The volume of a hexagonal pyramid is defined as the space occupied by the pyramid in three dimensions. Volume is a three-dimensional measure, so we use cubic units to represent it. The volume of any pyramid is found by multiplying the area of the base times the height of the pyramid and dividing by three. That means we have to find the area of a hexagon to calculate the volume of these types of pyramids.

Here, we will learn about the formula that we can use to calculate the volume of hexagonal pyramids. Then, we will apply this formula to solve some practice problems.

## Formula to find the volume of a hexagonal prism

A pyramid is a three-dimensional figure that is composed of a base and side faces that meet at a single point above. The volume of these figures is found by multiplying the area of their base by their height and dividing by three. Therefore, we have the following formula:

$latex \text{Volume} = \frac{1}{3}\text{Base} \times \text{Height}$

Hexagonal pyramids have a hexagon as their base, so we have to find an expression for the area of a hexagon. The area of a hexagon is calculated using its apothem and the length of one of its sides. However, it is also possible to find the area of hexagons simply by using the length of one of their sides. For this, we use the following formula:

$latex A= \frac{3\sqrt{3}}{2}{{l}^2}$

where *l* represents the length of one of the sides of the hexagon.

If we substitute this expression in the formula for the volume of a pyramid, we have:

$latex V=\frac{1}{3}\times \frac{3\sqrt{3}}{2}{{l}^2}\times h$

$latex V=\frac{\sqrt{3}}{2}{{l}^2}\times h$ |

where *l* is the length of one of the sides of the hexagonal base and *h* is the length of the height of the pyramid.

## Volume of a hexagonal pyramid – Examples with answers

The following examples can be used to practice using the hexagonal pyramid volume formula. Try to solve the problems yourself before looking at the answer.

**EXAMPLE 1**

If a pyramid has a height of 4 m and a hexagonal base with sides of 1 m, what is its volume?

##### Solution

We have the following information:

- Height, $latex h=4$
- Hexagon sides, $latex l=1$

We have to use these values in the volume formula:

$latex V=\frac{\sqrt{3}}{2}{{l}^2}\times h$

$latex V=\frac{\sqrt{3}}{2}{{(1)}^2}\times (4)$

$latex V=\frac{\sqrt{3}}{2}(4)$

$latex V=3.46$

The volume is equal to 3.46 m³.

**EXAMPLE 2**

What is the volume of a pyramid that has a height of 5 m and a hexagonal base with sides of 2 m?

##### Solution

We have the following values:

- Height, $latex h=5$
- Hexagon sides, $latex l=2$

We use the volume formula with this information:

$latex V=\frac{\sqrt{3}}{2}{{l}^2}\times h$

$latex V=\frac{\sqrt{3}}{2}{{(2)}^2}\times (5)$

$latex V=\frac{\sqrt{3}}{2}(20)$

$latex V=17.32$

The volume is equal to 17.32 m³.

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**EXAMPLE 3**

If a hexagonal pyramid has sides that are 3 m long and 8 m high, what is its volume?

##### Solution

We have the following:

- Height, $latex h=8$
- Hexagon sides, $latex l=3$

Using the volume formula with this information, we have:

$latex V=\frac{\sqrt{3}}{2}{{l}^2}\times h$

$latex V=\frac{\sqrt{3}}{2}{{(3)}^2}\times (8)$

$latex V=\frac{\sqrt{3}}{2}(72)$

$latex V=62.35$

The volume is equal to 62.35 m³.

**EXAMPLE 4**

What is the height of a pyramid that has a volume of 50 m³ and a hexagonal base with sides 3 m long?

##### Solution

We have the following information:

- Volume, $latex V=50$
- Hexagon sides, $latex l=3$

Here, we have the measure of the volume, but we have to find the length of the height. Therefore, we use the volume formula and solve for *h*:

$latex V=\frac{\sqrt{3}}{2}{{l}^2}\times h$

$latex 50=\frac{\sqrt{3}}{2}{{(3)}^2}\times h$

$latex 50=\frac{\sqrt{3}}{2}(9)\times h$

$latex 50=7.79 h$

$latex h\approx 6.42$

The length of the height is equal to 6.42 m.

**EXAMPLE 5**

If a pyramid has a volume of 64 m³ and a hexagonal base with sides of 4 m, what is the length of its height?

##### Solution

We have the following:

- Volume, $latex V=64$
- Hexagon sides, $latex l=4$

Similar to the previous exercise, we use the volume formula and solve for *h*:

$latex V=\frac{\sqrt{3}}{2}{{l}^2}\times h$

$latex 64=\frac{\sqrt{3}}{2}{{(4)}^2}\times h$

$latex 64=\frac{\sqrt{3}}{2}(16)\times h$

$latex 64=13.86 h$

$latex h\approx 4.62$

The length of the height is equal to 4.62 m.

## Volume of a hexagonal pyramid – Practice problems

Solve the following practice problems using the formula for the volume of hexagonal pyramids. If you need help with this, you can look at the solved examples above.

## See also

Interested in learning more about geometric pyramids? Take a look at these pages:

- Volume of a Square Pyramid – Formulas and Examples – Mechamath
- Surface Area of a Square Pyramid – Formulas and Examples – Mechamath
- Surface Area of a Hexagonal Pyramid – Formulas and Examples – Mechamath
- Volume of a Pentagonal Pyramid – Formulas and Examples – Mechamath
- Surface Area of a Pentagonal Pyramid – Formulas and Examples – Mechamath
- Parts of a Geometric Pyramid – Mechamath

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