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# 11.5: Pyramids

Difficulty Level: At Grade Created by: CK-12
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Practice Surface Area and Volume of Pyramids

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What if you wanted to know the volume of an Egyptian Pyramid? The Khafre Pyramid is the second largest pyramid of the Ancient Egyptian Pyramids in Giza. It is a square pyramid with a base edge of 706 feet and an original height of 407.5 feet. What was the original volume of the Khafre Pyramid? After completing this Concept, you'll be able to answer this question.

### Guidance

A pyramid has one base and all the lateral faces meet at a common vertex. The edges between the lateral faces are lateral edges. The edges between the base and the lateral faces are called base edges. If we were to draw the height of the pyramid to the right, it would be off to the left side.

When a pyramid has a height that is directly in the center of the base, the pyramid is said to be regular. These pyramids have a regular polygon as the base. All regular pyramids also have a slant height that is the height of a lateral face. Because of the nature of regular pyramids, all slant heights are congruent. A non-regular pyramid does not have a slant height.

##### Surface Area

Using the slant height, which is usually labeled l\begin{align*}l\end{align*}, the area of each triangular face is A=12bl\begin{align*}A=\frac{1}{2} bl\end{align*}.

Surface Area of a Regular Pyramid: If B\begin{align*}B\end{align*} is the area of the base and P\begin{align*}P\end{align*} is the perimeter of the base and l\begin{align*}l\end{align*} is the slant height, then SA=B+12Pl\begin{align*}SA=B+\frac{1}{2} Pl\end{align*}.

If you ever forget this formula, use the net. Each triangular face is congruent, plus the area of the base. This way, you do not have to remember a formula, just a process, which is the same as finding the area of a prism.

##### Volume

Recall that the volume of a prism is Bh\begin{align*}Bh\end{align*}, where B\begin{align*}B\end{align*} is the area of the base. The volume of a pyramid is closely related to the volume of a prism with the same sized base.

###### Investigation: Finding the Volume of a Pyramid

Tools needed: pencil, paper, scissors, tape, ruler, dry rice or sand.

1. Make an open net (omit one base) of a cube, with 2 inch sides.
2. Cut out the net and tape up the sides to form an open cube.
3. Make an open net (no base) of a square pyramid, with lateral edges of 2.45 inches and base edges of 2 inches. This will make the overall height 2 inches.
4. Cut out the net and tape up the sides to form an open pyramid.
5. Fill the pyramid with dry rice. Then, dump the rice into the open cube. How many times do you have to repeat this to fill the cube?

Volume of a Pyramid: If B\begin{align*}B\end{align*} is the area of the base and h\begin{align*}h\end{align*} is the height, then the volume of a pyramid is V=13Bh\begin{align*}V=\frac{1}{3} Bh\end{align*}.

The investigation showed us that you would need to repeat this process three times to fill the cube. This means that the pyramid is one-third the volume of a prism with the same base.

#### Example A

Find the slant height of the square pyramid.

Notice that the slant height is the hypotenuse of a right triangle formed by the height and half the base length. Use the Pythagorean Theorem.

82+24264+576640l=l2=l2=l2=640=810\begin{align*}8^2+24^2&=l^2\\ 64+576&=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 surface area of the four triangular faces are 4(12bl)=2(16)(810)=25610\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 162=256\begin{align*}16^2 = 256\end{align*}. The total surface area is 25610+2561065.54\begin{align*}256 \sqrt{10}+256 \approx 1065.54\end{align*}.

#### Example C

Find the volume of the pyramid.

V=13(122)12=576 units3\begin{align*}V=\frac{1}{3} (12^2)12=576 \ units^3\end{align*}

Watch this video for help with the Examples above.

#### Concept Problem Revisited

The original volume of the pyramid is 13(7062)(407.5)67,704,223.33 ft3\begin{align*}\frac{1}{3} (706^2)(407.5)\approx 67,704,223.33 \ ft^3\end{align*}.

### Vocabulary

A pyramid is a solid with one base and lateral faces that meet at a common vertex. The edges between the lateral faces are lateral edges. The edges between the base and the lateral faces are 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.

Surface area is a two-dimensional measurement that is the total area of all surfaces that bound a solid. Volume is a three-dimensional measurement that is a measure of how much three-dimensional space a solid occupies.

### Guided Practice

1. Find the area of the regular triangular pyramid.

2. If the lateral surface area of a square pyramid is 72 ft2\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 area of the regular hexagonal pyramid below.

4. Find the volume of the pyramid.

5. Find the volume of the pyramid.

6. A rectangular pyramid has a base area of 56 cm2\begin{align*}56 \ cm^2\end{align*} and a volume of 224 cm3\begin{align*}224 \ cm^3\end{align*}. What is the height of the pyramid?

1. The area of the base is A=14s23\begin{align*}A=\frac{1}{4} s^2 \sqrt{3}\end{align*} because it is an equilateral triangle.

BSA=14823=163=163+12(24)(18)=163+216243.71\begin{align*}B & =\frac{1}{4} 8^2 \sqrt{3}=16 \sqrt{3}\\ SA& = 16 \sqrt{3}+\frac{1}{2} (24)(18)=16 \sqrt{3}+216 \approx 243.71\end{align*}

2. In the formula for surface area, the lateral surface area is 12Pl\begin{align*}\frac{1}{2} Pl\end{align*} or 12nbl\begin{align*}\frac{1}{2} nbl\end{align*}. We know that n=4\begin{align*}n = 4\end{align*} and b=l\begin{align*}b = l\end{align*}. Let’s solve for b\begin{align*}b\end{align*}.

12nbl12(4)b22b2b2b=72 ft2=72=72=36=6\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*}

Therefore, the base edges are all 6 units and the slant height is also 6 units.

3. To find the area of the base, we need to find the apothem. If the base edges are 10 units, then the apothem is 53\begin{align*}5 \sqrt{3}\end{align*} for a regular hexagon. The area of the base is 12asn=12(53)(10)(6)=1503\begin{align*}\frac{1}{2} asn = \frac{1}{2} \left ( 5\sqrt{3} \right )(10)(6)=150 \sqrt{3}\end{align*}. The total surface area is:

SA=1503+12(6)(10)(22)=1503+660919.81 units2\begin{align*}SA& = 150 \sqrt{3}+\frac{1}{2}(6)(10)(22)\\ & = 150 \sqrt{3}+660 \approx 919.81 \ units^2\end{align*}

4. In this example, we are given the slant height. For volume, we need the height, so we need to use the Pythagorean Theorem to find it.

72+h2h2h=252=576=24\begin{align*}7^2+h^2&=25^2\\ h^2&=576\\ h&=24\end{align*}

Using the height, the volume is 13(142)(24)=1568 units3\begin{align*}\frac{1}{3} (14^2 )(24)=1568 \ units^3\end{align*}.

5. The base of this pyramid is a right triangle. So, the area of the base is 12(14)(8)=56 units2\begin{align*}\frac{1}{2} (14)(8)=56 \ units^2\end{align*}.

V=13(56)(17)317.33 units3\begin{align*}V=\frac{1}{3} (56)(17) \approx 317.33 \ units^3\end{align*}

6. The formula for the volume of a pyramid works for any pyramid, as long as you can find the area of the base.

2244=56h=h\begin{align*}224&=56h\\ 4& = h\end{align*}

### Practice

Fill in the blanks about the diagram below.

1. x\begin{align*}x\end{align*} is the ___________.
2. The slant height is ________.
3. y\begin{align*}y\end{align*} is the ___________.
4. The height is ________.
5. The base is _______.
6. The base edge is ________.

Find the area of a lateral face and the volume of the regular pyramid. Leave your answer in simplest radical form.

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

1. A regular tetrahedron has four equilateral triangles as its faces. Find the surface area of a regular tetrahedron with edge length of 6 units.
2. Using the formula for the area of an equilateral triangle, what is the surface area of a regular tetrahedron, with edge length s\begin{align*}s\end{align*}?

For questions 13-15 consider a square with diagonal length 102 in\begin{align*}10\sqrt{2} \ in\end{align*}.

1. What is the length of a side of the square?
2. If this square is the base of a right pyramid with height 12, what is the slant height of the pyramid?
3. What is the surface area of the pyramid?

A regular tetrahedron has four equilateral triangles as its faces. Use the diagram to answer questions 16-19.

1. What is the area of the base of this regular tetrahedron?
2. What is the height of this figure? Be careful!
3. Find the volume. Leave your answer in simplest radical form.
4. Challenge If the sides are length s\begin{align*}s\end{align*}, what is the volume?

A regular octahedron has eight equilateral triangles as its faces. Use the diagram to answer questions 20-22.

1. Describe how you would find the volume of this figure.
2. Find the volume. Leave your answer in simplest radical form.
3. Challenge If the sides are length s\begin{align*}s\end{align*}, what is the volume?

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### Vocabulary Language: English Spanish

TermDefinition
lateral edges Edges between the lateral faces of a prism.
lateral faces The non-base faces of a prism.
Cone A cone is a solid three-dimensional figure with a circular base and one vertex.
Pyramid A pyramid is a three-dimensional object with a base that is a polygon and triangular faces that meet at one vertex.
Vertex A vertex is a point of intersection of the lines or rays that form an angle.
Volume Volume is the amount of space inside the bounds of a three-dimensional object.
Cavalieri's Principle States that if two solids have the same height and the same cross-sectional area at every level, then they will have the same volume.
Base Edge The base edge is the edge between the base and the lateral faces of a prism.
Slant Height The slant height is the height of a lateral face of a pyramid.
Apothem The apothem of a regular polygon is a perpendicular segment from the center point of the polygon to the midpoint of one of its sides.

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