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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

License: CC BY-NC 3.0

Ashanti went to an alternative therapies market with her aunt and ended up buying a beautiful pyramid crystal paperweight. She’s interested in figuring out the total surface area of her new ornament and has measured and drawn the dimensions below. How can Ashanti figure out the surface area?

License: CC BY-NC 3.0

In this concept, you will learn to find the surface area of pyramids.

Surface Area

A pyramid has sides that are triangular faces and a base. The base can be any shape. Surface area is the total of the areas of each face in a solid figure. Imagine you could wrap 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 calculate surface area is to use a net. A net is a two-dimensional diagram of a three-dimensional figure. Imagine you could unfold a pyramid so that it is completely flat. Here is what the net of a pyramid would look like.

License: CC BY-NC 3.0

This net is of a square pyramid. You can imagine folding up the sides to create a pyramid. 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 in the net: the sides and the base. The side faces of a pyramid are always triangles, so you use the area formula for triangles to calculate their area: \begin{align*}A = \frac{1}{2} bh\end{align*}A=12bh. For triangles, you will need the height, or slant height in the case of a pyramid.

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. You use whichever area formula is appropriate for the shape.

Here are some common area formulas:

\begin{align*}\text{Rectangle}: A = lw\end{align*}Rectangle:A=lw

\begin{align*}\text{Square}: A = s^2\end{align*}Square:A=s2

\begin{align*}\text{Triangle}: A = \frac{1}{2} bh\end{align*}Triangle:A=12bh

In the net of the pyramid above, you can see that it is a square pyramid. Imagine that it has a slant height of 4 cm and a side length of 6 cm. You can use these measurements to find the area of each face of the pyramid.

First, find the area of the square base.

\begin{align*}\begin{array}{rcl} A &=& s^2\\ A &=& 6^2\\ A &=& 36 \end{array}\end{align*}AAA===s26236

Next, find the area of the triangle side.

\begin{align*}\begin{array}{rcl} A &=& \frac{1}{2} bh\\ A &=& \frac{1}{2} (4)(6)\\ A &=& 12 \end{array}\end{align*}AAA===12bh12(4)(6)12

Then, add up all of the areas. Remember you have one square base and four triangle sides.

\begin{align*}\begin{array}{rcl} A &=& \text{base} + 4 \times \text{side}\\ A &=& 36 + (4 \times 12)\\ A &=& 84 \end{array}\end{align*}AAA===base+4×side36+(4×12)84

The answer is 84.

The surface area of the square based pyramid is 84 cm2.

There is also a formula is to calculate the surface area of a pyramid.  Let \begin{align*}B\end{align*}B represent the area of the base.

\begin{align*}SA = \frac{1}{2} \times \text{perimeter} \times \text{slant height} + B\end{align*}SA=12×perimeter×slant height+B

Remember, pyramids can have bases of different shapes, so the area formula you use to find \begin{align*}B\end{align*}B varies. You will find the base’s area first and then put it into the formula in place of \begin{align*}B\end{align*}B.

Sometimes, you will have to find a linear measurement. This means that you will be given the surface area and one other dimension. Then you will need to work backwards to figure out the measurement for the missing dimension.

Let’s look at an example.

The base of a square pyramid has sides of 4 cm each and a surface area of 96 cm2. What is the slant height of the pyramid?

First, find the area of the square base.

\begin{align*}\begin{array}{rcl} A &=& s^2\\ A &=& 4^2\\ A &=& 16 \end{array} \end{align*}AAA===s24216

Next, find the perimeter of the square base.

\begin{align*}\begin{array}{rcl} P &=& 4s\\ P &=& 4(4)\\ P &=& 16 \end{array}\end{align*}PPP===4s4(4)16

Then, substitute what you know in the formula for the surface area.

\begin{align*}\begin{array}{rcl} SA &=& \frac{1}{2} \times \text{perimeter} \times \text{slant height} + B \\ 96 &=& \frac{1}{2} \times 16 \times \text{slant height} + 16 \\ 96 &=& 8 \times \text{slant height} + 16 \end{array}\end{align*}

Then, solve for the slant height by subtracting 16 and dividing both sides by 8.

\begin{align*}\begin{array}{rcl} 96 &=& 8 \times \text{slant height} + 16 \\ 96 - 16 &=& 8 \times \text{slant height} + 16 - 16 \\ 80 &=& 8 \times \text{slant height} \\ \frac{80}{8} &=& \frac{8 \times \text{slant height}} {8} \\ \text{slant height} &=& 10 \end{array}\end{align*}

The answer is 10.

The slant height of the pyramid is 10 cm.

Examples

Example 1

Earlier, you were given a problem about Ashanti and her crystal pyramid.

Ashanti needs to use the dimensions below to calculate the surface area of her pyramid.

License: CC BY-NC 3.0

First, find the area of the triangular base.

\begin{align*}\begin{array}{rcl} A &=& \frac{1}{2}bh\\ A &=& \frac{1}{2}(16)(13.86)\\ A &=& 110.88 \end{array}\end{align*}

Next, find the perimeter of the triangular base.

\begin{align*}\begin{array}{rcl} P &=& 3s\\ P &=& 3(16)\\ P &=& 48 \end{array}\end{align*}

Then, use the surface area formula to calculate the surface area.

\begin{align*}\begin{array}{rcl} SA &=& \frac{1}{2} \times \text{perimeter} \times \text{slant height} + B \\ SA &=& \frac{1}{2} \times 48 \times 13.86 + 110.88 \\ SA &=& 443.52 \end{array}\end{align*}

The answer is 443.52.

The surface area of the pyramid is 443.52 cm2.

Example 2

What is the surface area of the pyramid below?

License: CC BY-NC 3.0

First, find the area of the square base.

\begin{align*}\begin{array}{rcl} A &=& s^2\\ A &=& 8^2\\ A &=& 64 \end{array}\end{align*}

Next, find the perimeter of the square base.

\begin{align*}\begin{array}{rcl} P &=& 4s\\ P &=& 4(8)\\ P &=& 32 \end{array}\end{align*}

Then, use the surface area formula to calculate the surface area.

\begin{align*}\begin{array}{rcl} SA &=& \frac{1}{2} \times \text{perimeter} \times \text{slant height} + B \\ SA &=& \frac{1}{2} \times 32 \times 13.6 + 64 \\ SA &=& 281.6\end{array}\end{align*}

The answer is 281.6.

The surface area of the pyramid is 281.6 in2.

Example 3

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

First, find the area of the square base.

\begin{align*}\begin{array}{rcl} A &=& s^2\\ A &=& 8^2\\ A &=& 64 \end{array}\end{align*}

Next, find the perimeter of the square base.

\begin{align*}\begin{array}{rcl} P &=& 4s\\ P &=& 4(8)\\ P &=& 32 \end{array}\end{align*}

Then, use the surface area formula to calculate the surface area.

\begin{align*}\begin{array}{rcl} SA &=& \frac{1}{2} \times \text{perimeter} \times \text{slant height} + B \\ SA &=& \frac{1}{2} \times 32 \times 9 + 64 \\ SA &=& 208 \end{array}\end{align*}

The answer is 208.

The surface area of the pyramid is 208 in2.

Example 4

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, find the area of the rectangular base.

\begin{align*}\begin{array}{rcl} A &=& lw\\ A &=& (6)(4)\\ A &=& 24 \end{array}\end{align*}

Next, find the perimeter of the rectangular base.

\begin{align*}\begin{array}{rcl} P &=& 2l+2w\\ P &=& 2(6)+2(4)\\ P &=& 20 \end{array}\end{align*}

Then, use the surface area formula to calculate the surface area.

\begin{align*}\begin{array}{rcl} SA &=& \frac{1}{2} \times \text{perimeter} \times \text{slant height} + B \\ SA &=& \frac{1}{2} \times 20 \times 3 + 24 \\ SA &=& 54 \end{array}\end{align*}

The answer is 54.

The surface area of the pyramid is 54 in2.

Example 5

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

First, find the area of the square base.

\begin{align*}\begin{array}{rcl} A &=& s^2\\ A &=& 6^2\\ A &=& 36 \end{array}\end{align*}

Next, find the perimeter of the square base.

\begin{align*}\begin{array}{rcl} P &=& 4s\\ P &=& 4(6)\\ P &=& 24 \end{array}\end{align*}

Then, use the surface area formula to calculate the surface area.

\begin{align*}\begin{array}{rcl} SA &=& \frac{1}{2} \times \text{perimeter} \times \text{slant height} + B \\ SA &=& \frac{1}{2} \times 24 \times 5 + 36 \\ SA &=& 96 \end{array}\end{align*}

The answer is 96.

The surface area of the pyramid is 96 cm2.

Review

Answer the questions below each of the diagrams.

License: CC BY-NC 3.0

1. What is the name of the figure represented by this net?

2. What is the length of each of the sides of the base?

3. What is the surface area of the figure?

License: CC BY-NC 3.0

4. What is the name of this figure?

5. What is the shape of the base?

6. How many faces does this figure have?

7. What is the surface area of this figure?

License: CC BY-NC 3.0

8. What is the name of the figure represented by this net?

9. What is the length of each of the sides of the base?

10. What is the surface area of the figure?

License: CC BY-NC 3.0

11. What is the name of the figure represented by this net?

12. What is the length of each of the sides of the base?

13. What is the surface area of the figure?

14. True or false. The B in the formula for surface area of a pyramid stands for bottom.

15. True or false. You need to know the slant height to figure out the perimeter of a pyramid.

Review (Answers)

To see the Review answers, open this PDF file and look for section 8.8.

Resources

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Vocabulary

TermDefinition
Cone A cone is a solid three-dimensional figure with a circular base and one vertex.
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 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 is the total area of all of the surfaces of a three-dimensional object.
Vertex A vertex is a point of intersection of the lines or rays that form an angle.
Volume Volume is the amount of space inside the bounds of a three-dimensional object.
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 The base edge is the edge between the base and the lateral faces of a prism.
Slant Height The slant height is the height of a lateral face of a pyramid.
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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