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)

12 + 7 $\sqrt 3$

Answer 2

Let AB = AC = a and $\angle 120^\circ$. $\therefore$ Area of triangle = $\frac{1}{2} a^2 \, sin \, 120^\circ$ where, a = AD + BD = $\sqrt 3 \, tan \, 30^\circ + \sqrt 3 \, cot \, 15^\circ$ \hspace17mm = 1 + $\frac{ \sqrt 3}{ tan \, (45^\circ - 15^\circ)}$ $\Rightarrow a = 1 + \sqrt 3 \bigg( \frac{1 + tan \, 45^\circ \, tan \, 30^\circ}{ tan \, 45^\circ - tan \, 30^\circ}\bigg)$ = 1 + $\sqrt 3 \bigg( \frac{ \sqrt 3 + 1}{ \sqrt 3 + 1} \bigg) = 4 + 2 \sqrt 3$ $\therefore$ Area of triangle = $\frac{1}{2} (4 + 2 \sqrt 3)^2 \bigg( \frac{ \sqrt 3}{2}\bigg) = (12 + 7 \sqrt 3)$ sq units.