A triangle has vertices A, B, and C. Vertex A has an angle of #pi/8 #, vertex B has an angle of #( pi)/6 #, and the triangle's area is #27 #. What is the area of the triangle's incircle?

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Answer 1 · 2018-06-22T09:51:52

#color(maroon)("Area of incircle " = pi r^2 = pi * (1.71)^2 = 9.186#

#hat A = pi/8, hat B pi/6, hat C = (17pi) / 24), A_t = 27#

#A_t = (1/2) ab sin C = (1/2(bc sin A = (1/2) ac sin B#

#ab =( 2A_t) /sin C = (2 * 27) / sin ((17pi)/24) = 68.07#

#bc = (2 * 27) / sin (pi/8) = 141.11#

#ca = (2 * 27) / sin (pi/6) = 108#

#a = sqrt(abc)^2 / (bc) =sqrt (68.07 * 141.11 * 108) / 141.11 = 7.22#

#b = sqrt(abc)^2 / (ac) =sqrt (68.07 * 141.11 * 108) / 108 = 9.43#

#c = sqrt(abc)^2 / (ab) =sqrt (68.07 * 141.11 * 108) / 68.07 = 14.96#

#"Semiperimeter " = s = (a = b + c) / 2 = (7.22 + 9.43 + 14.96) / 2 = 15.81#

#color(chocolate)("Incircle radius " = r = A_t / s = 27 / 15.81 = 1.71#

#color(maroon)("Area of incircle " = pi r^2 = pi * (1.71)^2 = 9.186#