# 8.5: Surface Area of Pyramids and Cones

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

## Introduction

*The Award Platform*

The students ordered an award platform from a company for Olympics. This platform would go on top of the cement area that the students already made and would be the place for the first place finisher in each event to stand and receive his/her award.

“It isn’t a trapezoid like the area we created,” Jose commented when the platform arrived at school.

“I know. They didn’t have a trapezoid. We got this instead,” Carmen explained.

“I think it will still work, and we can paint it,” Jose said.

“How much paint will we need?” Carmen asked.

“I’m not sure,” Jose said scratching his head.

**This figure is a different kind of cone called a truncated cone. You can figure out the surface area of a figure like this one, but you will need a special formula. As you work through this lesson look for the formula. You will have a chance to use it at the end of the lesson.**

*What You Will Learn*

By the end of this lesson you will be able to complete the following skills.

- Recognize surface area of pyramids and cones as the sum of the areas of faces using nets.
- Find surface areas and linear dimensions of pyramids with any polygonal base.
- Find surface areas and linear dimensions of cones and truncated cones.
- Solve real - world problems involving surface area of pyramids and cones.

*Teaching Time*

I. **Recognize Surface Area of Pyramids and 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

**are solid shapes that exist in three-dimensional space.**

*cones***A pyramid has sides that are triangular faces and a base. The base can be any shape. Like pyramids, cones have a base and a point at the top. However, cones always have a circular base.**

Here is an example of a cone and a pyramid.

You can see that the pyramid is a rectangular pyramid. We call it a rectangular pyramid because the base is in the shape of a rectangle. Remember that the base can be any polygon. The cone is a cone with a circular base and a vertex.

In the last lesson you learned how to calculate surface area of prisms and cylinders. Now we can figure out the surface area of cones and pyramids.

Let’s review surface area.

*Surface area***is the total of the areas of each face 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.

We will look at different ways to calculate surface area. One way 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.

This net is of a square pyramid. You can imagine folding up the sides to create a pyramid. 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: \begin{align*}A = \frac{1}{2} bh\end{align*} A=12bh. For triangles, we 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. We use whichever area formula is appropriate for the shape.** Here are some common area formulas:

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

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

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

In the pyramid of the net above, we 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. We can use these measurements to find the area of each face of the pyramid.

**The base has a side length of 6 cm, so we use the formula for finding the area of a base.**

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

**The area of the base of the pyramid is 36 sq. cm.**

**Next, we need to find the area of each triangular side. To find the area of one side, we use the formula for finding the area of a triangle.**

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

**This is the area of one triangle. We have four total, so we can multiply this value by 4.**

\begin{align*}12 \times 4 = 48 \ sq. cm\end{align*}

**Next, we add up all of the areas.**

\begin{align*}48 + 36 = 84 \ sq. cm\end{align*}

**The surface area of this square pyramid is 84 sq. cm.**

**We can also find the surface area of a cone.** Cones have different nets. Imagine you could unroll a cone. Here is what the net of a cone would look like.

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 \begin{align*}A = \pi r^2\end{align*} A=πr2, where \begin{align*}r\end{align*}r 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 slant height to its radius, or \begin{align*}\frac{s}{r}\end{align*} sr.**

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

\begin{align*}A=rs \pi\end{align*}

Let’s see how this would work with an example.

Example

*Find the surface area of the following cone.*

**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.**

\begin{align*}& \mathbf{Bottom \ face \ (circle)} && \mathbf{Side}\\ & A = \pi r^2 && A = \pi rs\\ & \pi (5^2) && \pi (5) (11.7)\\ & 25 \pi && \pi (58.5)\\ & 78.5 && 58.5 \pi\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.**

\begin{align*}&\text{bottom face} \qquad \qquad \text{side} \qquad \qquad \text{surface area}\\ & \quad 78.5 \qquad \quad \ + \quad \ 183.69 \quad = \quad \ 262.19 \ in^2.\end{align*}

We used the formula \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 \begin{align*} \pi rs\end{align*}.

**When we add these together, we get a surface area of 262.19 square inches for this cone.**

There is another way to calculate surface area besides using a net. Let’s take a look at using a formula to find the surface area of a cone or pyramid.

II. **Find Surface Areas and Linear Dimensions of Pyramids with any Polygonal Base**

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 \begin{align*}SA\end{align*}, surface area. Let’s start with pyramids.

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*}

**Now let’s look at how we can understand this formula.**

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

Let’s look at an example.

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*}. We also know we will need to use the area formula for squares to find \begin{align*}B\end{align*}, the base’s area.**

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

Now that we have the area of the base, we have all the information that we need. We can 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 13.6\right] + 64\\ SA &= (16 \times 13.6) + 64\\ SA &= 217.6 + 64\\ SA &= 281.6 \ in^2.\end{align*}

**The surface area of the pyramid is 281.6 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 13.86 \right] + 110.88\\ SA &= (24 \times 13.86) + 110.88\\ SA &= 332.64 + 110.88\\ SA &= 443.52 \ cm^2\end{align*}

**The surface area of this triangular pyramid is 443.52 square centimeters.**

**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. This may seem challenging, but if you think of it as a puzzle, then you will be right on track.**

Let’s look at an example.

Example

*The base of a square pyramid has sides of 4 cm each and a surface area of \begin{align*}96 \ cm^2\end{align*}. What is the slant height of the pyramid?*

**This time we know the surface area, but we need to find the slant height.** Let’s find the perimeter and \begin{align*}B\end{align*} first so that we can put these into the formula. The base is a square with sides of 4 centimeters, so the perimeter must be \begin{align*}4 \times 4 = 16 \ cm\end{align*}. Now we use the square area formula to find \begin{align*}B\end{align*}.

\begin{align*}B &= s^2\\ B &= 4^2\\ B &= 16 \ cm^2\end{align*}

**\begin{align*}B\end{align*} is 16 square centimeters. Let’s put these values in for the appropriate variables in the formula and solve for \begin{align*}s\end{align*}, the slant height.**

\begin{align*}SA &= \frac{1}{2} \ \text{perimeter} \times \text{slant height} + B \\ 96 &= \frac{1}{6} (16)s + 16\\ 96 &= 8s + 16\\ 96 - 16 &= 8s\\ 80 &= 8s\\ 80 \div 8 &= s\\ 10 \ cm &= s\end{align*}

**A square pyramid with a base of 4 centimeters on each side and a surface area of 96 square centimeters must have a slant height of 10 centimeters.**

*Write these formulas for finding the surface area of a pyramid in your notebook. Be sure to note that you will need to find the area of the base \begin{align*}(B)\end{align*} before putting values into the formula for surface area.*

III. **Find Surface Areas and Linear Dimensions of Cones and Truncated Cones**

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 it 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.**

*Write this formula down in your notebook.*

Now let’s try it out.

Example

*What is the surface area of the cone?*

**We know that the radius of this cone is 3.5 inches and the slant height is 19 inches. We simply put these values in for \begin{align*}r\end{align*} and s in the formula and solve for \begin{align*}SA\end{align*}, surface area.**

\begin{align*}SA &= \pi r^2 + \pi rs\\ SA &= \pi (3.5^2) + \pi (3.5) (19)\\ SA &= 12.25 \pi + 66.5 \pi\\ SA &= 78.75 \pi\\ SA &= 247.28 \ in^2.\end{align*}

**This cone has a surface area of 247.28 square inches.**

**As with the surface area formula for pyramids, we can also use this formula to solve for either the radius or the slant height of a cone. As long as the problem gives us two of the three pieces of information we need, we can put them into the formula and solve for the unknown variable. Let’s try it.**

Example

*A cone with a radius of 15 m has a surface area of \begin{align*}292.5 \pi\end{align*}. What is its slant height?*

**This time we have been given the radius and the surface area and we need to solve for the slant height, \begin{align*}s\end{align*}.** All we have to do is put the values in for the appropriate variables and solve for \begin{align*}s\end{align*}. Notice that the surface area is given as a function of pi, that’s what it means when the number has the pi symbol next to it. That’s no problem simply put the whole number into the formula for \begin{align*}SA\end{align*}.

\begin{align*}SA &= \pi r^2 + \pi rs\\ 292.5 \pi &= \pi (15^2) + \pi (15)s\\ 292.5 \pi &= 225 \pi + 15 \pi s\end{align*}

We cannot add these two terms because they are not like terms.

\begin{align*}292.5 \pi - 225 \pi = 15 \pi s\end{align*}

Subtract \begin{align*}225 \pi\end{align*} from both sides.

\begin{align*}67.5 \pi = 15 \pi s\end{align*}

Divide both sides by \begin{align*}15 \pi\end{align*}. \begin{align*}\pi\end{align*} cancels out on both sides.

\begin{align*}4.5 \ m = s\end{align*}

**The slant height of the cone must be 4.5 meters.**

That may seem a little tricky, but if you think of it just like solving equations it will be much simpler. Now let’s look at how we can find the surface area of part of a cone.

First, let’s think about what a ** truncated cone** would look like.

**A truncated cone is one where the point of the cone is cut off leaving two circular bases and a side face.**Here is what one looks like.

Notice that with a truncated cone, that we will have two different circular bases-a top radius and a bottom one. We will have to find the area of both bases plus the area of the sector to find the surface area.

**The formula for finding the surface area of a truncated cone is:**

\begin{align*}SA= \pi [s(R+r)+ R^2+r^2]\end{align*}

Notice that s stands for slant height and the capital \begin{align*}R\end{align*} stands for the larger radius and the lowercase \begin{align*}r\end{align*} stands for the smaller radius.

Example

*What is the surface area of a truncated cone with a slant height of 6 cm, a radius of 8 cm and a radius of 6 cm?*

**To find the surface area of this figure, we fill the dimensions into the formula and solve.**

\begin{align*}SA &= \pi[6(8+6)+8^2+6^2]\\ SA &= \pi[84+64+36]\\ SA &= 3.14(184)\\ SA &= 577.76 \ cm^2 \end{align*}

Now we can look at applying what we have learned to some real – life problems.

IV. **Solve Real – World Problems Involving Surface Area of Pyramids and 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

*Trey is decorating conical party hats for his party by wrapping them in colored tissue paper. Each hat has a radius of 4.2 centimeters and a slant height of 8.6 centimeters. If he wants to wrap 6 party hats, how much paper will he 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 (4.2^2) + \pi (4.2) (8.6)\\ SA &= 17.64 \pi + 36.12 \pi\\ SA &= 53.76 \pi\\ SA &= 168.81 \ cm^2\end{align*}

**Trey will need 168.81 square centimeters of tissue paper to cover one hat, when we approximate pi as 3.14.**

**But we’re not done yet! Remember, he 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 he needs: \begin{align*}168.81 \times 6 = 1,012.86\end{align*}. Trey will need 1,012.86 square centimeters of paper to cover all 6 hats.**

Now let’s apply what we have learned to the problem from the introduction.

## Real-Life Example Completed

*The Award Platform*

**Here is the original problem once again. Reread it and then solve for the surface area of the truncated cone.**

The students ordered an award platform from a company for Olympics. This platform would go on top of the cement area that the students already made and would be the place for the first place finisher in each event to stand and receive his/her award.

“It isn’t a trapezoid like the area we created,” Jose commented when the platform arrived at school.

“I know. They didn’t have a trapezoid. We got this instead,” Carmen explained.

“I think it will still work, and we can paint it,” Jose said.

“How much paint will we need?” Carmen asked.

“I’m not sure,” Jose said scratching his head.

*Remember to use the formula to solve for the surface area of the figure.*

*Solution to Real – Life Example*

**We can use the following formula to find the surface area of a truncated cone.**

\begin{align*}\pi [s(R+r)+ R^2+r^2]\end{align*}

**Now we can take the given information and substitute it into the formula.**

\begin{align*}& 3.14[3(3 + 2) + 3^2 + 2^2]\\ & 3.14 [3(5) + 9 + 4]\\ & 3.14 [28]\\ & SA = 87.92 \ ft^2\end{align*}

## Vocabulary

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

- Pyramid
- a three – dimensional solid figure with any polygon as a base and all triangular side faces meeting at one vertex.

- Cone
- a three – dimensional solid figure with a circular base, a sector for a side that meets in a single vertex.

- Surface area
- the measurement of the outer covering of a solid figure.

- Net
- a diagram that represents what a solid figure would look like in two-dimensions, flattened out as a pattern.

- Sector
- a piece of a circle.

- Truncated cone
- when a cone is divided by a plane so that the point is removed from the cone leaving only the bottom section.

## Time to Practice

Directions: Look at each figure and answer the following questions about each.

- What is the name of the figure represented in this net?
- What is the diameter of this figure?
- What is the slant height of the figure?
- What is the surface area of the figure?

- What is the name of the figure represented by this net?
- What is the length of each of the sides of the base?
- What is the surface area of the figure?

- What is the name of this figure?
- What is the shape of the base?
- How many faces does this figure have?
- What is the surface area of this figure?

- What is the name of this figure?
- What is the shape of the base?
- What is the diameter of the base?
- What is the surface area of this figure?

Directions: Use what you have learned to solve each problem.

- What is the slant height of a pyramid whose square base has sides of 12 feet and surface area is 624 square feet?
- A cone with a radius of 6.2 meters has a surface area of \begin{align*}131.44 \pi\end{align*} square meters. What is the slant height of the cone?
- Alice and Mike made a teepee in their backyard out of sheets. The teepee had a radius of 3.7 feet and a slant height of 9.3 feet. How many square feet of sheets did they use to make the teepee?
- For a class project, Omar is making a model of the pyramids in Egypt. He needs to cover 3 square pyramids in gold foil. The base of each pyramid has sides of 10 inches, and each pyramid has a slant height of 18 inches. How much gold foil will Omar need to cover all three pyramids?
- Christy is decorating a conical piñata with construction paper. If the piñata has a radius of 9.5 centimeters and a slant height of 23 centimeters, how much construction paper will Christy need to use?

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Jul 25, 2013## Last Modified:

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