Question

Given an isoceles triangle, whose one angle is ${120}^{o}C$ and radius of its incircle is $\sqrt{3},$ 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. $4+2\sqrt{3}$

Answer 1

Answer: (B)

$12+7\sqrt{3}$

Let the angles be $A,B,C$

Given:$\angle A={120}^{\circ }$

and since it is an isosceles Triangle, other two angles must be equal

We have $B=C$

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

$\Rightarrow 2B={180}^{\circ }-{120}^{\circ }={60}^{\circ }$

since $B=C$ so $C=\frac{1}{2}\times {60}^{\circ }={30}^{\circ }$

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

Also,we know that $r=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

$\Rightarrow \sqrt{3}=4R\sin \frac{{120}^{\circ }}{2}\sin \frac{{30}^{\circ }}{2}\sin \frac{{30}^{\circ }}{2}$

$\Rightarrow \sqrt{3}=4R\sin {60}^{\circ }\sin {15}^{\circ }\sin {15}^{\circ }$

We know that $\sin {15}^{\circ }=\frac{\sqrt{3}-1}{2\sqrt{2}}$

$\Rightarrow \sqrt{3}=4R\times \frac{\sqrt{3}}{2}\times \frac{\sqrt{3}-1}{2\sqrt{2}}\times \frac{\sqrt{3}-1}{2\sqrt{2}}$

$\Rightarrow \sqrt{3}=2\sqrt{3}R\left(\frac{3+1-2\sqrt{3}}{4\times 2}\right)$

$\Rightarrow 1=R\times \frac{4-2\sqrt{3}}{4}$

$\Rightarrow 1=R\times \frac{2-\sqrt{3}}{2}$

$\Rightarrow R=\frac{2(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}=\frac{4+2\sqrt{3}}{4-3}=4+2\sqrt{3}$

Now,$a=R\sin A\Rightarrow (4+2\sqrt{3})\sin {60}^{\circ }=(4+2\sqrt{3})\times \frac{\sqrt{3}}{2}=2\sqrt{3}+3$

$b=R\sin B=(4+2\sqrt{3})\sin {30}^{\circ }=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}$

$c=R\sin C=(4+2\sqrt{3})\sin {30}^{\circ }=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}$

$\Delta =rs$ where $s=bc$

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

$=\sqrt{3}(4+3+2\sqrt{3})$

$=\sqrt{3}(7+2\sqrt{3})$

$=7\sqrt{3}+12$