Question

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 `1`. What is the area of the triangle's incircle?

Answer 1

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

Explanation:

Three angles are `A = pi / 8, B = pi / 12, C = pi - pi/8 - pi/12 = (19pi)/24`

Given Area of triangle A_t = 1#

`bc = A_t / ((1/2) sin A) = (2 * 1) / sin (pi / 8) = 5.2263`

`ca = A_t / ((1/2) sin B) = 2 / sin (pi / 12) = 7.7274`

`ab = A_t / ((1/2) sin C) = 2 / sin ((19pi)/24) = 3.2854`

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

`a = (abc) / (bc) = 11.5188 / 5.2263 = 2.204`

`b = (abc) / (ca) = 11.5188 / 7.7274 = 1.4906`

`c = (abc) / (AB) = 11.5188 / 3.2854 = 3.5061`

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

Inradius `I_r = A_t / (s/2) = 1 / 3.6`

Area of Incircle `A_i = pi (I_r)^2 = pi * (1/3.6)^2 = color(green)(0.2424)`