A triangle has vertices A, B, and C. Vertex A has an angle of #pi/3 #, vertex B has an angle of #(5 pi)/12 #, and the triangle's area is #27 #. What is the area of the triangle's incircle?

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Answer 1 · 2018-07-29T10:56:38

#color(indigo)("Area of inscribed circle " = A_i = pi r^2 ~~ 16.0643#

#"Area of "Delta = A_t = (1/2) a b sin C = (1/2) b c sin A = (1/2) c a sin B#

#"Given " hat A = pi/3, hat B = (5pi)/12, hat C = pi/4#, A_t = 27#

#a b = (2 A_t) / sin C = 54 / sin (pi/4) ~~ 76.3675#

#b c = (2 A_t) / sin A = 54 / sin (pi/3) = 62.3538#

#c a = (2 A_t) / sin B = 54 / sin ((5pi)/12) = 55.9049#

#a = (a b c) / (b c) = sqrt(76.3675 * 62.3538 * 55.9049) / 62.3538 = 7.8948#

#b = (a b c) / (c a) = sqrt(76.3675 * 62.3538 * 55.9049) / 55.9049 = 9.2291#

#c = (a b c) / (a b) = sqrt(76.3675 * 62.3538 * 55.9049) / 76.3675= 6.7562#

#"Semi-perimeter " = s = (a + b + c) / 2 = 23.8801 / 2 = 11.9401#

#"Radius of inscribed circle " = r = A_t / s = 27 / 11.9401#

#color(indigo)("Area of inscribed circle " = A_i = pi r^2 = pi * (27/ 11.9401)^2 ~~ 16.0643#