Question

A sector of a circle of radius 12 cm has the angle 120°. It is rolled up so that the two bounding radii are formed together to form a cone. Find the volume of the cone.

Answer 1

Hint: First, we shall draw a figure and it will give us insights about the cone and the sector. Then, we will find the radius of the newly made cone from the sector. We will make use of the length of the sector to find this radius. On rolling, the radius of the sector will become the slant height of the cone. Once we are able to get the radius and the slant height, we can use Pythagoras theorem to find the height of the cone. Then we will use the height and radius to find the volume of the cone given as $\dfrac{1}{3}\pi {{r}^{2}}h$, where r is the radius of the base circle and h is the height of the cone.

Complete step-by-step answer:
Let us draw the figure of the sector from the circle and the view of the cone.

AOB is the sector of the circle with centre O and radius of 12 cm. The angle of the sector is 120°.

If we roll the sector, the length of the arc of the sector will be equal to circumference of the base of the cone and the radius of the sector will be the slant height of the cone.
We know that for 120° or $2\pi $ radians is $2\pi r$.
Then, we will find the length of the arc of the sector by comparison.
$\begin{align}
  & \dfrac{2\pi }{\theta }=\dfrac{2\pi r}{{{l}_{s}}} \\
 & {{l}_{s}}=r\theta \\
\end{align}$
Where $\theta $ is the angle of the sector in radians.
The angle of the sector given to us is 120°.
We know that 180° is $\pi $ radians.
$\begin{align}
  & \dfrac{180}{120}=\dfrac{\pi }{\theta } \\
 & \theta =\dfrac{2\pi }{3} \\
\end{align}$
Therefore, we will substitute $\theta =\dfrac{2\pi }{3}$ and r = 12 in ${{l}_{s}}=r\theta $ to find the length of the arc of the sector.
 $\begin{align}
  & \Rightarrow {{l}_{s}}=\left( 12 \right)\left( \dfrac{2\pi }{3} \right) \\
 & \Rightarrow {{l}_{s}}=8\pi \\
\end{align}$
This length will be equal to the circumference of the base circle. Let the radius of the base circle be r.
$\begin{align}
  & \Rightarrow 2\pi r=8\pi \\
 & \Rightarrow r=4 \\
\end{align}$
Therefore, base radius is 4 cm.
As we can see in the figure, a right-angled triangle is formed with the height, base radius and slant height as the three sides.
We will apply the Pythagoras theorem on this triangle to find the height of the cone.
${{l}^{2}}={{h}^{2}}+{{r}^{2}}$, where l is the slant height, h is the height and r is the radius of the cone.
$\begin{align}
  & \Rightarrow {{\left( 12 \right)}^{2}}={{h}^{2}}+{{\left( 4 \right)}^{2}} \\
 & \Rightarrow 144={{h}^{2}}+16 \\
 & \Rightarrow {{h}^{2}}=132 \\
 & \Rightarrow h=\sqrt{132} \\
\end{align}$
We know that the volume of a cone is given by the relation $\dfrac{1}{3}\pi {{r}^{2}}h$.
$\begin{align}
  & \Rightarrow V=\dfrac{1}{3}\pi {{\left( 4 \right)}^{2}}\left( \sqrt{132} \right) \\
 & \Rightarrow V=\dfrac{16\pi \sqrt{132}}{3} \\
\end{align}$
Hence, the volume of the cone is $\dfrac{16\pi \sqrt{132}}{3}$.

Note: Students are advised to be vigilant that there are two radii, one of the sector and other of the base circle of the cone. The radius of the sector changes to slant height on rolling. While substitution, students should be careful not to substitute sector radius in place of base radius and vice versa.