# Frustum of a Prism

Frustum:

When a plane section is taken of a right prism parallel to its end (i.e., perpendicular to its axis), the section is known as a cross-section of the prism and the two positions of the prism are still prisms. If, however, the plane section taken is not parallel to the ends, the portion of the prism between the plane section and the base is called frustum.

Volume of a frustum of a Prism:

In figure $ABCEFGHI$ represents a frustum of a prism whose cutting plane $EFGH$ is inclined at an angle $\theta$ to the horizontal. In this case, the frustum can be taken as a prism with base $ABEF$ and height $BC$.

(i)         Volume of the frustum           $= \,{\text{area}}\,{\text{ABEF }} \times {\text{ BC}}$

$\therefore$            ABEF is a trapezium whose area is
$= \frac{{{\text{AF}} + {\text{BE}}}}{2} \times {\text{AB}}$
$\therefore$            Volume of frustum                 $= \frac{{{\text{AF}} + {\text{BE}}}}{2} \times {\text{AB}} \times \,{\text{BC}}$
$= \frac{{{h_1} + {h_2}}}{2} \times {\text{AB}} \times \,{\text{BC}}$
i.e.        $Volume\,of\,frustum\, = \,average\,height\, \times \,area\,of\,the\;base$

Lateral Surface Area of a Prism:

If the cutting plane is inclined at an angle $\theta$ to the horizontal then from figure we have
$\frac{{{\text{AB}}}}{{{\text{EF}}}} = \cos \theta$
or         $\frac{{{\text{AB}}}}{{{\text{EF}}}} \times \frac{{{\text{BC}}}}{{{\text{FG}}}} = \cos \theta$
${\text{(as BC = FG)}}$
or         $\frac{{{\text{Area of the base}}}}{{{\text{Area of section EFGH}}}} = \cos \theta$
or         ${\text{Area of section EFGH}} = \frac{{{\text{Area of the base}}}}{{\cos \theta }}$

Hence
Total surface area = area of the base + area of the section + lateral surface area

Note:   Lateral surface area of the frustum is the combination of rectangle and trapeziums whose area can be calculated separately.

Example:

A hexagonal right prism, whose base is inscribed in a circle of radius 2m, is cut by a plane inclined at an angle of $45^\circ$ to the horizontal. Find the volume of the frustum and the area of the section when the heights of the frustum are 8m and 6m respectively.

Solution:
Area of cross-section $= \frac{{n{R^2}}}{2}\sin \frac{{360^\circ }}{n}$
Here $n = 6,\,\,R = 2$
$\therefore$            Area of base                $= \frac{{6 \times 4}}{2} \times \sin 60^\circ$
$= 12 \times 0.87$
$= 10.4\,{\text{sq}}{\text{.m}}$

$\therefore$            Volume of frustum     $= \frac{{8 + 6}}{2} \times 10.4$
$= 73.1\,{\text{cu}}{\text{. m}}$

Area of the section      $= \frac{{10.4}}{{\cos 45^\circ }} = \sqrt 2 \times 10.4$
$= 1.414 \times 10.4$
$= 14.8\,{\text{sq}}{\text{. m}}$