A triangle has vertices A, B, and C. Vertex A has an angle of #pi/8 #, vertex B has an angle of #(pi)/12 #, and the triangle's area is #12 #. What is the area of the triangle's incircle?

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Answer 1 · 2018-06-30T11:11:38

#color(indigo)("Area of inscribed circle " = A_i = pi r^2 = 2.9092#

#"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/12, hat B = pi/8, hat C = (19pi)/24, A_t = 12#

#a b = (2 A_t) / sin C = 24 / sin ((19pi)/24) = 39.42#

#b c = (2 A_t) / sin A = 24 / sin (pi/12) = 92.73#

#c a = (2 A_t) / sin B = 24 / sin (pi/8) = 62.72#

#a = (a b c) / (b c) = sqrt(39.42 * 92.73 * 62.72) / 92.73= 5.16#

#b = (a b c) / (c a) = sqrt(39.42 * 92.73 * 62.72) / 62.72 = 7.63#

#c = (a b c) / (a b) = sqrt(39.42 * 92.73 * 62.72) / 39.42 = 12.15#

#"Semi-perimeter " = s = (a + b + c) / 2 = 24.94 / 2 = 12.47#

#"Radius of inscribed circle " = r = A_t / s = 12 / 12.47 = 0.9623#

#"Area of inscribed circle " = A_i = pi r^2 = pi * 0.9623^2 = 2.9092#