PROBLEMS OF CONSTRUCTION. Proposition 11. Problem. 442. To inscribe a square in a given circle. I to each other. Join AB, BC, CD, DA. Then ABCD is the square required. rts, in the same and the every (Cons.) B .. the sides AB, BC, etc., are equal, A equal Ls at the centre intercept equal chords (196), s BAD, ADC, etc., are rt. Zs, inscribed in a is a rt. (240). .. the figure ABCD is a square. (126) Q.E.F. 443. COR. 1. If tangents be drawn to the circle at the points A, B, C, D, the figure so formed will be a circumscribed square. 444. COR. 2. To inscribe, and circumscribe, regular polygons of 8 sides, bisect the arcs AB, BC, CD, DA, and proceed as before. By repeating this process, regular inscribed and circumscribed polygons of 16, 32, . and, in general, of 2n sides, may be drawn. EXERCISES. 1. What is the area of a square inscribed in a circle whose area is 48 feet? 2. What is the area of a regular hexagon inscribed in a circle whose area is 560 square feet? Proposition 12. Problem. 445. To inscribe a regular hexagon in a given circle. Given, the ABC, with centre O. Required, to inscribe a regular hexagon in it. Cons. With any pt. A on the Oce as D a centre, and AO as a radius, describe an arc cutting the Oce in B. Join AB, BO, OA. Then AB is a side of the hexagon required. Proof. Since the ▲ OAB is equilateral, .. it is equiangular. E .. ZAOB is of 2 rt. /s, or of 4 rt. Zs. .. the arc AB is of the Oce. (Cons.) (113) (103) .. the chord AB is a side of the regular inscribed hexagon. (400) .. the figure ABCDEF, completed by drawing the chords BC, CD, DE, EF, FA, each equal to the radius OA, is the regular inscribed hexagon required. Q.E.F. 446. COR. 1. If the alternate vertices of the regular hexagon be joined, we obtain the inscribed equilateral triangle ACE. 447. COR. 2. If tangents be drawn to the circle at the points A, B, C, D, E, F, the figure so formed will be a regular circumscribed hexagon. 448. COR. 3. To inscribe, and circumscribe, regular polygons of 12 sides, bisect the arcs AB, BC, CD, etc., and proceed as before. By repeating this process, regular inscribed and circumscribed polygons of 24, 48, etc., sides may be drawn. Proposition 13. Problem. 449. To inscribe a regular decagon in a given circle. Given, the ABCE. Required, to inscribe in it a regular dec agon. Cons. Draw any radius OA, and divide it in extreme and mean ratio at D, so that OA: OD OD: AD. (352) B With centre A and OD as a radius, describe an arc cut ting the Oce at B, and join AB. Then AB is a side of the required inscribed decagon. .. the As OAB and BAD are similar, having A common and the including sides proportional (314). being opp. the equal sides of an isosceles ▲ (111). •'. ZABO + ≤ BAO+ ≤ AOB = 5 / AOB = 2 rt. ≤ s. = ... ZAOB of 2 rt. /s, or of 4 rt. S. ... the arc AB is of the Oce. .. the chord AB is a side of the regular inscribed deca gon. (400) .. the figure ABCE. formed by applying AB ten times to the Oce, is the regular inscribed decagon required. Q.E.F. 450. Cor. 1. If the alternate vertices of the required decagon be joined, a regular pentagon is inscribed. 451. COR. 2. If tangents be drawn at the points at which the circumference is divided, the figure so formed will be a regular circumscribed decagon. 452. COR. 3. To inscribe, and circumscribe, regular polygons of 20 sides, bisect the arcs AB, BC, etc., and proceed as before. By repeating this process, regular inscribed and circumscribed polygons of 40, 80, etc., sides may be drawn. EXERCISES. 1. The side of an inscribed square is equal to the radius of the circle multiplied by 12. 2. The side of an inscribed equilateral triangle is equal to the radius multiplied by √3. 3. The apothem of an inscribed square is equal to half the radius multiplied by 12. 4. The apothem of an inscribed equilateral triangle is equal to half the radius. 5. The apothem of a regular inscribed hexagon is equal to half the radius multiplied by 3. 6. The area of a circumscribed square is double the area of the inscribed square. 7. Required the area of an equilateral triangle inscribed. in a circle whose radius is 4. 8. Required the area of a square inscribed in a circle whose radius is 5. 9. Required the area of a regular hexagon inscribed in a circle whose radius is 8. 10. Required the area of a circle whose circumference is 100. 11. Required the area of a circle inscribed in a square whose area is 36. cle. Proposition. 14. Problem. 453. To inscribe a regular pentadecagon in a given cir Given, the ABC. Required, to inscribe in it a regular polygon of 15 sides. Cons. Lay off HB equal to a side of a regular inscribed hexagon. (445) Lay off HA equal to a side of a regular inscribed decagon. Join AB. (449) H Then AB is a side of the required inscribed pentadeca = of Oce of Oce of Oce. ... the chord AB is a side of the regular inscribed pentadecagon. (400) .. the figure ABCD..., formed by applying AB 15 times to the Oce, is the regular inscribed pentadecagon required. Q.E. F. 454. COR. 1. If tangents be drawn at the points at which the circumference is divided, a regular circumscribed pentadecagon is obtained. 455. COR. 2. To inscribe, and circumscribe, regular polygons of 30 sides, bisect the arcs AB, BC, etc., and proceed as before. By repeating this process, regular inscribed and circumscribed polygons of 60, 120, etc., sides may be drawn. 456. SCH. These are the only polygons that could be constructed by the ancient geometers, by the use of the rule and compass. About the beginning of the present century, Gauss, the great German mathematician, proved that whenever 2" + 1 is a prime number, and an integer, a regular polygon of that number of sides could be inscribed in the circle, by the rule and compass. Therefore, it is possible to inscribe regular polygons of |