The volume of a hexagonal pyramid is found by multiplying the area of the hexagonal base by the height of the pyramid and dividing by three. To calculate the surface area, we have to add up the areas of all the faces of the pyramid.

Here, we will look at the formulas that we can use to calculate the volume and surface area of hexagonal pyramids. Then, we will apply this formula to solve some practice problems.

## How 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.

## How to find the surface area of a hexagonal pyramid

The surface area is found by adding up the areas of all the faces of the three-dimensional figure. These pyramids have one hexagonal face and six lateral triangular faces.

The area of the hexagonal base is found using the formula for the area of a hexagon:

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

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

On the other hand, the area of triangular faces is found using the formula for the area of any triangle:

$latex A=\frac{1}{2}b\times h$

where *b* is the length of the triangle’s base and *h* is the length of the height of the triangle.

In a hexagonal pyramid, the base of the triangular faces is equal to the sides of the hexagonal base. Furthermore, considering that the six triangular faces are equal, we have the following formula for the surface area of hexagonal pyramids:

$latex A_{s}=\frac{3\sqrt{3}}{2}{{l}^2}+6(\frac{1}{2}b\times h)$

$latex A_{s}=\frac{3\sqrt{3}}{2}{{l}^2}+3lh)$

## Volume and area of a hexagonal pyramid – Examples with answers

**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

If a pyramid has a hexagonal base with sides 1 m long and its triangular faces are 3 m high, what is its surface area?

##### Solution

We have the lengths $latex l=1$ and $latex h=3$. Therefore, we use the formula for surface area with these values:

$latex A_{s}=\frac{3\sqrt{3}}{2}{{l}^2}+3lh$

$latex A_{s}=\frac{3\sqrt{3}}{2}{{(1)}^2}+3(1)(3)$

$latex A_{s}=\frac{3\sqrt{3}}{2}(1)+9$

$latex A_{s}=2.6+9$

$latex A_{s}=11.6$

The surface area is equal to 11.6 m².

**EXAMPLE **3

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.

**EXAMPLE **4

If a pyramid has a hexagonal base with sides 2 m long and triangular faces 5 m high, what is its surface area?

##### Solution

We have the values $latex l=2$ and $latex h=5$. Using these values in the formula for surface area, we have:

$latex A_{s}=\frac{3\sqrt{3}}{2}{{l}^2}+3lh$

$latex A_{s}=\frac{3\sqrt{3}}{2}{{(2)}^2}+3(2)(5)$

$latex A_{s}=\frac{3\sqrt{3}}{2}(4)+30$

$latex A_{s}=10.4+30$

$latex A_{s}=40.4$

The surface area is equal to 40.4 m².

**EXAMPLE **5

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 **6

What is the surface area of a hexagonal pyramid with sides 4 m long and triangular faces 6 m high?

##### Solution

In the question, we have the lengths $latex l=4$ and $latex h=6$. Using these values in the formula, we have:

$latex A_{s}=\frac{3\sqrt{3}}{2}{{l}^2}+3lh$

$latex A_{s}=\frac{3\sqrt{3}}{2}{{(4)}^2}+3(4)(6)$

$latex A_{s}=\frac{3\sqrt{3}}{2}(16)+72$

$latex A_{s}=41.6+72$

$latex A_{s}=113.6$

The surface area is equal to 113.6 m².

**EXAMPLE **7

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 **8

If a pyramid has a hexagonal base with sides that are 5 m long and its triangular faces are 10 m high, what is its surface area?

##### Solution

We have the lengths $latex l=5$ and $latex h=10$. Therefore, we use the formula for surface area with these values:

$latex A_{s}=\frac{3\sqrt{3}}{2}{{l}^2}+3lh$

$latex A_{s}=\frac{3\sqrt{3}}{2}{{(5)}^2}+3(5)(10)$

$latex A_{s}=\frac{3\sqrt{3}}{2}(25)+150$

$latex A_{s}=65+150$

$latex A_{s}=215$

The surface area is equal to 215 m².

**EXAMPLE **9

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 and area of a hexagonal pyramid – Practice problems

#### What is the height of a pyramid that has a volume of 79.5 m^{3} and a hexagonal base with sides of 3 m?

Write the answer using a single decimal place.

## See also

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

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