Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/2`, vertex B has an angle of `( pi)/8`, and the triangle's area is `3`. What is the area of the triangle's incircle?

Answer 1

Area of incircle is `1.251`

Explanation:

It is a right angled triangle whose angles are `pi/2`, `pi/8` and third angle would be `pi-pi/8-pi/2=(3pi)/8`

Now let hypotenuse be `h`, then other two sides are `hsin(pi/8)` and `hsin((3pi)/8)` and area of triangle `Delta` is `3`. So we have

`1/2h^2sin((3pi)/8)sin(pi/8)=1/2h^2xx0.9239xx0.3827=3`

or `h=sqrt((3xx2)/(0.9239xx0.3827))=sqrt16.9695=4.12`

and sides are `4.12xx0.9239=3.81` and `4.12xx0.3827=1.58`

and half perimeter `s` is given by `s=1/2xx(4.12+3.81+1.58)=1/2xx9.51=4.755`

As radius of incircle is `Delta/s=3/4.755=0.631`

and its area is `0.631^2pi=3.1416xx0.631^2=1.251`