Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/2`, vertex B has an angle of `( 3pi)/8`, and the triangle's area is `16`. What is the area of the triangle's incircle?

Answer 1

6.681 (to four significant figures)

Explanation:

This is a right angled triangle with `a` as the hypotenuse. The other two sides obey
`b = a sin/_ B`
`c = a cos/_B`

The area `A` of the triangle is given by
`A = 1/2 bc = 1/2 a^2 sin/_B cos/_B`

Alternatively, by adding up the areas of the three triangles formed by joining the vertices to the in-center (each of which have a common height - namely, the radius `r` of the incircle), we get

`A = 1/2 (a+b+c)r`

Equating the two expressions for the area, we get

`r = {bc}/(a+b+c)= a {sin/_B cos/_B }/{1+sin/_B + cos/_B}`

So that the area of the in-circle is

`pi r^2 = pi a^2 {sin^2/_B cos^2/_B }/(1+sin/_B + cos/_B)^2 = 2 pi A {sin/_B cos/_B }/(1+sin/_B + cos/_B)^2 = pi A {sin(2/_B ) }/(1+sin/_B + cos/_B)^2`
Substituting the numerical values, we get an area of 6.681 units