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, $\tan 60^{\circ}=\frac{I E}{A E}$
$\sqrt{3}=\frac{\sqrt{3}}{A E}$
$ \Rightarrow AE =1= AF$
$\tan 150^{\circ}=\frac{I E}{B E}$
$\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}}{B E}$
$B E=\frac{\sqrt{3}(\sqrt{2}+1)}{(\sqrt{2}-1)}$
$=\frac{\sqrt{3}(\sqrt{3}+1)^{2}}{2}=\frac{\sqrt{3}(3+1+2 \sqrt{3})}{2}$
$B E=\sqrt{3}(2+\sqrt{3})=3+2 \sqrt{3} $
$A B=1+3+2 \sqrt{3}=4+2 \sqrt{3}$
Now, the area of $\triangle ABC$ 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} $
$=\sqrt{3}(4+3+4 \sqrt{3})=7 \sqrt{3}+12$