Squaring the ring: quadrature of circles and annuli

Right half. Upper margin.

I shall give problems from 113 books, one or two from each, and in geometry I shall give the stretched ring, etc.

Below here are 4 half-portions removed, namely a b c d, against 4 whole portions of the sub-double circle, etc.

Part of a circular ring and inscribed squares, with:

b a – 3 f 1 – g e – 4 h 2 – d c
I remove the half-portions 1 2 3 4, which are worth the 4 whole ones e f g h.

On the right; two concentric circles, a semicircle with a lune, with:

n – m – a – b – c – f

n is equal to m and to f.

In these two circles, one double the other, upon the same center, the excess of the greater is equal to the smaller. It follows that the portion c, made from the quarter of the greater, is worth the two portions a b made from the quarter of the smaller. Therefore the remainder of the greater, n, is equal to the remainder of the smaller, m.

On the left. Small figure partly struck through, with:

e a - c – f b – d

The curvilinear a b c d e f, a parallel [ring] of circles double one another, is squared by means of b f a c e.

Three circles, with:

4 - a; 1 e - d e b; a
* Because here one circle is worth 4 times the other circle, and the portions do likewise, therefore the portion a of the greater circle, being 1/6 of its circumference, is worth 4 of the 6 portions of the smaller circle. Therefore two portions remain to the smaller circle, and five remain to the greater.

* And because the whole is quadruple the other whole, the quarter of the greater is worth the whole of the smaller. Now I will take from the quarter of that greater one portion, which with its arc contained 1/6 of its whole greater circle, that portion being worth 4 of the portions of the smaller circle; there remains a equal to b. (There follow two figures marked a, b.)

Left half.

Large circular ring with a sector and two detached portions, with: f – h 12 11 g – d o n – d c b a – h – e

Note well.

Note that, since it is proved that no square can be given equal to any curvilinear surface unless it be squarable in itself by taking away and restoring its parts or their value, you shall do the like with this ring.

This ring is squarable in itself by taking the half and removing 6 portions from outside and putting back inside 6 of the smaller ones, which the smaller circle gives, and so much the more as the smaller circle is smaller than the greater circle.

Along the left margin.

If from two equal things equal things be taken, the remainder will be equal.

Therefore, taking away a b, a portion of the sector b c e, and taking away c d, a portion of the ring g f h, it is necessary – the ring being equal to the sector – that the remainder of the portion be equal to the remainder of the ring; and by this we have proved the ring to be squarable, being equal to the rectilinear triangle b c e.

And know that the curvature taken from the triangle b c e is of the same curvature as the circumferential line of the ring.

The sector below is the thirteenth part of the whole circle, and likewise is the capacity of the ring.

* Again I make you understand that if you wish to remove a very small...

The sector must always, by rule, be a sector of the greater circle of the ring, if the portions of the sector and of the ring are to be equal and similar to one another.

To the left of the sector;

n o d c, of one curved side, taken from the rectilinear n b c, enables me to remove that rectilinear n b c from the rectilinear triangle h.

To the left of the sector;

n o d c, of one curved side, taken from the rectilinear n b c, enables me to remove that rectilinear n b c from the rectilinear triangle h.

On the right, a circular ring and a double sector, with:

b – t c a S – m t n – v – o – p

The smaller circle a b c t is worth the excess of the greater circle. I will take from the greater the portion n m v, and there remains the triangle n m p, equal to the remainder of the excess.

I will do the same, with the same triangle n m p, to the smaller circle; and whatever is lacking to the said triangle I will take from the remainder of the smaller circle, etc.

On the root.

If you want the root of one, set a line of 2; and if you want the root of 2, set 3 in a line; and this process extends to infinity by always adding one to the number whose root you want. And this arises from one [proposition] of Euclid, here figured in the margin, because such a proportion is from circle to circle as from square to square touching that circle with its sides.

To the right of the preceding: a line divided into two parts, another into three, and a rectangle turned into a square, with:

n - b c a

Metà destra. Margine superiore.

Parte di corona circolare e quadrati iscritti, con:

A destra; due cerchi concentrici, semicerchio con falcata, con:

A sinistra. Piccola figura parzialmente depennata, con

Tre cerchi con:

Metà sinistra.

Lungo il margine sinistro.

A sinistra del settore;

A sinistra del settore;

A destra, corona circolare e doppio settore, con;

A destra del precedente, figure di una linea divisa in due parti, altra divisa in tre, rettangolo trasformato in quadrato, con:

Right half. Upper margin.

Part of a circular ring and inscribed squares, with:

On the right; two concentric circles, a semicircle with a lune, with:

On the left. Small figure partly struck through, with:

Three circles, with:

Left half.

Along the left margin.

To the left of the sector;

To the left of the sector;

On the right, a circular ring and a double sector, with:

To the right of the preceding: a line divided into two parts, another into three, and a rectangle turned into a square, with:

Squaring the ring: quadrature of circles and annuli

On this page

Excess of the greater circle equals the smaller

Squaring a circular ring in itself

Ring equal to a rectilinear triangle

A rule for successive roots after Euclid