When one mentions ‘pyramid’ we normally think of Egypt and its colossal pyramids. Pyramids are one of the oldest structures in history. The intriguing thing is that such stupendous pyramids were found not only in Egypt but also in other parts of the world like Mexico, Greece, Spain, China etc. In a pyramid, the majority of the weight of the structure is focused close to the base. The great pyramid of Khufu, Egypt is one of the seven wonders of the ancient world. These ancient pyramids were usually made up of limestone. The first pyramids are known to be built by the people of the Mesopotamian civilisation.

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Since this article focuses on the mathematical aspect of pyramids, we’ll not get too much into the history of pyramids. The geometric definition of pyramids would go like this: A solid structure with only one base and lateral faces that converge into a single point called the apex of the pyramid. Thus the lateral faces are all always triangles, unlike a prism that has rectangular lateral faces.

A hexagonal pyramid is the one that has a hexagonal base. A hexagon is a 6 sided polygon. Since the base is hexagonal, a hexagonal pyramid would have 6 lateral faces. Therefore the number of faces in a hexagonal pyramid would be $6\ +\ 1(base)$ = $7$. Since the base is hexagonal, the base has $6$ vertices. There would be one more vertex at the apex, so the total number of vertices would be $6\ +\ 1$ = $7$. The hexagonal base as $6$ edges and there will be $6$ more edges that join the base vertices to the apex, therefore a hexagonal pyramid would have $6\ +\ 6$ = $12\ edges$.

Shown above is a picture of a right regular hexagonal pyramid. A right pyramid is the one in which the apex is directly above the centre of the base. In other words, a line joining the apex to the centre of the base and the line joining centre of the base to any vertex of the base, form a right angle. A regular pyramid is the one in which the base is a regular hexagon. A regular hexagon has all sides equal in length.

Given below are the formulas associated with a hexagonal pyramid.

$\text{Area of base for regular pyramid}$ = $6\ \times$ $\frac{1}{2}$ $\times\ s\ \times\ a$

$\therefore$ **Area of base **= $3sa$

Here, $s$ is the length of one side of the regular hexagonal base and $a$ is the length of the apothem of the hexagon. Apothem refers to the distance between the centre of the hexagon to its vertex.

If the hexagon is irregular, then we need to find the area of each of the triangles and add them up to get the area of the base.

This can be seen in the figure below:

Area of each of those coloured portions is calculated using the side length and the length of each apothem and added to get the area of the hexagon.

$Volume$ = $\frac{1}{3}$ $\times\ \text{area of base}\ \times\ \text{height}$

Whatever is the base of the pyramid, we first need to calculate its area using any suitable method and then multiply it by the height of the pyramid and then take a third of it to get the volume occupied by the pyramid. As shown in the figure above the height of a pyramid is the perpendicular distance between the apex and the base of the pyramid. If $B$ represents the area of the base of the pyramid and $h$ is the height, then the volume can be given by:

$V$ = $\frac{1}{3}$ $Bh$

For an irregular hexagonal base, once we split it into triangles if we know the length of each of the sides of each of the triangles, then we can use the Heron’s formula to find the area of each triangle. The Heron’s formula is like this:

$Area$ = $\sqrt{s(s\ -\ a)(s\ -\ b)(s\ -\ c)}$

Where, $s$ = $\frac{a\ +\ b\ +\ c}{2}$

Here, $a,\ b$ and $c$ are the lengths of the sides of the triangle.

$Volume$ = $\frac{1}{3}$ $Bh$

Now, we saw earlier that for a regular pyramid, the area of base $B$ can be given by the formula:

$B$ = $3sa$

Here, s is the length of each side of the regular base and a is the apothem. So plugging this

formula into the volume formula we have:

$V$ = $\frac{1}{3}$ $(3sa)h$

Simplifying that we have:

$V$ = $sah$

Since the figure is a regular hexagon each of the central angles would measure: $\frac{360}{6}$ = $60^{\circ}$

In a regular hexagon, all the diagonals are congruent and bisect each other. Therefore, in the triangle $OAB$, $OA$ = $OB$. That makes the triangle OAB an isosceles triangle. The altitude $OM$ of the triangle $OAB$ is also the apothem of the hexagon. Since triangle OAB is an isosceles triangle, the altitude would also be the angle bisector as well as the median of the triangle. Therefore the angle $MOB$ would be $\text{30 degrees}$ and the length $MB$ would be half of the length $AB$. Thus,

$MB$ = $\frac{1}{2}$ $\times\ 10$ = $5ft$

Now using some trigonometry we have:

$tan(\theta)$ = $\frac{\text{opposite side}}{\text{adjacent side}}$

For the triangle $OMB$,

$tan(O)$ = $\frac{MB}{OM}$

Substituting the known values we have:

$tan(30^\circ)$ = $\frac{5}{a}$

(note: $OM$ = $apothem$ = $a$)

Solving that for a we have:

$a$ = $\frac{5}{tan(30^{\circ})}$

Evaluating that we have:

$a$ = $\frac{5}{0.5774}$ = $8.66\ ft$

Therefore now we know that for our hexagonal pyramid:

$s$ = $10\ ft$ (given)

$h$ = $12\ ft$ (given)

$a$ = $8.66\ ft$ (calculated)

Thus the volume would be:

$V$ = $sah$

$V$ = $10\ \times\ 8.66\ \times\ 12$

$V$ = $1039.23\ ft^3$ $\leftarrow$ Answer!

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