Let AB be a side of a polygon circumscribing a circle whose centre is 0. Let 8 = sum of sides of polygon. r = radius of inscribed circle. Draw OX=r perpendicular to AB. Thus the radius is determined by one side and the adjacent angles. 227. The perimeter and area of a regular polygon in terms of the radius of the inscribed circle. Take the figure of last article. Then the polygon being regular, and containing n sides, (say) n AOB = 4 right-angles, .. LAOB = 2π/n. 228. The perimeter and area of a regular polygon in terms of the radius of the circumscribing circle. Let AB be a side of a regular polygon of n sides. Let the centre of the circumscribing circle be 0. Draw OX perpendicular to AB. EXAMPLES X. [The following notation is employed :-area of ABC=A; inradius =r; circumradius=R; cosine-radius=p; eradii=71, 72, 73; ex-cosine radii=P1, P2, P3; half-sum of sides=s; and points are lettered as in the last chapter.] 13. 14. A = - Rr (sin A + sin B + sin C). r2 = ▲ tan 4 tan B tan C. 8 = 4R cos A cos B cos C. abc r = 4R (8 − a) (s - b) (8 −c). 2r+2R = a cot A+ b cot B + c cot C. 4R sin A sin B sin C a cos A+ b cos B+ c cos C. cot w=cosec A cosec B cosec C + cot A cot B cot C. 16. 17. 18. 19. cosec2 w = 112 35. 36. P2P3+ P3P1+ P1P2 = · 3R2 + R2 (cos2 A + cos2 B + cos2 C) sec A sec B sec C. 1 1 bc + ca + ab r2r3 r3r1 7172 A2 4r (r1 + r1⁄2 + r3) = 2bc + 2ca + 2ab — a2 – b2 — c2. ar 2r3 brar crir2 = = r2+r3 r3+r1 r1 + r2 √(r2r3 + r3r1 + r1ra) ' 37. √(ar) + √(br2) + √(cr3) −√(abc/r) 8√(Rs) sin (45° – 4) sin (45° – 4 B) sin (45° – 4C). 38. 2rp (r+r2+r3)= 2ps2 - abc. 40. r1 cot A = r2, cot B = r; cot 1⁄2 C = r cot A cot B cot C. 47. Given the inradius r, the circumradius R, and the area ▲ of a triangle, show that its sides are the roots of the equation x3 − 2x2▲/r + x (r2 + 4rR + ▲2/r2) = 4AR. 48. Given the half-sum of sides s, the half-sum of squares on sides o2, and the area ▲ of a triangle, show that the radii of its escribed circles are the roots of the equation x3 /8 + x8 = x2 (s2 — σ2)/A + A. 49. If the sides of a triangle are the roots of the equation x3 + px = qx2 + v, its cosine-radius is v/(q2 – 2p) and the rectangle contained by its inradius and circumradius is v/2q. 50. If the squares on the sides of a triangle are the roots of the equation a3 + px = qx2 + v2, its cosine-radius is v/q and its cir 51. In an equilateral triangle, the incircle coincides with the nine-points circle, the cosine-circle with the Lemoine circle, and the centres of the escribed circles with those of the ex-cosine |