Question

One angle of isosceles $△$ is ${120}^o$ and radius of its $=\sqrt{3}$.Then the area of the triangle is 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 A $7+12\sqrt{3}$

Given, the triangle is isosceles. One angle is ${120}^{\circ }$

$\therefore$ The other two angles will be ${30}^{\circ }$ each. Let $x$ be the length of the two sides which are equal, and $a$ is the length of the third side.

From Sine rule, we have,

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

$\frac{x}{\frac{1}{2}}=\frac{a}{\frac{\sqrt{3}}{2}}$

$a=x\sqrt{3}$ ---------- (i)

$Area=\frac{1}{2}{x}^2\sin {120}^{\circ }$

$Area=\frac{\sqrt{3}}{4}{x}^2$

We know, $RadiusofIncircle=\frac{Area}{Semi-perimeter}$

$⟹\sqrt{3}=\frac{Area}{s}=\frac{\frac{\sqrt{3}}{4}{x}^2}{\frac{2x+a}{2}}=\frac{\sqrt{3}{x}^2}{2(2x+a)}$

$\sqrt{3}=\frac{\sqrt{3}{x}^2}{2(2x+x\sqrt{3})}$ ----------from (i)

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

$Area=\frac{\sqrt{3}}{4}{x}^2=\frac{\sqrt{3}}{4}4(2+\sqrt{3}{)}^2$

$\therefore Area=7\sqrt{3}+12$

Option A