Ex. 524. Find the diameter of a carriage wheel that makes 264 revolutions in going half a mile. Ex. 525. The sides of three octagons are 6 feet, 7 feet, 8 feet, respectively. Find the side of a regular octagon equivalent to the sum of the three given octagons. Let x be the side of the regular octagon required. Two regular polygons of the same number of sides are to each other as the squares of their sides. § 446 Ex. 381 = 12.206. 12.206 ft. Ans. Ex. 526. A circular pond 100 yards in diameter is surrounded by a walk 10 feet wide. Find the area of the walk. = 50 yd. The radius of the pond is of 100 yd. Area of walk is π.Χ. 25,600 § 463 π.χ 22,500 =π × 3100 3.1416 x 3100 = 9738.96. 9738.96 sq. ft. Ans. Ex. 527. The span (chord) of a bridge in the form of a circular arc is 120 feet, and the highest point of the arch is 15 feet above the piers. Find the radius of the arc. Let chord AB be 120 ft., and CD be 15 ft. Ex. 528. Three equal circles are described each tangent to the other two. If the common radius is R, find the area contained between the circles. The A 00'0" is equilateral, each side being 2 R, and the area is R2 √3. The area of the sector cut out of each O by the A is of the O, or πR2. § 462 Ex. 404 & is the .. the area of the three sectors is πR2. But the area contained between the area of the ▲ diminished by the area of the three sectors; that is, Area contained between the circles = R2√3 − ‡ π R2 – † R2 (2√3 − π). = Ex. 529. Given p, P, the perimeters of regular polygons of n sides inscribed in and circumscribed about a given circle. Find p', P', the perimeters of regular polygons of 2 n sides inscribed in and circumscribed about the given circle. Let AB be one side of the circumscribed polygon, and let CD, || to AB, be one side of the similar inscribed polygon. To construct inscribed and circumscribed A regular polygons of double the number of sides, and to compute their perimeters p', P' in terms of the perimeters p, P of the given polygons. Construction. Draw OE to AB, and draw AO, BO, CE, DE. Draw CFL to AO, and DH 1 to BO. F E H B Then FH is a side of the circumscribed polygon required, and CE and DE sides of the inscribed polygon required. Proof. C and D lie in AO and BO, respectively. § 441 § 245 .. CE = DE. .. CE and DE are sides of the required inscribed polygon. CF and DH are tangent to the O. § 241 § 442 .. FH is a side of the required circumscribed polygon. § 253 § 440 But AE is contained as often in P as FH is in P'. Ax. 1 · § 332 :. Ax. 1 D 2 pP P+P P' the rt. A CKE and ENF are similar. § 356 LECK = FEN. § 110 CK EN .*. § 351 But and p § 340 § 340 Ax. 1 Ex. 530. Given the radius R and the apothem r of an inscribed regular polygon of n sides. Find the radius R' and the apothem r' of an isoperimetrical regular polygon of 2 n sides. Let AB be one side of the given regular polygon. To construct an isoperimetrical regular polygon of double the number of sides, and to compute its radius R' and apothem in terms of the radius R and apothem r of the given polygon. Construction. Draw OD 1 to AB and draw AD, BD, OA, OB. Draw EF. Draw OE to AD, and OF 1 to BD. D +G MISCELLANEOUS EXERCISES. Ex. 531. If two adjacent angles of a quadrilateral are right angles, the bisectors of the other two angles are perpendicular. In the quadrilateral ABCD, let A and D be D rt., and let BE bisect the ABC and CE bisect the BCD. To prove that BE is 1 to CE. Proof. But E B Ex. 532. If two opposite angles of a quadrilateral are right angles, the bisectors of the other two angles are parallel. In the quadrilateral ABCD, let A and C be rt. 4, and let BE bisect the ABC and DF bisect the CDA. Ex. 533. The two lines that join the middle points of the opposite sides of a quadrilateral bisect each other. Let ABCD be a quadrilateral, and EG and FH be lines joining the middle points of the opposite sides. To prove that EG and FH bisect each other. Then EFGH is a with EG and FH for diagonals. .. EG and FH bisect each other. D F E B Ex. 74 $ 184 Q. E. D. |