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# 10.5: Surface Area of Pyramids and Cones

Difficulty Level: At Grade Created by: CK-12

## Introduction

Painting a Pyramid

After about a week of working at the wrapping station, both Candice and Trevor were beginning to feel at ease. Neither one of them thought that there was a package that was too difficult to wrap. They had wrapped lots of regular shaped boxes and also some odd shaped things too.

Then one Saturday morning, an interesting dilemma came their way. A girl came up to the booth with a box in one hand.

“I wonder if you can help me,” she said. “I bought my brother a model pyramid and it needs paint, but the paint isn’t included. I need to know how much paint I’ll need.”

“Well, to figure that out, you need the dimensions of the pyramid to calculate the amount of surface area that the paint will need to cover. Let me see the box and I’ll see if I can help you,” Candice said taking the box from the girl.

The box said that once the pyramid is built, it will have a base of 7.7 cm, a height of 6.2 cm and a slant height of 7.3 cm.

“Okay, I know what to do,” Candice said smiling, as she took out a piece of a paper and a calculator.

Do you know what to do? How can you use these measurements to find the surface area of the pyramid?

In this lesson you will learn how to find the surface area of a pyramid. Learn as you go through this lesson and at the end you can check out how Candice finds the surface area of the pyramid model.

What You Will Learn

In this lesson, you will learn how to demonstrate the following skills.

• Recognize surface area of pyramids or cones as the sum of the areas of faces using nets.
• Find surface areas of pyramids using formulas.
• Find surface areas of cones using formulas.
• Solve real-world problems involving surface areas of pyramids or cones.

Teaching Time

I. Recognize Surface Area of Pyramids or Cones as the Sum of the Areas of Faces Using Nets

In this lesson we will learn to find the surface area of pyramids and cones. Pyramids and cones are solid shapes that exist in three-dimensional space. A pyramid has sides that are triangular faces and a base. The base can be any shape. Let’s look at some pyramids.

Like pyramids, cones have a base and a point at the top. However, cones always have a circular base.

Surface area is the sum of all of the areas of the faces in a solid figure. Imagine you could wrap a pyramid or cone 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, we 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 using 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. It might look something like this net.

With a net, we can see each face of the pyramid more clearly. To find the surface area, we 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 we use the area formula for triangles to calculate their area: A=12bh\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. We use whichever area formula is appropriate for the shape. Here are some common formulas to find area.

rectangle:A=lwsquare:A=s2triangle: A=12bh\begin{align*}\text{rectangle}: \quad \qquad A = lw\!\\ \text{square}: \qquad \qquad A = s^2\!\\ \text{triangle}: \quad \qquad \ A = \frac{1}{2} bh\end{align*}

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

The base of the pyramid is a square with sides of 6 centimeters, so we should use the area formula for a square: A=s2\begin{align*}A = s^2\end{align*}. Now we can find the area of the base.

AA=62=36 sq.cm\begin{align*}A & = 6^2 \\ A&=36 \ sq.cm \end{align*}

Next, we need to find the area of one of the triangles. There are four triangles, but we can work by finding the area of one of the triangles. We use the formula A=12bh\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.

AAA=12(6)(4)=12(24)=12 sq.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 we can take this number and multiply it by four and then add it to the area of the square.

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

Cones have different nets. Imagine you could unroll a cone.

The shaded circle is the base. Remember, cones always have circular bases. The unshaded portion of the cone represents its side. Technically we don’t call this a face because it has a round edge.

To find the surface area of a cone, we need to calculate the area of the circular base and the side and add them together.

The formula for finding the area of a circle is A=πr2\begin{align*}A = \pi r^2\end{align*}, where r\begin{align*}r\end{align*} is the radius of the circle. We use this formula to find the area of the circular base.

The side of the cone is actually a piece of a circle, called a sector. The size of the sector is determined by the ratio of the cone’s radius to its slant height, or rs\begin{align*}\frac{r}{s}\end{align*}.

To find the area of the sector, we take the area of the portion of the circle.

A=πr2sr\begin{align*}A = \pi r^2 \cdot \frac{s}{r}\end{align*}

This simplifies to πrs\begin{align*}\pi rs\end{align*}.

To find the area of the cone’s side, then, we multiply the radius, the slant height, and pi.

This may seem a little tricky, but as you work through a few examples you will see that this becomes easier as you go along.

Example

What is the surface area of the figure below?

Now that we have the measurements of the sides of the cone, let’s calculate the area of each. Remember to use the correct area formula.

bottom faceA=πr2π(5)225π78.5sideA=πrsπ(5)(11)π(55)55π172.7\begin{align*}& \text{bottom face} && \text{side}\\ & A = \pi r^2 && A = \pi rs\\ & \pi (5)^2 && \pi (5) (11)\\ & 25 \pi && \pi (55)\\ & 78.5 && 55 \pi\\ & && 172.7\end{align*}

We know the area of each side of the cone when we approximate pi as 3.14. Now we can add these together to find the surface area of the entire cone.

bottom faceside surface area78.5  + 172.7  =   251.2 in.2\begin{align*}& \text{bottom face} \qquad \quad \text{side} \qquad \quad \ \text{surface area}\\ & 78.5 \qquad \quad \ \ + \quad \ 172.7 \ \ = \ \ \ 251.2 \ in.^2\end{align*}

Now we can look at what we did to solve this problem. We used the formula A=πr2\begin{align*}A = \pi r^2\end{align*} to find the area of the circular base. Then we found the area of the side by multiplying πrs\begin{align*}\pi rs\end{align*}. When we add these together, we get a surface area of 251.2 square inches for this cone.

When you learned how to find the surface areas of other three-dimensional figures, we discussed using a formula to simplify our work. Cones and pyramids aren’t any different. Let’s look at simplifying our work with formulas.

II. Find Surface Areas of Pyramids using Formulas

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

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

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

The first part of the formula, 12 perimeter×slant height\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 A=12bh\begin{align*}A = \frac{1}{2} bh\end{align*}. In the formula, b\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, so we can just call this the slant height of the pyramid.

Therefore “12 perimeter×slant height\begin{align*}\frac{1}{2} \ \text{perimeter} \times \text{slant height}\end{align*}” is really the same as 12bh\begin{align*}\frac{1}{2} bh\end{align*}.

The B\begin{align*}B\end{align*} in the formula represents the base’s area. Remember, pyramids can have bases of different shapes, so the area formula we use to find B\begin{align*}B\end{align*} varies. We find the base’s area first and then put it into the formula in place of B\begin{align*}B\end{align*}. Let’s give it a try.

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 8×4=32 inches\begin{align*}8 \times 4 = 32 \ inches\end{align*}. We also know we will need to use the area formula for squares to find B\begin{align*}B\end{align*}, the base’s area.

BBB=s2=(8)2=64 in.2\begin{align*}B & = s^2\\ B & = (8)^2\\ B & = 64 \ in.^2\end{align*}

Now we have all the information that we need. We can put it into the formula and solve for SA\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 surface area of the pyramid is 112 square inches.

Example

Find the surface area of the figure below.

First of all, what kind of pyramid is this? It is a triangular pyramid because its base is a triangle. That means we need to use the area formula for triangles to find \begin{align*}B\end{align*}. The base’s sides are all the same length, so we can calculate the perimeter by multiplying \begin{align*}16 \times 3 = 48\end{align*}. Now let’s find \begin{align*}B\end{align*}.

\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*}

Now we’re ready to put all of the information into the formula. Let’s see what happens.

\begin{align*}SA & = \frac{1}{2} \ \text{perimeter} \times \text{slant height} + B \\ SA & = \left [ \frac{1}{2} (48) \times 6 \right ] + 110.88\\ SA & = (24 \times 6) + 110.88\\ SA & = 144 + 110.88\\ SA & = 254.88 \ cm^2\end{align*}

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

Write this formula down in your notebook.

Try a few of these on your own.

10G. Lesson Exercises

Find the surface area of each pyramid.

1. A square pyramid with side of 6 cm, slant height of 5 cm
2. A rectangular pyramid with a length of 6 in, a width of 4 in and a slant height of 3 in

III. Find Surface Areas of Cones Using Formulas

Cones have a different formula because they have a circular base. But the general idea is the same. The formula is a short cut to help us combine the area of the circular base and the area of the cone’s side. Here’s what the formula looks like.

\begin{align*}SA = \pi r^2 + \pi rs\end{align*}

The first part of the formula, \begin{align*}\pi r^2\end{align*}, is simply the area formula for circles. This represents the base area. The second part, as we have seen, represents the area of the cone’s side. We simply put the pieces together and solve for the area of both parts at once. Let’s try it out.

Example

What is the surface area of the cone?

We know that the radius of this cone is 3 inches and the slant height is 9 inches. We simply put these values in for \begin{align*}r\end{align*} and \begin{align*}s\end{align*} in the formula and solve for \begin{align*}SA\end{align*}, surface area.

\begin{align*}SA & = \pi r^2 + \pi rs\\ SA & = \pi (3^2) + \pi (3)(9)\\ SA & = 9 \pi + 27 \pi\\ SA & = 36 \pi\\ SA & = 113.04 \ in.^2\end{align*}

This cone has a surface area of 113.04 square inches.

Example

A cone has a radius of 2.5 meters and a slant height of 7.5 meters. What is its surface area?

This time we have not been given a picture of the cone, so we’ll need to read the problem carefully. It tells us the radius and the slant height of the cone, though, so we can put these numbers in for \begin{align*}r\end{align*} and \begin{align*}s\end{align*} and solve.

\begin{align*}SA & = \pi r^2 + \pi rs\\ SA & = \pi (2.5^2) + \pi (2.5)(7.5)\\ SA & = 6.25 \pi + 18.75 \pi\\ SA & = 25 \pi\\ SA & = 78.5 \ in.^2\end{align*}

This cone has a surface area of 78.5 square inches.

Take a few minutes to write this formula down in your notebooks.

Now try a few of these on your own.

10H. Lesson Exercises

Find the surface area of each cone.

1. Radius = 4 in, slant height = 6 in
2. Radius = 3.5 in, slant height = 5 in
3. Radius = 5 m, slant height = 7 m

Take a few minutes to check your work with a friend.

You’re right! Just be sure to put the numbers in the correct place in the formula!

IV. Solve Real-World Problems Involving Surface Areas of Pyramids or Cones

We have learned two ways to find the surface area of pyramids and cones: drawing a net or using a formula. We can use either of these methods to solve word problems involving surface area. Nets may be especially useful if the problem does not provide an image of the figure. If you choose to use a formula, be sure you know whether the problem deals with a cone or a pyramid, and what the shape of the pyramid’s base is. Let’s practice using what we have learned.

Example

Jessica is decorating conical party hats for her party by wrapping them in colored tissue paper. Each hat has a radius of 4 centimeters and a slant height of 8 centimeters. If she wants to wrap 6 party hats, how much paper will she need?

This problem involves a cone. It does not include a picture, so it may help to draw a net. In your drawing, label the radius and the slant height of the cone. We can also use the formula. We simply put the radius and slant height in for the appropriate variables in the formula and solve for \begin{align*}SA\end{align*}.

\begin{align*}SA & = \pi r^2 + \pi rs\\ SA & = \pi (4)^2 + \pi (4) (8)\\ SA & = 16 \pi + 32 \pi\\ SA & = 48 \pi\\ SA & = 150.72 \ cm^2\end{align*}

Jessica will need 150.72 square centimeters of tissue paper to cover one hat.

But we’re not done yet! Remember, she wants to cover 6 party hats. We need to multiply the surface area of one hat by 6 to find the total amount of paper she needs: \begin{align*}150.72 \times 6 = 904.32\end{align*}.

Jessica will need 904.32 square centimeters of paper to cover all 6 hats.

Now let’s take what we have learned in this lesson and apply it to our work solving the problem from the introduction.

## Real–Life Example Completed

Painting a Pyramid

Here is the original problem once again. Reread it and underline any important information.

After about a week of working at the wrapping station, both Candice and Trevor were beginning to feel at ease. Neither one of them thought that there was a package that was too difficult to wrap. They had wrapped lots of regular shaped boxes and also some odd shaped things too.

Then one Saturday morning, an interesting dilemma came their way. A girl came up to the booth with a box in one hand.

“I wonder if you can help me,” she said. “I bought my brother a model pyramid and it needs paint, but the paint isn’t included. I need to know how much paint I’ll need.”

“Well, to figure that out, you need the dimensions of the pyramid to calculate the amount of surface area that the paint will need to cover. Let me see the box and I’ll see if I can help you,” Candice said taking the box from the girl.

The box said that once the pyramid was built, it would have a base of 7.7 cm, a height of 6.2 cm and a slant height of 7.3 cm.

“Okay, I know what to do,” Candice said smiling as she took out a piece of a paper and a calculator.

First, Candice needs to find the surface area of each of the four lateral faces. She can use this formula.

\begin{align*}A & = \frac{1}{2} bh\\ A & = 4 \left [ \frac{1}{2} (7.7)(7.3) \right ]\\ A & = 112.42 \ sq. cm\end{align*}

Now she can find the area of the base.

\begin{align*}A & = s^2\\ A & = 7.7(7.7)\\ A & = 59.29 \ sq. cm\end{align*}

The surface area of the pyramid is 171.71 sq. cm.

“You need to find paint that stretches across 172 sq. cm. That will give him just enough. Try the hobby shop. They will help you,” Candice said writing the figure on a piece of paper as she handed it to the girl.

The girl smiled and thanked her. Candice felt good about her ability to use math to figure out the problem.

## Vocabulary

Here are the vocabulary words that are found in this lesson.

Pyramid
a three-dimensional object with a polygon for a base and triangles for sides that join at a single vertex.
Cone
a three-dimensional object with a circle as a base and a side that wraps around the base connecting at one vertex at the top.
Surface area
the measurement of the outer covering or surface of a three-dimensional figure.

## Technology Integration

1. http://www.mathplayground.com/mv_surface_area_cones.html – This is a Brightstorm video on finding the surface area of a cone.
2. http://www.mathplayground.com/mv_surface_area_pyramids.html – This is a Brightstorm video on finding the surface area of pyramids.

## Time to Practice

Directions: Find the surface area of each 4-sided 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*}

Directions: Find the surface area of each cone. Remember that \begin{align*}sh\end{align*} means slant height and \begin{align*}r\end{align*} means radius.

11. \begin{align*}r = 4 \ in, \ sh = 5 \ in\end{align*}

12. \begin{align*}r = 5 \ m, \ sh = 7 \ m\end{align*}

13. \begin{align*}r = 3 \ cm, \ sh = 6 \ cm\end{align*}

14. \begin{align*}r = 5 \ mm, \ sh = 8 \ mm\end{align*}

15. \begin{align*}r = 8 \ in, \ sh = 10 \ in\end{align*}

16. \begin{align*}r = 11 \ cm, \ sh = 14 \ cm\end{align*}

17. \begin{align*}r = 12 \ in, \ sh = 16 \ in\end{align*}

18. \begin{align*}r = 3.5 \ cm, \ sh = 6 \ cm\end{align*}

19. \begin{align*}r = 4.5 \ mm, \ sh = 7 \ mm\end{align*}

20. \begin{align*}r = 5.5 \ cm, \ sh = 9 \ cm\end{align*}

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