Question

In an isosceles triangle, if one angle is ${120}^o$ and radius of its incircle is $\sqrt{3}$, then the area of the triangle in square units is

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

Answer 1

Answer: (C)

$12+7\sqrt{3}$

Given that ABC is an isosceles triangle. In the figure, $\angle A={120}^o$. I is the incenter. AD is the angle bisector of $\angle A$ and also the height of the triangle.
Area of the triangle is $\frac{1}{2}BC.AD$
From $△IBD$, $BD=\frac{1}{2}BC=\frac{r}{\tan {15}^o}$...(1)
In triangle $ABD$, $AD=BD\tan {30}^o$
Hence, area of triangle is $\frac{1}{2}BC.AD=BD.BD\tan {30}^o=B{D}^2\tan {30}^o$
Using (1), we get area$={\left(\frac{r}{\tan {15}^o}\right)}^2\tan {30}^o=r^2\frac{\tan {30}^o}{{\tan }^2{15}^o}$
Using $r^2=3,\tan {30}^o=\frac{1}{\sqrt{3}},{\tan }^2{15}^o=7-4\sqrt{3}$, we get area$=3\frac{1/\sqrt{3}}{7-4\sqrt{3}}=\frac{\sqrt{3}}{7-4\sqrt{3}}=\frac{\sqrt{3}(7+4\sqrt{3})}{49-48}=7\sqrt{3}+12$