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 `18`. What is the area of the triangle's incircle?

Answer 1

`A=pi*(6/(2+sqrt(2)))^2`

Explanation:

Since `B=pi/4` we get `b=c` so we can compute `b^2/2=18` so `b=6` and by the Pythagorean Theorem we get `a=sqrt(2)*6` and our inradius is `a=36/(12+6sqrt(2))=6/(2+sqrt(2))` by the formula `r=A/s` `s=(a+b+c)/2` and `A` denotes the area.