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 $45$ . What is the area of the triangle's incircle?

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Answer 1 · 2018-06-30T11:00:18

$color(magenta)("Area if incircle " A_c = pi r^2 = 24.24 " sq units"$

http://mathibayon.blogspot.com/2015/01/derivation-of-formula-for-radius-of-incircle.html#.Wzdgo9Iza70

$"Given : " hat A = pi/2, hat B = pi/4, hat C = pi - pi/2 - pi/4 = pi/4, A_t = 45$

It's as isosceles right triangle with sides in the ratio 1 : 1 : sqrt 2

$A_t = (1/2) b c = (1/2) b^2 " as b = c and a = b sqrt2$

$b^2 = 45 * (2/1) = 90, b = 9.49$

$a = b * sqrt 2 = sqrt90 * sqrt2 = sqrt180 = 13.42$

$"Semi perimeter " s = (a + b + c) / 2 = (13.42 + 9.49 + 9.49) / 2 = 16.2$

$"Radius of incircle " r = A_t / s = 45/16.2 = 2.7778$

$color(magenta)("Area if incircle " A_c = pi r^2 = pi * 2.7778^2 = 24.24 " sq units"$

A triangle has vertices A, B, and C. Vertex A has an angle of pi/2 pi/2 , vertex B has an angle of ( pi)/4 ( pi)/4 , and the triangle's area is 45 45 . What is the area of the triangle's incircle?