Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/12`, vertex B has an angle of `(5pi)/8`, and the triangle's area is `25`. What is the area of the triangle's incircle?

Answer 1

`color(purple)("Area of inscribed circle " A_r = pi r^2 = pi * 0.065^2 = 0.0133`

Explanation:

The distances from the incenter to each side are equal to the inscribed circle's radius. The area of the triangle is equal to 12×r×(the triangle's perimeter), 1 2 × r × ( the triangle's perimeter ) , where r is the inscribed circle's radius.

First to find the perimeter of the triangle.

`hat A = pi/12, hat B = (5pi)/8, hat C = (7pi)/24, A_t = 25`

Area of triangle `A_t = (1/2)ab sin C = (1/2)bc sin A = (1/2)ca sin B`

`ab = (25 * 2) / sin ((7pi)/24) = 63.02`

Similarly, `bc = (25 * 2) / sin (pi/12) = 193.19`

`ca = (25 * 2) / sin ((5pi/8) = 54.12`

`ab * bc * ca = (abc)^2 = 63.02 * 193.19 * 54.12`

`abc = sqrt(63.02 * 193.19 * 54.12) ~~ 810.98`

`a = (abc) / (bc) = 810.98 / 193.19 = 4.2`

Likewise, `b = (abc) / (ca) = 810.98 / 54.12 = 14.98`

`c = (abc) / (ab) = 810.98 / 63.02 = 12.87`

`"Perimeter of the triangle " p = a + b + c = 4.2 + 14.98 + 12.87 = 32.05`

`"Radius of incircle " r = A_t / (12 * p) = 25 / (12 * 32.05) = 0.065`

`color(purple)("Area of inscribed circle " A_r = pi r^2 = pi * 0.065^2 = 0.0133`