Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/2`, vertex B has an angle of `( pi)/4`, and the triangle's area is `25`. What is the area of the triangle's incircle?

Answer 1

The area of the incircle is `=13.48u^2`

Explanation:

The area of the triangle is `A=25`

The angle `hatA=1/2pi`

The angle `hatB=1/4pi`

The angle `hatC=pi-(1/2pi+1/4pi)=1/4pi`

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(50/(sin(pi/2)*sin(1/4pi)*sin(1/4pi)))`

`=10`

Therefore,

`a=10sin(1/2pi)=10`

`b=10sin(1/4pi)=7.07`

`c=10sin(1/4pi)=7.07`

The radius of the incircle is `=r`

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

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

`=50/(24.14)=2.07`

The area of the incircle is

`area=pi*r^2=pi*2.07^2=13.48u^2`