52. A circular grass-plot, 100 ft. in diameter, is surrounded by a walk 4 ft. wide. Find the area of the walk. 53. Two plots of ground, one a square and the other a circle, contain each 70686 sq. ft. How much longer is the perimeter of the square than the circumference of the circle? (= 3.1416.) 54. A wheel revolves 55 times in travelling diameter in inches ? 1045 ft. What is its If r represents the radius, a the apothem, s the side, and ✯ the area, prove that 55. In a regular octagon, s = r √ 2 − √2, a = { r √2 + √2, and k = 2r2 √2. (§ 375) 56. In a regular dodecagon, s = r √ 2 − √3, a = {r √2 + √3, and k = 3r2. 57. In a regular octagon, 8 = 2 a ( √2 − 1), r = a √ 4 − 2 √2, and k = 8 a2 (√2 − 1). 58. In a regular dodecagon, 8 = 2a (2√3), r = 2a V2 - √3, and k = 12 a2 (2 - √3). 59. In a regular decagon, a = ‡r V10 +2 √5. (§ 359.) (Find the apothem by § 273.) 60. What is the number of degrees in an arc whose length is equal to that of the radius of the circle? (= 3.1416.) (Represent the number of degrees by x.) 61. Find the side of a square equivalent to a circle whose diameter is 3. (3.1416.) 62. Find the radius of a circle equivalent to a square whose side is 10. (3.1416.) 63. Given one side of a regular hexagon, to construct the hexagon. 64. Given one side of a regular pentagon, to construct the pentagon. (Draw a of any convenient radius, and construct a side of a regular inscribed pentagon.) 65. In a given square, to inscribe a regular octagon. (Divide the angular magnitude about the centre of the square into eight equal parts.) 66. In a given equilateral triangle to inscribe a regular hexagon. 67. In a given sector whose central angle is a right angle, to inscribe a square. Note. For additional exercises on Book V., see p. 231. APPENDIX TO PLANE GEOMETRY. MAXIMA AND MINIMA OF PLANE FIGURES. PROP. I. THEOREM. 377. Of all triangles formed with two given sides, that in which these sides are perpendicular is the maximum. Given, in A ABC and A'BC, AB= A'B, and AB 1 BC. To Prove area ABC area A'BC. Proof. Draw A'D L BC; then, A'B> A'D. .. AB> A'D. ($46) (1) Multiplying both members of (1) by 1⁄2 BC, BC AB> BC × A'D. .. area ABC > area A'BC. (§ 312) 378. Def. Two figures are said to be isoperimetric when they have equal perimeters. PROP. II. THEOREM. 379. Of isoperimetric triangles having the same base, that which is isosceles is the maximum. Given ABC and A'BC isoperimetric A, having the same base BC, and ▲ ABC isosceles. Proof. Produce BA to D, making AD= AB, and draw line CD. Then, BCD is a rt. ; for it can be inscribed in a semicircle, whose centre is A and radius AB. (§ 195) Draw lines AF and A'G 1 to CD; take point E on CD so that A'E = A'C, and draw line BE. Then since ABC and A'BC are isoperimetric, But, AB+ AC = A'B+ A'C= A'B+ A'E. .. A'B+ A'E = AB + AD = BD. A'B+A'E > BE. .. BD > BE. ... CD>CE. (Ax. 4) (§ 51) Now AF and A'G are the Is from the vertices to the bases of isosceles ACD and A'CE, respectively. ... CF CD, and CG CE. = .. CF> CG. = Multiplying both members of (1) by 1⁄2 BC, 1⁄2 BC × CF > BC × CG. .. area ABC> area A'BC. (§ 94) (1) (?) 380. Cor. Of isoperimetric triangles, that which is equilateral is the maximum. For if the maximum ▲ is not isosceles when any side is taken as the base, its area can be increased by making it isosceles. Then, the maximum ▲ is equilateral. PROP. III. THEOREM. (§ 379) 381. Of isoperimetric polygons having the same number of sides, that which is equilateral is the maximum. Given ABCDE the maximum of polygons having the given perimeter and the given number of sides. To Prove ABCDE equilateral. Proof. If possible, let sides AB and BC be unequal. Let AB'C be an isosceles ▲ with the base AC, having its perimeter equal to that of AABC. .. area AB'C> area ABC. Adding area ACDE to both members, area AB'CDE> area ABCDE. (§ 379) But this is impossible; for, by hyp., ABCDE is the maxi mum of polygons having the given perimeter. Hence, AB and BC cannot be unequal. In like manner we have BC= CD = DE, etc. Then, ABCDE is equilateral. ード PROP. IV. THEOREM. 382. Of isoperimetric equilateral polygons having the same number of sides, that which is equiangular is the maximum. Given AB, BC, and CD any three consecutive sides of the maximum of isoperimetric equilateral polygons having the same number of sides. If possible, let ZABC be > BCD, and draw line AD. In Fig. 1. Let E be the middle point of BC; and draw line EF, meeting AB produced at F, making EF BE. Produce FE to meet CD at G. = Then in A BEF and CEG, by hyp., BE = CE. Also, L BEF=2 CEG. (?) |