Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/8`, vertex B has an angle of `( pi)/6`, and the triangle's area is `3`. What is the area of the triangle's incircle?

Answer 1

The area of the incircle is `=1.02` square units.

Explanation:

The area of the triangle is `A=3`

The angle `hatA=1/8pi`

The angle `hatB=1/6pi`

The angle `hatC=pi-(1/8pi+1/6pi)=17/24pi`

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(6/(sin(pi/8)*sin(1/6pi)*sin(17/24pi)))`

`=6.29`

Therefore,

`a=6.29sin(1/8pi)=2.41`

`b=6.29sin(1/6pi)=3.15`

`c=6.29sin(17/24pi)=4.99`

Observe in the figure above that joining incenter with the three vertices `A,B` and `C` results in three triangles, whose bases are `a,b` and `c` and height is `r`, the radius of incenter and therefore the sum of the area of these triangles is area of `DeltaABC`.

Hence we have, `1/2*r*(a+b+c)=A`

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

`=6/(10.55)=0.57`

The area of the incircle is

`area=pi*r^2=pi*0.57^2=1.02` square units.