1 1 1 1000
2 4 32 500
250
16 125
8
1 12 1/2
6 1/4
25 3 1/8
1 1/16
25 (⁰
32 25
64
25
128

o c – q b f d e a – m L – n K i h – g

If you make a semidiameter of 200 braccia, it will be 1/6 of its whole circle six-sided, which will be 6 chords of 200 braccia each, over which you will make its arc, which you will divide into a thousand equal parts, by first dividing it in half, which contains 500 of them; and then divide one of these parts in half, which will make 250; then divide the half, and you will have 125; then the other half, which will be [half again] 2); and so continue up to 14 times, and you will have made, over a chord of 200 braccia, an arc divided into more than a thousand parts, that is d e; and from one of these parts draw the sides as far as the centre of such a circle, which is 3) the semidiameter of the circle that we set at 200 braccia, and therefore this is known. Then take half the length of that diameter, which makes 4) a pyramid, and you will have 100 square parts, of which the root is 10, and the root of the square that the great circle would hold in itself would be 400 5). Thus make a small circle in the small square that has 10 braccia of side, and make with the mind a great circle in the square that contains the great circle that has 200 braccia per side, and you will have the circles proportioned, of which 6) the smaller is born of the thousandth part of the great circle, that is of the pyramid which was the thousandth part of the great circle, and the other is the whole circle which turns 6000 bases of the said pyramid, and so comes to be composed of 6000 of the said pyramids.

Then straighten the curvature of the base of the aforesaid pyramid and take away half of that base together with the whole pyramid; and then straighten its base and take half of the circle and draw off the curvature of the base of the pyramid, and so there will remain the half of the circle equal to the whole rectilinear pyramid. Then square it as you see l m K n.

g – b a; d c – f e

n – f e d - c b a - o

It is more useful, once you have made a pyramid equal to a circle, to find two portions equal to each other, of which one is in proportion to the circle as the other is to the pyramid, and then, having thrown them away, there will remain the remainder of the circle equal to the remainder of the pyramid. But why do I go entangling myself? The method stated above does the same, because I make a pyramid that is equal to a semicircle, and I occupy with its base that semicircle on the straight side, and so I remove from them whatever touches. Thus the remainder is equal, because what touches is like and equal, and because I shall make the base of the pyramid of so little curvature that it will seem straight. I shall straighten it 7) by taking half the sagitta of that arc and adding it to the pyramid, as is shown below. Thus study to find the half of a portion, which, although it does not have rectilinear fronts, will be of great utility for the approximation 8) of such a squaring.

b a – m q o – p – n c; e h r – f g – r – c - d

And the triangles e h and r f are equal.

a b c is a concave triangle from a quarter of a circle, and like it is o n m; and from both is removed the part o p q.

With these 2 figures no squaring will be given by taking and returning with their own same parts, but the parts will be borrowed and foreign.