Question

A metallic sector of central angle 240° and radius 15cm is rolled up to form a cone. What is the volume of the cone so formed?

Answer 1

Hint:
The sector is basically a portion of a circle which is enclosed by two radii and an arc. A sector divides the circle into two regions, namely major and minor sectors. The length of the arc of the sector with angle θ, is given by:
Arc length $ = l = \dfrac{{\left( {\theta \pi r} \right)}}{{180}}$
The volume of the cone defines the capacity of the cone. The volume of the cone (V) which has a radius of its circular base as r, height from the vertex to the base as h is given as:
Volume of the cone = $V = \dfrac{1}{3}\pi {r^2}h$
The slant height of the cone (l) (specifically right circular) is the distance from the vertex or apex to the point on the outer line of the circular base of the cone i.e. $l = \sqrt {{r^2} + {h^2}} $
For volume of the cone use, this metallic sector is rolled to form a cone. Therefore the arc becomes the circumference of the base of the cone. Radius of the metallic sector becomes slant height of the cone.

Complete step by step solution:
According to the question,

Radius of the metallic sector = 15cm
Angle of the sector = 240°
We know that
Arc length = $l = \dfrac{{\left( {\theta \pi r} \right)}}{{180}}$
$\begin{gathered}
  l = \dfrac{{240 \times \pi \times 15}}{{180}} \\
  l = 20\pi {\text{cm}} \\
\end{gathered} $
Now, as we know that the length of the arc will become the circumference of the base of cone.
Length of arc = Circumference of base of the cone
$\begin{gathered}
  20\pi = 2\pi r \\
  r = 10cm \\
\end{gathered} $
Hence, the radius of the cone = 10cm

Slant height of cone = Radius of the sector = 15cm
In $\Delta $OPQ, to find the height of the cone we use the formula, ${h^2} + {r^2} = {l^2}$
$\begin{gathered}
  {h^2} + {10^2} = {15^2} \\
  {h^2} = 225 - 100 \\
  {h^2} = 125 \\
  h = 5\sqrt 5 \\
\end{gathered} $
The height of the cone = $5\sqrt 5 $
∴Volume of the cone $ = \dfrac{1}{3}\pi {r^2}h$
$\begin{gathered}
   = \dfrac{1}{3}\pi \times {10^2} \times 5\sqrt 5 \\
   = \dfrac{{500\sqrt 5 \pi }}{3}c{m^3} \\
\end{gathered} $

Note:
 The formula for the volume of a regular cone or right circular cone and the oblique cone is the same.To find arc, we can use the formula Angle$ = \dfrac{{arc}}{{radius}}$, angle must be in radian. To change into radian from degree, multiply with $\dfrac{\pi }{{180}}$ .