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# 11.5: Volume of Pyramids and Cones

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Find the volume of pyramids and cones.

## Review Queue

1. Find the volume of a square prism with 8 inch base edges and a 12 inch height.
2. Find the volume of a cylinder with a diameter of 8 inches and a height of 12 inches.
3. Find the surface area of a square pyramid with 10 inch base edges and a height of 12 inches.

Know What? The Khafre Pyramid is a pyramid in Giza, Egypt. 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?

## Volume of a Pyramid

The volume of a pyramid is closely related to the volume of a prism with the same sized base.

Investigation 11-1: Finding the Volume of a Pyramid

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

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 2.5 inches and base edges of 2 inches.
4. Cut out the net and tape up the sides to form an open pyramid.
5. Fill the pyramid with dry rice and dump the rice into the open cube. Repeat this until you have filled the cube?

Volume of a Pyramid: V=13Bh\begin{align*}V=\frac{1}{3} Bh\end{align*}.

Example 1: Find the volume of the pyramid.

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

Example 2a: Find the height of the pyramid.

Solution: In this example, we are given the slant height. Use the Pythagorean Theorem.

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

Example 2b: Find the volume of the pyramid in Example 2a.

Solution: V=13(142)(24)=1568 units3\begin{align*}V=\frac{1}{3} (14^2)(24)=1568 \ units^3\end{align*}.

Example 3: Find the volume of the pyramid.

Solution: The base of the pyramid is a right triangle. 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*}

Example 4: 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?

Solution:

V22412=13Bh=1356h=h\begin{align*}V &= \frac{1}{3} Bh\\ 224 &= \frac{1}{3} \cdot 56h\\ 12 &= h\end{align*}

## Volume of a Cone

Volume of a Cone: V=13πr2h\begin{align*}V=\frac{1}{3} \pi r^2 h\end{align*}.

This is the the same relationship as a pyramid’s volume with a prism’s volume.

Example 5: Find the volume of the cone.

Solution: First, we need the height. Use the Pythagorean Theorem.

52+h2hV=152=200=102=13(52)(102)π370.24\begin{align*}5^2+h^2 &=15^2\\ h &= \sqrt{200}=10\sqrt{2}\\ V &= \frac{1}{3}(5^2)\left(10\sqrt{2}\right) \pi \approx 370.24\end{align*}

Example 6: Find the volume of the cone.

Solution: We can use the same volume formula. Find the radius.

V=13π(32)(6)=18π56.55\begin{align*}V=\frac{1}{3} \pi (3^2)(6)=18 \pi \approx 56.55\end{align*}

Example 7: The volume of a cone is 484π cm3\begin{align*}484 \pi \ cm^3\end{align*} and the height is 12 cm. What is the radius?

Solution: Plug in what you know to the volume formula.

484π12111=13πr2(12)=r2=r\begin{align*}484 \pi &= \frac{1}{3} \pi r^2 (12)\\ 121 &= r^2\\ 11 &= r\end{align*}

## Composite Solids

Example 8: Find the volume of the composite solid. All bases are squares.

Solution: This is a square prism with a square pyramid on top. First, we need the height of the pyramid portion. Using the Pythagorean Theorem, we have, h=252242=7\begin{align*}h=\sqrt{25^2-24^2}=7\end{align*}.

VprismVpyramid=(48)(48)(18)=41472 cm3=13(482)(7)=5376 cm3\begin{align*}V_{prism} &= (48)(48)(18)=41472 \ cm^3\\ V_{pyramid} &= \frac{1}{3} (48^2)(7)=5376 \ cm^3\end{align*}

The total volume is 41472+5376=46,848 cm3\begin{align*}41472 + 5376 = 46,848 \ cm^3\end{align*}.

Know What? 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*}.

## Review Questions

• Questions 1-13 are similar to Examples 1-3, 5 and 6.
• Questions 14-22 are similar to Examples 4 and 7.
• Questions 23-31 are similar to Example 8.

Find the volume of each regular pyramid and right cone. Round any decimal answers to the nearest hundredth. The bases of these pyramids are either squares or equilateral triangles.

Find the volume of the following non-regular pyramids and cones. Round any decimal answers to the nearest hundredth.

1. base is a rectangle

A regular tetrahedron has four equilateral triangles as its faces. Use the diagram to answer questions 14-16. Round your answers to the nearest hundredth.

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.

A regular octahedron has eight equilateral triangles as its faces. Use the diagram to answer questions 17-21. Round your answers to the nearest hundredth.

1. Describe how you would find the volume of this figure.
2. Find the volume.
3. The volume of a square pyramid is 72 square inches and the base edge is 4 inches. What is the height?
4. If the volume of a cone is 30π cm3\begin{align*}30 \pi \ cm^3\end{align*} and the radius is 5 cm, what is the height?
5. If the volume of a cone is 105π cm3\begin{align*}105 \pi \ cm^3\end{align*} and the height is 35 cm, what is the radius?
6. The volume of a triangle pyramid is 170 in3\begin{align*}170 \ in^3\end{align*} and the base area is 34 in2\begin{align*}34 \ in^2\end{align*}. What is the height of the pyramid?

1. Find the volume of the base prism.
2. Find the volume of the pyramid.
3. Find the volume of the entire solid.

The solid to the right is a cube with a cone cut out.

1. Find the volume of the cube.
2. Find the volume of the cone.
3. Find the volume of the entire solid.

The solid to the left is a cylinder with a cone on top.

1. Find the volume of the cylinder.
2. Find the volume of the cone.
3. Find the volume of the entire solid.

1. (82)(12)=768 in3\begin{align*}(8^2)(12) = 768 \ in^3\end{align*}
2. (42)(12)π=192π603.19\begin{align*}(4^2)(12) \pi = 192 \pi \approx 603.19\end{align*}
3. Find slant height, l=13\begin{align*}l = 13\end{align*}. SA=100+12(40)(13)=360 in2\begin{align*}SA = 100 + \frac{1}{2} (40)(13)=360 \ in^2\end{align*}

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