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 $36$ . What is the area of the triangle's incircle?

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Answer 1 · 2016-07-02T19:38:28

Area of Incircle $=pir^2=36pi(sqrt2-1)^2=36pi(3-2sqrt2)$

We will use the usual notation to solve the problem.

In $DeltaABC, A=pi/2, B=pi/4 rArr C=pi/4 rArr b=c$

$Delta=1/2*b*c*sinA rArr36=1/2*b*b*sin(pi/2)rArr36=b^2/2$

$:. b=c=6sqrt2$ $:. a^2=b^2+c^2=b^2+b^2=2b^2=144rArra=12$

In right $DeltaABC,$ in which, $a$ is hypo., it is a well=known result that,

$r=(b+c-a)/2=(6sqrt2+6sqrt2-12)/2=12/2(sqrt2-1)=6(sqrt2-1)$

$:.$ Area of Incircle $=pir^2=36pi(sqrt2-1)^2=36pi(3-2sqrt2)$

Hope this will help!

A triangle has vertices A, B, and C. Vertex A has an angle of pi/2 pi/2 , vertex B has an angle of ( pi)/4 ( pi)/4 , and the triangle's area is 36 36 . What is the area of the triangle's incircle?