Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/12`, vertex B has an angle of `(3pi)/8`, and the triangle's area is `21`. What is the area of the triangle's incircle?

Answer 1

Area of Incircle `A_i = color(green)(6.6)`

Explanation:

`hat A = pi/12, hatB = 3pi/8, hat C = pi - pi/12 - (3pi)/8 = (13pi)/24, A_t = 21`

`(1/2) bc sin A = (1/2) ca sin B = (1/2) a b sin C = A_t`

`bc = (2 * A_t) / sin A = (2 * 21) = sin (pi/12) = 162.28`

`ca = (2*21) / sin ((3pi)/8) = 45.46`

`ab = (2*21)/sin ((13pi)/24) = 42.36`

`abc = sqrt(ab* bc * ca) = sqrt (42.36 * 45.46 * 162.28) = 559`

`a = (abc) / (bc) = 559 / 162.28 = 3.44`

Similarly, `b = 559 / 45.46 = 12.3`

`c = 559 / 42.36 = 13.2`

Semi perimeter of triangle ABC

`p/2 = (a + b + c) / 2 = (3.44 + 12.3 + 13.2) / 2 = 14.47`

Radius of Incircle `r_i = A_t / (p/2) = 21 / 14.47 = 1.45`

Area of Incircle `A_i = pi r_i^2 = pi * 1.45^2 = color(green)(6.6)`