Proposition 17. Problem. 398. Two similar polygons being given, to construct a similar polygon equal to their sum. Given, two homologous sides P and Q of two similar polygons R and S. Required, to construct a similar polygon equivalent to their sum. Cons. Draw AB = P. AC C At A construct the rt. A, draw A к P P On BC, homologous to P and Q, construct a polygon T similar to R and S, as in (354). Then T is the polygon required. 394. SCH. To construct a polygon similar to two given. similar polygons, and equivalent to their difference, we find the side of a square equivalent to the difference of the squares on P and Q (389), and on this side, homologous to P and Q, construct a polygon similar to the given polygons R and S (354). This will be the polygon required (379), 395. To find two straight lines proportional to two given polygons. Given, two polygons R and S. Required, to find two st. lines proportional to R and S. Cons. Find two squares equiva -lent to the given polygons R and S (386); let P and Q be the sides of these squares. Construct the rt. / A, draw AB = P, and AC = Q. Join BC, and draw AD 1 to BC. Then BD and DC are the lines required. Proof. Since AD is a L from the rt. A on the hypotenuse BC, AB: AC = BD : DC, the sqs. of the sides about the rt. Z are proportional to the adj. segments of the hypotenuse (324). But AB = P, and AC = Q. (Cons.) ... P2: Q2 = BD : DC, and ... BD, DC are proportional to the areas of the given polygons. Q. E. F. EXERCISE. Bisect a quadrilateral by a straight line drawn from one of its vertices. Let ABCD be the quad.; bisect BD in E, let E lie between AC and B; through E draw EF to AC to meet BC in F; join CE, EA, AF, then AFCD = }ABCD, Proposition 19. Problem. 396. To construct a square which shall have to a given square the ratio of two given lines. Given, the square S, and the ratio P : Q. Required, to construct a square which shall be to S as P is to Q. Cons. Draw the st. line AK; take ADP and DB = Q. Upon AB as a diameter, describe a E Then CE is the side of the required square. S Proof. Since CD is a from the rt. /C to the hypote nuse AB, But .·. CA' : CB2 – AD : DB. СА : СВ = CE: CH, (324 a st. line | to a side of a ▲ cuts the other two sides proportionally (298), 397. SCH. To construct a polygon similar to a given polygon S, and having to it the given ratio of P to Q, we find, as in (396), a side x so that x2 shall be to (where s is a side of S) as P is to Q, and upon a as a side homologous to s, construct the R S S polygon R similar to S (354); this will be the polygon required, (379) Proposition 20. Problem. 398. To construct a polygon equivalent to a given polygon P, and similar to a given polygon Q. Given, two polygons P and Q. Required, to construct a polygon equivalent to P and similar to Q. Cons. Find Ρ and q, the sides of squares equivalent respectively to P and Q. (386) Take any side of Q as AB, and find a fourth proportional A'B' to q, P, and AB. (347) p P A B Α' Upon A'B', homologous to AB, construct Q' similar ... Q' is equivalent to P and similar to Q. Q. E. F. APPLICATIONS. 1. To find the area of an equilateral triangle in terms of its side a.* Let h denote the alt., and S the area, of the A. 2. To find the area of a triangle in terms of its sides and the radius of the circumscribing circle. Let a, b, c denote the sides and h the alt. of the ▲, R the radius of the circumscribing O. and Therefore, the area of a triangle is equal to the product of the three sides divided by four times the radius of the circumscribing circle. 3. To find the area of a triangle in terms of its sides. 5. Find the area of an equilateral triangle if a side = 1 foot. Ans. 0.433 sq. ft. 6. Find (1) the area of the triangle whose sides are 3, 4, and 5 feet, and (2) the radius of the circumscribing circle. Ans. (1) 6 sq. ft. ; (2) 2.5 ft. * Rouché et Comberousse, p. 286. |