Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/8`, vertex B has an angle of `( 3pi)/8`, and the triangle's area is `16`. What is the area of the triangle's incircle?

Answer 1

The area of the incircle is `=6.52u^2`

Explanation:

The area of the triangle is `A=16`

The angle `hatA=1/8pi`

The angle `hatB=3/8pi`

The angle `hatC=pi-(1/8pi+3/8pi)=4/8pi=pi/2`

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(32/(sin(pi/8)*sin(3/8pi)*sin(1/2pi)))`

`=9.81`

Therefore,

`a=9.81sin(1/8pi)=3.64`

`b=9.81sin(3/8pi)=9.06`

`c=9.81sin(1/2pi)=9.51`

The radius of the incircle is `=r`

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

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

`=32/(22.2)=1.44`

The area of the incircle is

`area=pi*r^2=pi*1.44^2=6.52u^2`