A triangle has vertices A, B, and C. Vertex A has an angle of pi/8 π8, vertex B has an angle of (pi)/12 π12, and the triangle's area is 1 1. What is the area of the triangle's incircle?

1 Answer
Feb 1, 2018

Area of triangle’s Incircle A_i = color(green)0.2424Ai=0.2424

Explanation:

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Three angles are A = pi / 8, B = pi / 12, C = pi - pi/8 - pi/12 = (19pi)/24A=π8,B=π12,C=ππ8π12=19π24

Given Area of triangle A_t = 1#

bc = A_t / ((1/2) sin A) = (2 * 1) / sin (pi / 8) = 5.2263bc=At(12)sinA=21sin(π8)=5.2263

ca = A_t / ((1/2) sin B) = 2 / sin (pi / 12) = 7.7274ca=At(12)sinB=2sin(π12)=7.7274

ab = A_t / ((1/2) sin C) = 2 / sin ((19pi)/24) = 3.2854ab=At(12)sinC=2sin(19π24)=3.2854

sqrt(ab * bc * ca)= (abc) = sqrt(5.2263 * 7.7274 * 3.2854) = 11.5188abbcca=(abc)=5.22637.72743.2854=11.5188

a = (abc) / (bc) = 11.5188 / 5.2263 = 2.204a=abcbc=11.51885.2263=2.204

b = (abc) / (ca) = 11.5188 / 7.7274 = 1.4906b=abcca=11.51887.7274=1.4906

c = (abc) / (AB) = 11.5188 / 3.2854 = 3.5061c=abcAB=11.51883.2854=3.5061

Semi perimeter s = (a + b + c) / 2 = (2.204 + 1.4906 + 3.5061) / 2 ~~ 3.6s=a+b+c2=2.204+1.4906+3.506123.6

Inradius I_r = A_t / (s/2) = 1 / 3.6Ir=Ats2=13.6

Area of Incircle A_i = pi (I_r)^2 = pi * (1/3.6)^2 = color(green)(0.2424)Ai=π(Ir)2=π(13.6)2=0.2424