Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/2`, vertex B has an angle of `( pi)/4`, and the triangle's area is `12`. What is the area of the triangle's incircle?

Answer 1

Area of Incircle `A_i = pi r_i^2 = color (purple)(6.4683)` sq units

Explanation:

`hat A = pi/2, hat B = pi/4, hat C = pi - pi/2 -pi/4 = pi/4, A_t = 12`

#It’s an isosceles right triangle with sides b and c equal

`A_t = (1/2)ab sin C = (1/2) ca sin B`

`ab = ca = (2 * A_t) / sin C = (2*12)/ sin (pi/4) = 24sqrt2`

`bc = (2 * 12) / sin (pi/2) = 24`

`abc = sqrt(ab* bc* ca) = sqrt (24sqrt2 * 24sqrt2 * 24) = 96 sqrt3`

`a = (96 sqrt3) / 24 = 4 sqrt3`

`b = c = (96 sqrt3) / (24 sqrt2 )= 2 sqrt(6)`

Radius of Incircle `r_i = (b + c - a)/2 = (4 sqrt6 - 4 sqrt3) /2 = 1.4349`

Area of Incircle `A_i = pi r_i^2 = pi * 1.4349^2 = color (purple)(6.4683)` sq units