Question

The least perimeter (in cm) of an isosceles triangle in which a circle of radius $\sqrt{3}$ cm can be inscribed, is

• A. $1$
• B. $6\sqrt{3}$
• C. $18$
• D. $2$

Answer 1

The correct option is C $18$


Let the sides be $ax,ax$ and $a.$
$ax+ax>a$
$\Rightarrow x>\frac{1}{2}$

$\Delta =\frac{a}{2}\times \sqrt{a^2{x}^2-\frac{a^2}{4}}$
$r=\frac{\Delta }{\frac{ax+ax+a}{2}}=\frac{a}{2}\sqrt{\frac{2x-1}{2x+1}}=\sqrt{3}$

Perimeter, $p=ax+ax+a=a(2x+1)$
$=2\sqrt{3}\times \sqrt{\frac{2x+1}{2x-1}}(2x+1)=2\sqrt{3}\frac{(2x+1{)}^{3/2}}{\sqrt{2x-1}}$

$\frac{dp}{dx}=0$
$\Rightarrow \frac{3}{2}\times 2(2x+1{)}^{1/2}(2x-1{)}^{1/2}-\frac{1}{2}\times 2(2x-1{)}^{-1/2}(2x+1{)}^{3/2}=0$
$\Rightarrow (2x-1{)}^{-1/2}(2x+1{)}^{1/2}\left[3\right(2x-1)-(2x+1\left)\right]=0$
$\Rightarrow x=1$

${\frac{dp}{dx}|}_{x=1^{+}}=+ve$ and ${\frac{dp}{dx}|}_{x=1^{-}}=-ve$
$\Rightarrow x=1$ is a point of minima.
$\therefore p_{min}=(2\sqrt{3}\times \sqrt{3})3=18$