A triangle has vertices A, B, and C. Vertex A has an angle of $pi/8$ , vertex B has an angle of $(7pi)/12$ , and the triangle's area is $28$ . What is the area of the triangle's incircle?

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Answer 1 · 2018-06-30T11:21:17

$color(orange)("Area of inscribed circle " = A_i = pi r^2 = 11.2446$

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

$"Given " hat A = pi/8, hat B = (7pi)/12, hat C = (7pi)/24, A_t = 28$

![http://mathibayon.blogspot.com/2015/01/ derivation-of-formula-for-radius-of-incircle.html

$a b = (2 A_t) / sin C = 56 / sin ((7pi)/24) = 70.59$

$b c = (2 A_t) / sin A = 56 / sin (pi/8) = 146.34$

$c a = (2 A_t) / sin B = 56 / sin (7pi)/12) = 57.98$

$a = (a b c) / (b c) = sqrt(70.59 * 146.34 * 57.98) / 146.34= 5.29$

$b = (a b c) / (c a) = sqrt(70.59 * 146.34 * 57.98) / 57.98 = 13.35$

$c = (a b c) / (a b) = sqrt(70.59 * 146.34 * 57.98) / 70.59 = 10.96$

$"Semi-perimeter " = s = (a + b + c) / 2 = 29.6 / 2 = 14.8$

$"Radius of inscribed circle " = r = A_t / s = 28 / 14.8$

$"Area of inscribed circle " = A_i = pi r^2 = pi * (28/14.8)^2 = 11.2446$

A triangle has vertices A, B, and C. Vertex A has an angle of pi/8 pi/8 , vertex B has an angle of (7pi)/12 (7pi)/12 , and the triangle's area is 28 28 . What is the area of the triangle's incircle?