This section is from the book "The American House Carpenter", by R. G. Hatfield. Also available from Amazon: The American House Carpenter.

Let A and B (Fig. 396) be the given circles. In the right-angled triangle a bc make a b equal to the diameter of the circle B, and cb equal to the diameter of the circle A; then the hypothenuse a c will be the diameter of a circle C, which will be equal in area to the two circles A and B, added together.

Fig. 396.

Any polygonal figure, as A (Fig. 397), formed on the hypothenuse of a right-angled triangle, will be equal to two similar figures,* as B and C, formed on the two legs of the triangle.

Fig. 397.

Let A (Fig. 398) be the given rectangle. Extend the side ab and make bc equal to be; bisect ac in f, and upon f, with the radius fa, describe the semi-circle agc; extend eb till it cuts the curve in g; then a square bghd, formed on the line bg, will be equal in area to the rectangle A.

* Similar figures are such as have their several angles respectively equal, and their sides respectively proportionate.

Another method. Let A (Fig. 399) be the given rectangle. Extend the side ab and make ad equal to ac; bisect ad in e; upon e, with the radius ea, describe the semi-circle afd; extend gb till it cuts the curve in f; join a and f; then the square B, formed on the line af, will be equal in area to the rectangle A. (See Arts. 352 and 353.)

Fig. 398.

Let ab (Fig. 398) equal the base of the given triangle, and be equal half its perpendicular height (see Fig. 392); then proceed as directed at Art. 538.

Fig. 399.

Let A and B (Fig. 400) be the given lines. Make a b equal to A; from a draw a c at any angle with ab; make ac and ad each equal to B; join c and b; from d draw de parallel to cb; then ae will be the third proportional required. That is, ae bears the same proportion to B as B does to A.

Fig. 400.

Let A, B, and C (Fig. 401) be the given lines. Make ab equal to A; from a draw ac at any angle with a b; make a c equal to B and a e equal to C; join r and b; from e draw e f parallel to cb; then a f will be the fourth proportional required. That is, af bears the same proportion to C as B does to A.

Fig. 401.

To apply this problem, suppose the two axes of a given ellipsis and the longer axis of a proposed ellipsis are given. Then, by this problem, the length of the shorter axis to the proposed ellipsis can be found; so that it will bear the same proportion to the longer axis as the shorter of the given ellipsis does to its longer. (See also Art. 559.)

Continue to: