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!