Question

Given an isosceles triangle, whose one side is ${120}^0$ and radius of its in-circle is $\sqrt{3}$.Then the area of the triangle in square unit is

• A. $(7+12\sqrt{3})$
• B. $(12-7\sqrt{3})$
• C. $(12+7\sqrt{3})$
• D. $4\pi$

Answer 1

Answer: (C)

$(12+7\sqrt{3})$

In $△IBD,\tan {15}^0=\frac{\sqrt{3}}{\frac{a}{2}}$
We know that $\tan {15}^0=2-\sqrt{3}$
$\Rightarrow 2-\sqrt{3}=\frac{2\sqrt{3}}{a}$
$\Rightarrow a=\frac{2\sqrt{3}}{2-\sqrt{3}}$
On Rationalising the denominator, we get
$a=\frac{2\sqrt{3}(2+\sqrt{3})}{4-3}=2\sqrt{3}(2+\sqrt{3})$
In $△ABD,\tan {30}^0=\frac{AD}{\frac{a}{2}}$
$\Rightarrow AD=\frac{a}{2}\tan {30}^0=\frac{a}{2\sqrt{3}}$
$\therefore$ Area of $△ABC=\frac{1}{2}.a.AD$
$=\frac{1}{2}.a\frac{a}{2\sqrt{3}}$
$=\frac{a^2}{4\sqrt{3}}$
$=\frac{{\left(2\sqrt{3}(2+\sqrt{3})\right)}^2}{4\sqrt{3}}$
$=\frac{3(7+4\sqrt{3})}{\sqrt{3}}$
$=\sqrt{3}(7+4\sqrt{3})$
$=12+7\sqrt{3}$.sq.unit.