Question

Inradius of a circle which is inscribed in an isosceles triangle one of whose angle is $\frac{2\pi }{3}$, is $\sqrt{3}$, then the area of the triangle is

• A. $4\sqrt{3}$
• B. $12–7\sqrt{3}$
• C. $12+7\sqrt{3}$
• D. None of these

Answer 1

The correct option is C

$12+7\sqrt{3}$

Explanation For Correct Option:

Finding the Area Of The Triangle:

The given triangle is an isosceles triangle.

Let $A=\frac{2\mathrm{\pi }}{3}=120°,b=c$

Then $B=C=30°\left[\text{by property of isosceles triangle}\right]$
From sine rule

$\frac{\sin 120°}{a}=\frac{\sin 30°}{b}$

$$\begin{gathered} \Rightarrow \frac{\frac{\sqrt{3}}{2}}{a}=\frac{\frac{1}{2}}{b}\left[\sin 30°=\frac{1}{2}\&\sin 120°=\frac{\sqrt{3}}{2}\right] \\ \Rightarrow a=\sqrt{3}b...\left(i\right) \end{gathered}$$

Area of triangle

$$\begin{gathered} ∆=\frac{bc\sin A}{2} \\ =\frac{\sqrt{3}{b}^2}{4}.....\left(ii\right)[\mathrm{from}(i\left)\right] \end{gathered}$$

Inradius of a circle

$r=\frac{∆}{s}\left[\therefore s=\frac{a+b+c}{2}\right]$

$\begin{array}{rcl}& \Rightarrow & \frac{∆}{s}=\sqrt{3}[\text{given}r=\sqrt{3}]\\ & \Rightarrow & \frac{\sqrt{3}{b}^2}{4s}=\sqrt{3}[\text{from}(ii\left)\right]\\ & \Rightarrow & b^2=4s=2a+4b\left[\therefore s=\frac{a+b+c}{2},b=c\right]\\ & \Rightarrow & b^2=2\sqrt{3}b+4b\\ & \Rightarrow & b=2\sqrt{3}+4\end{array}$

Substituting the value of $b$ in equation $\left(ii\right)$

$$\begin{gathered} ∆=\frac{\sqrt{3}{b}^2}{4} \\ =\frac{\sqrt{3}{(2\sqrt{3}+4)}^2}{4} \\ =12+7\sqrt{3} \end{gathered}$$

Hence, option (C) is the correct answer.