Question

A model of a building is in the from of a cone surmounting a cylinder. The radius of the cylinder is $6$ cm The height of the model is $6\sqrt{3}$ cm. The vertical angle of the cone is ${120}^{\circ }$. Find the volume of the model (in $c{m}^2$).

• A. $148\sqrt{3}\pi$
• B. $156\sqrt{3}\pi$
• C. $162\sqrt{3}\pi$
• D. $168\sqrt{3}\pi$

Answer 1

The correct option is D $168\sqrt{3}\pi$

In $\Delta OAB$

$tan{60}^0=\frac{AB}{OA}\Rightarrow \sqrt{3}=\frac{6}{OA}=OA=\frac{6}{\sqrt{3}}=\frac{6\sqrt{3}}{3}=2\sqrt{3}$

$\therefore$ $h_2=6\sqrt{3}-2\sqrt{3}=4\sqrt{3}$

Now, Volume of model $=$ $\frac{1}{3}\pi r^2{h}_1+\pi r^2{h}_2$

$=$ $\pi r^2(\frac{h_1}{3}+h_2)$

$=$ $\pi \times 6\times 6(\frac{2\sqrt{3}}{3}+4\sqrt{3})=\pi \times 36\times \left(\frac{2\sqrt{3}+12\sqrt{3}}{3}\right)$

$=$ $\pi \times 36\times \frac{14\sqrt{3}}{3}=168\sqrt{3}\pi {cm}^3$