A triangle has vertices A, B, and C. Vertex A has an angle of #pi/2 #, vertex B has an angle of #( pi)/3 #, and the triangle's area is #25 #. What is the area of the triangle's incircle?

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Answer 1 · 2018-06-30T11:42:33

#color(green)("Area of inscribed circle " = A_i = pi r^2 = 12.1354#

#"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/2, hat B = pi/3, hat C = pi/6, A_t = 25#

#a b = (2 A_t) / sin C = 50 / sin (pi/6) = 100#

#b c = (2 A_t) / sin A = 50 / sin (pi/2) = 50#

#c a = (2 A_t) / sin B = 50 / sin (pi/3) = 57.98#

#a = (a b c) / (b c) = sqrt(100 * 50 * 57.98) / 50= 10.7684#

#b = (a b c) / (c a) = sqrt(100 * 50 * 57.98) / 57.98 = 9.2863#

#c = (a b c) / (a b) = sqrt(100 * 50 * 57.98) / 100 = 5.3842#

#"Semi-perimeter " = s = (a + b + c) / 2 = 25.4389 / 2 = 12.72#

#"Radius of inscribed circle " = r = A_t / s = 25 / 12.72#

#"Area of inscribed circle " = A_i = pi r^2 = pi * (25/12.72)^2 = 12.1354#