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 `18`. What is the area of the triangle's incircle?

Answer 1

The area of the incircle is `=8.74u^2`

Explanation:

The area of the triangle is `A=18`

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(36/(sin(1/2pi)*sin(1/3pi)*sin(1/6pi)))`

`=9.12`

Therefore,

`a=9.12sin(1/2pi)=9.12`

`b=9.12sin(1/3pi)=7.90`

`c=9.12sin(1/6pi)=4.56`

The radius of the incircle is `=r`

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

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

`=36/(21.58)=1.67`

The area of the incircle is

`area=pi*r^2=pi*1.67^2=8.74u^2`