Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/8`, vertex B has an angle of `( pi)/3`, and the triangle's area is `18`. What is the area of the triangle's incircle?

Answer 1

`7.414\ \text{unit}^2`

Explanation:

Given that in `\Delta ABC`, `A=\pi/8`, `B={\pi}/3`

`C=\pi-A-B`

`=\pi-\pi/8-{\pi}/3`

`={13\pi}/24`

from sine in `\Delta ABC`, we have

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

`\frac{a}{\sin(\pi/8)}=\frac{b}{\sin ({\pi}/3)}=\frac{c}{\sin ({13\pi}/24)}=k\ \text{let}`

`a=k\sin(\pi/8)=0.383k`

`b=k\sin({\pi}/3)=0.866k`

`c=k\sin({13\pi}/24)=0.991k`

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

`=\frac{0.383k+0.866k+0.991k}{2}=1.12k`

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

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

`18=\sqrt{1.12k(1.12k-0.383k)(1.12k-0.866k)(1.12k-0.991k)}`

`18=0.1644k^2`

`k^2=109.4506`

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

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

`r=\frac{18}{1.12k}`

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

`=\pi r^2`

`=\pi (18/{1.12k})^2`

`=\frac{324\pi}{1.2544k^2}`

`=\frac{811.4445}{109.4506}\quad (\because k^2=109.4506)`

`=7.414\ \text{unit}^2`

Answer 2

Area of Incircle `color(indigo)(A_i = pi r^2 = pi * (1.5357)^2 ~~ 7.4087`

Explanation:

`hat A = pi/8, hat B = pi/3, hat C = (13pi)/24, A_t = 18`

`A_t = 1/2 ab sin C = 1/2 bc sin A = 1/2 ca sin B`

`ab = (2 A_t) / sin C = 36 / sin ((13pi)/24) ~~ 36.3106`

`bc = (2 A_t) / sin A = 36 / sin ((pi)/8) ~~ 94.0725`

`ca = (2 A_t) / sin B = 36 / sin ((pi)/3) ~~ 41.5692`

`a = sqrt (ab * bc * ca) / (bc) = sqrt ((ab * ca) / (bc))`

`a = sqrt((36.3106*41.5692)/94.0725) ~~ 4`

Similarly, `b = sqrt((36.3106 * 94.0725)/41.5692) ~~ 9.0649`

`c = sqrt((94.0725*41.5692)/36.3106) ~~ 10.3777`

Semi perimeter `s = (a + b + c) / 2 = 23.4426/2 = 11.7213`

Incircle radius `r = A_t / s = 18 / 11.7213 ~~ 1.5357`

Area of Incircle `A_i = pi r^2 = pi * (1.5357)^2 ~~ 7.4087`