Question

One angle of an isosceles triangle is ${120}^0$.If $r=\sqrt{3},$ then $△=$

• 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 $12+7\sqrt{3}$

Let $x$ are the two equal sides of the isosceles $△$ and $a$ is the third side. One of the angle given is ${120}^o$, So the remaining two equal angle will be ${30}^o$. So, from sine law;

$\frac{x}{\sin {30}^o}=\frac{x}{\sin {30}^o}=\frac{a}{\sin {120}^o}$

Now, $△=\frac{1}{2}.x.x.\sin {120}^o=\frac{\sqrt{3}}{4}{x}^2$

Inradius, $r=\frac{\Delta }{s}=\frac{\sqrt{3}/4x^2}{\frac{a+2x}{2}}$

$\Rightarrow \sqrt{3}=\frac{\sqrt{3}{x}^2}{2(x\sqrt{3}+2x)}$

$\Rightarrow x=2(2+\sqrt{3})$

$\therefore \Delta =\frac{\sqrt{3}}{4}{x}^2=\frac{\sqrt{3}}{4}\left(2\right(2+\sqrt{3}){)}^2=7\sqrt{3}+12sq.unit$

Hence, option $C$ is correct.