Question

Given an isosceles triangle, whose one angle is ${120}^o$ and radius of its incircle $=\sqrt{3}$.

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}$

Let the angles be $A,B,C$

it is given that $A={120}^o$

and since it is an isosceles Triangle, other two angles must be equal $(B=C)$

$A+B+C={120}^o+2B={180}^o\Rightarrow B={30}^o$

since $B=C$ so, $C={30}^o$

we know that radius $r$ of incircle $=\sqrt{3}$

also ,

we know that $r=4Rsin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}\Rightarrow \sqrt{3}=4R\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)$

After, Simplification and rationalization we get

$R=4+2\sqrt{3}$

now,

$a=RsinA\Rightarrow (4+2\sqrt{3})\frac{\sqrt{3}}{2}=2\sqrt{3}+3$

similarly,

$b=2+\sqrt{3}$

$c=2+\sqrt{3}$

$△=rs=\left[\sqrt{3}\right]\left[\right(2+\sqrt{3}\left)\right(2+\sqrt{3}\left)\right]=\sqrt{3}(2+\sqrt{3}{)}^2=7\sqrt{3}+12$

therefore, Answer is $C$