What if you were given a solid three-dimensional figure with one base and lateral faces that meet at a common vertex? How could you determine how much two-dimensional and three-dimensional space that figure occupies? After completing this Concept, you'll be able to find the surface area and volume of a pyramid.

### Watch This

### Guidance

A **pyramid** is a solid with one ** base** and

**that meet at a common**

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

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

*lateral edges.*

*base edges.*

A **regular pyramid** is 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.

##### Surface Area

**Surface area** is a two-dimensional measurement that is the total area of all surfaces that bound a solid. The basic unit of area is the square unit. For pyramids, we will need to use the slant height, which is labeled \begin{align*}l\end{align*}

**Surface Area of a Regular Pyramid:** If \begin{align*}B\end{align*}

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

##### Volume

To find the **volume** of any solid you must figure out how much space it occupies. The basic unit of volume is the cubic unit.

**Volume of a Pyramid:** \begin{align*}V=\frac{1}{3} Bh\end{align*}

#### Example A

Find the slant height of the square pyramid.

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*}

#### Example B

Find the surface area of the pyramid from Example A.

The total surface area of the four triangular faces is \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*}.

#### Example C

Find the volume of the pyramid.

\begin{align*}V=\frac{1}{3} (12^2 )12=576 \ units^3\end{align*}

-->

### Guided Practice

1. Find the surface area of the ** regular** triangular pyramid.

2. If the lateral surface area of a regular 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?

3. Find the height and then volume of the pyramid.

4. Find the volume of the pyramid with a right triangle as its base.

5. A rectangular pyramid has a base area of \begin{align*}56 \ cm^2\end{align*} and a volume of \begin{align*}224 \ cm^3\end{align*}. What is the height of the pyramid?

**Answers:**

1. “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*}

2. 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 feet\end{align*}

3. In this example, we are given the *slant* height. Use the Pythagorean Theorem.

\begin{align*}7^2+h^2 &=25^2\\ h^2 &= 576\\ h &= 24\end{align*}

\begin{align*}V=\frac{1}{3} (14^2)(24)=1568 \ units^3\end{align*}.

4. The base of the pyramid is a right triangle. The area of the base is \begin{align*}\frac{1}{2} (14)(8)=56 \ units^2\end{align*}.

\begin{align*}V=\frac{1}{3} (56)(17) \approx 317.33 \ units^3\end{align*}

5. Use the formula for volume and plug in the information we were given. Then solve for the height.

\begin{align*}V &= \frac{1}{3} Bh\\ 224 &= \frac{1}{3} \cdot 56h\\ 12 &= h\end{align*}

### Explore More

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 ________.

For questions 7-8, 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 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. Round your answer to the nearest hundredth.

Find the surface area and volume of the regular pyramid. Round your answers to the nearest hundredth.

- 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 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 volume of a square pyramid is 72 square inches and the base edge is 4 inches. What is the height?
- The volume of a triangle pyramid is \begin{align*}170 \ in^3\end{align*} and the base area is \begin{align*}34 \ in^2\end{align*}. What is the height of the pyramid?

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 11.5.