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.!

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)/6 ( pi)/6 , and the triangle's area is 45 45 . What is the area of the triangle's incircle?