Question

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?

Answer 1

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

Explanation:

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.!