A triangle has vertices A, B, and C. Vertex A has an angle of #pi/2 #, vertex B has an angle of #( pi)/6 #, and the triangle's area is #45 #. What is the area of the triangle's incircle?

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Answer 1 · 2016-09-20T18:34:49

#"The Area of the Incircle="pir^2=15sqrt3pi(2-sqrt3).#

We solve this problem with the help of Trigo.

In the usual notation for the #DeltaABC#, by the Sine-Rule , .

#a/sinA=b/sinB=c/sinC=2R#.

#rArr a/1=b/(1/2)=c/(sqrt3/2)=2R#.

#rArr a=2R, b=R, c=sqrt3R#.

Now, the Area of #DeltaABC=1/2bcsinA rArr 45=1/2*R*sqrt3R#.

#:. R^2=90/sqrt3=30sqrt3#.

Also, the Area of #DeltaABC=Delta=rs, where,#

#2s=a+b+c=2R+R+sqrt3R=sqrt3(sqrt3+1)R", so that,"#

#Delta=rs rArr 4Delta^2=r^2(2s)^2 rArr 4(45^2)=r^2(sqrt3(sqrt3+1)R)^2#

#rArr r^2=(4*45^2)/{3(sqrt3+1)^2*30sqrt3}=45/((2+sqrt3)sqrt3),#

#or, r^2=15sqrt3(2-sqrt3)#. And, therefore, finally,

#"The Area of the Incircle="pir^2=15sqrt3pi(2-sqrt3).#

Enjoy Maths.!