Question

The in-radius of a circle inscribed in an isosceles triangle whose one angle is $\frac{2\pi }{3}$ 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

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

since, the triangle is an isosceles triangle.
and $A={120}^{\circ }$; $b=c$
$\Rightarrow B=C={30}^{\circ }$
From sine rule: $\frac{a}{\sin {120}^{\circ }}=\frac{b}{\sin {30}^{\circ }}$
$\Rightarrow a=\sqrt{3}b$ ...(1)

Now, $△=\frac{bc\sin A}{2}=\frac{\sqrt{3}{b}^2}{4}$
given that, $r=\sqrt{3}$
$\Rightarrow \frac{△}{s}=\sqrt{3}$
$\Rightarrow \frac{\sqrt{3}{b}^2}{4s}=\sqrt{3}$
$\Rightarrow b^2=4s=2a+4b$
$\Rightarrow b^2=2\sqrt{3}b+4b$ [ from (1) ]
$\Rightarrow b=2\sqrt{3}+4$
$△=\frac{\sqrt{3}{b}^2}{4}=12+7\sqrt{3}$
Ans: C