# 7.4: Base, Lateral and Surface Areas of Pyramids

**At Grade**Created by: CK-12

## Learning Objectives

- Identify pyramids.
- Find the surface area of a pyramid using a net or a formula.

## Pyramids

A **pyramid** is a 3-dimensional figure with a *single* base and 3 or more *non-parallel* sides that meet at a *single point* above the base called the **apex**. The sides of a pyramid are **triangles**.

A **pyramid** has only one ____________________________, while a **prism** has 2.

The sides of a **pyramid** are ______________________________.

A **pyramid** has 3 or more _________________________.

What is the name of the point on the very top of the pyramid? _______________

Below are examples of **pyramids**. Notice that each pyramid is named after the shape of its **base** (so the pyramid with a *square* **base** is called a *square pyramid*, and so on):

A **regular pyramid** is a pyramid that has a *regular polygon* for its **base** and whose **sides** are all *congruent triangles.*

A **regular pyramid** has sides that are ___________________________ triangles.

The base of a **regular pyramid** is a regular ________________________.

## Surface Area of a Pyramid Using Nets

You can deconstruct a pyramid into a net:

As you can see in the diagram, the easiest way to construct a net of a pyramid is to draw the base in the center, with each triangular side extending outwards from the base edges.

To find the **surface area** of the figure using the net, first find the *area* of the **base**. In this case, the base is a square:

\begin{align*}A& = s^2 \\ &= (12)(12)\\ &= 144 \ \text{square units}\end{align*}

Now find the *area* of each **side** of the pyramid. Each side is an *isosceles triangle.* Use the Pythagorean Theorem to find the *height* of the triangles.

The *height of each triangle* is called the **slant height** of the pyramid. The **slant height** of the pyramid is the **altitude** of one of the triangles.

The height of the isosceles triangle side of the pyramid is labeled \begin{align*}n\end{align*} above, so we will call the **slant height** \begin{align*}n\end{align*} for this problem.

The **slant height** is the same as the _________________________ of the triangular side.

The triangular **side** of the pyramid is drawn again below:

We use the Pythagorean Theorem (with only half of the triangle, as shown on the right above) to find the *height* of the triangle:

\begin{align*}6^2 + n^2 &= (11.66)^2\\ 36 + n^2 &= 136\\ n^2 &= 100\\ n &= 10\end{align*}

Now find the **area** of each *triangular side* (using the formula for area of a triangle):

\begin{align*}A &= \frac{1}{2} bh \qquad \text{where}\ b = 12 \ \text{and}\ h = n = 10\\ &= \frac{1}{2} (12)(10)\\ &= 60 \ \text{square units}\end{align*}

Since there are 4 triangular sides of the pyramid:

\begin{align*}\text{Area of all 4 triangles} &= 4(60)\\ &= 240 \ \text{square units}\end{align*}

Finally, *add* the *total area of the 4 triangles* to the *area of the base:*

\begin{align*}\text{Total area} &= \text{Area of the triangular sides} + \text{Area of the base}\\ &= 240 + 144\\ &= 384\ \text{square units}\end{align*}

Review the steps for finding the **surface area of a regular pyramid:**

- Find the area of the
**base**of the pyramid. - Find the
**slant height**of the triangular side of the pyramid. - Find the
**area**of each triangular**side**and multiply it by the number of sides. - Add the area of the sides to the area of the base for total area!

**Example 1**

*Use the net below to find the total area of the regular hexagonal pyramid with an apothem of 5.19. The dimensions are given in centimeters.*

The **base** of the pyramid is a **hexagon**, which is a 6-sided figure. The **area** of the hexagonal base is given by the formula for the area of a regular polygon.

Since each **side** of the hexagon measures 6 cm, the perimeter is \begin{align*}6 \cdot 6\end{align*} or 36 cm.

The **apothem**, or perpendicular distance to the center of the hexagon, is 5.19 cm.

(*Step #1*)

\begin{align*}\text{Area of base} &= \frac{1}{2} (\text{perimeter})(\text{apothem})\\ &= \frac{1}{2} ( \underline{\;\;\;\;\;\;\;\;\;\;\;} )( \underline{\;\;\;\;\;\;\;\;\;\;\;} ) \\ &= 93.42 \ \text{square cm}\end{align*}

Use the Pythagorean Theorem to find the **slant height** of each **lateral triangle**

(*Step #2*)

\begin{align*}3^2 + n^2 &= (14)^2\\ 9 + n^2 &= 196\\ n^2 &= 187\\ n & \approx 13.67 cm\end{align*}

Now find the **area** of *each triangle:*

(*Step #3*)

\begin{align*}A &= \frac{1}{2} bh \qquad \text{where}\ b = 6 \ \text{and}\ h = n = 13.67\\ &= \frac{1}{2} ( \underline{\;\;\;\;\;\;\;\;\;\;\;} )( \underline{\;\;\;\;\;\;\;\;\;\;\;} )\\ &= 41 \ \text{ square cm}\end{align*}

Together, the **area** of all 6 triangles that make up the **lateral sides** of the pyramid are:

\begin{align*}\text{Area} &= 6 \cdot \text{(area of each triangle)}\\ &= 6 \cdot 41\\ &= 246 \ \text{square cm}\end{align*}

Add the *area of the 6 lateral triangular sides* to the *area of the hexagonal base*

(*Step #4*)

\begin{align*}\text{Total area} &= \text{Area of the triangular sides} + \text{Area of the base}\\ &= 246 + 93.42\\ &= 339.42 \ \text{square cm}\end{align*}

**Reading Check:**

1. *What is the slant height of a pyramid?*

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

2. *What are the 4 steps for finding the surface area of a regular pyramid? Describe them in your own words.*

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

## Surface Area of a Regular Pyramid

To get a general formula for the **area** of a **regular pyramid**, look at the net for this square pyramid:

The **slant height** of each **lateral triangle** is labeled \begin{align*}l\end{align*} (the lowercase letter \begin{align*}L\end{align*}), and the side of the regular polygon base is labeled \begin{align*}s\end{align*}. For each **lateral triangle**, the area is:

\begin{align*}A = \frac{1}{2}\ sl\end{align*}

There are \begin{align*}n\end{align*} triangles in a regular pyramid (for example, \begin{align*}n = 3\end{align*} for a triangular pyramid, \begin{align*}n = 4\end{align*} for a square pyramid, \begin{align*}n = 5\end{align*} for a pentagonal pyramid, and so on.)

The total area, \begin{align*}L\end{align*}, of the lateral triangles is:

\begin{align*}L &= n \cdot \text{(area of each lateral triangle)}\\ L &= n \cdot \left ( \frac{1}{2}\ sl \right )\end{align*}

If we re-arrange the above equation we get:

\begin{align*}L = \frac{1}{2}\ nsl\end{align*}

Notice that \begin{align*}(n \cdot s)\end{align*} is just the perimeter, \begin{align*}P\end{align*}, of the regular polygon base. So we can substitute \begin{align*}P\end{align*} into our equation to get the following postulate:

\begin{align*}L = \frac{1}{2}\ Pl\end{align*}

To get the *total* area of the pyramid, add the area of the base, \begin{align*}B\end{align*}, to the equation above:

\begin{align*}A = \frac{1}{2}\ Pl + B\end{align*}

## Area of a Regular Pyramid

The **surface area** of a **regular pyramid** is:

\begin{align*}A = \frac{1}{2} Pl + B\end{align*}

Where \begin{align*}P\end{align*} is the **perimeter** of the regular polygon that forms the pyramid’s **base**, \begin{align*}l\end{align*} is the **slant height** of the pyramid, and \begin{align*}B\end{align*} is the **area** of the **base**.

**Example 2**

*A tent without a bottom has the shape of a hexagonal pyramid with a slant height \begin{align*}l\end{align*} of 30 feet. The sides of the hexagonal perimeter of the figure each measure 8 feet. Find the surface area of the tent that exists above ground.*

For this problem, \begin{align*}B\end{align*} is zero because the tent has no bottom. So simply calculate the **lateral area** of the figure:

\begin{align*}A &= \frac{1}{2}\ Pl + B\\ &= \frac{1}{2}\ Pl + 0\end{align*}

Now substitute the given information to find the perimeter and the given slant height:

\begin{align*}A &= \frac{1}{2} (30)(6 \cdot 8)\\ &= 720 \ \text{square feet}\end{align*}

In the formula for the **surface area** of a **regular pyramid:**

\begin{align*}A = \frac{1}{2}\ Pl + B\end{align*}

\begin{align*}A\end{align*} stands for the _____________________ ________________ of the pyramid,

\begin{align*}P\end{align*} stands for the ____________________________ of the base,

\begin{align*}l\end{align*} stands for the ________________ __________________ of the pyramid, and

\begin{align*}B\end{align*} stands for the ___________________ of the _________________.

**Example 3**

*A pentagonal pyramid has a slant height \begin{align*}l\end{align*} of 12 cm. The sides of the pentagonal perimeter of the figure each measure 9 cm. The apothem of the figure is 6 cm.*

*Find the surface area of the figure.*

If the **base** is a *pentagon* (which has 5 sides), and each side is 9 cm, then the **perimeter** of the base is (\begin{align*}5 \cdot 9\end{align*}) or 45 cm.

First find the **lateral area** of the figure:

\begin{align*}L &= \frac{1}{2}\ Pl\\ &= \frac{1}{2} (45)(12)\\ &= 270 \ cm^2\end{align*}

Now use the formula for the area of a regular polygon to find the **area** of the **base**:

\begin{align*}A &= \frac{1}{2} \ \text{(perimeter)(apothem)}\\ &= \frac{1}{2} (45)(6)\\ &= 135 \ cm^2\end{align*}

Finally, add these values together to find the *total* **surface area:**

\begin{align*}\text{Total area} &= \text{Lateral area} + \text{Base area}\\ &= \underline{\;\;\;\;\;\;\;\;\;\;\;\;} cm^2 + \underline{\;\;\;\;\;\;\;\;\;\;\;\;} cm^2\\ &= 405 \ cm^2\end{align*}

**Total surface area of a pyramid** \begin{align*} = \ \text{Lateral} \underline{\;\;\;\;\;\;\;\;\;\;\;\;} + \underline{\;\;\;\;\;\;\;\;\;\;\;\;} \ \text{area}\end{align*}

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