Question

In radius of a circle which is inscribed in a isosceles triangle one of whose angle is $\frac{2\pi }{3}$ is $\sqrt{3}$, then area of triangle (in sq units) is

• A. $4\sqrt{3}$
• B. $12-7\sqrt{3}$
• C. $12+7\sqrt{3}$
• D. None of the above

Answer 1

The correct option is C $12+7\sqrt{3}$

Let $AB=AC=aand\angle A={120}^{\circ }$
$\therefore$ Area of triangle $=\frac{1}{2}{a}^{2}sin{120}^{\circ }$
Where $a=AD+BD=\sqrt{3}tan{30}^{\circ }+\sqrt{3}cot{15}^{\circ }=1+\frac{\sqrt{3}}{tan({45}^{\circ }-{15}^{\circ })}$



$\Rightarrow a=1+\sqrt{3}\left(\frac{1+tan{45}^{\circ }tan{30}^{\circ }}{ta{45}^{\circ }-tan{30}^{\circ }}\right)$
$=1+\sqrt{3}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)=4+2\sqrt{3}$
$\therefore$ Area of triangle
$=\frac{1}{2}(4+2\sqrt{3}{)}^2\left(\frac{\sqrt{3}}{2}\right)=(12+7\sqrt{3})squnits$