# Frustum of a Cone

If a cone is cut by a plane parallel to its base, the portion of a solid between this plane and the base is known as the frustum of a cone.

The volume denoted by $ABCD$ in the figure is the frustum of the cone $ABE$.

Volume of the Frustum of a Cone

Since we know that cone is a limit of a pyramid, therefore the frustum of a cone will be the limit of the frustum of a pyramid. But the volume of a pyramid is

$V = \frac{1}{3}h\left[ {{A_1} + {A_2} + \sqrt {{A_1}{A_2}} } \right]$
Where              ${A_1} = \pi {R^2}$
${A_2} = \pi {r^2}$
$\therefore$                        $= \frac{h}{3}\left( {\pi {R^2} + \pi {r^2} + \pi \sqrt {{R^2}{r^2}} } \right)$
$= \frac{{\pi h}}{3}\left( {{R^2} + {r^2} + Rr} \right)$

Example:

A cone 12cm high is cut 8cm from the vertex to form a frustum with a volume of 190cu.cm. Find the radius of the cone.

Solution:

Given that
Height of cone            $= 12{\text{cm}}$
Height of the frustum       $h = 12 – 8 = 4{\text{cm}}$
Volume of the frustum     $= 190{\text{cu}}.{\text{cm}}$

Now the volume of the frustum cone
$= \frac{{\pi h}}{3}\left( {{R^2} + {r^2} + Rr} \right)$
or         $190 = \frac{{3.1415 \times 4}}{3}\left[ {\left( {\frac{2}{3}{r^2}} \right) + {r^2} + \frac{2}{3}{r^2}} \right] = 4.19 \times \frac{{19}}{9}{r^2}$
or         ${r^2} = \frac{{190 \times 9}}{{19 \times 4.19}} = \frac{{1710}}{{79.61}} = 21.48$
$r = \sqrt {21.40} = 4.63$

Hence the required radius of the cone $= 4.63$

Curved Surface Area of the Frustum of a Cone

Since a cone is the limiting case of a pyramid, therefore the lateral surface of the frustum of a cone can be deduced from the slant surface of the frustum of a pyramid, i.e. the curved (lateral) surface of the frustum of the cone.

$= \frac{1}{2}\left( {{\text{sum of circumferences of bases}}} \right) \times \left( {{\text{slant height}}} \right)$
$= \frac{1}{2}\left( {2\pi R + 2\pi r} \right)l\,\,\, = \pi \left( {R + r} \right)l$
$l$, being the slant height of the frustum, $R$ and $r$ being the two radii of bases.

Note:
(1)        Total surface area of the frustum of a cone
$= \pi {R^2} + \pi {r^2} + \pi \left( {R + r} \right)l$
(2)        To find the slant height of the cone, use the Pythagorean theorem.

Example:

A bucket is in the shape of the frustum of a right circular cone, as shown in the figure below. Find the volume and the total surface area of the bucket.

Solution:

Slant height                 $l = \sqrt {{h^2} + {r^2}} \,\,\, = \sqrt {{{\left( {17} \right)}^2} + {{\left( {2.5} \right)}^2}} \,\,\, = 17{\text{cm}}$
Lateral surface area     $= \pi \left( {R + r} \right)l$
$= \pi \left( {7.5 + 5} \right)\left( {17} \right)\,\,\,\, = 670\,{\text{sq}}{\text{.cm}}$
$\therefore$            Base areas                   $= \pi {R^2} + \pi {r^2}\,\,\,\, = \pi \left( {{R^2} + {r^2}} \right)$
$= \pi \left( {56.25 + 25} \right)\,\,\,\, = 177 + 79$
$\therefore$            Total surface area        $= \pi {R^2} + \pi {r^2} + \pi \left( {R + r} \right)l$
$= 177 + 79 + 670$
$= 926\,{\text{sq}}{\text{.cm}}$

Volume           $= \frac{{\pi h}}{3}\left( {{R^2} + {r^2} + Rr} \right)$
$= \frac{{17h}}{3}\left( {{{\left( {7.5} \right)}^2} + {{\left( 5 \right)}^2} + \left( {7.5} \right)\left( 5 \right)} \right)$
$= 2114\,{\text{cu}}{\text{.cm}}$