Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/12`, vertex B has an angle of `(pi)/2`, and the triangle's area is `12`. What is the area of the triangle's incircle?

Answer 1

Area of incircle is `3.808`

Explanation:

Let us consider a right angled triangle in general and an incircle within it as shown below.

![https://www.quora.com/What-is-the-radius-of-the-incircle-of-the-3-4-5-right-triangle]()

Observe that centre of incircle `O` makes three triangles with sides `a`, `b` and `c` and as area of triangle is `1/2xx"base"xx"height"` `-` and hence area of these triangles is `(ar)/2`, `(br)/2` and `(cr)/2`. As area of complete triangle is `1/2axxb`, we have

`r/2(a+b+c)=(ab)/2` and as area of triangle is `12`, we have `a+b+c=24/r` or `r=24/(a+b+c)`.

Here we have two angles `pi/2` and `pi/12` and third angle is `pi-pi/2-pi/12=(5pi)/12` and using sine law, we have

`a/sin(pi/12)=b/sin((5pi)/12)` or `a/b=sin(pi/12)/sin((5pi)/12)`

i.e. `a/b=0.25882/0.96593=0.26795` and as `ab=24`

Hence `b^2=(ab)xxa/b=24xx0.26795`

and `b=sqrt6.4308=2.536` and `a=24/2.536=9.464`

and `c=sqrt(2.536^2+9.464^2)=sqrt95.9986=9.8`

Hence `r=24/(2.536+9.464+9.8)=24/21.8=1.101`

and area of incircle is `pi(1.101)^2=3.808`