Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/3`, vertex B has an angle of `(5 pi)/12`, and the triangle's area is `16`. What is the area of the triangle's incircle?

Answer 1

Area of I’m circle `color(green)(A_i = 9.19) sq. Units`

Explanation:

Given `A = pi /3, B = (5pi) / 12, C = pi - A - B = pi / 4`

Area of triangle `A_t = 16`

`a*b = A_t / ((1/2) sin C) = 16 / (0.5 sin (pi/3)) ~~ 36.95`

Similarly,
`b*c = 16 / (0.5 sin A) = 16 / (0.5 * sin ((5pi)/12)) ~~ 33.13`

`c*a = 16 / (0.5 sin B) = 16 / (0.5 * sin (pi/4)) ~~ 45.25`

`(ab * bc * ca) = (abc)^2 = 36.95 * 33.13 * 45.25 = 55392.95`

`abc = sqrt(55392.95) ~~ 235 36`

`a = (abc) / (bc) = 235.36 / 33.13 ~~ 7.1`

`b = (abc) / (ca) = 235.36 / 45.25 ~~ 5.2`

`c = (abc) / (ab) = 235.36 / 36.95 ~~ 6.37`

Semi perimeter `S/ 2 = (a + b + c) / 2 = (7.1 + 5.2 + 6.37) / 2 = 9.34`

In radius `r = A_t / (S/2) = 16 / 9.34 = 1.71`

Area of in circle `A_i = pi r^2 = pi * (1.71)^2 = color(green)(9.19)`