A triangle has vertices A, B, and C. Vertex A has an angle of pi/12 , vertex B has an angle of pi/12 , and the triangle's area is 8 . What is the area of the triangle's incircle?

Mar 20, 2016

${A}_{\odot} = \pi \cdot {r}_{\Delta}^{2} = {\left[\frac{8}{8 + \frac{4}{\sqrt{2 - \sqrt{3}}}}\right]}^{2} \cdot \pi$

${A}_{\odot} \approx {\left[\frac{8}{8 + 7.73}\right]}^{2} \pi \approx .259 \pi$

Explanation:

Firs by AA the triangle is isosceles triangle with sides $a = b$ see geometric figure. Thus using the sine law and the cross product we can write:
1) $\frac{a}{\sin} \alpha = \frac{b}{\sin} \beta = \frac{c}{\sin} \gamma$
2) ${A}_{\Delta} = \frac{1}{2} | a | \cdot | c | \cdot \sin \left(\alpha\right) = \frac{1}{2} \left(a c\right) \sin \left(\alpha\right)$
3) $a = b \text{ and } \alpha = \beta = \frac{\pi}{12}$

From 1) we have $a = \sin \frac{\alpha}{\sin} \gamma c$
Substitute into 2) $2 {A}_{\Delta} = {\sin}^{2} \frac{\alpha}{\sin} \beta {c}^{2}$
Now letting A_Delta=8; sinalpha=sin(pi/12); sinbeta= sin(5/6pi)
$16 = {\sin}^{2} \frac{\frac{\pi}{12}}{\sin} \left(\frac{5}{6} \pi\right) {c}^{2}$
$c = \sqrt{16 \cdot \sin \frac{\frac{5}{6} \pi}{\sin} ^ 2 \left(\frac{\pi}{12}\right)} = \frac{4}{\sin} \left(\frac{\pi}{12}\right) \sqrt{\sin \left(\frac{5}{6} \pi\right)}$
c=(4*2)/sqrt((2-sqrt(3)))color(red)(sqrt(sin(5/6pi))
color(red)(sqrt(sin(5/6pi)) = color(blue)(cos(pi/2-5/6pi)=cos(pi/3)=1/2)
$c = \frac{4 \cdot 2}{\sqrt{\left(2 - \sqrt{3}\right)}} \textcolor{b l u e}{\frac{1}{2}} = \frac{4}{\sqrt{2 - \sqrt{3}}}$
$a = \sin \frac{\frac{\pi}{12}}{\cos} \left(\frac{\pi}{3}\right) \cdot \frac{4}{\sqrt{2 - \sqrt{3}}} = 4 = b$
Now the task is to find the radius of the Incenter, Inradius ${r}_{\Delta}$
${r}_{\Delta} = {A}_{\Delta} / {P}_{\Delta}$ Where
${A}_{\Delta}$ = Area and ${P}_{\Delta}$ = Perimeter
r_Delta = 8/(8+(4)/sqrt(2-sqrt(3))
An the ${A}_{\odot} = \pi \cdot {r}_{\Delta}^{2} = {\left[\frac{8}{8 + \frac{4}{\sqrt{2 - \sqrt{3}}}}\right]}^{2} \pi$
${A}_{\odot} \approx {\left[\frac{8}{8 + 7.73}\right]}^{2} \pi \approx .259 \pi$