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

Answer 1

The area of the triangle's incircle is `(10-4sqrt6)pi`.

Explanation:

Let `S` be the area of the incircle, calculated with the formula

`S = pir^2`, where `r` is the inradius of the triangle.

The simplest formula to find the inradius is

`r= "Area"/"Semiperimeter" = Delta/s`

where `Delta` is the area of the triangle, equal to `2` in this case, and `s` is the semiperimeter:

`s=(a+b+c)/2`

See this answer of mine for a generalised proof of this. As all triangles are tangential, the formula applies here aswell.

The angle of vertex `C` can easily be calculated:

`C = pi - A - B = (5pi)/12`

Our triangle is visualised below.

As it is a right triangle, its area will be equal to

`Delta = (AB xx BC)/2 = (c*a)/2=2 => ac=4`

Using the trigonometric functions, we see that

`sin A = "opposite"_A/"hypotenuse"=a/b`

`sin C = "opposite"_C/"hypotenuse" = c/b`

`=> (ac)/b^2 = sinAsinC => ac = b^2*sinAsinC = 4`

Using the sum and difference formulas, we find out that

`sin C = sin color(black)((5pi)/12) = (sqrt6+sqrt2)/4`

`sin A = sin color(black)(pi/12) = (sqrt6-sqrt2)/4`

Hence

`1/16 b^2 (sqrt6+sqrt2)(sqrt6-sqrt2) = 4`

`1/16 b^2 (sqrt6^2-sqrt2^2) = 4`

`1/16 b^2 *4 =4 => b^2 = 16 => color(blue)(b=4)`

Then,

`color(blue)(a = bsinA = sqrt6-sqrt2)`

`color(blue)(c = bsinC = sqrt6+sqrt2)`

`:. s = (sqrt6-sqrt2+4+sqrt6+sqrt2)/2 =sqrt6+2`

`r = Delta/s = 2/(sqrt6+2) = 2/(sqrt6+2) * color(purple)((sqrt6-2)/(sqrt6-2))=(2sqrt6-4)/2`

`color(red)( :. r=sqrt6-2)`

`S = pir^2 = pi(sqrt6-2)^2 = pi(6-4sqrt6+4)=(10-4sqrt6)pi`

Answer 2

`0.6346\ \text{unit}^2`

Explanation:

Given that in `\Delta ABC`, `A=\pi/12`, `B=\pi/2`

`C=\pi-A-B`

`=\pi-\pi/12-\pi/2`

`={5\pi}/12`

from sine in `\Delta ABC`, we have

`\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}`

`\frac{a}{\sin(\pi/12)}=\frac{b}{\sin (\pi/2)}=\frac{c}{\sin ({5\pi}/12)}=k\ \text{let}`

`a=k\sin(\pi/12)=0.259k`

`b=k\sin(\pi/2)=k`

`c=k\sin({5\pi}/12)=0.966k`

`s=\frac{a+b+c}{2}`

`=\frac{0.259k+k+0.966k}{2}=1.1125k`

Area of `\Delta ABC` from Hero's formula

`\Delta=\sqrt{s(s-a)(s-b)(s-c)}`

`2=\sqrt{1.1125k(1.1125k-0.259k)(1.1125k-k)(1.1125k-0.966k)}`

`2=0.125k^2`

`k^2=16`

Now, the in-radius (`r`) of `\Delta ABC`

`r=\frac{\Delta}{s}`

`r=\frac{2}{1.1125k}`

Hence, the area of inscribed circle of `\Delta ABC`

`=\pi r^2`

`=\pi (2/{1.1125k})^2`

`=\frac{4\pi}{1.2376k^2}`

`=\frac{10.15336}{16}\quad (\because k^2=16)`

`=0.6346\ \text{unit}^2`