
|
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10583 </p> |
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10584 <p class="main"> |
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10585 <lb></lb> |
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10586 </p> |
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10587 <p class="runhead"> |
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10588 Distinctio octava. De Corporibus regularibus. |
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10589 </p> |
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10590 <p class="main"> |
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10591 <lb></lb> |
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10592 suo lato. Tu hai che, se il lato de una basa è .4., el catetto è .R.12. e la superficie di quella basa è<lb></lb> |
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10593 .R.48. E, si hai che .200. è la superficie de .20. base, peró parti .200. per .20., ne ven .10. e .10. è superficie de<lb></lb> |
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10594 una. basa. E, perché la proportione de superficie a superficie è dopia ala proportione de un lato de una su-<lb></lb> |
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10595 perficie a uno lato del’ altra superficie, quando sonno simili, per la .17a. del .6o. de Euclide. Peró dirai: se<lb></lb> |
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10596 .R.48. de superficie me dá de lato .4., che me dará .10. de superficie. Reca .4. a .R.R., fa .256. e re-<lb></lb> |
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10597 ca .10.R., fa .100. Ora dí: se .48. dá de lato .256., che dará .100. Multiplica .100.via.256., fa .25600., parti in<lb></lb> |
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10598 .48., ne ven .533 1/3. e .R.R.533 1/3. sirá per lato el .20. base triangulari che la superficie sua sia .300. Facta. <lb></lb> |
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10599 29a. |
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10600 </p> |
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10601 <p class="main"> |
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10602 E gli é un .20. base triangulari equilatero la cui superficie è bracia .200. Domando del diametro<lb></lb> |
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10603 dela spera dove sia descripto. Tu hai, per la precedente, che ’l .20. base che la sua super-<lb></lb> |
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10604 ficie sia .200. che il suo lato è .R.R.533 1/3. e, per la .2a. del .20. base, hai che il lato, che è .4.,<lb></lb> |
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10605 de diametro .40. piú .R.320. E, perché tu hai il lato che .R.R., reca .4. a .R.R., fa .256. e re-<lb></lb> |
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10606 ca .40. a .R., fa .1600. e poi reca a .R.320., fa .102400. e hai .1600. piú .R.102400. Ora dí cosí: se<lb></lb> |
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10607 .256. de lato dá de diametro .1600. piú .R.102400., che dará .533 1/3. Multiplica .533 1/3. via.1600., fa .853333 1/3.,<lb></lb> |
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10608 parti per .256., ne ven .3333 1/3. Ora reca a .R.533 1/3., fa .284444 4/9.; multiplicalo con .102400., fará .29127311111 <lb></lb> |
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10609 1/9.<lb></lb> |
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10610 el quale parti per .256., recato a .R., fa .65536., ne vene .429188 26068/32489. e hai .3333 1/3. piú .R.429188 <lb></lb> |
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10611 26068/32489. Adon-<lb></lb> |
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10612 ca dí che ’l diametro dela spera dove è discripto ditto corpo base .20. triangoli che la sua superficie<lb></lb> |
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10613 sia .200., sia la .R. dela summa che fa la .R.R.429188 26068/32489. posta sopra la .R.3333 1/3. Facta. et cetera. 30a.<lb></lb> |
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10614 E gli é un .20. base triangulare equilatero che il lato de ciascuna sua basa è .4. Domando<lb></lb> |
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10615 quanto sirá quadrato. Tu hai, per la .2a. del corpo .20. base, che, se ’l suo lato è .4., che ’l dia-<lb></lb> |
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10616 metro dela spera dove è descripto è .R. dela summa che fa la .R.320. posta sopra .40. |
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10617 </p> |
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10618 <p class="main"> |
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10619 Adonca dividi in doi parti equali .40. piú .R.320. Cosí reca .2. a .R., fa .4., parti .40. per .4., ne ven<lb></lb> |
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10620 .10. e reca .4. a .R., fará .16., parti .320. per .16., ne vene .20. che hai .10. piú .R.20. che è mezzo diametro.<lb></lb> |
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10621 Ora trova il catetto de una basa del .20. base che sai che uno lato è .4. Multiplica in sé, fa .16. e multiplica la<lb></lb> |
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10622 .1/2. dela basa, che è .2., fa .4., trallo de .16., resta .12. e .R.12. sia il suo catetto. Del quale trova il centro<lb></lb> |
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10623 che è neli .2/3. Cosí dividi .12. per .9., ne ven .1 1/3. el quale multiplica per .4., fa .5 1/3. che è la .R. de 2/3. de .12.; <lb></lb> |
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10624 trallo de <lb></lb> |
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10625 .10.<lb></lb> |
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10626 piú .R.20., resta .4 2/3. piú .R.20. el quale hai a multiplicare nela superficie de .20. base. Tu hai, nella terza, che la<lb></lb> |
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10627 superficie di tal corpo è .R.19200. Del quale piglia el .1/3. commo .R.: reca .3. a .R., fa .9., parti .19200. per .9.,<lb></lb> |
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10628 ne ven .2133 1/3., il quale multiplica con .4 2/3., fa .9955 5/9. Ora reca a .R.2133 1/3., fa .4551111 1/9. che, <lb></lb> |
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10629 multiplicato con <lb></lb> |
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10630 .20.<lb></lb> |
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10631 fa .910222222 2/9. Adonca dirai che gli é quadrato ditto corpo .20. base, che per lato è .4. la .R. dela<lb></lb> |
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10632 summa che fa la .R. de .91022222 2/9. posta sopra de .9955 5/9. et cetera. 31a.<lb></lb> |
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10633 E gli é un .20. base triangolari equilatero che è quadrato .400.bracia. Dimando quanto è il suo<lb></lb> |
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10634 lato. Per la precedente, tu hai che il lato de .20. base, ch’ é .4., dá de quadratura .9955 5/9. piú<lb></lb> |
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10635 .R.91022222 2/9. Adonca .9955 5/9. piú .R.91022222 2/9. dá de lato .R.18. Peró reca .16. a .R.<lb></lb> |
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10636 cuba, fa .4096. Ora dí: se .9955 5/9. piú .R.91022222 2/9. de quadratura dá de lato .4096.,<lb></lb> |
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10637 che dará .400. Recalo a .R., fa .160000., il quale multiplica con .4096., fa .655360000. Volse partire per .9955 5/9.<lb></lb> |
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10638 piú .R. de .910522222 2/9. Trova el partitore, cioé cosí perché gli é binomio multiplica .9955 5/9. piú .R.91022222 2/9.<lb></lb> |
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10639 via. 9955 5/9. men .R.91022222 2/9., fa .8190864 16/81. e questo è partitore. Poi multiplica .9955 5/9. via <lb></lb> |
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10640 .655360000., fa<lb></lb> |
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10641 .6524472888888 8/9. a partire per .8190864 16/81. Reduci a una natura, harai .5.28483304000000.;<lb></lb> |
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10642 a partir per .663460000. ne vene .796556 26084/66346. Ora reca .R.655360000. fa .429496729600000000.;<lb></lb> |
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10643 e questo multiplica con .91022222 2/9. fa .390937467665368888888888888 8/9., el quale e se vol partir per .8190864 <lb></lb> |
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10644 16/81.,<lb></lb> |
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10645 recato a .R. il qual fará .4401787123 29/81.900000., per lo qual parti quello de sopra, che sirá<lb></lb> |
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10646 recato a otantunesimi, commo che éne il partitore .256400074527574057777777789000., che ne vene .582490855067 <lb></lb> |
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10647 239590987174489/440178712329900. Adonca dirai che ’l .20. base triangolare equilatero che è<lb></lb> |
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10648 quadrato .400.bracia. sia per lato la .R. cuba dela .R. del remanente de .796556 26084/66346., tratone la .R. de<lb></lb> |
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10649 .582490855067 239590987174489/440178712329900. Fatta. 32a. |
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10650 </p> |
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10651 <p class="main"> |
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10652 E gli é uno corpo de .8. base triangolari equilatero che il lato suo è .4. Domando del diametro<lb></lb> |
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10653 dela spera dove se descrive. De questo corpo facilmente se ánno le sue mesure, peró non<lb></lb> |
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10654 de detti niente e per questo è remaso derietro, ma, perché ci sienno tutti li .5. corpi regola-<lb></lb> |
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10655 ri, non lo voglio lasciare. Dico che, per havere il diametro dela spera dove è scripto lo<lb></lb> |
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10656 .8. base che il suo lato è .4., che multiplica .4. in sé, fa .16., el qual redopia commo numero, fa .32. e .R.32 sirá el<lb></lb> |
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10657 diametro dela spera dove se descrive ditto corpo et cetera. 33a. |
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10658 </p> |
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10659 <p class="main"> |
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10660 E, se ’l ti fosse detto e gli é una spera che ’l suo diametro è .10. e contene uno corpo de .8. base<lb></lb> |
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10661 triangulari equilatero. Dimando che sirá il suo lato. Multiplica il diametro dela spera in sé, ch’ é<lb></lb> |
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10662 .10., fa .100., dividelo per .2., ne nene .50. e la .R. dirai che sia per lato. 34a.<lb></lb> |
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10663 E gli é una spera il cui diametro è .20.bracia. e contene uno .8. base triangulari equilateri. Domando dela<lb></lb> |
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10664 sua superficie. Trova prima uno lato delo .8. base, commo per la precedente s’ é ditto, cioé multiplica .20. in<lb></lb> |
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10665 <lb></lb> |
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10666 <pb></pb></p> |
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10667 <p class="folio"> |
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10668 folio 72r |
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10669 </p> |
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10670 <p class="main"> |
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10671 <lb></lb> |
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10672 </p> |
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10673 <p class="runhead"> |
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10674 Distinctio octava. De Corporibus regularibus. 72 |
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10675 </p> |
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10676 <p class="main"> |
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10677 <lb></lb> |
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10678 sé, fa .400., dividilo per .2., ne ven .200. e tanto è la possanza de uno de’ lati del triangolo d’ una basa.<lb></lb> |
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10679 Trova il suo catetto, cioé dividi .200. in .2. parti equali commo .R., ne ven .50., cavalo del .200., resta<lb></lb> |
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10680 .150. e .R.150. è il catetto de una dele base. Ora multiplicalo con la mitá dela basa, ch’ é .R.50., via<lb></lb> |
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10681 .150., fa .7500. el quale multiplica per .8. base recato a .R., che è .64., via .7500., fa .480000. e la .R.480000.<lb></lb> |
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10682 sirá la superficie domandata . 35a. |
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10683 </p> |
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10684 <p class="main"> |
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10685 E gli é uno .8. base triangulari equilatero che è quadrato .400.bracia. Domando che sia dia-<lb></lb> |
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10686 metro dela spera che ’l circunscrive aponto. Tu hai, per la passata, che .1333 1/3. de qua-<lb></lb> |
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10687 dratura dá de diametro .20. Reca .20. a .R. cuba, fa .8000. Peró dí: se .1333 1/3. de quadratu-<lb></lb> |
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10688 ra dá de diametro .8000., che dará .400. Multiplica .400. via .8000. fa .3200000. il qual parti per .1333 1/3.,<lb></lb> |
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10689 fa terzi dele parti, harai .9600000., a partire per .4000., che ne vene .2400. e la .R. cuba de .2400.<lb></lb> |
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10690 sirá el diametro dela spera che lo contene . 36a. |
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10691 </p> |
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10692 <p class="main"> |
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10693 E gli é un .4. base triangulari .abcd. ch’ é .ab.20.ac.18.ad.16.bd.15.bc.14.cd.13. Di-<lb></lb> |
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10694 mandase la quantitá del suo assis .ag. Tu dei far cosí: trovar el catetto dela basa .bcd.<lb></lb> |
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10695 cadente sopra la linea .bc., ch’ é .14.bd.15.cd.13. e trovarai il catetto .ce. esser .12. e casca<lb></lb> |
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10696 presso al .c.5. Ora trova el catetto dela facia .abc. cadente pure su la linea .bc., che tro-<lb></lb> |
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10697 varai essere .R.305 31/49. e casca presso al .c.4 2/7. Or piglia la differenza ch’ é da .4 2/7. a .5., ch’ é .5/7., multiplicala in <lb></lb> |
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10698 sé,<lb></lb> |
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10699 fa .25/49., trallo dela possanza dela potumissa .ad. che è da rescontro, che è .256., resta .255 24/39. Ora<lb></lb> |
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10700 reca .de., ch’ é .12., a .R., fa .144. Ora tu hai uno triangolo che uno lato è .R.305 31/49. e l’ altro .R.<lb></lb> |
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10701 .255 24/49. e l’ altro è .R.144. Trova el suo catetto che cade sopra la basa che è .R.144., cioé giongni<lb></lb> |
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10702 insiemi .R.255 24/49. e .R.144. commo numero, fa .399 24/49. tranne .305 31/49., resta .93 42/49. El quale parti per lo <lb></lb> |
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10703 dopio delo .de.<lb></lb> |
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10704 che sirá .24. Fanne quarantanovesimi, harai .4599. a partire per .1176. che ne vene .3 51/56. Tanto éne<lb></lb> |
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10705 .gh., el quale multiplica in sé, fa .15 421/3136., trallo de .255 24/19., restará .R.240 615/1336. e la .R. de .240. e quello <lb></lb> |
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10706 rotto<lb></lb> |
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10707 sirá longo l’ assis .ag. Facta. E, se tu voli .dg., giongni .25/49. sopra .15 921/3136. che fa .R.15 2521/3136. e la .R.<lb></lb> |
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10708 di questa summa éne aponto .dg. et cetera. 37a. |
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10709 </p> |
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10710 <p class="main"> |
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10711 E gli é uno .4. base triangulari .abcd. che è quadrato .252.bracia. e la sua basa éne .bcd.<lb></lb> |
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10712 E lo .bd.15.bc.14.cd.13. Domandase quanto sia l’ assis .ag. Fa cosí. Vedi quanto è<lb></lb> |
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10713 la superficie dela basa .bcd. che la troverai .84. Poi multiplica la quadratura del .4. base, che è<lb></lb> |
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10714 .252., per .3., fa .756. qual parti per la superficie dela basa, ch’ é .84., ne vene .9. e tanto dirai che sia<lb></lb> |
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10715 l’ axis. Ala prova: multiplica la superficie dela basa per .9., fa .756. E giá tu sai che ogni piramide éne uno ter-<lb></lb> |
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10716 zo del suo chelindro, adoca parti .756. per .3., ne vene .252., sí commo dicemo esser quadro el .4. base. Facta.<lb></lb> |
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10717 E gli é un .4. base triangulari .abcd. la cui basa .bcd. E .bd. è .15.bc.14.cd.13. el suo<lb></lb> |
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10718 assis .ag. e .bg.10.cg.9. Domando quanto sirá .dg. Tu hai el .4. base triangulari<lb></lb> |
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10719 .abcd. ch’ é .bc.14.bd.15.cd.13.bg.10.cg.9. Ora trova il catetto cadente dal pon-<lb></lb> |
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10720 to .d. sopra la basa .bc. in ponto .e. che è .12. e cade apresso .c. .5. E giá tu hai .bgc. noto.<lb></lb> |
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10721 Ora trova il suo catetto cadente pure sopra .bc. Trovarai che cade presso .c. 6 9/28. e lo catetto vene<lb></lb> |
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10722 esser .R.41 31/784. Trallo delo .de., ch’ é .12., restará .12. men .R.41 31/784., il qual multiplica in sé, fa .185 31/784. <lb></lb> |
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10723 men<lb></lb> |
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10724 .R.2363. 8 608/784. al quale giongni la posanza dela differenza ch’ é dal caso delo .fg. al catetto .de. ch’ é .1 9/28., il qual <lb></lb> |
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10725 multiplica in<lb></lb> |
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10726 sé, fa .1 585/784., giongnilo con .185 31/784., fará .186 616/74. Tanto sia la linea .dg., cioé .R. del rimanente de<lb></lb> |
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10727 .186 616/784. trattone la .R. de .23638 608/784. fatta aponto et cetera. 39a.<lb></lb> |
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10728 E gli é un .4. base triangulari .abcd., ch’ é .bd.15.bc.14.cd.13. e l’ assis suo<lb></lb> |
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10729 .ag.8.bg.10.cg.9.dg. éne .R. del rimanente de .186 616/74. trattone .R.23638 608/784. Domando che sia .ab.ac.<lb></lb> |
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10730 .ad. Volse far prima per .ab., cioé multiplicare .bg., ch’ é .10., in sé, fa .100.; poi multiplicare .ag., ch’ é<lb></lb> |
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10731 .8., in sé, fa .64., giogni con .100., fa .164. la cui .R. éne la linea .ab. Ora, per la linea .ac., multiplica .cg.,<lb></lb> |
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10732 ch’ é .9., in sé, fa .81., poi multiplica l’ axis .ag., ch’ é .8., in sé, fa .64., giogni con .81., fa .145. la cui .R. éne .ac. Ora<lb></lb> |
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10733 se vole fare per .ad. Tu hai che la posanza delo .dg. é .R. del rimanente de .186. 616/784. tratone la .R.23638.<lb></lb> |
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10734 608/784. Giogni la posanza del’ axis .ag., ch’ é .64., fa .250. 616/784. men .R.23638 608/784. e tanto sia .ad., cioé .R. del<lb></lb> |
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10735 rimanente de .250 619/784. trattone .R. 23638 608/784. Fatta et cetera. 40a.<lb></lb> |
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10736 E gli é un .4. base triangoli .abcd. e la basa sua .ebcd. e un lato è .10. l’ altro .9., il .3o.12. e l’ a-<lb></lb> |
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10737 xis è .8. Domando quanto sirá quadro. Trova prima la superficie dela basa .bcd., ch’ é .bc.12.<lb></lb> |
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10738 cd.9.bd.10. Bisogna trovare el catetto cadente sopra .bc. ch’ é .12., cioé cosí: multiplica .12.<lb></lb> |
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10739 in sé, fa .144. e .9. in sé, fa .81.; giongni insiemi fa .225. Poi multiplica .10. in sé, fa .100., trallo de .225. resta<lb></lb> |
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10740 .125. il quale parti per lo dopio de .bc., ch’ é .24., ne ven .5 5/14. e tanto casca il catetto apresso .c., ch’ é il menor<lb></lb> |
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10741 lato. Peró multiplica .5 5/24. in sé, fa .27 73/576. Ora multiplica .cd., ch’ é .9., in sé, fa .81., tratone .27 73/576.<lb></lb> |
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10742 resta .53 503/576. la cui .R. sia il ditto catetto il qual multiplica con la .1/2. de .bc. ch’ é .6., recalo a .R., fa .36. e .36. via<lb></lb> |
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10743 .53.503/576. fa .R. 1939 252/576. e tanto éne la superficie dela basa .bcd., la qual multiplica con l’ axis, ch’ é .8., recato<lb></lb> |
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10744 a .R. ch’ é .64., fará .R. 124124., de questo piglia el .1/3. commo .R., cioé el .1/9., ne ven aponto .R. 13780 4/8. è<lb></lb> |
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10745 <lb></lb> |
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10746 <pb></pb></p> |
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10747 <p class="folio"> |
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10748 folio 72v |
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10749 </p> |
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10750 <p class="main"> |
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10751 <lb></lb> |
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10752 </p> |
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10753 <p class="runhead"> |
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10754 Distinctio octava. De Corporibus regularibus. |
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10755 </p> |
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10756 <p class="main"> |
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10757 <lb></lb> |
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10758 tanto sia quadrato ditto .4. base. Fatta et cetera. 41a. |
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10759 </p> |
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10760 <p class="main"> |
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10761 E gli é un .4. base triangulari equilatero .abcd. che è quadro bracia .100. Domando deli soi<lb></lb> |
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10762 lati. Fa cosí. Trova un .4. base che te sia noto il suo axis e li soi lati, cioé dirai che<lb></lb> |
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10763 sia un .4. base che ’l suo axis sia .R.16., sirá li lati suoi ciascuno .R.14., perché la pos-<lb></lb> |
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10764 sanza del’ axis è sexquialtera ala possanza del lato suo, quando il .4. base è equilatero<lb></lb> |
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10765 Adonca trova il catetto de una basa cosí. Piglia la .1/2. de .R.24., ch’ é .R.6., trallo de .24., resta .18. e<lb></lb> |
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10766 .R.18. sia il catetto el qual multiplica con la .1/2. dela basa .bc., ch’ é .R.6., fa .R.108. e questo multiplica con l’ a-<lb></lb> |
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10767 xis, ch’ é .R.16., fa .R.1728., piglian el .1/3., cioé reca .3. a .R., fa .9., parti .1728. in .9., ne ven .192. e .R.192. sia qua-<lb></lb> |
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10768 dro il .4. base il cui axis sia .4. Peró reca .4. a .R.cuba., fa .64. E, perché .192. éne .R., reca .64. a .R.,<lb></lb> |
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10769 fa .4096. Ora dirai: se .192. me dá .4096., che me dará .100. Reca a .R., fa .10000. el quale multiplica con<lb></lb> |
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10770 .4096. fa .40960000. Partilo per .192., ne ven .R.213333 1/3. e la .R. dela .R.cuba. overo la .R.cuba. de<lb></lb> |
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10771 la .R. quadra, che tanto fa, de .213333 1/3. éne l’ axis. E tu voli il suo lato. E, commo è ditto di sopra, la<lb></lb> |
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10772 possanza del’ axis è sexquialtera ala possanza del suo lato. Peró trova doi numeri in proportione<lb></lb> |
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10773 sexquialtera, ch’ é .2. e .3. Reca .2. a .R.cuba., fa .8., reca .3. a .R.cuba., fa .27. Peró dirai: se .8. me dá .27.,<lb></lb> |
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10774 che me dará .213333 1/3. Multiplica .27. via .213333 1/3., fa .5760000., partilo per .8., ne vene .720000. e la .R.cu.<lb></lb> |
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10775 de .720000. síra il lato de tal .4. base et cetera pulcra per viam proportionis. 42a.<lb></lb> |
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10776 E gli é un .4. base triangulari equilatero .abcd. che è quadro .bracia.100. Taglio del’ axis li<lb></lb> |
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10777 .2/3. con una linea piana. Domando che levará de quadratura, cioé quanti braci solidi. Tu hai,<lb></lb> |
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10778 per la precedente, che ’l suo axis è .R.R.cuba. de .213333 1/3., donca piglia el .1/3., cioé cosí: reca<lb></lb> |
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10779 .3. a .R.R.cuba., fa .729. Ora parti la .R.R.cuba. de .213333 1/3. per la .R.R.cuba. de .729., ne<lb></lb> |
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10780 vene .R.R.cuba. de .292 1396/2187. el qual radopia commo .R.R.cu., cioé reca .2. a .R.R.cu., fa .64. e .64. via<lb></lb> |
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10781 .292 1396/2887. fa .R.R.cuba.18728 1864/2187. Questo éne .2/3. delo axis e tu voli il suo quadrato, peró dirai commo<lb></lb> |
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10782 di sopra. Se .R.4096. me dá .R.192., che me dará la .R.18728 1864/2187. Multiplica .192. via .18728 1864/2187., fa<lb></lb> |
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10783 .3595939 1407/2187. el qual parti per .4096., ne ven .877 8196096/8957952. e la .R. de .877 8196096/8957952. tanto <lb></lb> |
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10784 se <lb></lb> |
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10785 leva de<lb></lb> |
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10786 quadratura del .4. base. 43<lb></lb> |
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10787 E gli é un .4. base triangulari equilatero .abcd. la cui basa è .bcd.bd. é .15.bc.14.cd.<lb></lb> |
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10788 .13. e ’l suo axis .ag.9. cadente dentro dale linea .bcd. Taglio del’ axis .ag.3.bracia. apresso<lb></lb> |
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10789 .a. Domando che se levará de quadratura del .4. base .abcd. Fa cosí. Vedi quanto è qua-<lb></lb> |
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10790 drata la sua basa .bcd. che è .84. Multiplica .81. via l’ axis, ch’ é .9., fa .756., partilo per .3., ne ven<lb></lb> |
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10791 .252. e tanto è quadrato tutto el .4. base che ’l suo axis è .9. E tu voli un .4. base che ’l suo axis sia .3.<lb></lb> |
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10792 che è .1/3. de .ag. che è .9. e in quella proportione che è diviso l’ axis sonno divisi li lati dela basa .bcd.<lb></lb> |
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10793 Peró piglia el .1/3. de .bd., ch’ é .15., sirá .5., piglia el .1/3. de .14., sirá .4 2/3.; piglia el .1/3. de .1/3., sirá .4 1/3. e piglia<lb></lb> |
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10794 el .1/3 del catetto .de., ch’ é .12., sirá .4. el qual multiplica con la .1/2. de .4 2/3., ch’ é .2 1/3., fa .9 1/3. e questo <lb></lb> |
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10795 multiplica con <lb></lb> |
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10796 l’ axis<lb></lb> |
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10797 ch’ é .3., fa .28. del qual piglia .1/3., ch’ é .9 1/3. e tanto se leva de quadratura del .4. base a noi proposto et cetera.<lb></lb> |
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10798 .44. |
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10799 </p> |
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10800 <p class="main"> |
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10801 E gli é un .4. base triangulari .abcd. che l’ axis suo .ag. è .10. ed é quadrato .280. Taglione<lb></lb> |
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10802 con una linea piana .40. Domando quanto levará del’ axis .ag. Fa cosí. Tu sai che gli é qua-<lb></lb> |
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10803 drato .280. e l’ axis suo è .10. Reca .10. a .R.cu., fa .1000. Donca .280. dá de axis .1000., che<lb></lb> |
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10804 dará .40. Multiplica .40. via .1000., fa .40000. el qual parti per la quadratura del .4. base, ch’ é .280.<lb></lb> |
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10805 ne ven .142 5/7. e la .R.cu.142 5/7. tagliará del’ axis .ag. togliendo .40.bracia. dela quadra dal capo. da lato. a .et cetera.<lb></lb> |
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10806 E gli é un cubo che è .4.bracia. per lato, voglio dela superficie sua fare superficie de una spera. Doman-<lb></lb> |
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10807 do quanto sirá il diametro de ditta spera. Fa cosí. Vedi prima quanto è la superficie de tutto el cubo il<lb></lb> |
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10808 quale á .6. facce e .16. per facia. Adonca .6. via .16. fa .96. E tu voi una spera che la superficie sua<lb></lb> |
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10809 sia .96., peró multiplica .96. per .14., fa .1344. el quali parti per .11., ne vene .122 2/11. E tu hai a pigliare la<lb></lb> |
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10810 .1/2. commo .R., peró reca .2. a .R., fa .4., parti .122 2/11. in .4., ne ven .30 6/11. e .R.30 6/11. sirá il diametro de ditta <lb></lb> |
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10811 spera. |
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10812 </p> |
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10813 <p class="main"> |
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10814 Tu<lb></lb> |
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10815 sai che l’ .1/4. de .96. ch’ é .24., è la superficie del magior cerchio il cui diametro è diametro dela spera. 46<lb></lb> |
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10816 E gli é una spera che ’l suo diametro è .7. Voglio dela superficie sua fare superficie de .1.cubo. Domando de’ soi la-<lb></lb> |
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10817 ti. Trova la superficie dela spera, che sai che ’l suo diametro è .7. El magior circulo che l’ abia è .22. el qual<lb></lb> |
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10818 multiplica con lo diametro, ch’ é .7., fa .154. e tanto è la superficie dela spera. E tu sai che ’l cubo á .6. facce, peró parti <lb></lb> |
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10819 .154.<lb></lb> |
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10820 in .6., ne ven .25 2/3. e la .R.25 2/3. sirá el cubo per facia et cetera. 47.<lb></lb> |
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10821 E gli é .1.cubo che è quadrato .64.; vo-<lb></lb> |
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10822 glione fare una spera. Domando che sirá suo diametro. Tu dei sapere che ogni spera è .11/21. del suo cubo, commo di<lb></lb> |
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10823 sopra fo ditto. Donca multiplica .21. via .64., fa .1344. el qual parti per .11., ne ven .122 2/11. e la radici .cuba.<lb></lb> |
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10824 di questo sirá el diametro dela spera adimandata. Si tu voi saper perché tu multiplichi per .21. e<lb></lb> |
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10825 partisti per .11., te lo dico: perché, se la spera á essere quadra .64., cioé tanto quanto il .cu., questo .64. fo li .11/21. de <lb></lb> |
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10826 un<lb></lb> |
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10827 certo cubo che, quando fo scantonato, rimase .64. e peró dirai .64. de che numero foron li .11/21., sia che si vo-<lb></lb> |
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10828 glia quel numero che fo multiplicato per .11. e partito per .21. Donca a multiplicare .64. per .21. e repartire per .11. ne <lb></lb> |
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10829 retornará<lb></lb> |
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10830 quella tal quantitá de che .64. ne fo li .11/21. e quella tal quantitá sirá el tenuto de un cubo el cui lato aponto, quando<lb></lb> |
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10831 sia scantonato, vene a essere diametro dela spera et cetera. 48<lb></lb> |
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10832 <lb></lb> |
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10833 <pb></pb></p> |
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10834 <p class="folio"> |
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10835 folio 73r |
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10836 </p> |
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10837 <p class="main"> |
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10838 <lb></lb> |
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10839 </p> |
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10840 <p class="runhead"> |
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10841 Distinctio octava. De Corporibus regularibus. 73 |
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10842 </p> |
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10843 <p class="main"> |
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10844 <lb></lb> |
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10845 E gli é una spera il cui diametro è .7. Voglio dela sua quadratura fare un cubo. Doman-<lb></lb> |
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10846 do quanto siran suoi lati. Quadra la spera che ’l diametro è .7., reca a .R.cu., fa .343. |
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10847 </p> |
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10848 <p class="main"> |
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10849 E, perché la spera è .11/21. del suo cubo e lo suo .cu. è .343., peró piglia li .11/21. de .343., ne ven<lb></lb> |
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10850 .179 2/3., peró sirá el lato del cubo .R.cu. de .179 2/3. Fatta. Posiam fare per altra via. Dire<lb></lb> |
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10851 cosí, sí commo di sopra, che il diametro dela spera ch’ é .R.cu.122 2/11. dá de lato de cubo .R.cu. de .64.,<lb></lb> |
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10852 adonca che dará el diametro ch’ é .7. Reca .7. a .R. cuba, fa .343., multiplica .64. via .343., fa .21952., qual parti per<lb></lb> |
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10853 .122 2/11., ne ven .179 2/3. commo prima e .R. cuba .179 2/3. sirá per facia il cubo. 49<lb></lb> |
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10854 E gli é una piramide, o voi dire cono, che la basa sua é circulare e ’l suo diametro è equale<lb></lb> |
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10855 ali lati e ’l suo axis è .4. Voglio dela sua quadratura fare una spera. Dimando che sirá<lb></lb> |
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10856 suo diametro. Bisogna prima quadrare la piramide, che sai che l’ axis è .4. e la<lb></lb> |
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10857 possanza del’ axis ala possanza del suo lato é in proportione sexquitertia e la possanza de-<lb></lb> |
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10858 l’ axis é .16., adonca la possanza del lato éne .21 1/3., la qual multiplica per .11., fa .234 2/3., partilo in .14., ne ven<lb></lb> |
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10859 .16 16/21. e tanto è la supeficie dela basa, la qual multiplica con l’ axis, ch’ é .4., fa .64 1/21. E, perché questo è che-<lb></lb> |
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10860 lindro e noi volemo la piramide e ogni piramide è el .1/3. del suo chelindro, peró dividi .67 1/21. per .3., ne ven<lb></lb> |
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10861 .22 2/63., tanto sia quadrata la piramide. E tu hai, per la passata, che la quadratura dela spera<lb></lb> |
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10862 ch’ é .179 2/3., te dá .343., che te dará .22 22/63. Multiplica .22 22/63. via .343., fa .7665 2/9., el qual parti per .179 <lb></lb> |
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10863 2/3., ne <lb></lb> |
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10864 ven .42.<lb></lb> |
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10865 .2/3. e la .R.cu. de .42 2/3. sirá il diametro dela spera che cerchiamo aponto, et cetera. .50. |
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10866 </p> |
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10867 <p class="main"> |
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10868 E gli é una spera il cui diametro è .7. Voglio dela sua quadratura fare una piramide<lb></lb> |
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10869 che i lati suoi sienno equali al diametro dela circunferentia dela basa. Domandase del suo<lb></lb> |
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10870 axis. Trova la quadratura dela spera, che sirá .179 2/3., ut supra. E de questo se vol fa-<lb></lb> |
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10871 re una piramide. E giá tu hai de sopra che la piramide dela quale l’ axis è .4. dá de qua-<lb></lb> |
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10872 dratura .22 22/63. Ora reca .4. a .R.cu., fa .64., adonca .22 22/63. te dá la quadratura .64., che te dará<lb></lb> |
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10873 .179.2/3. de quadratura. Multiplica .64. via .179 2/3., fa .11498 2/3., il quale parti in .22 22/63., ne vene .514 1/2. e la <lb></lb> |
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10874 .R.cu.<lb></lb> |
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10875 de .514 1/2. sirá l’ axis dela piramide. |
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10876 </p> |
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10877 <p class="main"> |
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10878 E gli é una spera il cui diametro è .14., levone, con una linea piana, tanto che taglia del dia-<lb></lb> |
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10879 metro .4.bracia. Domando quanto che levará dela superficie de ditta spera e quanto sirá la<lb></lb> |
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10880 linea dividente. Havemo ditto nelli corpi sperici che la superficie dela spera è .4. tan-<lb></lb> |
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10881 to che la superficie del magior circulo dela spera. E disesse che, a multiplicare il diametro dela<lb></lb> |
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10882 spera nella circunferentia del magior cerchio, produciva la superficie de tutta la spera. Adonca, multiplicando<lb></lb> |
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10883 .14. via .44., fa .616. per la superficie di questa spera. Ora, per trovar quella parte che leva quella linea che ta-<lb></lb> |
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10884 glia del diametro .4., multiplica .4. nel resto del diametro, ch’ é .10., fa .40. e .R.40. è la .1/2. dela linea dividen-<lb></lb> |
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10885 te e tutta è .R.160. Ora tu hai el diametro .ad. ch’ é .14. e la linea dividente .bc.R.160., che è corda e<lb></lb> |
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10886 sega il diametro in ponto .e., ái che .be.R.40., ch’ é la .1/2. de .bc., e lo .ae., che è saetta, é .4., multiplicalo in sé, fa<lb></lb> |
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10887 .16., giongni lo con .40., fa .56. Donca .ab. è .R.56., el qual, se vol dopiare commo .R., fa .R.224. Multiplica in<lb></lb> |
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10888 sé, fa .224., pigliane li .11/14., per quello dici Archimede, ne ven .176. e tanto sia la superficie di quella portion<lb></lb> |
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10889 menore che l’ axis è .4. e lo diametro dela sua basa è .R.160. Facta et cetera. 52<lb></lb> |
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10890 E gli é una spera il cui diametro .14. meno una linea piana longa .9. segante l’ axis ad angolo re-<lb></lb> |
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10891 cto. Domando quanto tagliará del’ axis. Tu hai la spera .abcd.ad. è diametro, o vo-<lb></lb> |
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10892 li axis, e .bc. lo sega in ponto .e. e, perché lo sega ad angol recto, é divisa per equali in ponto .e. Adon-<lb></lb> |
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10893 ca .be. è .4 1/2., che è una parte de .9. che, multiplicato 4 1/2. in sé fa .20 1/4. Ora dí cosí: famme del dia-<lb></lb> |
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10894 metro, ch’ é .14., doi parti che, multiplicato una in l’ altra, facia .20 /14. Opera, troverai che l’ una parte menore<lb></lb> |
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10895 sia .7. men .R.28 3/4., l’ altra magiore sia .7. piú .R.28 3/4. Siché dirai che .ea. sia .7. men .R.28 3/4. Fatta et cetera.<lb></lb> |
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10896 53<lb></lb> |
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10897 E gli é una spera il cui diametro è .14.; meno una linea piana segando l’ axis ad angolo recto,<lb></lb> |
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10898 che la ditta linea è .R.96. Domando quanto levará dela superficie dela spera. La spera<lb></lb> |
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10899 è .abcd. e il diametro suo è .ad., che è .14. e la linea dividente, ch’ é .bc. è .R.96. Pigliane la<lb></lb> |
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10900 .1/2., ch’ é .R.24., ch’ é .be., multiplica in sé, fa .24. Ora fa del diametro, ch’ é .14., doi parti che<lb></lb> |
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10901 multiplicata l’ una via l’altra facia .24., per la ragion che pone el philosopho nella .34a. del terzo. Opera e trova-<lb></lb> |
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10902 rai che l’ una parte, cioé la menore .ae., sia .7. men .R.25., ch’ é .2. e l’ altra, ch’ é .ed., sia .R.25. piú .7., ch’ é .12. |
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10903 </p> |
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10904 <p class="main"> |
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10905 Adonca taglia del’ axis .2.bracia. E tu voi saper la superficie. Peró multiplica .ae., ch’ é .2., in sé, fa .4. e .R.24., ch’ é<lb></lb> |
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10906 be., in sé, fa .24.; giongni insiemi, fa .28. e .R.28. sia .ba., la qual radopia commo .R., fa .112. e di<lb></lb> |
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10907 questo piglia li .11/14. che ne ven .88. e tanti braci leva dela superficie per la portion .bac., per la ragion so-<lb></lb> |
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10908 pra ditta d’ Archimede, che vole che la superficie de ciascuna portione de spera sia equale al cerchio<lb></lb> |
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10909 el cui semidiametro sia la linea che si move dala sumitá del cono e vene ala circunferentia de-<lb></lb> |
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10910 la basa de ditta portione. E peró multiplicasti .ae., ch’ é .2., in sé e .be. in sé e giongnisti insiemi<lb></lb> |
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10911 e .R.28.fo.ab., che ditta linea dopia, fa .R.112. e questo sia diametro de ditto tondo. Multiplica<lb></lb> |
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10912 in sé, fará .112., pigliane li .11/14., commo festi, ne vene .88. per la superficie de ditto cerchio che è equa-<lb></lb> |
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10913 <lb></lb> |
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10914 <pb></pb></p> |
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10915 <p class="folio"> |
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10916 folio 73v |
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10917 </p> |
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10918 <p class="main"> |
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10919 <lb></lb> |
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10920 </p> |
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10921 <p class="runhead"> |
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10922 Distinctio octava. De corporibus regularibus |
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10923 </p> |
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10924 <p class="main"> |
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10925 <lb></lb> |
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10926 le ala superficie dela portione dela spera .bac. E cosí responde in tutte portioni, o sien magiori o sien<lb></lb> |
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10927 menori de mezza spera ut inpe acuratissime probat. Et cetera. .53. |
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10928 </p> |
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10929 <p class="main"> |
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10930 E gli é una spera il cui diametro è .14. e la sua superficie .616. Levone con una linea pia-<lb></lb> |
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10931 na .100.bracia. de superficie. Domandose quanto se levará del diametro, cioé axis. Tu hai<lb></lb> |
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10932 che la spera .abcd., che il diametro.ad. è .14. E la linea dividente .bc. tira; .ab. e dí che la sia<lb></lb> |
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10933 .1.co., redopia, fa .2.co., multiplica in sé, fa .4. ce .che sonno equali a .100.bracia., per la ragione asi-<lb></lb> |
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10934 gnata in la precedente. Ma prima piglia li .11/14. de .4.ce. ne ven .3 1/7.ce. e questo sirá equale a .100. Parti, ave-<lb></lb> |
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10935 rai la cosa valere aponto .R.31.9/11. e tanto sirá .ab. Ora multiplica .ad., ch’ é .14., in sé, fa .196., tranne la possan-<lb></lb> |
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10936 za de .ab., ch’ é .31 9/11., resta .164 2/11. e la .R. di questo è .bd. Ora multiplica .R.31 9/11. via .R.164 2/11., fa .5223.<lb></lb> |
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10937 118/121., el quale parti per lo diametro del tondo, cioé dela spera, ch’ é .14., ne ven .R. 26 15484/23716. Quando <lb></lb> |
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10938 .14.parti<lb></lb> |
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10939 sia recato a .R., che fa .196., e tanto sia .be., che vene a essere catetto sopra .ad. del triangulo .abd.<lb></lb> |
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10940 Cava ora la possanza del .be. dela possanza del .ab., cioé .26 15484/23716., resta .5 21560/130438. per la possanza de-<lb></lb> |
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10941 lo .ae. Donca dirai che taglia del diametro.ad. la quinta .ae., che è .R.5 21560/130438. Fatta et cetera. 54.<lb></lb> |
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10942 E gli é una spera il cui diametro è .14.; taglione con una linea piana .5.bracia. Domando<lb></lb> |
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10943 che levará dela quadratura del corpo. Fa cosí. Vedi prima quanto è la linea dividen-<lb></lb> |
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10944 te .bc. a questo modo. Tu sai che .ae. è .5. e .de.9., ch’ é il resto del diametro, peró multiplica .5.<lb></lb> |
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10945 via .9., fa .45. e la .R.45. é .be., ch’ é la .1/2. dela dividente. E tu voi .ab., peró multiplica .ae.,<lb></lb> |
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10946 ch’ é .5., in sé, fa .25., gongnilo con la possanza .be., fa .70. e .R.70. sia .ab., la qual redopia commo .R.,<lb></lb> |
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10947 fa .R.280., donca piglia li .11/14. de .280., ne vene .148 4/7. E tanto e va dela superficie de ditta spera per la<lb></lb> |
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10948 portione .bac. E tu voli del corpo, peró multiplica .148 4/7. per la .1/2. del diametro, ch’ é .7., fa .1040. e questo parti<lb></lb> |
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10949 per .3., ne ven .346 2/3., del qual se vol trare el cono .bcf., cioé cosí. Tu hai la linea .bc., che è .R.180. e<lb></lb> |
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10950 perché s’ é ditto che .be., ch’ é la .1/2. del .bc., é .R.45., siché dopia é .R.180. E la possanza de .bc., ch’ é dia-<lb></lb> |
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10951 metro dela basa de ditto cono, si’ é .180., pigliane li .11/14., ne ven .142 6/7. e questo multiplica per .fe., ch’ é l’ alteza<lb></lb> |
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10952 de ditto cono, ch’ é .285 5/7., pigliane el .1/3., ne ven .95 5/21., trallo de .346 2/3., che di sopra havesti, re-<lb></lb> |
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10953 sta .251 3/7. e tanti braci leva del solido de ditta spera. 55. |
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10954 </p> |
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10955 <p class="main"> |
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10956 E gli é una spera il cui diametro è .14.bracia. Segolo in doi luoghi con doi linee piane eque-<lb></lb> |
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10957 distanti. La prima taglia del diametro .3.bracia, l’ altra ne taglia .6. Dimando quan-<lb></lb> |
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10958 ta superficie sirá tra l’ una linea e l’ altra. Fa cosí. Vedi prima quanto é la linea che<lb></lb> |
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10959 taglia .6.bracia. de diametro. Tu sai che ’l diametro .ad. è .14. e la linea .f.g. lo sega in<lb></lb> |
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10960 ponto .h., che .ah. è .6.hd.8. Multiplica .6. via .8., fa .48. e .R.48. sia .fh. Ora multiplica .6. in sé, fa<lb></lb> |
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10961 .36., giongnilo ala possaza de .fh., cioé con .48., fa .84. e .R.84. sia .af. Redopiata fa .R.336. Per<lb></lb> |
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10962 quello che di sopra s’ é ditto, multiplica in sé, fa .336., pigliane li .11/14., ne ven .264. e questo tieni a mente, per-<lb></lb> |
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10963 ché l’ é la superficie dela portione .fag. Poi trova la superficie dela portione .bac. per la me-<lb></lb> |
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10964 desima via. Cioé trova .ab. E tu sai per la linea .bc. che sega .ad. in ponto e che .ae. è .3. e lo<lb></lb> |
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10965 .ed.11. Multiplica .3. via .11., fa .33. e .R.33. é .be. Poi quadra .ae., ch’ é .3., fa .9.; quadra .be., fa .33.; giongni<lb></lb> |
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10966 insiemi fa .42. e .R.42. é .ab.; dopialo, fa .R.168., multiplica in sé, fa .168., pigliane li .11/14., ne ven .132.<lb></lb> |
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10967 e questo cava de quello che sopra serbasti, cioé de .264., resta .132. e tanti bracci se levará de super-<lb></lb> |
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10968 ficie de ditta spera fra le doi linee et cetera. 56. |
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10969 </p> |
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10970 <p class="main"> |
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10971 E gli é una spera il cui diametro è .14. ad.; meno doi linee piane equidistante l’ una<lb></lb> |
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10972 dal’ altra; l’ una sega el diametro in ponto .e., l’ altra in ponto .h.; .ae. è .3.ah. è .6. e la su-<lb></lb> |
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10973 perficie ch’ é tra l’ una e l’ altra è .132. Domando che si levará de quadratura del cor-<lb></lb> |
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10974 po solido fra una linea e l’ altra. Tu sai che ’l diametro dela spera è .14.; piglia la<lb></lb> |
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10975 mitá, ch’ é .7. e multiplica .7. via la superficie dela portione .fag., fa .1848., pigliane .el.1/3., ne vene<lb></lb> |
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10976 .616. e de questo cava el cono. Tu hai, per la precedente, che .fh. è .R.48., el quale redopia, fa .R.<lb></lb> |
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10977 .192., multiplica in sé, fa .192., pigliane li .11/14., ne vene .150 6/7., multiplica per .1., che è l’ alteza, cioé da .h.<lb></lb> |
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10978 fin al centro, fa pur .150 6/7., pigliane el .1/3., ne ven .50 2/7., trallo de .616., resta .565 5/7. e tanto si leva de qua-<lb></lb> |
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10979 dratura del corpo sperico per la portione .fag. Dela qual cava la portione .bac., a modo che di<lb></lb> |
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10980 sopra in l’ altre festi, e restaratte per la ditta quadratura .349. et cetera. E cosí tu per te in ciascuna si-<lb></lb> |
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10981 mili farai. E queste, a tuo documento, sienno bastanti et cetera. 57<lb></lb> |
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10982 Quadrupla è la superficie di qualunque spera ala superficie dil magior cerchio in<lb></lb> |
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10983 essa contenuto, che cosí si prova. Sia una sphera qual voli e sia de poi una superficie qua-<lb></lb> |
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10984 drupla al magior cerchio che in essa capi e sia el cerchio .a., qual dico ch’ é equal a tut-<lb></lb> |
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10985 ta la spoglia dela sphera. Se non è equale, donca è magior overo minore. Or<lb></lb> |
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10986 sia prima magiore per l’ aversario la superficie dela spera che ’l cerchio .A. Donca habiamo<lb></lb> |
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10987 doi quantitá inequali, l’ una la spera, l’ altra el cerchio .A. Possibile adonca prendere doi linee<lb></lb> |
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10988 <lb></lb> |
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10989 <pb></pb></p> |
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10990 <p class="folio"> |
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10991 folio 74r |
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10992 </p> |
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10993 <p class="main"> |
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10994 <lb></lb> |
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10995 </p> |
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10996 <p class="runhead"> |
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10997 Distinctio octava. De Corporibus regularibus. 74 |
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10998 </p> |
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10999 <p class="main"> |
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11000 <lb></lb> |
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11001 recte inequali in tal modo che la magiore habia minore proportione ala minore che la superficie dela spera al cer-<lb></lb> |
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11002 chio .A. E sienno ditte linee .Bc.b. maior e .c. minor. Poi, fra queste, per la .9a. del .6o., trova la linea .d. media pro-<lb></lb> |
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11003 portionale. Poi sia una superficie piana che seghi ditta spera in doi parti e passi per lo centro de decta spera e sia decta<lb></lb> |
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11004 superficie piana el cerchio .efgh., el qual, de necessitá, sirá el magior cerchio che sia in spera. Poi in questo cerchio <lb></lb> |
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11005 imagina<lb></lb> |
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11006 una figura multiangola inscripta e una altra figura pur multiangola circunscripta simile ala prima inscripta, cioé<lb></lb> |
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11007 che ‘lati del’ una a ’lati del’ altra sienno proportionali e gli angoli di l’ una iguali ali angoli del’ altra, secondo<lb></lb> |
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11008 la diffinitione prima del .6o. E poni che il lato dela figura circunscripta al lato dela figura inscripta habia minor pro-<lb></lb> |
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11009 portione che .b. al .d. linee. Queste cose cosí disposte, arguesci in questa forma lo intento dicendo: la superficie dela<lb></lb> |
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11010 multiangola inscripta è minore che la superficie del cerchio. A, perché el cerchio .A. quadruplo al cerchio .efgh. nel qual<lb></lb> |
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11011 ditta multiangola è inscripta, perché gli é il maxino nele spera contenuto, perché la segha in centro, donca ma-<lb></lb> |
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11012 gior proportione ará la superficie dela spera ala inscrita multiangola che al cerchio .A. Donca la superficie multiangola al <lb></lb> |
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11013 ditto <lb></lb> |
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11014 cerchio cir-<lb></lb> |
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11015 cunscrita è minore che la proportione dela spera, perché ex ypotesi dala circunscrita ala inscrita é minore proportione <lb></lb> |
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11016 che da-<lb></lb> |
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11017 la superficie dela spera al cerchio .A., peroché foron posti i lati de ditte figure multiangole in tal modo che ’l la-<lb></lb> |
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11018 to dela circunscripta al lato dela inscripta havesse minor proportione che .b. al .d. E, per consequente, asa mol-<lb></lb> |
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11019 to minore che dal .b. al .c. e poi molto piú che dala spera al cercio .A. Diché sequita che, duplata la pro-<lb></lb> |
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11020 portione dal lato dela circunscripta al lato dela inscripta, sirá anche minore che duplata la proportione dal<lb></lb> |
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11021 .b. al .d., cioé dal .b. al .c., perché la proportione dal .b. al .c. è doppia ala proportione dal .b.d., per la .10a. diffinitione del<lb></lb> |
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11022 .5o. Ma la proportione della circunscripta piana ala inscripta piana é doppia ala proportione del lato dela cir-<lb></lb> |
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11023 cunscripta ala inscripta, per la .18a. del .6o., perché sonno figure simili. Ora ymagina doi figure solide multi-<lb></lb> |
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11024 angole facte de medesimi lati dela inscritta e circunscritta e che una sia dentro ala spera e l’ altra di fora, che li angoli <lb></lb> |
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11025 de-<lb></lb> |
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11026 le intrinseca tochino equaliter la pancia dentro e quella di fore le base tochino ditta spera aponto. Poi recogli tut-<lb></lb> |
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11027 ta la superficie dela figura solida circunscripta e anche recogli tutta la superficie dela figura solida inscripta. Sirá ancora <lb></lb> |
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11028 mi-<lb></lb> |
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11029 nore la proportione de tutta la superficie dela solida circunscripta a tutta la superficie dela solida iscripta che dal .b.<lb></lb> |
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11030 al .c., cioé che la doppia dal .b. al .d. è molto piú minore ancora che quella dela spera al cerchio .A. E sia<lb></lb> |
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11031 la intrinseca ala extrinseca de quante voi base, che pare siranno a numero lá dentro e lá di fore, havenga che non si<lb></lb> |
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11032 grandi parti. Ora, questo notato, dirai la superficie dela solida intrinseca tutta mai sirá quanto la superficie di tuta la <lb></lb> |
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11033 spera, <lb></lb> |
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11034 perché<lb></lb> |
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11035 abia quante face si voglia mai s’ aguaglia ala spera. Donca dirai che la superficie tutta dela solida circunscripta ará<lb></lb> |
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11036 magiore proportione al cerchio .A. che non haverá la spera al cercio .A., perché molto piú magiore é la superficie dela<lb></lb> |
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11037 multiangola circunscripta che la spera e, per consequente, magior proportione averá la superficie dela multiangola <lb></lb> |
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11038 circunscrip-<lb></lb> |
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11039 ta ala multiangola inscripta che la spera a essa multiangola inscripta. Le qual cose fin qui serba in mente. Poi ar-<lb></lb> |
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11040 gumenta cosí: o la superficie tutta dela multiangola inscripta è magior o minor o uguale al cerchio .A. E sia qual vo-<lb></lb> |
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11041 glia, sequita lo inconveniente contra la ypotesi. Or sia prima la superficie tutta dela multiangola intrisica magior che ’l <lb></lb> |
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11042 cerchio<lb></lb> |
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11043 .A., donca minore sia la proportione dala spera a essa che dala spera al cerchio .A. E pur magior nondimen sia an-<lb></lb> |
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11044 cor la proportione dela multiangola extrinsica ala multiangola intrinsica che dala spera ala intrinsica et ex unti magior si-<lb></lb> |
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11045 <lb></lb> |
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11046 rá la proportione dela extrinsica al’ itrinsica che dala spera al cerchio .A. E giá prima avamo minore per la ypotesi ecco<lb></lb> |
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11047 contra al presupposto secondo non é ita, che la spera sia piú che .4. tanto dela superficie del maximo cerchio in lei <lb></lb> |
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11048 contenuto. El me-<lb></lb> |
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11049 desimo molto piú magior sequiri la proportione del’ estrinseca ala intrinsica che dala spera al cerchio .A., quando la in-<lb></lb> |
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11050 trinsica fosse equale al cerchio .A., over minor, arguendo commo finora é facto. Donca non è piú che .4. tanto. E an-<lb></lb> |
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11051 co, per le medesime vie, poi provare che non è manco che .4. tanto del ditto cerchio, ponendo ogni<lb></lb> |
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11052 cosa commo di sopra, excepto che ’l cerchio .A. sia magior che la spera, mediante ditte linee proportionali<lb></lb> |
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11053 e figure multiangole intrinseca e extrinseca et cetera. Nota che tal proportione è da una superficie multiangola a un’ al-<lb></lb> |
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11054 tra multiangola simili, descripte intra doi cerchi, qual è dal quadrato de’ diametri de’ lor cerchi uno al’ altro, per la<lb></lb> |
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11055 prima del .12o. Le simili s’ intendeno, commo è ditto in principio del .6o., d’ angoli iguali e lati che li contengono pro-<lb></lb> |
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11056 portonali a’ sui relativi. E ancora de ponto la superficie dela spera, per ditto archi, quanto la superficia d’ un qua-<lb></lb> |
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11057 drangolo ortogonio che sia contenuto sotto linee equali al diametro dela spera e ala circunferentia del ma-<lb></lb> |
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11058 ximo cerchio in lei contento. Commo sia il diametro dela spera .7., la circunferentia del magiore cerchio sia .22. Don-<lb></lb> |
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11059 ca il quadrangolo a lei equale sia longo .22. e largo .7. e possiede .154. che è il medesimo che habiam mostro.<lb></lb> |
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11060 E questo havene perché tal quadrangolo sia composto dela superficie de ponto di .4. cerchi. maximi dela spera commo <lb></lb> |
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11061 de<lb></lb> |
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11062 sopra è detto. Del modo a far li stagiuoli e tavola de scemi in ogni luogo.<lb></lb> |
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11063 Parrá forse al pratico geometra che, in questo trattato de geometria, al suo bisogno, io sia stato di<lb></lb> |
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11064 minuto. Conciosiaché al suo principal desiderio, qual è de sapere ben con tuta diligentia me-<lb></lb> |
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11065 surare el tenuto d’ una botte e anche el suo scemo. E anche, per tutti ’luoghi, fabricar-<lb></lb> |
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11066 se i stagiuoli a sue occurentie dele prime non l’ abia suvenuto. Diché, qui sequente, el bastante a ció intendo<lb></lb> |
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11067 dire. Sappi che, de tutte le mesure geometriche, quella dela botte è difficile e del scemo difficili-<lb></lb> |
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11068 sima, peroché, se finora d’ una figura semplici plana circulare non s’ á per li antichi e moderni philosophy aponto<lb></lb> |
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11069 sua quadratura e manco dele sue portioni, molto magiormente sia ascosta la demensione circulare<lb></lb> |
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11070 <lb></lb> |
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11071 <pb></pb></p> |
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11072 <p class="folio"> |
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11073 folio 74v |
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11074 </p> |
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11075 <p class="main"> |
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11076 <lb></lb> |
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11077 </p> |
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11078 <p class="runhead"> |
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11079 Distinctio octava. |
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11080 </p> |
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11081 <p class="main"> |
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11082 <lb></lb> |
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11083 sperica e di quelle che a spherico tendano, quali non solo importano piano ma ancora soliditá. E ben-<lb></lb> |
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11084 ché di sopra in questo mostrassemo di quadrare i cerchi e sue parti, nella distinctione .4a. al capitolo primo a carti-<lb></lb> |
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11085 .26., questo fo facto solo con lo presuposito de Archimede che, in lo suo .3o. libro, che fa de quadratura circuli,<lb></lb> |
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11086 ci mostra la proportione del suo diametro ala circunferentia esser magior che tripla sexquictava e minore che<lb></lb> |
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11087 tripla sexquiseptima, onde per questo finora ci siamo fermati a questo: che ’l diametro multiplicato via .3 1/7. facia la<lb></lb> |
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11088 sua circunferentia, che con precisione non é da mettere, ma poco varia: commo dici Photolomeo è quantitá<lb></lb> |
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11089 non sensibile. Se tu domandi perché piú á .3 1/7. che .3 1/8. ci siamo fermati. Si fa perché menore è la differenza<lb></lb> |
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11090 da .3. tanto e .1/7. ala circunferentia, che non è da .3. tanto e .1/8. e per questo s’ é facto. E, se ’l si havesse aponto la<lb></lb> |
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11091 proportione del diametro ala circunferentia aponto, ancora quadraremo el cerchio e sue parti. La qual finora non s’ é<lb></lb> |
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11092 trovata. Esser porria che giá fosse nato colui che ci habia a dar modo aponto quadrarlo. La cui pos-<lb></lb> |
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11093 sibilitá per niun philosofo se denega. Onde AR. dici che scientia de quadratura circuli est scibilis et dabilis<lb></lb> |
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11094 quamvis nondum sit scita neque data. E sopra di questo molti se sonno affatigati, tutti li matemaci, ma-<lb></lb> |
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11095 xime Raymundo. Havenga che nel .1489. nella citá de Roma dove publice legiavamo Maestro Pier-<lb></lb> |
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11096 Lione da Spoleti, medico che lí se stava in casa del Reverendisimo Cardinal de San Marco a sua Revederendisima Santita <lb></lb> |
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11097 (me <lb></lb> |
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11098 pre-<lb></lb> |
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11099 sente e tutti a una mensa per sua humanitá) mostró un libro in .4o. foglio, de circa carti .150., impresso ultra-<lb></lb> |
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11100 montes, compilato per un certo vescovo di quelle parti dove lui diciva haverlo studiato tutto. E che<lb></lb> |
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11101 altro non trattava che de quadratura circuli con moltissime figure e diciva che la concludiva. La qual<lb></lb> |
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11102 cosa non poddi mai vedere el libro dapoi non ebbi in libertá. Or commo sia stiamoci ancora un poco<lb></lb> |
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11103 a’ patti vechi e modi usati, finche ’l certo si trova. Al proposito nostro tornando. Dico che le botti tut-<lb></lb> |
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11104 te, a bene mesurarle, convengonse prenderle a modo piramide corta. E per imaginatione sempre tagliarla<lb></lb> |
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11105 nel mezzo al cochiume e mesurarne la mitá. E poi, quel che fa, dopiare e harai tutto el tenuto. La<lb></lb> |
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11106 qual cosa farai cosí, commo succintamente di sopra, nel trattato de’ corpi regolari, in una domanda te<lb></lb> |
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11107 disti e anco fra li casi geometrici in la distitone .8a. al caso .2o. in ordine. Prima prendi l’ altezza del cochiume e l’ al-<lb></lb> |
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11108 tezza d’ uno deli fondi e la distantia dal’ un fondo al’ altro, che, cosí facto, harai una piramide corta<lb></lb> |
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11109 la quale, a ben mesurare, te conven finire fin al suo cono, commo te mostrai sopra nella distinctione .6a.<lb></lb> |
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11110 al capitolo .2o. carti .44. E poi d’ essa trovare l’ area corporale al modo lí dato, qual, trovata commo è ditto,<lb></lb> |
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11111 doppiala e harai el tenuto quadrato de tutta la botte, secondo la mesura che tu arai adoptata, cioé<lb></lb> |
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11112 se arai facto a piedi, arai piedi quadrati e, se arai li toi stagiuoli divisi a bracia, arai a bracia quadri et cetera. Ma que-<lb></lb> |
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11113 sto modo presuppone sempre nel vaso uniformitá la qual cosa, al piú dele volte, nelle botti, non si tro-<lb></lb> |
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11114 va, per esser mal facte e con doghe non equali et cetera. E peró bisogna a te adoperare oltra l’ arte ancora el tuo in-<lb></lb> |
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11115 gegno in guardare la botte con diligentia dentro e di fore e considerare piú apontino si pó quello che<lb></lb> |
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11116 si perde overo avanza ala uniformitá et cetera. E questo non è possibile con penna insegnartelo, peró a te lo<lb></lb> |
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11117 lascio meglio saparai et cetera. Ora, per li scemi, sonno per molti fabricate assai tavole di quali uno fo chia-<lb></lb> |
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11118 mato garofano, l’ altro modo paulo e alcune son chiamate tavole del .60., altre del .12., altre del .10., altre<lb></lb> |
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11119 de AR. e son quelle che hano molti rotti, bench’ io creda che mai AR. le vedesse, ma li vulgari li ha-<lb></lb> |
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11120 no dato questo nome per piú auctentico lor fondamento. Or commo se sia le tavole ognun le pó da sé fa-<lb></lb> |
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11121 re experimentado, commo altri sempre hano facto, perché de loro, commo è ditto, non c’ é scientia certa e peró, a far-<lb></lb> |
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11122 tele da te, tirrai questo modo. Fa de haver una botte meglio lavorata sia possibile per bon maestro e,<lb></lb> |
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11123 s’ altramente non la trovasse conporrate con un bottaro, quando e gli é per farne una, che vi duri in tuo ser-<lb></lb> |
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11124 vitio alquanto piú fatiga in giustarla e usali qualche cortesia e farallo. E poi quella situa a livello in luo-<lb></lb> |
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11125 go piano e falla impire d’ aqua. Facto questo prenderai li tuoi stagiuoli segnati e divisi a tuo modo, ma<lb></lb> |
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11126 comunamente se dividano in .12. parti equali e ognuna se chiama ciuocolo per la Toscana, verso el Borgo<lb></lb> |
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11127 nostro, Castello, Arezzo, Cortona e poscia altri li chiama braciuoli et cetera. E con questi stagiuoli prendi a livel-<lb></lb> |
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11128 vello l’ altezza del fondo denanze che pongo l’ abi conmenzato e pregiato giá con quello di drieto. E segna<lb></lb> |
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11129 una riga sul ditto fondo, aponto in mezzo, che sia diametro. El qual diametro dividi sutilmente secondo li ponti del tuo<lb></lb> |
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11130 stagiuolo in bracia e {+} et cetera. E poi, supra ditta linea, de mano in mano, commenzando di sopra, andarai con un tri-<lb></lb> |
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11131 vellino, over veriuola, piú sotil se possa, forando e quinde uscirá l’ aqua, qual fa che aponto tutta la recol-<lb></lb> |
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11132 ga in qualche vaso sotto la botte prima asettato. E mesurala con la mesura del campione del paese dove<lb></lb> |
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11133 te trovi a foro per foro e segna nel foglio da parte li bocali, petitti, fogliette et cetera. E cosí va facendo<lb></lb> |
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11134 fin in fondo, a foro per foro. E arai, metiamo, che fin al primo foro de sopra sirá insita .4. bocali e fin al .2o.9.<lb></lb> |
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11135 e fin al terzo .22. E cosí discorrendo in tutti. E quanti piú spessi farai li fori, cioé in bracci e<lb></lb> |
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11136 mezzi bracci e quarti bracci e octavi bracci et cetera, meglio te sia. E, questo facto, tieni questa ta-<lb></lb> |
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11137 vola scripta apresso te, con li ponti denanze, cioé, braccia, once, ponti et cetera. E al’ incontro<lb></lb> |
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11138 porrai el numero de’ bocali, o barili, et cetera, che te sirá sempre tuo maestro a tutte altre botti<lb></lb> |
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11139 che con li medesimi stagiuoli mesurasse, peroché sempre le tavole di scemi prosupponga-<lb></lb> |
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11140 no la mesura con che loro foron fabricate. E de qua nasci che li mesuratori ale volte non s’ a-<lb></lb> |
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11141 <lb></lb> |
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11142 <pb></pb></p> |
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11143 <p class="folio"> |
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11144 folio 75r |
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11145 </p> |
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11146 <p class="main"> |
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11147 <lb></lb> |
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11148 </p> |
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11149 <p class="runhead"> |
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11150 Distinctio octava. 75 |
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11151 </p> |
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11152 <p class="main"> |
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11153 <lb></lb> |
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11154 cordano nelli scemi, havenga che usino la medesima tavola, perché la tavola sirá facta a bracia<lb></lb> |
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11155 e ‘l stagiuolo di colui sirá a piedi magiori o menori del bracio. E quando poi torrai el scemo<lb></lb> |
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11156 d’ alcuna botte con lo tuo stagiuolo vedi quanti ponti prende piú aponto possi e, se piglia un<lb></lb> |
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11157 sol ponto recorri ala tavola giá facta e vedi scontra un ponto la quantitá segnata de’ boccali o de’ ba-<lb></lb> |
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11158 rili et cetera e dirai: uno ponto me dá tanto et cetera. E cosí, se havesse .2. ponti, dirai che te dia .9. e, se .3. <lb></lb> |
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11159 ponti,<lb></lb> |
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11160 dirai che te dia .22. et cetera, commo qui habiamo ditto, et cetera. Ma, perché le botti non sonno tutte longhe a <lb></lb> |
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11161 un<lb></lb> |
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11162 modo né tutte alte a un modo, peroché in Toscana e di lá per tutto fin Roma sonno botti alte corte e gran-<lb></lb> |
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11163 di. E in Romagna, commo a Rimino e Cesena et cetera sonno botti longhe a basse, per rispecto che le porta-<lb></lb> |
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11164 no in carri. Non guardare a questo, che la tua mesura fatta in un luogo te serve per tutto, ma te con-<lb></lb> |
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11165 verrá proportionarla per tutto dove vai con quella del paese che fosse, se non te converria farne un’ altra.<lb></lb> |
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11166 de nuovo per quello. Siché bisogna sappi far ragione, se la tua te dá metadelle o bocali e tu vo-<lb></lb> |
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11167 lesse raspondere a petitti. Convente esser noto el tuo vaso che parte sia di quelle, over quello che parte<lb></lb> |
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11168 sia del tuo e poi respondere e mai falle. Peró tutte le mesure de una qualitá e natura sonno fra<lb></lb> |
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11169 loro proportionate in modo che, per la piccola, se trova la grande e, per la grande, si trova la<lb></lb> |
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11170 piccola, quando sien simili, cioé tutti cerchi o tutti quadri o tutti triangoli et cetera. E cosí riesci ne-<lb></lb> |
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11171 li corpi aponto. E tu, per te, piú amplamente in ció te stenderai et cetera.<lb></lb> |
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11172 <lb></lb> |
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11173 Tavola dela .2a. parte principale de tutta l’ opera dove se trat-<lb></lb> |
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11174 ta de Geometria in tutti li modi Theorica e pratica.<lb></lb> |
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11175 <lb></lb> |
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11176 Divisione de tutto el trattato de geometria. K. pa. |
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11177 </p> |
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11178 <p class="main"> |
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11179 Divisione e continentia dela prima distinctione. K. pa. |
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11180 </p> |
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11181 <p class="main"> |
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11182 Dele cose che sonno necessarie al buono agrimensore e pratico geometra,<lb></lb> |
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11183 quali per numero son .5., commo apare al primo capitolo dela ditta di. K. pa.<lb></lb> |
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11184 Del sentimento necessario senza el quale non è possibile in geometria bene operare<lb></lb> |
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11185 de tutte le conclusioni e demostrationi del primo libro de Euclide. cao. 2 K. 2<lb></lb> |
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11186 Substantia efficacissima del secondo libro de Euclide. cao. 3 K. 4<lb></lb> |
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11187 De tutte le conclusioni e demostrationi dignissime del sexto libro. de Eucli. cao. 4 K. 5<lb></lb> |
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11188 Del modo a mesurare secondo el degno strumento fiorentino aplica-<lb></lb> |
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11189 bile a ciascun altro in tutte parti. cao. 5 K. 7<lb></lb> |
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11190 Del modo a mesurare tutte le superficie quadrate. cao. 6 K. 7<lb></lb> |
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11191 Del modo a saper mesurare tutte sorte de’ triangoli. cao. 7. K. 8<lb></lb> |
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11192 Del modo a saper trovare tutti li catetti, over perpendiculari, in le figure triangulari. E com-<lb></lb> |
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11193 me secondo un vulgar modo s’ usi in sul terreno a mesurare assai commendabile in<lb></lb> |
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11194 la pratica usuale. La seconda distinctione.] cao. 8 K. 10<lb></lb> |
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11195 Del modo a saper trovare la quantitá de una linea menata da un ponto dato<lb></lb> |
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11196 de fore, over dentro, d’ alcun triangolo.] cao. po. K.13<lb></lb> |
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11197 Dela protractione extrinseca dela ypotumissa nel triangolo ortogo-<lb></lb> |
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11198 nio regola optima.] cao. 2 K. 15<lb></lb> |
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11199 <lb></lb> |
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11200 La terza distinctione. |
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11201 </p> |
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11202 <p class="main"> |
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11203 Del dilicato modo a saper solvere varie e diverse questioni proposte sopra le figure<lb></lb> |
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11204 quadrilatere, cioé de .4. lati, che sienno rettangole, per via algebratica. cao. po. K. 16<lb></lb> |
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11205 Del modo a saper trovar l’ area, over superficie, dele figure quadrilatere, ditte dal vul-<lb></lb> |
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11206 go Rombi e da Eucli. helmuaym, che sonno figure de .4. lati equali e non hano<lb></lb> |
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11207 alcun angolo retto, ma solo li .2. oppositi equali e obtusi e li altri. 2. acuti, pur equali<lb></lb> |
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11208 cao. 2 K. 21<lb></lb> |
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11209 Del modo a saper trovare l’ area dele figure de .4. lati, dal vulgo chiamate Rom-<lb></lb> |
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11210 boide e da Euclide Simili helmuaym, che solo hano li lati oppositi<lb></lb> |
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11211 equali e fra loro equidistanti e manca de angoli retti, cioé che non é rettangola. cao. 3 K. 22<lb></lb> |
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11212 Del modo a trovar l’ area dele superficie dele figure quadrilatere decte dal vulgo Ca-<lb></lb> |
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11213 potagliato, cioé Caput abscisum e per Eucli. chiamate helmuariphe: e son-<lb></lb> |
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11214 no quelle che hano doi lati oppositi equali e non equedistanti. E li altri doi<lb></lb> |
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11215 equedistanti e non equali e non hano alcun angolo retto.] cao. 4 K. 23<lb></lb> |
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11216 Del modo a mesurare le figure quadrilatere, ditte da’ vulgari mezzo capo<lb></lb> |
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11217 tagliato e da Eucli. similmente helmuariphe chiamate, che hano doi<lb></lb> |
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11218 lati oppositi equidistanti non equali e li altri doi non sonno equali fra loro<lb></lb> |
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11219 né equedistanti. E hano solo doi angoli retti: uno ala basa e l’altro al capo. cao. 5 K. 24<lb></lb> |
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11220 Del modo a saper trovar l’ area dele figure quadrilatere, ditte dal vulgo<lb></lb> |
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11221 Capo tagliato declinante, dele quali el capo e la basa sonno equedistanti non<lb></lb> |
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11222 <lb></lb> |
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11223 <pb></pb></p> |
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11224 <p class="folio"> |
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11225 folio 75v |
|
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11226 </p> |
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11227 <p class="main"> |
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11228 <lb></lb> |
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11229 </p> |
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11230 <p class="runhead"> |
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11231 Distinctio octava |
|
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11232 </p> |
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11233 <p class="main"> |
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11234 <lb></lb> |
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11235 equali. E li altri doi non equidistanti che fano in sula basa doi sorte an-<lb></lb> |
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11236 goli: cioé l’ uno obtuso e l’ altro acuto. cao. 5 k. 24<lb></lb> |
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11237 Del modo a trovar l’ area dele figure Multilatere, cioé figure de piú<lb></lb> |
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11238 de .4. lati, como sonno pentagoni, exagoni, eptagoni et cetera. cao. 6 k. 25<lb></lb> |
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11239 La quarta distinctione. |
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11240 </p> |
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11241 <p class="main"> |
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11242 Delucidatione de tutte le conclusioni e demostrationi del terzo libro de<lb></lb> |
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11243 Euclide. Summamente necessari a chi vole haver notitia del’ area su-<lb></lb> |
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11244 perficiale deli cerchi e de tutte le sue parti commo immediate da poi<lb></lb> |
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11245 ordinatamente se dirá. cao. po. k. 26<lb></lb> |
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11246 Del modo a saper ritrovar l’ area dele figure circulari e dele sue parti, como se-<lb></lb> |
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11247 micirculi, portioni magiori e minori e settori de cerchi. E del mo-<lb></lb> |
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11248 do a trovar li archi per le corde, con la tavola de corda et arcu, secondo<lb></lb> |
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11249 Ptolomeo in suo Almegesto. cao. 2 k 30<lb></lb> |
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11250 Del modo a saper mesurare le superficie in monti e valli. cao. 3 k 35<lb></lb> |
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11251 La quinta distinctione. |
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11252 </p> |
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11253 <p class="main"> |
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11254 Del modo a saper dividere le superficie triangolari in quante parti se voglia<lb></lb> |
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11255 tirando le linee da qualunche angolo se voglia e cosí da qualunche pon-<lb></lb> |
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11256 to se voglia segnato nelli soi lati, over de fore, in tutti li modi. cao. po. k. 36<lb></lb> |
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11257 Del modo a sapere dividire le figure de .4. lati in quante parte faccia mestieri. cao. 2 k. 38<lb></lb> |
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11258 Del modo a saper dividere giustamente le figure multilatere, commo<lb></lb> |
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11259 sonno pentagoni, exagoni et cetera, in quante voi parti. cao. 3 k. 42<lb></lb> |
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11260 Del modo a dividere le figure circulari in piú parti con diversi tagli.] cao. 4 k. 43<lb></lb> |
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11261 La sexta distinctione. |
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11262 </p> |
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11263 <p class="main"> |
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11264 Dele cose necessarie a chi vol saper mesurare tutte le sorti de’ corpi. Com-<lb></lb> |
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11265 mo sonno Cubi, Seratili, Colonne tonde e laterate e cosí le lor pyrami-<lb></lb> |
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11266 di sane e scapezze, con le loro dechiaratione e diffinitoni poste dal philosopho.<lb></lb> |
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11267 Eucli. nel suo .11o. libro, senza la cui notitia non è possibile del’ area corpora-<lb></lb> |
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11268 le d’ alcum solido havere intelligentia. cao. po. 2. 44<lb></lb> |
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11269 Del modo a saper mesurare un solido rettangolo e cosí li cubi. cao. 2 k. 44<lb></lb> |
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11270 Dechiaratione del corpo Seratile e del modo a mesurarlo con quello de colonne<lb></lb> |
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11271 tonde e laterate e cosí de tutte pyramide, cioé tonde laterate e scapez-<lb></lb> |
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11272 ze e ancora dele loro portioni. cao. 3.k. 45<lb></lb> |
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11273 Dela mesura dele superficie Spherali e dela loro capacitá corporale<lb></lb> |
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11274 e ancora de tutte le sue portioni, o sieno magiori over menori de 1/2 spera. cao. 4 k. 49<lb></lb> |
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11275 La septima distinctione. |
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11276 </p> |
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11277 <p class="main"> |
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11278 Deli strumenti e modi diversi con li quali se costuma mesurare solo con l’ aspe-<lb></lb> |
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11279 cto, cioé col vedere, ogni distantia de longheza, alteza, largheza e profonditá, pur-<lb></lb> |
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11280 ché l’ ochio la possa scorgere, senza movarse de luogo. E anche del mo-<lb></lb> |
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11281 do a mesurare quelle che per alcun impedimento non se potessero vedere, per la<lb></lb> |
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11282 notitia dele visibili lor certa mesura proportionalmente arguire. cao. po. k. 50<lb></lb> |
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11283 De diversi casi che col vedere pó occorrere a mesurare. Exemplificativi<lb></lb> |
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11284 a ciascun altro che te acadesse de Torri, Pozzi, Campagne, Valli, Fiu-<lb></lb> |
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11285 mi, Fossi et cetera. Dove l’ Homo habilmente non se potesse acostare. Commo in-<lb></lb> |
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11286 terven per guerra a muraglie, fortezze, bastioni o ali ostacoli che impacessero.] cao. 2. k. 50<lb></lb> |
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11287 La octava distinctione Solo per numero de’ casi distinta. |
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11288 </p> |
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11289 <p class="main"> |
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11290 De diversi casi gentilissimi indifferentemente posti ala pratica geometrica<lb></lb> |
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11291 spectanti. E applicabili a tutte occurentie in ciascuna operatione commo<lb></lb> |
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11292 legendo intenderai, distincti solamente fra loro per numero de petitioni, per tutta<lb></lb> |
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11293 ditta distinctione compresi, che per numero sonno .100., de dignissim piaceri. k. 53<lb></lb> |
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11294 Trattato particulare de tutti li .5. corpi regolari e sue mesure quali sonno de<lb></lb> |
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11295 grandissimo artificio e la lor scientia é la subtilissima che esser possi in tut-<lb></lb> |
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11296 te le discipline mathematici. E da ciascun phylosopho prosuposta. Similmen-<lb></lb> |
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11297 te per numero de’ casi distincto che sonno .59. k. 69<lb></lb> |
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11298 Prova e demostratione apertissima secondo Archimede commo la superfi-<lb></lb> |
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11299 cie di ciascuna spera é quatro tanto che l’ area del suo maximo cer-<lb></lb> |
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11300 chio in essa contenuto. k. 74<lb></lb> |
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11301 Modo a far tavole de scemi e stagiuoli a mesurar botti in tutti luoghi cao. ultimo k.<lb></lb> |
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11302 <lb></lb> |
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11303 <pb></pb></p> |
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11304 <p class="folio"> |
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11305 folio 76r |
|
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11306 </p> |
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11307 <p class="main"> |
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11308 <lb></lb> |
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11309 </p> |
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11310 <p class="runhead"> |
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11311 Distinctio octava 76 |
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11312 </p> |
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11313 <p class="main"> |
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11314 <lb></lb> |
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11315 Pervenuti (Idio laudato e il Seraphyco Patriarca de sancta povertá. Padre<lb></lb> |
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11316 e fondatore del nostro sacro ordine Meser San Francesco benedecto) al desidera-<lb></lb> |
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11317 to fine delo intento nostro in questa utilissima opera commenzato, altro, ale vostre lau-<lb></lb> |
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11318 dabil exercitanti caritá, in remuneratione de tante fatighe, l’ umil servo di quelle non di-<lb></lb> |
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11319 manda, se non per lui l’ altissimo li piacia pregare che con quel felici fine che ogni buon cristiano<lb></lb> |
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11320 desidera a sé lo chiami e interim in lo curriculo de questa calamitosa vita, con sua gratia, lo go-<lb></lb> |
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11321 verni e guidi. E non manco per lo degno e Reverendo Piovano de santo Apostolo de Vinegia. Me-<lb></lb> |
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11322 ser pre Ysidero Bagnuoli. El simile per lo Mco. e Nobile dela excelsa Republica de Vinegia<lb></lb> |
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11323 Patritio e in le scientie Mathematici fondatissimo. E de tutti virtuosi colonna firmissima Me-<lb></lb> |
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11324 ser Marco Sannuto quondam Mci. domini Francisci. Perché mediante loro aiuto e favore tan-<lb></lb> |
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11325 ta commoditá de’ volumi al’ universo è consequita. Con spesa e diligentia. E opifitio del pru-<lb></lb> |
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11326 dente homo Paganino de’ Paganini da Brescia. Nella excelsa citá de Vinegia, con gran del<lb></lb> |
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11327 suo excelso Dominio, che per anni .x. proximi null’altro in quello la possi restampare né altrove<lb></lb> |
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11328 stampata in quello portarla sotto pena in ditta gratia contenuta. Negli anni de nostra Salute<lb></lb> |
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11329 M.cccc.lxliiij. adí .10. de novembre. Sotto el felicissimo Governo del D.D. de’ venitiani Au-<lb></lb> |
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11330 gustino Barbadico Serenissimo Principe di quello. Frater Lucas de Burgo sancti Se-<lb></lb> |
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11331 pulchri Ordinis minorum. Et sacre theologie humilis professor suo parvo ingenio ignaris<lb></lb> |
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11332 compatiens hanc summam Arithmetice et Geometrie Proportionunque et proportionalitum edi-<lb></lb> |
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11333 dit. Ac impressoribus assistens die noctuque proposse manu propia castigavit.<lb></lb> |
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11334 <lb></lb> |
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11335 <lb></lb> |
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11336 LAUS DEO<lb></lb> |
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11337 Registrum Geometrie.<lb></lb> |
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11338 <lb></lb> |
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11339 A<lb></lb> |
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11340 Tractatus<lb></lb> |
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11341 Ogni linea<lb></lb> |
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11342 Tutti i trian<lb></lb> |
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11343 Denanze in<lb></lb> |
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11344 e simile] C<lb></lb> |
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11345 a’ quali agion-<lb></lb> |
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11346 peró ciascuna<lb></lb> |
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11347 il lato brie<lb></lb> |
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11348 tetragono-<lb></lb> |
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11349 to fo .on.<lb></lb> |
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11350 E<lb></lb> |
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11351 havere<lb></lb> |
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11352 Sia il cerchio<lb></lb> |
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11353 E se l’ area<lb></lb> |
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11354 Al triangolo-<lb></lb> |
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11355 re se ’l qua] G<lb></lb> |
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11356 con la linea<lb></lb> |
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11357 Distinctio .7a. de<lb></lb> |
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11358 Se tu fosse<lb></lb> |
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11359 e .ab. che<lb></lb> |
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11360 me adunque] I<lb></lb> |
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11361 obscura<lb></lb> |
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11362 .3. che me dará<lb></lb> |
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11363 questo e quel<lb></lb> |
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11364 gia con la qua.<lb></lb> |
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11365 <lb></lb> |
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11366 B<lb></lb> |
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11367 def. |
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11368 </p> |
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11369 <p class="main"> |
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11370 E se ’l tri<lb></lb> |
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11371 Havendo mo<lb></lb> |
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11372 Onde oxigo<lb></lb> |
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11373 Ancora pos] D<lb></lb> |
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11374 sia il diame-<lb></lb> |
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11375 agiognerai<lb></lb> |
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11376 Se uno cer-<lb></lb> |
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11377 Dico che gli é<lb></lb> |
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11378 Per la .20a.] F<lb></lb> |
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11379 se multiplicarai<lb></lb> |
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11380 quadrilatero<lb></lb> |
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11381 detta propor-<lb></lb> |
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11382 Se uno solido<lb></lb> |
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11383 il seratile] H<lb></lb> |
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11384 Per la mitá<lb></lb> |
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11385 ditti cerchi<lb></lb> |
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11386 e perché lo la<lb></lb> |
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11387 dal’ una al’ al<lb></lb> |
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11388 po grosso] K<lb></lb> |
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11389 .2 2/9.<lb></lb> |
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11390 va.37348000.<lb></lb> |
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11391 se fa .400. |
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11392 </p> |
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11393 <p class="main"> |
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11394 E gli é una<lb></lb> |
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11395 rette inequali<lb></lb> |
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11396 <lb></lb> |
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11397 </p> |
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11398 </archimedes> |