Question

A sphere is inscribed in a cone of radius $3\sqrt{3}$ and slant height $6\sqrt{3}$. The radius of the sphere is:

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

Answer 1

The correct option is A $3$

$DE=DO=$ radius of sphere

$△ABC$ is an equilateral $△$, with $DO\perp BC$. The $△DOB$ is a $30-60-90$ triangle & as $BO=OC=3\sqrt{3}$

$\frac{OD}{OB}=tan{30}^0\Rightarrow \frac{OD}{3\sqrt{3}}=\frac{1}{\sqrt{3}}\Rightarrow OD=3$

$\therefore$ Radius of sphere is $3$.