Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/2`, vertex B has an angle of `( pi)/8`, and the triangle's area is `16`. What is the area of the triangle's incircle?

Answer 1

Area of in circle `A_i = 6.7` sq units

Explanation:

`A_t = (1/2) ab sin C =(1/2)b c sin A = (1/2) c a sin B`

`hat A = pi/2, hat B = pi/8, hat C = (3pi)/8`

`a b = (2 * 16) / sin ((3pi)/8) = 34.64`

`c a = (2*16)/ sin (pi/8) = 83.62`

`b c= (2*16)/ sin(pi/2) = 32`

`a = (abc) / (b c) = sqrt(34.64*83.62*32) / 32 = 9.51`

`b = (abc) / (c a) = sqrt(34.64* 83.62*32) / 83.62 = 3.64`

`c = sqrt(34.64*83.62*32)/34.64 = 8.79`

Semi perimeter `s = (a + b + c) / 2`

`s = (9.51 + 3.64 + 8.79) / 2 = 10.97`

Radius of in circle `r = A_t / s = 16 / 10.97 = 1.46`

Area of in circle `A_i = pi r^2 = pi * 1.46^2 = 6.7`