Topics
11. Modeling in Two and Three dimensions
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United States of America
Grades 9-12

11.03 Pyramids and cones

Lesson
Practice

Introduction

We were introduced to solving problems involving the volume of cones in 8th grade, and will use the volume of a cone to relate to the volume of a pyramid. This lesson also extends to solving problems involving the surface area of cones and pyramids.

Volume of cones

While a cylinder was formed by a pair of congruent circles joined by a curved surface, a cone is formed from a single circle with a curved surface that meets itself at the apex.

The volume of a cone is exactly one-third the volume of a cylinder formed from the same base with the same perpendicular height. That is, the volume of a cone is given by

\displaystyle V = \frac{1}{3}Bh
\bm{B}
area of the base
\bm{h}
perpendicular height of the apex from the base
Right cone

A cone in which the apex lies directly above the center of the base

A cone with a dashed line segment drawn from the vertex of the cone to the center of the base. The angle the segment makes with the base is shown to be a right angle.
Oblique cone

A cone in which the apex does not lie directly above the center of the base

A cone with a dashed line segment drawn from the vertex of the cone to the center of the base. The angle the segment makes with the base is not a right angle.

Examples

Example 1

Consider the diagram shown below:

Recall that the formula for the volume of a prism or cylinder is V=Bh, where B is the area of the base and h is the height of the prism or cylinder.

Use the diagram to explain why the formula for the volume of a cone is V=\dfrac{1}{3}Bh.

Worked Solution
Apply the idea

The diagram shows the same cylinder being filled three times, up to three different lines. If we relate the formula for the volume of a cone to the formula for the volume of a cylinder, we can see that the cone in the diagram is only filling a third of the cylinder with each pour, so it makes sense that the volume of a cone is one-third the volume of a cylinder.

Reflect and check

If the formula for the volume of a prism or cylinder is the same, there may be another solid that is one-third the volume of a prism.

Example 2

A carrot has grown such that it can be approximated by a cone with radius 1 \text{ in} and height 12 \text{ in}. If the carrot weighs \frac{3}{4} \text{ lb}, find the approximate density of the carrot.

Worked Solution
Create a strategy

To find the density, we first want to calculate the volume of the cone. Before we can do that, we want to start by calculating the area of the circular base.

Apply the idea

We can calculate the area of the base as follows:

\displaystyle B\displaystyle =\displaystyle \pi r^2Area of a circle
\displaystyle =\displaystyle \pi \cdot 1^2 Substitute the radius
\displaystyle =\displaystyle \piEvaluate the exponent and multiplication

So the area of the circular base is \pi \text{ in}^2.

We can now use this to find the approximate volume of the carrot:

\displaystyle V\displaystyle =\displaystyle \dfrac{1}{3}BhVolume of a cone
\displaystyle {}\displaystyle =\displaystyle \dfrac{1}{3}\cdot\pi \cdot12Substitute known values
\displaystyle {}\displaystyle =\displaystyle 4\piEvaluate the multiplication

So the approximate volume of the carrot is 4\pi \text{ in}^3.

Finally, we can approximate the density:

\displaystyle \text{Density}\displaystyle =\displaystyle \frac{\text{Mass}}{\text{Volume}}Density formula
\displaystyle =\displaystyle \dfrac{\frac{3}{4}}{4\pi}Substitute known values
\displaystyle =\displaystyle \frac{3}{16 \pi}Evaluate the division

The density of the carrot is approximately 0.06 \, \text{lb}/\text{in}^3.

Reflect and check

Since this is only an approximation of the density we could also express it as a decimal value, rounded to some number of decimal places. In this case, the density is approximately 0.0597 \, \text{ lb}/\text{in}^3 to four decimal places.

Note that even when approximating, it is generally good practice to leave values in exact form throughout the work and only round at the final expression.

Example 3

A cone is sawed in half to create the following solid:

What is the volume of the solid?

Worked Solution
Create a strategy

Calculate the volume of the original cone, then take half the volume. We need the height of the cone, so use the Pythagorean theorem with the height of the slanted lateral face as the hypotenuse and the radius of the base as the length of the triangle's short leg.

Apply the idea

First, we'll find the height of the cone:

\displaystyle a^2+b^2\displaystyle =\displaystyle c^2Pythagorean theorem
\displaystyle 5^2+b^2\displaystyle =\displaystyle 11^2Substitute a=5 and c=11
\displaystyle 25+b^2\displaystyle =\displaystyle 121Evaluate the exponents
\displaystyle b^2\displaystyle =\displaystyle 96Subtract 25 from both sides
\displaystyle b\displaystyle =\displaystyle \sqrt{96}Evaluate the square root of both sides

The height of the cone is \sqrt{96 } \text{ cm}.

Now, we need the area of the base of the original cone. We know that the radius of the base is 5 \text{ cm}, so the area of the base is \pi \left(5 ^2 \right) = 25 \pi \text{ cm}^2.

We can calculate the volume of the original cone as follows:

\displaystyle V\displaystyle =\displaystyle \dfrac{1}{3}BhVolume of a cone
\displaystyle =\displaystyle \dfrac{1}{3} \left( 25 \pi \right) \left( \sqrt{96} \right)Substitute B= 25 \pi and h = \sqrt{96}
\displaystyle \approx\displaystyle 256.50997Evaluate the multiplication

Finally, since the cone was sawed in half, we can take half of the total cone of the original cone to be approximately 128.25 \text{ cm} ^3.

Idea summary

The volume of a cone can be found by taking one-third the volume of a cylinder with the same base area and height. The formula for the volume of a cone is given by:

\displaystyle V = \frac{1}{3}Bh
\bm{B}
area of the base
\bm{h}
perpendicular height of the apex from the base

Volume of pyramids

Pyramid

A figure formed from a polygonal base and a set of triangular faces. The triangular faces connect to one side of the base and all join together at the apex.

A pair of three dimensional solids. One has a rectangular base and triangular sides which meet at a point, the other has a triangular base and triangular sides which meet at a point.
Regular pyramid

A pyramid formed from a regular polygonal base. The triangular faces of a regular pyramid are congruent.

A pyramid with a square base.
Rectangular pyramid

A pyramid with a rectangular base

A pyramid with a rectangular base.
Right pyramid

A pyramid in which the apex lies directly above the center of the base

A pyramid with a triangular base. The height is shown to be perpendicular to the base.
Triangular pyramid

A pyramid with a triangular base

A pyramid with a triangular base.

Exploration

Drag the sliders to fold the pyramids in to the cube and change the size of the figure.

Loading interactive...

  1. How many pyramids fit into the prism?

  2. If the volume of a prism is found using the formula V=Bh, how do you think we can find the volume of a pyramid?

Similarly to the volume of a cylinder and a cone, the volume of a pyramid is also calculated by taking one-third the volume of a prism that has the same height and base area.

Examples

Example 4

The Pyramid of Giza is a square pyramid, that is 280 Egyptian royal cubits high and has a base length of 440 Egyptian royal cubits. What is the volume of the Pyramid of Giza?

A pyramid.
Worked Solution
Create a strategy

First we find the area of the base, then we can use that to find the volume. Since this solid is a square pyramid, the base is a square.

Apply the idea

Finding the area of the base, we have:

\displaystyle B\displaystyle =\displaystyle \text{side length}^2Area of a square
\displaystyle =\displaystyle 440^2Substitute the side length
\displaystyle =\displaystyle 193600Simplify

We can now use this to calculate the volume:

\displaystyle V\displaystyle =\displaystyle \dfrac{1}{3}BhVolume of a pyramid
\displaystyle =\displaystyle \dfrac{1}{3} \cdot 193600 \cdot 280Substitute known values
\displaystyle =\displaystyle 18069333 \frac{1}{3}Simplify

So the volume of the Pyramid of Giza is 18 \, 069 \, 333 \dfrac{1}{3} cubic Egyptian royal cubits.

Example 5

A small square pyramid of height 4 \text{ cm} was removed from the top of a large square pyramid of height 8 \text{ cm} forming the solid shown. Find the exact volume of the solid.

Worked Solution
Create a strategy

Subtract the volume of the smaller pyramid from the volume of the larger pyramid.

Apply the idea

Start by calculating the volume of the larger pyramid. We have

\displaystyle V\displaystyle =\displaystyle \dfrac{1}{3}BhVolume of a pyramid
\displaystyle =\displaystyle \dfrac{1}{3} \left(64 \right) \left(8 \right)Substitute B=64 for the area of the base and h=8 for the height of the larger pyramid
\displaystyle =\displaystyle \dfrac{512}{3}Evaluate the multiplication

The volume of the larger pyramid is exactly \dfrac{512}{3} \text{ cm}^3.

Now we can calculate the volume of the smaller pyramid. We have

\displaystyle V\displaystyle =\displaystyle \dfrac{1}{3}BhVolume of a pyramid
\displaystyle =\displaystyle \dfrac{1}{3} \left(16 \right) \left(4 \right)Substitute B=16 for the area of the base and h=4 for the height of the smaller pyramid
\displaystyle =\displaystyle \dfrac{64}{3}Evaluate the multiplication

The volume of the smaller pyramid is exactly \dfrac{64}{3} \text{ cm}^3.

The volume of the solid after removing the smaller pyramid is \dfrac{512}{3} \text{ cm}^3 - \dfrac{64}{3} \text{ cm}^3 = \dfrac{448}{3} \text{ cm}^3.

Reflect and check

We could choose to leave the solution in exact form, since the directions do not specify rounding requirements, or we could simplify the solution and round to a measure we find appropriate.

Idea summary

The volume of a pyramid can be found by taking one-third the volume of a prism with the same base area and height. The formula for the volume of a pyramid is given by:

\displaystyle V = \frac{1}{3}Bh
\bm{B}
area of the base
\bm{h}
perpendicular height of the apex from the base

Surface area of cones and pyramids

Similar to prisms, we can consider the nets of pyramids and cones to determine their surface area and lateral surface area.

Exploration

Flatten the cone using the sliders and then straighten the arc and dissect the sector created by the lateral face of the cone. Use the sliders to explore what happens.

Loading interactive...

  1. The formula for the lateral surface area of a cone is given by LA=\pi r h_s, where r is the radius of the base of the cone and h_s is the slant height of the cone. How does this formula relate to the area of the rearranged lateral face on the cone?

The surface area of a cone is the sum of the area of the base and the area of the lateral face. The area of the lateral face is calculated as the product of the slant height of the cone and half the perimeter of the cone's base. For a right cone, the surface area can be calculated using the formula

\displaystyle SA = B + \pi rh_s
\bm{B}
area of the circular base
\bm{r}
radius of the base
\bm{h_s}
slant height of the cone
A diagram of a right cone showing the height of the cone as a dashed segment from the apex to the center of the circular base. From the center of the circle to a point on the circumference is labelled r which is perpendicular to the height of the cone. The distance of the segment from the apex to a point on the circle is labelled h sub s.

The slant height of a right cone is the distance from any point on the circumference of the base to the apex.

Examples

Example 6

Consider the diagram of the right square pyramid and its net shown:

Let b= the base of each triangle and l= the height of each triangle.

a

Explain a process for finding the surface area of the square pyramid.

Worked Solution
Apply the idea

To find the surface area of the square pyramid, we will need to calculate the sum of the area of the pyramid's base and the area of the pyramid's lateral faces.

First, we find the area of the base of the pyramid.

Then, we find the area of the lateral faces of the pyramid. Since the base of this pyramid is a square, which is a regular polygon, each of the lateral face triangles are congruent. So, we can calculate the area of one of the faces and multiply that by the number of faces on the pyramid.

Together, these areas make up the entire surface area of the square pyramid, which is the area of its net.

Reflect and check

We could use the given dimensions to write an expression for the surface area of the given pyramid.

For this pyramid, the area of the base, B, is B=b \times b or B=b^2.

Then the area of each triangle, A, is A=\frac{1}{2}bl. Since there are 4 faces, we multiply the area of the triangular face by 4.

Therefore, the total surface area of the square pyramid can be expressed as SA=b^2 + 4 \left(\frac{1}{2}bl \right) or SA=B + 4A.

b

Now, consider the regular right octagonal pyramid shown below, where a= the distance from the midpoint of a side of the base to the center of the base:

Draw the net of the right octagonal pyramid and describe how to find the surface area of the pyramid.

Worked Solution
Create a strategy

We can draw the net of the octagonal pyramid the same way as the net in part (a) is drawn, or we can imagine unraveling the lateral faces of the pyramid and drawing them side by side.

Apply the idea

For this pyramid, we will again need to calculate the area of the base and the lateral surface area.

We can find the area of the base by cutting the octagon into eight congruent triangles, and then finding the sum of the area of each triangle. For the lateral surface area, we can calculate the area of one triangle and multiply by eight triangles. Together, the base area and lateral surface area make up the surface area of the pyramid.

Reflect and check

We could use the given dimensions to write an expression for the surface area of the given pyramid. This would summarise our description algebraically

For this pyramid, the area of the base, B, can be calculated by finding the total area of the 8 congruent triangles formed by joining the corners of the base to its center B=8 \left(\dfrac{1}{2}b\times a\right) or B=4ab.

Then the area of each triangular face, A, is A=\frac{1}{2}bl. Since there are 8 faces, we multiply the area of the triangular face by 8.

Therefore, the total surface area of the octagonal pyramid can be expressed as: SA=4ab + 4bl or SA=B + 8A.

c

Write a formula for finding the surface area of a regular, right n-gon pyramid. Explain your reasoning.

Worked Solution
Create a strategy

For any pyramid, we know that the net we need the area from will have a base and lateral faces that we need the sum of.

Apply the idea

If we let B= the area of the base and A= the area of one of the lateral face triangles, we can say that the surface area, SA, is equivalent to the sum of the area of the base and the area of the lateral faces. For any n-gon, the number of lateral faces is equal to the number of sides of the polygon.

A formula for the surface area of any pyramid is SA= B+ nA.

Reflect and check

We can express the formula above in different forms depending on the dimensions given in a problem. For example, if we are given the following dimensions:

  • b - the base of each triangular face

  • l - the height of each triangular face

  • a - the distance from the midpoint of a side of the base to the center of the base

Then the formula can also be expressed in terms of these parameters. We have the area of the base, B, will be n triangles of base b and height a. Hence, B=n\left(\dfrac{1}{2}ab\right). And the area of each face is A=\dfrac{1}{2}bl.

So, the total surface area is SA=n\left(\dfrac{1}{2}ab\right)+n\left(\dfrac{1}{2}bl\right), which could also be expressed as SA=\dfrac{nb}{2}\left(a+l\right)

Example 7

What is the surface area of the following cone?

Worked Solution
Create a strategy

Calculate the slant height of the cone using the Pythagorean theorem, then use the formula for the surface area of a cone to calculate the surface area.

Apply the idea

The slant height of the cone can be found using the Pythagorean theorem as follows:

\displaystyle a^2+b^2\displaystyle =\displaystyle c^2Pythagorean theorem
\displaystyle 3^2 + 9^2\displaystyle =\displaystyle c^2Substitute a=3 and b=9
\displaystyle 90\displaystyle =\displaystyle c^2Evaluate the exponents and addition
\displaystyle \sqrt{90}\displaystyle =\displaystyle cEvaluate the square root of both sides

The slant height of the cone is \sqrt{90} \text{ cm}.

Now, we can calculate the surface area as follows:

\displaystyle SA\displaystyle =\displaystyle B+ \pi r h_sSurface area of a cone
\displaystyle =\displaystyle \left(\pi \cdot 3^2 \right) + \left( \pi \cdot 3 \cdot \sqrt{90} \right)Substitute r=3 and h_s = \sqrt{90}
\displaystyle =\displaystyle 117.7Evaluate the exponent, square root, multiplication, and addition

The surface area of the cone is 117.7 \text{ cm}^2.

Example 8

The Pyramid of Giza is a square pyramid, that is 280 Egyptian royal cubits high and has a base length of 440 Egyptian royal cubits. What is the surface area of the Pyramid of Giza?

A pyramid.
Worked Solution
Create a strategy

We will need the height of a lateral face of the pyramid to calculate the lateral surface area. First, find the height of one lateral face using the Pythagorean theorem, then calculate the surface area of the pyramid. The diagram that follows shows the dimensions of the Pyramid of Giza:

Apply the idea

First, we'll use the height and half the length of the base as the legs in the Pythagorean theorem:

\displaystyle a^2 + b^2\displaystyle =\displaystyle c^2Pythagorean theorem
\displaystyle 220^2+280^2\displaystyle =\displaystyle c^2Substitute a=220 andb=280
\displaystyle 126\,800\displaystyle =\displaystyle c^2Evaluate the exponents and addition
\displaystyle \sqrt{126\,800}\displaystyle =\displaystyle cEvaluate the square root of both sides

The height of one lateral face of the pyramid is exactly \sqrt{126800} Egyptian royal cubits.

We can use this to calculate the area of one of the faces of the Pyramid of Giza as follows:

\displaystyle A\displaystyle =\displaystyle \dfrac{1}{2}bhArea of a triangle
\displaystyle =\displaystyle \dfrac{1}{2} \left(440 \right) \left(\sqrt{126800} \right)Substitute b=440 and h= \sqrt{126800}
\displaystyle =\displaystyle \left(220 \right) \left(\sqrt{126800} \right)Evaluate the multiplication

The area of one of the faces of the Pyramid of Giza is \left(220 \right) \left(\sqrt{126 800} \right) Egyptian royal cubits^2so the area of all four lateral faces is \left(880 \right) \left(\sqrt{126800} \right) Egyptian royal cubits^2.

The area of the base of the pyramid is 440 \text{ Egyptian royal cubits} \times 440\text{ Egyptian royal cubits} = 193\,600 \text{ Egyptian royal cubits}^2

Finally, the surface area of the Pyramid of Giza is calculated as \left(880 \right) \left(\sqrt{126 800} \right) \text{ Egyptian royal cubits}^2 + 193600 \text{ Egyptian royal cubits}^2 \\= 506 \, 959.0911 \text{ Egyptian royal cubits}^2

The surface area of the Pyramid of Giza to the nearest square royal cubit is 506 \, 959 \text{ Egyptian royal cubits}^2.

Reflect and check

The actual length of an Egyptian royal cubit is about 20 \text{ in}, consider how the surface area compares to area of a football field. Note that we did not simplify the square root until the final calculation in order to preserve the precision of the length.

Idea summary

The surface area of a cone is the sum of the area of the base and the area of the lateral face. For a right cone, the surface area can be calculated using the formula

\displaystyle SA = B + \pi rh_s
\bm{B}
area of the circular base
\bm{r}
radius of the base
\bm{h_s}
slant height of the cone

The surface area of a pyramid is the sum of the area of the base and the area of each triangle. For a regular pyramid, the triangles are congruent, and so the surface area of a regular pyramid is given by

\displaystyle SA = B + nA
\bm{B}
area of the base polygon
\bm{n}
number of sides of the base polygon
\bm{A}
area of one triangular face

Outcomes

A.SSE.A.1

Interpret expressions that represent a quantity in terms of its context.

A.SSE.A.1.B

Interpret complicated expressions by viewing one or more of their parts as a single entity.

A.CED.A.4

Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations.

G.GMD.A.1

Give an informal argument for the formulas for the circumference of a circle, area of a circle, volume of a cylinder, pyramid, and cone. Use dissection arguments, Cavalieri's principle, and informal limit arguments.

G.GMD.A.3

Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems.

G.MG.A.1

Use geometric shapes, their measures, and their properties to describe objects.

G.MG.A.2

Apply concepts of density based on area and volume in modeling situations.

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