A pyramid is one of the important geometrical shapes. It is characteristically has a polygon base and the faces are isosceles triangles. Thus the top of the shape converges to a single point. The pyramids, which were built in ancient types at Egypt, are considered to be historical monuments and the largest one of those at Giza is considered as one of the Seven Wonders of the World.

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As mentioned earlier, a pyramid is shape with a polygon base and converging to a point which is the top vertex. If the base is a rectangle, then it is called as rectangular pyramid. Hence there are four triangular faces. The picture of a rectangular pyramid is shown below.

The faces are isosceles triangles with opposite ones are congruent. The base rectangle dimensions are denoted as $'l'$ and $'w'$. The normal to the base passing through the top vertex is called as axis of the pyramid. It is also the vertical height and denoted as $'h'$. With the help of geometrical concepts, let us study more about rectangular pyramids.

As we are aware, volume of a shape is the amount of space it occupies. The same fact applies to rectangular pyramids. If the shape is hollow, then its volume becomes as the capacity it can contain something like granules, liquid, gas etc.

The formula for volume of a three dimensional is fundamentally defined as the cross sectional area times the height, provided the shape has a uniform cross section throughout. But, in case of a rectangular pyramid, the cross sections are not uniform. The base section is a rectangle but the cross section truncates uniformly as we move up and finally reduces to point. Had the cross sections been same, then the volume would have been $l \times w \times h$, where ‘$l$’ and ‘$w$’ are the base dimensions and ‘$h$’ is the vertical height. Mathematically, it can be proved that, for the same height, the volume of a pyramid is $\frac{1}{3}$ of the prism with the same base dimensions. In the next section, we will establish this fact in case of rectangular pyramid.
With the application of integral calculus we will try to derive the formula for volume of a rectangular pyramid. Let us consider a rectangular pyramid placed with its top vertex at origin and the axis lying on $x$-axis. The following diagram shows the mid cross section perpendicular to the base.

Consider a thin cross section normal to the $x$-axis at a distance $'x'$ from origin with thickness as $'dx'$. The cross section is a rectangle whose height can be derived as $\frac{lx}{h}$ using the concept of similar triangles.

The above diagram shows the isometric view the cross section. Since it is a rectangle with a height of $\frac{lx}{h}$, the width can be expressed as $\frac{klx}{h}$ where $'k'$ is a constant. If $dV$ is the volume of this cross section, then

$dV$ = $(\frac{l}{h})(\frac{kl}{h})$ $dx$ = $(\frac{l}{h})(\frac{kl}{h})$ $x^2dx$

And the total volume is computed as,

$V$ = $\int_0^h$ $(\frac{l}{h})(\frac{kl}{h})$ $x^2 dx$ = $(\frac{1}{3})(\frac{l}{h})(\frac{kl}{h})$ $h^3$ = $(\frac{1}{3})$ $(l)(kl)\ h$ = $(\frac{1}{3})$ $lwh$, since $kl$ = $w$.

When a rectangular pyramid is cut chopped off by a plane parallel to the base, the remaining part of the pyramid is called as truncated rectangular pyramid or as frustum of rectangular pyramid.In the above diagram a truncated rectangular pyramid is shown. It is as if a regular pyramid is sliced off at a height $h'$from the base. In case of a truncated rectangular pyramid, how do we find the volume? It can be determined as $V_1 - V_2$, where $V_1$ = $(\frac{1}{3})$ $lwh$, the volume of the original pyramid and and $V_2$ = $(\frac{1}{3})$ $l' w' (h - h')$, the volume of the truncated part of the pyramid. But the difficulty is we may not know what $'h'$ is since the truncated part is only imaginary. However, $'h'$ can be found indirectly by realizing that,

$\frac{(h - h')}{h}$ = $\frac{l'}{l}$ = $\frac{w'}{w}$

To grasp well the concepts we discussed, let us study a few worked out examples.

The base dimensions of a rectangular pyramid are $14 in. \times 6 in.$ and its vertical height is $12 in.$ Calculate the volume of this rectangular pyramid.

In this case we can directly use the formula and arrive at the volume $V$ as,

$V$ = $(\frac{1}{3})$ $lwh$ = $(\frac{1}{3})$ $14 \times 6 \times 8$ = $224 cu.in.$

A lamp shade is in the shape of a truncated rectangular pyramid. The dimensions of the bottom section and the top section are $12in. \times 6in.$ and $8in. \times 4in.$ respectively. If its height is $6$ in. how much of space is there inside the lamp shade?

The amount of space inside the lamp shade is same as the volume of the lamp shade in the shape of a truncated rectangular pyramid. To the formula for volume of the truncated rectangular pyramid

$V$ = $(\frac{1}{3})$ $[lwh – l’w’(h – h’)$, let us first find ‘$h$’ from the equation $\frac{(h – h’)}{h}$ = $\frac{l'}{l}$**]**.

So, $\frac{(h – 6)}{h}$ = $\frac{8}{12}$ = $\frac{2}{3}$

Or, $3h – 18$ = $2h$ or $h$ = $18$ and hence $(h – h’)$ = $18 – 6$ = $12$

This means that the complete pyramid would have had a vertical height of $18in.$ and the height of the removed part of that would have been $6 in.$ Therefore, the volume of the truncated rectangular pyramid is = $(\frac{1}{3})$ $[12 \times 6 \times 18 – 8 \times 4 \times 6]$ = $368 cu.in.$

That is, the amount of space inside the lamp shade is $368 cu.in.$

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