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

Answer 1

`r= sqrt30/(2+sqrt2)`

Explanation:

Triangle ABC is a right triangle. Since one of its angle is `pi/4` it is an isosceles triangle also. If one of the equal sides is 'x' , its area would be `1/2 x*x= 1/2 x^2`

Thus `1/2 x^2 =15 -> = sqrt30`

The hypotenuse of the right triangle ABC would be `sqrt(30+30)= sqrt60`

The sum of the sides of triagle ABC would this be `sqrt30+sqrt30 +sqrt60= sqrt30 (2+sqrt2)`

If 'r' is the radius of the incircle then

`r= 2(Area of triangle)/(sum of the sides of the triangle)`
Thus `r= 30/(sqrt30 (2+sqrt2)`

`r= sqrt30/(2+sqrt2)`