Let there be removed within a circle the curved parts of various figures, and let the remainder be squarable.

First – e a – f b – c. a b c is worth b f c; a e is worth e f.

Second – d – b a – f – c. In the second, d is worth c of the first, and the a lent is worth b c removed. It follows that a d f of the second is worth a b c of the first, because a half portion is worth the two portions b c.

This is not a quadrature.

Third – a – b – c e – d. a c d is worth a b c. It is proved: because b is worth d, add a c to b to make the triangle a b c; and add a c to d, and you have made a c d, equal to a b c.

Fourth – a – c b. b c is worth a b.

Fifth – f d b a – o c. a b d f is worth b d c o.

Sixth – b a – d c – f e. The triangle a b c d is worth the a b, e f lunes.

Seventh – a b. a together with its adjoining lunes is worth b squared.

Eighth.

Ninth – b – a – c – d. b c d is worth b c a. It is proved, because a is worth d; if a is added to b c you have the rectilinear triangle b a c, and if you add d to that a c you have the curvilinear lune b c d, equal to the triangle b a c.

Tenth – a – f e – b – d c.

Eleventh – b a – d c – f e. a b e f is worth c d e f.

Twelfth – a b.

Thirteenth – o m f – n – g. The lune o n g is worth the triangle o m n, and is worth the lune o f n. It is proved to be so because m and f are equal between themselves, and each of them is worth the lune g. Now adding the portion m to the surface o n does as much as the portion f and as much as the lune g. This being so, our conclusion is verified.

Fourteenth – a b – e c – d. a b c d e lune.

Fifteenth.

Sixteenth.

Seventeenth.

Eighteenth.

Nineteenth.

Twentieth – b a c e d. Here from the square b a c d e are drawn the two lunes c d; the remainder a b c is worth the remainder of the aforesaid square.

Twenty-first – Twenty-second. These two surfaces lack the four greatest portions.

If within a square one removes a squarable curved area, the remainder will remain a squarable curved area.

a c – b d. a b is worth a c, and a c is worth b d.

o – c – b a. a c is worth b o. Therefore the portion a b c o is divided into four equal parts. It is proved because here there are three portions each half of the next: the portion a b c o is double the portion a b c, and that portion a b c is double the portion a b.

a c f h d e b g – Portion of a circle. a b c d e f g h is a portion divided into four equal parts, like those above.

Ant[onio] di Mic...

a b o p q s r t. The greatest portion a b o is worth the four portions b p q r. Now I will remove the greatest, a b o.

a o b c n r m.

Second – First – Second. The greatest portion a o is worth the four smallest o b r c. Remove therefore the greatest, and the three smallest b c r remain with n m, which is a squared quantity. Then return that surface to the first degree and remove b c r, and a o n m remains. Therefore I remove four portions and three portions remain, both in the first demonstration and in the second.

Francesco del Morano, shoemaker. Messer Alessandro, canon of Fiesole.

Sansavino. Whether the priest Alessandro Amadori is alive or not. Martino the brass-worker. Get the Peruzzo. Bellows(?). The plan of Pisa from Giorgio the stationer. Blue spectacles.

A column of arithmetical workings: 13, 1, 100, 110, 18, 118, 5, 13 1/2, 13, 765, 6, 20, 65, 300, 108, 000, 3, 100, 236, 195, 1300, 2360, 118 20, 000 236 2360, 22 2360, 2300 744 365, 305 365.