Casati, Paolo, Fabrica, et uso del compasso di proportione, dove insegna à gli artefici il modo di fare in esso le necessarie divisioni, e con varij problemi ...

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            icoſaedro, che conſta di 20 faccie triangolari equilatere. </s>
            <s xml:id="echoid-s4279" xml:space="preserve">Se-
              <lb/>
            condo trouato il lato del triangolo equilatero ſi cerchi la ſua
              <lb/>
            area, trouando ſ
              <unsure/>
            a perpendicolare, che da vn’angolo cade nel
              <lb/>
            mezzo del lato oppoſto: </s>
            <s xml:id="echoid-s4280" xml:space="preserve">il che ſi fà nella linea Geometrica,
              <lb/>
            applicando il lato del triangolo, ela metà di detto lato, à due
              <lb/>
            numeri, de’quali neceſſariamente vno è quadruplo dell’altro,
              <lb/>
            per eſſempio 48, e 12, e preſa la differenza 36 piglio l’inter-
              <lb/>
            uallo 36. </s>
            <s xml:id="echoid-s4281" xml:space="preserve">36, & </s>
            <s xml:id="echoid-s4282" xml:space="preserve">applico nella linea Aritmetica il lato del
              <lb/>
            triangolo al ſuo numero competente trouato nella prima
              <lb/>
            operatione, e poi veggo qual interuallo comprenda quella
              <lb/>
            diſtanza vltimamente preſa, che è illato d’vn quadrato, a cui
              <lb/>
            il quadrato del lato del triangolo è come 4 à 3, e queſto mol-
              <lb/>
            tiplicato per la metà del lato del triangolo dà l’area del trian-
              <lb/>
            golo. </s>
            <s xml:id="echoid-s4283" xml:space="preserve">Terzo, perche il corpo iſcritto nella sfera è vguale à
              <lb/>
            tante piramidi, che hanno la cima nel centro della sfera tra
              <lb/>
            di ſ
              <unsure/>
            oro vguali, per hauer le baſi, e gl’aſſi vguali, conuien tro-
              <lb/>
            uare la perpendicolare, che dal centro della sfera cade nel
              <lb/>
            piano del triangolo. </s>
            <s xml:id="echoid-s4284" xml:space="preserve">Ora ſe il piano del triangolo s’intenda
              <lb/>
            prolongato per ogni parte, taglia la sfera, e fà vn circolo, in
              <lb/>
            cui è iſcritto detto triangolo. </s>
            <s xml:id="echoid-s4285" xml:space="preserve">Prendaſi dunque il lato del
              <lb/>
            triangolo, e nella linea de’poligoni s’applichi all’interuallo
              <lb/>
            proprio del triangolo, econ vn’altro compaſſo ſi prenda il
              <lb/>
            raggio del ſuo circolo, cioè il lato dell’eſſagono: </s>
            <s xml:id="echoid-s4286" xml:space="preserve">e nella linea
              <lb/>
            Aritmetica applicato il lato del triangolo al numero, che gli
              <lb/>
            compete già trouato, veggaſi à qual numero cada il raggio
              <lb/>
            del circolo. </s>
            <s xml:id="echoid-s4287" xml:space="preserve">Cadendo dunque dal centro della sfera la per-
              <lb/>
            pendicolare nel centro di tal circolo, è noto il raggio del cir-
              <lb/>
            colo, & </s>
            <s xml:id="echoid-s4288" xml:space="preserve">è noto il raggio della sfera oppoſto all’angolo retto,
              <lb/>
            dunque applicati queſti due raggi alla linea Geometrica, ſi
              <lb/>
            troua la proportione de’l
              <unsure/>
            oro quadrati, & </s>
            <s xml:id="echoid-s4289" xml:space="preserve">alla differenza </s>
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