Question

In radius of a circle which is inscribed in a isosceles
triangle one of whose angle is $2\pi /3,is\sqrt{3}$, 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 these

Answer 1

Answer: (C)

Let AB = AC = a and $\angle 120^{\circ }$.
$\therefore$ Area of triangle = $\frac{1}{2}{a}^{2}sin120^{\circ }$
where, a = AD + BD = $\sqrt{3}tan30^{\circ }+\sqrt{3}cot15^{\circ }$
= 1 + $\frac{\sqrt{3}}{tan(45^{\circ }-15^{\circ })}$
$\Rightarrow a=1+\sqrt{3}\left(\frac{1+tan45^{\circ }tan30^{\circ }}{tan45^{\circ }-tan30^{\circ }}\right)$
= 1 + $\sqrt{3}\left(\frac{\sqrt{3}+1}{\sqrt{3}+1}\right)=4+2\sqrt{3}$
$\therefore$ Area of triangle = $\frac{1}{2}(4+2\sqrt{3}{)}^2(\frac{\sqrt{3}}{2})=(12+7\sqrt{3})$ sq units.