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 16 . What is the area of the triangle's incircle?

1 Answer
Jun 27, 2018

color(crimson)("Area of inscribed circle " = A_i = pi r^2 = 8.6186

Explanation:

"Area of "Delta = A_t = (1/2) a b sin C = (1/2) b c sin A = (1/2) c a sin B

"Given " hat A = pi/2, hat B = pi/4, hat C = pi/4, A_t = 16

It's an isosceles right triangle.

http://mathibayon.blogspot.com/2015/01/derivation-of-formula-for-radius-of-incircle.html#.WzLtzNIza70http://mathibayon.blogspot.com/2015/01/derivation-of-formula-for-radius-of-incircle.html#.WzLtzNIza70

a b = (2 A_t) / sin C = 32 / sin (pi/4) = 45.25

b c = (2 A_t) / sin A = 32 / sin (pi/2) = 32

c a = (2 A_t) / sin B = 32 / sin (pi/4) = 45.25

a = (a b c) / (b c) = sqrt(45.25 * 45.25 * 32) / 32 = 8

b = (a b c) / (c a) = sqrt(45.25 * 45.25 * 32) / 45.25 = 5.66

c = (a b c) / (a b) = sqrt(45.25 * 45.25 * 32) / 45.25 = 5.66

"Semi-perimeter " = s = (a + b + c) / 2 = 19.32 / 2 = 9.66

"Radius of inscribed circle " = r = A_t / s = 16 / 9.66 = 1.66

"Area of inscribed circle " = A_i = pi r^2 = pi * 1.66^2 = 8.6186