Question

Inradius of a circle which is inscribed in an isosceles triangle, one of whose angle is $2\pi /3$, is $\sqrt{3}$, then the area of the triangle is

• A. $4\sqrt{3}$
• B. $12-7\sqrt{3}$
• C. $12+7\sqrt{3}$
• D. none of these

Answer 1

Answer: (C)

From the figure shown here,
we have $\angle A=2\pi /3$ and $\angle B=\angle C$.
Now, $\frac{2\pi }{3}+2\angle B=\pi$
$\angle B=\frac{\pi }{6}$
$r=\sqrt{3}$

Now, $\tan 60^{\circ }=\frac{IE}{AE}$
$\sqrt{3}=\frac{\sqrt{3}}{AE}$
$\Rightarrow AE=1=AF$
$\tan 150^{\circ }=\frac{IE}{BE}$
$\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}}{BE}$
$BE=\frac{\sqrt{3}(\sqrt{2}+1)}{(\sqrt{2}-1)}$
$=\frac{\sqrt{3}(\sqrt{3}+1{)}^2}{2}=\frac{\sqrt{3}(3+1+2\sqrt{3})}{2}$
$BE=\sqrt{3}(2+\sqrt{3})=3+2\sqrt{3}$
$AB=1+3+2\sqrt{3}=4+2\sqrt{3}$
Now, the area of $△ABC$ is
$\frac{1}{2}(4+2\sqrt{3})(4+2\sqrt{3})\sin 120^{\circ }=\frac{1}{2}(4+2\sqrt{3}{)}^2\cdot \frac{\sqrt{3}}{2}$
$=\frac{\sqrt{3}}{4}(4+2\sqrt{3}{)}^2$
$=\sqrt{3}(4+3+4\sqrt{3})=7\sqrt{3}+12$