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?

1 Answer
Jun 30, 2018

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

Explanation:

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"#