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To make a Square equal to the Sum of given Squares.

262. Problem.

To make a square equal to the dif

ference of two given squares.

Solution. Construct, by art. 145, a right triangle, of which the hypothenuse BC (fig. 79) is equal to the side of the greater square, and the leg AB is equal to the side of the less square; and AC is the side of the required square.

Demonstration. For, by art. 257,

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263. Problem. To make a square equivalent to the sum of any number of given squares.

Solution. Take AB (fig. 132) equal to the side of one of the given squares. Draw BC, perpendicular to AB, and equal to the side of the second given square.

Join AC, and draw CD, perpendicular to AC, and equal to the side of the third given square.

Join AD, and draw DE, perpendicular to AD, and equal to the side of the fourth given square; and so on. The line which joins A to the extremity of the last side is the side of the required square.

Demonstration. For, by art. 256,

AC2 = AB2 + BC2,

AD2=AC2 + CD2 = AB2 + BC2+CD2, AE2=AD2+DE2=AB2+BC2+CD2+DE2; &c.

264. Scholium. If either of the squares BC2, CD2, &c. were to have been subtracted instead of being added, the problem might still have been solved by means of art. 262.

265. Problem. To make a square which is to a given square in a given ratio.

Ratio of Similar Polygons.

Solution. Divide any line, as EG (fig. 133), by art. 163, into two parts, at the point F, which are to each other in the given ratio of the given square to the required square.

Upon EG describe the semicircle EMG; draw FM perpendicular to EG.

Join ME and MG; take, on ME produced if necessary, MH AB the side of the given square.

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Draw HI parallel to EG, meeting ME in I, and MI is the side of the required square.

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Demonstration. Produce MF to P; and, as the triangle HMI is, by art. 108, right-angled at M, we have, by art. 260, MH2 M12 = HP: PI.

But, by art. 181,

HP : PI = EF : FG ;

whence, on account of the common ratio HP : P1, MH2: MI2 = EF: FG.

266. Theorem. Similar triangles are to each other as the squares of their homologous sides.

Demonstration. In the similar triangles ABC, A'B'C (fig. 109), we have, by art. 199,

CE: CE = AB : A'B',

which, multiplied by the proportion

gives

≥ AB : ≥ A'B' = AB : A'B',

ABX CE: A'B' × C'E' = AB2 : A'B'2,

and, by art. 252, 23]

the area of ABC: the area of A'B'C' — AB2 : A'B'2,

=

267. Corollary. Hence, by art. 198, şimilar triangles are to each other as the squares of their

Ratio of Similar and Regular Polygons.

homologous altitudes, and, by art. 200, as the squares of their perimeters.

268. Theorem. Similar polygons are to each other as the squares of their homologous sides.

Demonstration. In the similar polygons ABCD, &c., A'B'CD, &c. (fig. 108), the triangles ABC, A'B'C', which are similar, by art. 196, give, by art. 266, the proportion ABC: A'B'C' = AC2 : A'C2;

also the similar triangles ACD, A'C'D', give the proportion

ACD: A'CD' = AC2: A'C2;

hence, on account of the common ratio AC2 : A'C'2,

ABC: A'B'C' = ACD : A'C'D'.

In the same way may be obtained the continued proportion

ABC: A'B'C' ACD: A'C'D': ADE: A'D'E', =

&c.

Now the sum of the antecedents ABC, ACD, ADE, &c. is the polygon ABCD, &c., and the sum of the consequents A'B'C', ACD', A'D'E', &c. is the polygon A'B'C'D', &c.; so that, by art. 266,

ABCD, &c.: A'B'C'D', &c. ABC: A'B'C=AB2:A'B'2.

269. Corollary. Similar polygons are, therefore, to each other, by art. 197, as the squares of their perim

eters.

270. Corollary. As regular polygons of the same number of sides are, by art. 206, similar polygons, they are to each other as the squares of their homologous sides, and, by art. 232, as the squares of the

Ratio of Circles.

radii of their inscribed circles, and also as the squares of the radii of their circumscribed circles.

271. Theorem. Circles are to each other as the squares of their radii.

Demonstration. For, by art. 208, they are regular polygons of the same number of sides, and, as in art. 234, the radii of their inscribed and circumscribed circles are their own radii.

272. Problem. Two similar polygons being given, to construct a similar polygon equivalent to their sum or to their difference.

Solution. Let A and B be the homologous sides of the given polygons. Find, by art. 261, or by art. 262, the line

such that the square constructed upon X is equal to the sum or the difference of the squares constructed upon A and B. The polygon similar to the given polygons, constructed, by art. 194, upon the side X homologous to A or B, is the required polygon.

Demonstration. For, by art. 268, the similar polygons constructed upon A, B, and X, have the same ratio to each other as the squares upon A, B, and X.

273. Corollary. If A and B were the radii of two given circles, I would, by art. 271, be the radius of a circle equivalent to their sum or to their difference.

274. Corollary. By the process of art. 263, a polygon might be constructed equivalent to the sum of any number of given similar polygons, and similar to them, or a circle equivalent to the sum of any number of given circles; or, if either of the given polygons or circles is to be (added instead of

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being subtracted, the resulting polygon or circle may be obtained, as in art. 264.

275. Problem. To construct a polygon similar to a given polygon, and having a given ratio to it.

Solution. Let A be a side of the given polygon. Find, by art. 265, the side of a square which is to the square, constructed upon A, in the given ratio of the polygons. The polygon, constructed upon X, similar to the given polygon, is the required polygon.

Demonstration. For, by art. 268, the similar polygons constructed upon A and X, have the same ratio to each other as the squares upon A and X.

276. Corollary. In the same way, a circle may be constructed having a given ratio to a given circle, by taking for A and X the radii of the given and of the required circles.

277. Theorem. The area of any circumscribed polygon is half the product of its perimeter by the radius of the inscribed circle.

Demonstration. From the centre O (fig. 134) of the circle draw OA, OB, OC, &c. to the vertices of the circumscribed polygon ABCD, &c. Draw the radii OM, ON, OP, &c. to the points of contact of the sides.

If, now, the sides AB, BC, CD, &c. are taken for the bases of the triangles OAB, OBC, OCD, &c.; their altitudes, being the radii OM, ON, OP, &c., are all equal. The area of each of these triangles is, then, by art. 251, half the product of its base AB, BC, CD, &c. by the common altitude OM.

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