Question

Given an isosceles triangle, whose one angle is ${120}^{\circ }$ and radius of its incircle is $\sqrt{3}$ unit. Then the area of the triangle in sq. units is

• A. $7+12\sqrt{3}$
• B. $12-7\sqrt{3}$
• C. $12+7\sqrt{3}$
• D. $4\pi$

Answer 1

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