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Surface Area and Volume of Pyramids

Surface area and volume of solids with a base and lateral faces that meet at a common vertex.

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Surface Area of Pyramids

Let's Think About It

Janice and Julian want to make a tent that is in the shape of a square pyramid for their summer camping adventure. They want it to have a base of 6 feet wide and a slant height of 8 feet .  They need to find out the surface area of the tent so they know how much canvas to buy.  How can they calculate the surface area of their future tent?

In this concept, you will learn how to find the surface area of a pyramid. 

Guidance

A pyramid has sides that are triangular faces and a base. The base can be any shape. Here are three examples:

The surface area is the sum of all of the areas of the faces in a solid figure. Imagine wrapping a pyramid in wrapping paper, like a present. The amount of wrapping paper needed to cover the figure represents its surface area. To find the surface area, you must be able to calculate the area of each face and then add these areas together.

One way to find the surface area of a three-dimensional figure is by using a net. A net is a two-dimensional diagram of a three-dimensional figure. Imagine unfolding a pyramid so that it is completely flat. It would look something like this net.

With a net, you can see each face of the pyramid more clearly. To find the surface area, you need to calculate the area for each face of the net, the sides and the base.

The side faces of a pyramid are always triangles, so the area formula for triangles is used to calculate their areas: \begin{align*}A = \frac{1}{2} bh\end{align*}.

The area of the base depends on what shape it is. Remember, pyramids can have bases in the shape of a triangle, square, rectangle, or any other polygon. Use whichever area formula is appropriate for the shape. Here are some common formulas to find area.

\begin{align*}\text{rectangle} &: A = lw\\ \text{square} &: A = s^2\\ \text{triangle} &: A = \frac{1}{2} bh\end{align*}

Take a look at the pyramid below. Which formula should you use to find the area of the base?

The base of the pyramid is a square with sides of 6 centimeters, so you should use the area formula for a square: \begin{align*}A = s^2\end{align*}. Now find the area of the base.

\begin{align*}A & = 6^2 \\ A&=36 \ sq.cm \end{align*}

Next, find the area of one of the triangles. You use the formula \begin{align*}A= \frac{1}{2} bh\end{align*}. The base width is 6 cm and the slant height, the height of the side is 4 cm.

\begin{align*}A & = \frac{1}{2} (6)(4) \\ A & = \frac{1}{2}(24) \\ A & = 12 \ sq.cm \end{align*}

There are four triangles, so multiply this number by four and then add it to the area of the square.

\begin{align*}SA & =4(12)+ 36 \\ SA & =48+36 \\ SA & =84 \ sq.cm \end{align*}

Nets let you see each face of a pyramid so that you can calculate its area. However, you can also use a formula to represent the surface area of a pyramid. The formula is like a short cut because you can put all of the measurements in for the appropriate variable in the formula and solve for \begin{align*}SA\end{align*}, surface area.

Here is the formula for finding the surface area of a pyramid.

\begin{align*}SA = \frac{1}{2} \ \text{perimeter} \times \text{slant height} + B\end{align*}

The first part of the formula, \begin{align*}\frac{1}{2} \ \text{perimeter} \times \text{slant height}\end{align*}, is a quick way of finding the area of all of the triangular sides of the pyramid at once. Remember, the area formula for a triangle is \begin{align*}A = \frac{1}{2} bh\end{align*}. In the formula, \begin{align*}b\end{align*} stands for base. The perimeter of the pyramid’s bottom face represents all of the bases of the triangular faces at once, because it’s their sum. The height of each triangle is always the same; this is called the slant height of the pyramid.  Therefore “\begin{align*}\frac{1}{2} \ \text{perimeter} \times \text{slant height}\end{align*}” is really the same as \begin{align*}\frac{1}{2} bh\end{align*}.

The \begin{align*}B\end{align*} in the formula represents the area of the base. Remember, pyramids can have bases of different shapes, so the area formula is used to find \begin{align*}B\end{align*} varies. You find the area of the base first and then put it into the formula in place of \begin{align*}B\end{align*}.

Let's look at an example.

What is the surface area of the pyramid below?

This is a square pyramid. The four sides of the base are all 8 inches, so the perimeter of the base is \begin{align*}8 \times 4 = 32 \ inches\end{align*}. You will need to use the area formula for a square to find \begin{align*}B\end{align*}.

\begin{align*}B & = s^2\\ B & = (8)^2\\ B & = 64 \ in.^2\end{align*}

Now you have all the information that you need. Put it into the formula and solve for \begin{align*}SA\end{align*}, surface area.

\begin{align*}SA & = \frac{1}{2} \ \text{perimeter} \times \text{slant height} + B\\ SA & = \left [ \frac{1}{2} (32) \times 3 \right ] + 64\\ SA & = (16 \times 3) + 64\\ SA & = 48 + 64\\ SA & = 112 \ in.^2\end{align*}

The answer is the surface area of the pyramid is 112 square inches.

Guided Practice

Find the surface area of the figure below.

\begin{align*}B\end{align*}.This is a triangular pyramid because its base is a triangle. That means you need to use the area formula for triangles to find .  \begin{align*}B\end{align*} First, calculate the area of the base and the perimeter.  To do this, identify the shape of the base and use the appropriate formula for 

\begin{align*}B & = \frac{1}{2} bh\\ B & = \frac{1}{2} (16) (13.86)\\ B & = 8 (13.86)\\ B & = 110.88 \ cm^2\end{align*}

The sides of the base are all the same length, so you can calculate the perimeter by multiplying \begin{align*}16\ \times\ 3\ =\ 48\end{align*}.

Next, plug the values into the formula for surface area of a pyramid and multiply the values together inside the brackets:

\begin{align*}SA & = \frac{1}{2} \ \text{perimeter} \times \text{slant height} + B \\ SA & = \left [ \frac{1}{2} (48) \times 6 \right ] + 110.88\\ \end{align*} 

\begin{align*}SA & = (24 \times 6) + 110.88\\ SA & = 144 + 110.88\\\end{align*} 

Then, add the values together, making sure to include the appropriate unit of measurement:

\begin{align*}SA & = 144 + 110.88\\ SA & = 254.88 \ cm^2\end{align*}

The answer is the surface area of this triangular pyramid is 254.88 square centimeters.

Examples

Example 1

Find the surface area of a square pyramid with side of 6 cm, slant height of 5 cm.

First, calculate the area and perimeter of the square base: 

\begin{align*}B &= {s}^{2}\\ B &= {6}^{2}\\ B &= 36 {cm}^{2}\\ \\ P &= 4 \times\ 6\\ P &= 24\ cm \end{align*} 

Next, plug the values into the formula for surface area of a pyramid and multiply the values together inside the brackets:\begin{align*}SA & = \frac{1}{2} \ \text{perimeter} \times \text{slant height} + B \\ SA & = \left [ \frac{1}{2} (24) \times 5 \right ] + 36\\ \end{align*} 
\begin{align*}SA & = (12 \times 5) + 36\\ SA & = 60 + 36\\\end{align*}Then, add the values together for the answer, remembering the appropriate unit of measurement.\begin{align*}SA & = 60 + 36\\ SA & = 96 \ cm^2\end{align*} 

The answer is the surface area of the square pyramid is \begin{align*}96 \ cm^2\end{align*}.

Example 2

Find the surface area of a rectangular pyramid with a length of 6 in, a width of 4 in and a slant height of 3 in.

First, calculate the area and perimeter of the rectangular base:

 \begin{align*}B&\ =\ lw\\ B&\ =\ 6 \times 4\\ B&\ =\ 24\ {in}^{2}\\ \\ P&\ =\ 2l\ +\ 2w\\ P&\ =\ 2(6)\ +\ 2(4)\\ P&\ =\ 12\ +\ 8\\ P&\ =\ 20\ in\end{align*}

Next, plug the values into the formula for surface area of a pyramid and multiply the values together inside the brackets:
\begin{align*}SA & = \frac{1}{2} \ \text{perimeter} \times \text{slant height} + B \\ SA & = \left [ \frac{1}{2} (20) \times 3 \right ] + 24\\ \end{align*} 
\begin{align*}SA & = (10 \times 3) + 24\\ SA & = 30 + 24\\\end{align*}Then, add the values together for the answer.  Remember the appropriate unit of measurement.

\begin{align*}SA & = 30 + 24\\ SA & = 54 \ in^2\end{align*}

The answer is \begin{align*}54 \ in^2\end{align*}.

Example 3

Find the surface area of a square pyramid with side of 8 in and slant height of 9 in.

First, calculate the area and perimeter of the square base.  \begin{align*}B &= {s}^{2}\\ B &= {8}^{2}\\ B &= 64 {in}^{2}\\ \\ P &= 4 \times\ 8\\ P &= 32\ in \end{align*}Next, plug the values into the formula for surface area of a pyramid and multiply the values together inside the brackets:\begin{align*}SA & = \frac{1}{2} \ \text{perimeter} \times \text{slant height} + B \\ SA & = \left [ \frac{1}{2} (32) \times 9 \right ] + 64\\ \end{align*} 
\begin{align*}SA & = (16 \times 9) + 64\\ SA & = 144 + 64\\\end{align*}

Then, add the values together for the answer, remembering the appropriate unit of measurement:

\begin{align*}SA & = 144 + 64\\ SA & = 208 \ in^2\end{align*}

The answer is the surface area of the square pyramid is \begin{align*}208 \ in^2\end{align*}.

Follow Up

Remember Janice and Julian and their square pyramid tent? 

They are planning for their tent to have a base of 6 feet wide and a slant height of 8 feet.  How much canvas will they need?

First, calculate the area and perimeter of the square base:
\begin{align*}B &= {s}^{2}\\ B &= {6}^{2}\\ B &= 36\ {ft}^{2}\\ \\ P &= 4 \times\ 6\\ P &= 24\ ft \end{align*} 

Next, plug the values into the formula for surface area of a pyramid and multiply the values together inside the brackets:
\begin{align*}SA & = \frac{1}{2} \ \text{perimeter} \times \text{slant height} + B \\ SA & = \left [ \frac{1}{2} (24) \times 8 \right ] + 36\\ \end{align*} 

\begin{align*}SA & = (12 \times 8) + 36\\ SA & = 96 + 36\\\end{align*}

Then, add the values together for the answer, making sure to include the appropriate unit of measurement:

\begin{align*}SA & = 96 + 36\\ SA & = 132 \ ft^2\end{align*} 

The answer is the surface area of the tent is 132 square feet.  Janice and Julian will require 132 square feet of canvas to make their tent.

Explore More

Find the surface area of each square pyramid. Remember that \begin{align*}b\end{align*} means base and \begin{align*}sh\end{align*} means slant height.

1. \begin{align*}b = 4 \ in, \ sh = 5 \ in\end{align*}

2. \begin{align*}b = 4 \ in, \ sh = 6 \ in\end{align*}

3. \begin{align*}b = 6 \ in, \ sh = 8 \ in\end{align*}

4. \begin{align*}b = 5 \ in, \ sh = 7 \ in\end{align*}

5. \begin{align*}b = 7 \ m, \ sh = 9 \ m\end{align*}

6. \begin{align*}b = 8 \ m, \ sh = 10 \ m\end{align*}

7. \begin{align*}b = 9 \ cm, \ sh = 11 \ cm\end{align*}

8. \begin{align*}b = 11 \ m, \ sh = 13 \ m\end{align*}

9. \begin{align*}b = 6 \ in, \ sh = 9 \ in\end{align*}

10. \begin{align*}b = 10 \ cm, \ sh = 12 \ cm\end{align*}

Find the surface area of each rectangular pyramid.

11. \begin{align*}l = 10 \ in, w = 8 \ in, sh = 6 \ in\end{align*}

12. \begin{align*}l = 12 \ ft, w = 8 \ ft, sh = 10 \ ft\end{align*}

13. \begin{align*}l = 4 \ in, w = 2 \ in, sh = 3 \ in\end{align*}

14. \begin{align*}l = 16 \ m, w = 12 \ m, sh = 11 \ m\end{align*}

15. \begin{align*}l = 15 \ in, w = 10 \ in, sh = 12 \ in\end{align*}

Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 10.8. 

Vocabulary

Cone

Cone

A cone is a solid three-dimensional figure with a circular base and one vertex.
Net

Net

A net is a diagram that shows a “flattened” view of a solid. In a net, each face and base is shown with all of its dimensions. A net can also serve as a pattern to build a three-dimensional solid.
Pyramid

Pyramid

A pyramid is a three-dimensional object with a base that is a polygon and triangular faces that meet at one vertex.
Surface Area

Surface Area

Surface area is the total area of all of the surfaces of a three-dimensional object.
Vertex

Vertex

A vertex is a point of intersection of the lines or rays that form an angle.
Volume

Volume

Volume is the amount of space inside the bounds of a three-dimensional object.
Cavalieri's Principle

Cavalieri's Principle

States that if two solids have the same height and the same cross-sectional area at every level, then they will have the same volume.
Base Edge

Base Edge

The base edge is the edge between the base and the lateral faces of a prism.
Slant Height

Slant Height

The slant height is the height of a lateral face of a pyramid.
Apothem

Apothem

The apothem of a regular polygon is a perpendicular segment from the center point of the polygon to the midpoint of one of its sides.

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