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

Answer 1

Area of Incircle `=pi(6-4sqrt2).`

Explanation:

We follow the usual notation for `DeltaABC` in the following discussion.

Observe that `DeltaABC` is an isosceles right triangle with `b=c.` Its area `=1/2*b*c=1/2*b^2=2`(given). `rArr b^2=4 rArr b=c=2,` so that hypo. `a=sqrt(b^2+c^2)=sqrt(4+4)=2sqrt2.`

Thus, we have, `s=(a+b+c)/2=(2sqrt2+2+2)/2=(4+2sqrt2)/2=2+sqrt2.`

Now we use the Formula :`Delta=rs rArr 2=r(2+sqrt2),` giving inradius `r=2/(2+sqrt2)=2*(2-sqrt2)/(4-2)=(2-sqrt2).`

Hence, the Area of Incircle `=pir^2=pi(2-sqrt2)^2=pi(4-4sqrt2+2)=pi(6-4sqrt2).`