Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/2`, vertex B has an angle of `( pi)/3`, and the triangle's area is `56`. What is the area of the triangle's incircle?

Answer 1

The area of the incircle is `=27.4u^2`

Explanation:

The area of the triangle is `A=56`

The angle `hatA=1/2pi`

The angle `hatB=1/3pi`

The angle `hatC=pi-(1/2pi+1/3pi)=1/6pi`

The sine rule is

`a/sinA=b/sinB=c/sinC=k`

So,

`a=ksinA`

`b=ksinB`

`c=ksinC`

Let the height of the triangle be `=h` from the vertex `A` to the opposite side `BC`

The area of the triangle is

`A=1/2a*h`

But,

`h=csinB`

So,

`A=1/2ksinA*csinB=1/2ksinA*ksinC*sinB`

`A=1/2k^2*sinA*sinB*sinC`

`k^2=(2A)/(sinA*sinB*sinC)`

`k=sqrt((2A)/(sinA*sinB*sinC))`

`=sqrt(112/(sin(pi/2)*sin(1/3pi)*sin(1/6pi)))`

`=10.58/0.66=16.03`

Therefore,

`a=16.03sin(1/2pi)=16.03`

`b=16.03sin(1/3pi)=13.88`

`c=16.03sin(1/6pi)=8.02`

The radius of the incircle is `=r`

`1/2*r*(a+b+c)=A`

`r=(2A)/(a+b+c)`

`=112/(37.93)=2.95`

The area of the incircle is

`area=pi*r^2=pi*2.95^2=27.4u^2`