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

• A

  $4 \sqrt{3}$
• B

  $12-7 \sqrt{3}$
• Correct answer
  C

  $12+7 \sqrt{3}$
• D

  None of these

Solution

Let $A B=A C$ and $\angle A=120^{\circ}$ $\therefore$ Area of $\Delta=\frac{1}{2} a^2 \sin 120^{\circ}$ where, $a=A D+B D$ $=\sqrt{3} \tan 30^{\circ}+\sqrt{3} \cot 15^{\circ}$ $=1+\sqrt{3}\left(\frac{1+\tan 45^{\circ} \tan 30^{\circ}}{\tan 45^{\circ}-\tan 30^{\circ}}\right)$ $=1+\sqrt{3}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)$ $\therefore \quad a=4+2 \sqrt{3}$ $\therefore$ Area of $\Delta=\frac{1}{2}(4+2 \sqrt{3})^2\left(\frac{\sqrt{3}}{2}\right)=12+7 \sqrt{3}$

Hence, the correct answer is option: 'C'.