b c a - n; m. Every curvilinear surface that is equal to a given square will always likewise be squarable in itself. See here the whole surface n, [with] the holes of the hatched c: being equal to the figure m below, it is likewise squarable in itself, because, once 12 triangles of two curved sides, equal and similar such as a b (which are themselves squarable) are taken away, all the remainder will be entirely rectilinear, and consequently is squarable.

n - q o - m - p. By what is said above, a b c is squarable. I shall set it out below and take away q, which, having two similar curves, is likewise equal and squarable. (Figure). There remains to me (figure with o) o, which has two curves and two straight sides. I shall take the curve n from its place and set it into m, which m, having a curvature similar to n, will receive that n exactly; and so I shall cut off the triangle o, rectilinear and known, and join it to the quadrature of q, and it will stand thus (figure with q o i). This done, there remains to me p, not yet squarable, so I shall return to the square q o and finish it with the square i; and thus, having finished the whole quadrature of the a b c above, I shall draw off from it the square q and then the square o, and there will remain the square n, which with certainty is equal to p.

d c b - f e - a

a

c - b. a b remain equal to one another. c is known and so is b. Take b from c, which is the greater, and the remainder of c is squarable. And ...

a; a

S b - r - a - o. I have r S which are equal to b a. I add a o to b a and lose it in o r, and they remain as before and are similar to b S.

I remove the portion and add the triangle.

a; b

3; 5; 8, 6; 6; 5 (22.⁶; 10; 4; 8, 6; 6; 2; 1, 6 (³⁷; total 59, 6.