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?

1 Answer
Ratnaker Mehta
Jun 23, 2016

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).#

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