# 10.9: Surface Area of Cones

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

**Practice**Surface Area and Volume of Cones

Have you ever tried to figure out the covering of a cone? Take a look at this dilemma.

Jessica came to buy paper at the card store. She 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 Concept will teach you how to find the surface area of cones. Then you will know how to figure out this dilemma.**

### Guidance

In the last Concept, you learned about pyramids and surface area. Let's look at cones.

**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 \begin{align*}A = \pi r^2\end{align*}

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 \begin{align*}\frac{r}{s}\end{align*}

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

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

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

**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.**

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.**

\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.**

\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 \begin{align*}A = \pi r^2\end{align*}**we get a surface area of 251.2 square inches for this cone.**

**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*} πr2, 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.

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.**

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.

Find the surface area of each cone.

#### Example A

Radius = 4 in, slant height = 6 in

**Solution: \begin{align*}125.6 \ in^2\end{align*}**

#### Example B

Radius = 3.5 in, slant height = 5 in

**Solution: \begin{align*}93.42 \ in^2\end{align*}**

#### Example C

Radius = 5 m, slant height = 7 m

**Solution: \begin{align*}188.4 \ m^2\end{align*}**

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

Now back to Jessica and the hats.

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.**

### Vocabulary

Here are the vocabulary words in this Concept.

- 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.

### Guided Practice

Here is one for you to try on your own.

Figure out the surface area of a cone with a radius of 4.5 inches and a slant height of 8 inches.

**Answer**

To figure this out, we can use the formula for finding the surface area of a cone.

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

\begin{align*}SA = \pi (4.5)^2 + \pi (4.5)(8)\end{align*}

\begin{align*}SA = 63.585 + 113.04\end{align*}

\begin{align*}SA = 176.625\end{align*}

\begin{align*}SA = 176.63 \ in^2\end{align*}

**This is our answer.**

### Video Review

Here is a video for review.

- This is a video on surface area of cones.

### Practice

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.

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

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

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

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

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

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

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

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

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

10. \begin{align*}r = 6 \ cm, \ sh = 8 \ cm\end{align*}

11. \begin{align*}r = 7.5 \ cm, \ sh = 9 \ cm\end{align*}

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

13. \begin{align*}r = 16 \ cm, \ sh = 18 \ cm\end{align*}

14. \begin{align*}r = 13 \ cm, \ sh = 20 \ cm\end{align*}

15. \begin{align*}r = 15.5 \ cm, \ sh = 18.5 \ cm\end{align*}

### Image Attributions

Here you'll learn to find the surface area of cones using formulas.