Then, the side of an inscribed equilateral triangle equals the radius of the circle multiplied by √3. PROP. IX. PROBLEM 324. To inscribe a regular decagon in a given circle. M A D Given O AC. Required to inscribe a regular decagon in O AC. Construction. 1. Divide radius OA internally in extreme and mean ratio (§ 273), making ОА OM = OM AM 2. OM is a side of a regular inscribed decagon. (1) Proof. 3. Take chord AB OM, and draw lines OB, BM; in A OAB, ABM, ZA = ZA. = 4. Since OM= AB, (1) becomes OA AB AB AM 5. ▲ OAB, ABM are similar (§ 242); then / ABM = 20. 6. ▲ ABM is isosceles, being similar to ▲ OAB. 7. By Ax. 1, AB=BM=0M; then / OBM= ≤ 0. 8. Add results of (5) and (7), ZOBA=2 ≤0. (?) (2) 9. Find sum of ▲ of ▲ OAB; 2 ≤ OBA+20=180°. (?) 10. Substitute in this value of ≤OBA in (2), then 5 ≤ 0 = 180°; and 20 is of 4 rt. . 325. Note. If chords be drawn joining the alternate vertices of a regular inscribed decagon, there is formed a regular inscribed pentagon. 326. Denoting the radius of the O by R, we have AB=OM= R(√5 −1). 2 (§ 274) This is an expression for the side of a regular inscribed decagon in terms of the radius of the circle. Ex. 11. The diameter of a circle is 20. Inscribe a regular hexagon and find its area. Ex. 12. The side of a regular hexagon is 6; find its area. If this hexagon be an inscribed one, what is the area of the regular hexagon circumscribing the same circle ? Ex. 13. A regular polygon is inscribed in a circle. Give a general method for finding its area. Ex. 14. An equilateral triangle is inscribed in a circle. If its side is 10, what is its area, and how does it compare with the area of the equilateral triangle circumscribing the same circle? Ex. 15. If the altitude of the triangle in exercise 14 were 10, what would be its area and the area of the equilateral triangle circumscribing the circle ? Ex. 16. The centre of a regular hexagon bisects every line drawn through it which terminates in the sides of the hexagon. Ex. 17. The apothem of an inscribed equilateral triangle is one-half the radius of the circle. PROP. X. PROBLEM 327. To construct the side of a regular pentedecagon inscribed in a given circle. M C B N Given arc MN. Required to construct the side of a regular inscribed polygon of fifteen sides. Construction. If AB is a side of a regular inscribed hexagon (§ 321), and AC a side of a regular inscribed decagon (§ 326), arc BC is -16, or of the circumference. 109 Ex. 18. An equilateral triangle is inscribed in a circle. The distance from the intersection of the medians to the middle point of the base is 6. Find the radius of the circle. Ex. 19. If a circle be inscribed in an equilateral triangle, the altitude of the triangle passes through the centre of the circle, and is three times the radius of the circle. Ex. 20. The area of the square which circumscribes a circle is twice that of the inscribed square. Ex. 21. The area of the equilateral triangle circumscribing a circle is four times the area of the inscribed one. MEASUREMENT OF THE CIRCLE PROP. XI. THEOREM 328. The circumference of a circle is shorter than the perimeter of any circumscribed polygon. Given polygon ABCD circumscribed about a O. To Prove circumference of O shorter than perimeter ABCD. Proof. 1. Of the perimeters of the O, and of all possible circumscribed polygons, there must be some perimeter such that all the others are of the same or greater length. 2. But no circumscribed polygon can have this perimeter; for suppose polygon ABCD to have this perimeter, and draw line EF tangent to the O, meeting AB at E and AD at F. 3. We know that EF is < (AE+AF). (Ax. 6) 4. Then, the perimeter of polygon BEFDC is < the perimeter of polygon ABCD. 5. Hence, the circumference of the O is the perimeter of any circumscribed polygon. PROP. XII. THEOREM 329. If a regular polygon be inscribed in, or circumscribed about, a circle, and the number of its sides be indefinitely increased, I. Its perimeter approaches the circumference as a limit. II. Its area approaches the area of the circle as a limit. Given p, P perimeters, k, K areas, of two regular polygons of same number of sides, respectively inscribed in, and circumscribed about, a O, whose circumference is C, and area S. To Prove that, if the number of sides of the polygons be indefinitely increased, P and p approach the limit C, and K and k the limit S. Proof. 1. Let A'B' be side of polygon whose perimeter is P; draw radius OF to point of contact. 2. If OA' and OB' cut circumference at A and B, AB is side of polygon whose perimeter is p. OA' = p OF (§ 309) (§§ 314, 315) P OF (?); or P-p= × (OA' — OF). 5. p is circumference; OA' OF is < A'F. (Ax. 7, § 62) 6. Then, C (1) 7. If number of sides of each polygon be indefinitely increased, the polygons continuing to have same number of sides, the length of each side will be indefinitely diminished, and A'F will approach the limit 0. 8. OF being constant, P-p will approach limit 0. 9. Now circumference is < P, and > p. (§ 328, Ax. 7) 10. Then, P Cand C-p will approach limit 0, and P and 13. If the number of sides of each polygon be indefinitely increased, the polygons continuing to have the same number of sides, A'F will approach limit 0. 15. Then, K – k will approach limit 0; and S being < K, and >k, K and k will approach limit S. 330. If a regular polygon be inscribed in a circle, and the number of its sides be indefinitely increased, its apothem approaches the radius of the circle as a limit. - For, it was shown in § 329 that OA' OF approaches the limit 0, whence OF approaches the limit OA'. OF is the apothem of a regular polygon inscribed in a circle whose radius is OA'. PROP. XIII. THEOREM 331. The circumferences of two circles are to each other as their radii. оо Given C and C' the circumferences of two whose radii are R and R', respectively. |