# Frustum of a Pyramid

If a pyramid is cut through by a plane parallel to its base, the portion of the pyramid between that plane and the base is called the frustum of the pyramid.

__The Volume of the Frustum of a Pyramid__

A general formula for the volume of any pyramid can be derived in terms of the areas of the two bases and the height of the frustum.

Consider any frustum of a pyramid $$KE$$ in the figure with the lower base $${A_1}$$, upper base $${A_2}$$ and the altitude $$h$$. Complete the pyramid $$O – HD$$ of which the frustum $$KE$$ is a part.

Denoted by $$p$$, the volume of the small pyramid $$O – QM$$ whose altitude is $$a$$. Then the altitude of $$O – HD$$ is $$a + h$$.

Let $$V$$ and $$P$$ represent the volume of the frustum $$KE$$ and the pyramid $$O – HD$$, respectively.

$$\therefore $$ from the figure it is easily seen that $$V = P – p$$. Expressing this equality in terms of the dimensions, we may write:

\[\begin{gathered} V = \frac{1}{3}{A_1}\left( {h + a} \right) – \frac{1}{3}{A_2}a = \frac{1}{3}\left[ {{A_1}h + {A_1}a – {A_2}a} \right] \\ \Rightarrow V = \frac{1}{3}\left[ {{A_1}h + a\left( {{A_1} – {A_2}} \right)} \right]\, – – – \left( i \right) \\ \end{gathered}\]

The pyramid $$O – HD$$ may be considered as cut by the two parallel planes $$HD$$ and $$QM$$.

Hence

\[\frac{{{a^2}}}{{{{\left( {h + a} \right)}^2}}} = \frac{{{A_2}}}{{{A_1}}}\]

(If a pyramid is cut by two parallel planes, the areas of the sections are proportional to the squares of their distances from the vertex.)

Taking the square root of both sides, we have

\[\begin{gathered} \frac{a}{{h + a}} = \frac{{\sqrt {{A_2}} }}{{\sqrt {{A_1}} }} \\ \Rightarrow a\sqrt {{A_1}} = h\sqrt {{A_2}} + a\sqrt {{A_2}} \\ \end{gathered} \]

Transposing $$a\sqrt {{A_2}} $$ to the L.H.S. of this equation and factorizing,

\[\begin{gathered} a\left( {\sqrt {{A_1}} – \sqrt {{A_2}} } \right) = h\sqrt {{A_2}} \\ \Rightarrow a = \frac{{h\sqrt {{A_2}} }}{{\sqrt {{A_1}} – \sqrt {{A_2}} }} \\ \end{gathered} \]

Substituting the value of $$a$$ in (1), we have

\[\begin{gathered} V = \frac{1}{3}\left[ {{A_1}h + \frac{{h\sqrt {{A_2}} }}{{\sqrt {{A_1}} – \sqrt {{A_2}} }}\left( {{A_1} – {A_2}} \right)} \right] \\ \Rightarrow V = \frac{1}{3}\left[ {{A_1}h + \frac{{h\sqrt {{A_2}} }}{{\sqrt {{A_1}} – \sqrt {{A_2}} }}\left( {{A_1} + {A_2}} \right)\left( {{A_1} – {A_2}} \right)} \right] \\ \Rightarrow V = \frac{1}{3}\left[ {{A_1}h + h\sqrt {{A_2}} \left( {{A_1} + {A_2}} \right)} \right] \\ \Rightarrow V = \frac{1}{3}\left[ {{A_1}h + h\sqrt {{A_1}{A_2}} + {A_2}h} \right] = \frac{1}{3}h\left[ {{A_1} + {A_2} + \sqrt {{A_1}{A_2}} } \right] \\ \end{gathered} \]

i.e. the volume of a frustum of a pyramid is equal to one-third the product of the altitude and the sum of the upper base, the lower base and the square root of the product of the two bases.

__The Lateral Surface Area of the Frustum of a Pyramid__

Each of the faces, such as $$CDML$$, of the frustum of a pyramid is a trapezium, and the area of each trapezium will be half the sum of the parallel sides, $$CD$$ and $$ML$$, multiplied by the slant distance between them.

In the frustum of pyramid on a square base, let $$a$$ denote the length of each side of the base, $$b$$ the length of each side of the other end, and $$l$$ the height of the frustum.

Each face $$CDML$$ is a trapezium, and the lengths of the parallel sides are $$a$$ and$$b$$.

\[\begin{gathered} {\text{Area}}\,{\text{CDML}} = \frac{1}{2}\left( {a + b} \right)l \\ Lateral\,area = \frac{1}{2}\,\left( {sum\,of\,perimeters\,of\,bases \times slant\,height} \right) \\ \end{gathered} \]

__Example__**:**

A frustum of a pyramid has rectangular ends, and the sides of the base are **25dm** and **36dm**. If the area of the top face is **784sq.dm** and the height of the frustum is **60dm**, find its volume.

__Solution__**:**

Here

$${A_1} = 25 \times 36 = 900\,{\text{sq}}{\text{.dm}}$$, $${A_2} = 784\,{\text{sq}}{\text{.dm}}$$ \[\begin{gathered} V = \frac{1}{3}h\left[ {{A_1} + {A_2} + \sqrt {{A_1}{A_2}} } \right] = \frac{1}{3} \times 60\left[ {900 + 784 + \sqrt {900 \times 784} } \right] \\ V = 20 \times \left( {1684 + 840} \right) = 20 \times 2524 = 50480\,{\text{sq}}{\text{.}}\,{\text{dm}} \\ \end{gathered} \]

Ivo Satoru Yoshimatsu

August 21@ 5:38 amThere are some missing roots in the volume proof: when we rationalize the areas we substituted in the calculation.