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.

538. - To Construct A Square Equal To A Given Rectangle

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.

539. - To Form A Square Equal To A Given Triangle

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.

540. - Two Right Lines Being Given, To Find A Third Proportional Thereto

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.

541. - Three Right Lines Being Given, To Find A Fourth Proportional Thereto

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.)