# 11.3: Surface Area of Pyramids and Cones

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

## Learning Objectives

- Find the surface area of pyramids and cones.

## Review Queue

- A rectangular prism has sides of 5 cm, 6 cm, and 7 cm. What is the surface area?
- A cylinder has a diameter of 10 in and a height of 25 in. What is the surface area?
- A cylinder has a circumference of \begin{align*}72 \pi \ ft\end{align*}. and a height of 24 ft. What is the surface area?
- Draw the net of a square pyramid.

**Know What?** A typical waffle cone is 6 inches tall and has a diameter of 2 inches. What is the surface area of the waffle cone? (You may assume that the cone is straight across at the top)

## Parts of a Pyramid

**Pyramid:** A solid with one ** base** and the

**meet at a common**

*lateral faces*

*vertex.*
The edges between the lateral faces are *lateral edges.*

The edges between the base and the lateral faces are *base edges.*

**Regular Pyramid:** A pyramid where the base is a regular polygon.

All regular pyramids also have a ** slant height** which is the height of a lateral face. A non-regular pyramid does not have a slant height.

**Example 1:** Find the slant height of the square pyramid.

**Solution:** The slant height is the hypotenuse of the right triangle formed by the height and half the base length. Use the Pythagorean Theorem.

\begin{align*}8^2+24^2 &= l^2\\ 640 &= l^2\\ l = \sqrt{640} & = 8\sqrt{10}\end{align*}

## Surface Area of a Regular Pyramid

Using the slant height, which is labeled \begin{align*}l\end{align*}, the area of each triangular face is \begin{align*}A=\frac{1}{2} bl\end{align*}.

**Example 2:** Find the surface area of the pyramid from Example 1.

**Solution:** The four triangular faces are \begin{align*}4 \left(\frac{1}{2} bl \right)=2(16)\left(8\sqrt{10}\right)=256 \sqrt{10}\end{align*}. To find the total surface area, we also need the area of the base, which is \begin{align*}16^2 = 256\end{align*}. The total surface area is \begin{align*}256 \sqrt{10}+256 \approx 1065.54 \ units^2\end{align*}.

From this example, we see that the formula for a square pyramid is:

\begin{align*}SA &= (area \ of \ the \ base)+4(area \ of \ triangular \ faces)\\ SA &= B+n\left(\frac{1}{2} bl\right)\end{align*}

\begin{align*}B\end{align*} is the area of the base and \begin{align*}n\end{align*} is the number of triangles.

**Surface Area of a Regular Pyramid:** If \begin{align*}B\end{align*} is the area of the base, then \begin{align*}SA=B+\frac{1}{2} nbl\end{align*}.

The net shows the surface area of a pyramid. If you ever forget the formula, use the net.

**Example 3:** Find the area of the ** regular** triangular pyramid.

**Solution:** “Regular” tells us the base is an equilateral triangle. Let’s draw it and find its area.

\begin{align*}B=\frac{1}{2} \cdot 8 \cdot 4\sqrt{3}=16\sqrt{3}\end{align*}

The surface area is:

\begin{align*}SA=16\sqrt{3}+\frac{1}{2} \cdot 3 \cdot 8 \cdot 18=16\sqrt{3}+216 \approx 243.71\end{align*}

**Example 4:** If the lateral surface area of a square pyramid is \begin{align*}72 \ ft^2\end{align*} and the base edge is equal to the slant height. What is the length of the base edge?

**Solution:** In the formula for surface area, the lateral surface area is \begin{align*}\frac{1}{2} nbl\end{align*}. We know that \begin{align*}n = 4\end{align*} and \begin{align*}b = l\end{align*}. Let’s solve for \begin{align*}b\end{align*}.

\begin{align*}\frac{1}{2} nbl &= 72 \ ft^2\\ \frac{1}{2} (4) b^2 &= 72\\ 2b^2 &= 72\\ b^2 &= 36\\ b &= 6\end{align*}

## Surface Area of a Cone

**Cone:** A solid with a circular base and sides taper up towards a vertex.

A cone has a slant height, just like a pyramid.

A cone is generated from rotating a right triangle, around one leg, in a circle.

**Surface Area of a Right Cone:** \begin{align*}SA=\pi r^2+\pi rl\end{align*}.

Area of the base: \begin{align*}\pi r^2\end{align*}

Area of the sides: \begin{align*}\pi rl\end{align*}

**Example 5:** What is the surface area of the cone?

**Solution:** First, we need to find the slant height. Use the Pythagorean Theorem.

\begin{align*}l^2 &= 9^2+21^2\\ &= 81+441\\ l &= \sqrt{522} \approx 22.85\end{align*}

The surface area would be \begin{align*}SA=\pi 9^2+\pi (9)(22.85) \approx 900.54 \ units^2\end{align*}.

**Example 6:** The surface area of a cone is \begin{align*}36 \pi\end{align*} and the radius is 4 units. What is the slant height?

**Solution:** Plug in what you know into the formula for the surface area of a cone and solve for \begin{align*}l\end{align*}.

\begin{align*}36 \pi &= \pi 4^2+\pi 4l\\ 36 &= 16+4l \qquad When \ each \ term \ has \ a \ \pi, \ they \ cancel \ out.\\ 20 &= 4l\\ 5 &= l\end{align*}

**Know What? Revisited** The standard cone has a surface area of \begin{align*}\pi+ \sqrt{35}\pi \approx 21.73 \ in^2\end{align*}.

## Review Questions

- Questions 1-10 use the definitions of pyramids and cones.
- Questions 11-19 are similar to Example 1.
- Questions 20-26 are similar to Examples 2, 3, and 5.
- Questions 27-31 are similar to Examples 4 and 6.
- Questions 32-25 are similar to Example 5.

Fill in the blanks about the diagram to the left.

- \begin{align*}x\end{align*} is the ___________.
- The slant height is ________.
- \begin{align*}y\end{align*} is the ___________.
- The height is ________.
- The base is _______.
- The base edge is ________.

Use the cone to fill in the blanks.

- \begin{align*}v\end{align*} is the ___________.
- The height of the cone is ______.
- \begin{align*}x\end{align*} is a __________ and it is the ___________ of the cone.
- \begin{align*}w\end{align*} is the _____________ ____________.

For questions 11-13, sketch each of the following solids and answer the question. Your drawings should be to scale, but not one-to-one. Leave your answer in simplest radical form.

- Draw a right cone with a radius of 5 cm and a height of 15 cm. What is the slant height?
- Draw a square pyramid with an edge length of 9 in and a 12 in height. Find the slant height.
- Draw an equilateral triangle pyramid with an edge length of 6 cm and a height of 6 cm. What is the height of the base?

Find the slant height, \begin{align*}l\end{align*}, of one lateral face in each pyramid or of the cone. Round your answer to the nearest hundredth.

Find the area of a lateral face of the regular pyramid. Round your answers to the nearest hundredth.

Find the surface area of the regular pyramids and right cones. Round your answers to 2 decimal places.

- A
has four equilateral triangles as its faces.*regular tetrahedron*- Find the height of one of the faces if the edge length is 6 units.
- Find the area of one face.
- Find the total surface area of the regular tetrahedron.

- If the lateral surface area of a cone is \begin{align*}30 \pi \ cm^2\end{align*} and the radius is 5 cm, what is the slant height?
- If the surface area of a cone is \begin{align*}105 \pi \ cm^2\end{align*} and the slant height is 8 cm, what is the radius?
- If the surface area of a square pyramid is \begin{align*}40 \ ft^2\end{align*} and the base edge is 4 ft, what is the slant height?
- If the lateral area of a square pyramid is \begin{align*}800 \ in^2\end{align*} and the slant height is 16 in, what is the length of the base edge?
- If the lateral area of a regular triangle pyramid is \begin{align*}252 \ in^2\end{align*} and the base edge is 8 in, what is the slant height?

The traffic cone is cut off at the top and the base is a square with 24 in sides. Round answers to the nearest hundredth.

- Find the area of the entire square. Then, subtract the area of the base of the cone.
- Find the lateral area of the cone portion (include the 4 inch cut off top of the cone).
- Subtract the cut-off top of the cone, to only have the lateral area of the cone portion of the traffic cone.
- Combine your answers from #27 and #30 to find the entire surface area of the traffic cone.

## Review Queue Answers

- \begin{align*}2(5 \cdot 6) + 2(5 \cdot 7) + 2(6 \cdot 7) = 214 \ cm^2\end{align*}
- \begin{align*}2(15 \cdot 18) + 2(15 \cdot 21) + 2(18 \cdot 21) = 1926 \ cm^2\end{align*}
- \begin{align*}2 \cdot 25 \pi + 250 \pi = 300 \pi \ in^2\end{align*}
- \begin{align*}36^2 (2 \pi) + 72 \pi (24) = 4320 \pi \ ft^2\end{align*}