Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/12`, vertex B has an angle of `pi/8`, and the triangle's area is `4`. What is the area of the triangle's incircle?

Answer 1

`color(purple)("Area of incircle " A_r = pi r^2 = 0.00166 " sq units"`

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 = pi/8, hat C = (19pi)/24, A_t = 4`

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

`ab = (4 * 2) / sin ((19)/24) = 13.14`

Similarly, `bc = (4 * 2) / sin (pi/12) = 30.91`

`ca = (4 * 2) / sin (pi/8) = 20.91`

`ab * bc * ca = (abc)^2 = 13.14 * 30.91 * 20.91`

`abc = sqrt(13.14 * 30.91 * 20.91)`

`a = (abc) / (bc) = sqrt(13.14 * 30.91 * 20.91) / 30.91 = 2.98`

Likewise, `b = (abc) / (ca) = sqrt(13.14 * 30.91 * 20.91) / 20.91 = 4.41`

`c = (abc) / (ab) = sqrt(13.14 * 30.91 * 20.91) / 13.14 = 7.01`

`"Perimeter of the triangle " p = a + b + c = 2.98 + 4.41 + 7.01 = 14.4`

`"Radius of incircle " r = A_t / (12 * p) = 4 / (12 * 14.4) = 0.023`

`color(purple)("Area of incircle " A_r = pi r^2 = pi * 0.023^2 = 0.00166`