COMMON CORE MATHEMATICS CURRICULUM
Lesson 26
7•6
Lesson 26: Volume of Composite Three-Dimensional
Objects
Student Outcomes

Students compute volumes of three-dimensional objects composed of right prisms by using the fact that
volume is additive.
Lesson Notes
Lesson 26 is an extension of work done in the prior lessons on volume as well as an extension of work started in the final
lesson of Module 3 (Lesson 26). Students have more exposure to composite figures such as prisms with prism shaped
holes or prisms that have smaller prisms removed from their volumes. Furthermore, in applicable situations, students
compare different methods to determine composite volume. This is necessary when the entire prism can be
decomposed into multiple prisms or when the prism hole shares the height of the main prism.
Classwork
Example 1 (4 minutes)
Example 1
Find the volume of the following three-dimensional object composed of two
right rectangular prisms.
Volume of object = Volume of top prism + Volume of bottom prism
Volume of top prism:
𝑽𝒐𝒍𝒖𝒎𝒆𝒕𝒐𝒑 𝒑𝒓𝒊𝒔𝒎 = (𝟒 ∙ 𝟓 ∙ 𝟓) m3
= 𝟏𝟎𝟎 m3
Volume of bottom prism:
𝑽𝒐𝒍𝒖𝒎𝒆𝒃𝒐𝒕𝒕𝒐𝒎 𝒑𝒓𝒊𝒔𝒎 = (𝟔 + 𝟒)(𝟓 + 𝟓)(𝟑) m3
= 𝟑𝟎𝟎 m3
The volume of the object is (𝟏𝟎𝟎 + 𝟑𝟎𝟎) m3 = 𝟒𝟎𝟎 m3
There are different ways the volume of a composite figure may be calculated. If the figure is like the figure in Example 1,
where the figure can be decomposed into separate prisms and it would be impossible for the prisms to share any one
dimension, the individual volumes of the decomposed prisms can be determined and then summed. If, however, the
figure is similar to the figure in Exercise 1, there are two possible strategies. In Exercise 1, the figure can be decomposed
into two individual prisms, but a dimension is shared between the two prisms—in this case the height. Instead of
calculating the volume of each prism and then taking the sum, we can calculate the area of the entire base by
decomposing it into shapes we know and then multiplying the area of the base by the height.
Lesson 26:
Date:
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Lesson 26
COMMON CORE MATHEMATICS CURRICULUM
7•6
Exercise 1 (4 minutes)
Exercise 1
Find the volume of the following three-dimensional figure composed of two right
rectangular prisms.
Volume of object = Volume of back prism + Volume of front prism
Volume of back prism:
Volume of front prism:
𝑽𝒐𝒍𝒖𝒎𝒆𝒃𝒂𝒄𝒌 𝒑𝒓𝒊𝒔𝒎 = (𝟏𝟎 ∙ 𝟐 ∙ 𝟗) in3
= 𝟏𝟖𝟎 in3
𝑽𝒐𝒍𝒖𝒎𝒆𝒇𝒓𝒐𝒏𝒕 𝒑𝒓𝒊𝒔𝒎 = (𝟐 ∙ 𝟔 ∙ 𝟗) in3
= 𝟏𝟎𝟖 in3
The volume of the object is (𝟏𝟖𝟎 + 𝟏𝟎𝟖) in3 = 𝟐𝟖𝟖 in3
Exercise 2 (10 minutes)
Exercise 2
The right trapezoidal prism is composed of a right rectangular
prism joined with a right triangular prism. Find the volume of the
right trapezoidal prism shown in the diagram using two different
strategies.
Strategy #1
The volume of the trapezoidal prism is equal to the sum of the
volumes of the rectangular and triangular prisms.
MP.1
Volume of object = Volume of rectangular prism + Volume of triangular prism
Volume of rectangular prism:
Volume of triangular prism:
𝑽𝒐𝒍𝒖𝒎𝒆𝒓𝒆𝒄𝒕𝒂𝒏𝒈𝒖𝒍𝒂𝒓 𝒑𝒓𝒊𝒔𝒎 = 𝑩𝒉
𝑽𝒐𝒍𝒖𝒎𝒆𝒕𝒓𝒊𝒂𝒏𝒈𝒖𝒍𝒂𝒓 𝒑𝒓𝒊𝒔𝒎 = 𝑩𝒉
𝟏
= (𝒍𝒘)𝒉
= ( 𝒍𝒘) 𝒉
𝟏
𝟐
= (𝟑 𝒄𝒎 ∙ 𝟐 𝒄𝒎) ∙ 𝟏 𝒄𝒎
𝟏
𝟏
𝟏
𝟐
=
(
∙ 𝟑 𝒄𝒎 ∙ 𝟐 𝒄𝒎) ∙ 𝟏 𝒄𝒎
𝟐
𝟒
𝟐
= 𝟗 𝒄𝒎𝟑
𝟏
=𝟓
𝒄𝒎𝟑
𝟏𝟔
𝟏
𝟏
The volume of the object is (𝟗 + 𝟓 ) cm3 = 𝟏𝟒 cm3
𝟏𝟔
𝟏𝟔
Strategy #2
The volume of a right prism is equal to the area of its base time its height. The base consists of a rectangle and a triangle.
𝑩 = 𝑨𝒓𝒆𝒂𝒓𝒆𝒄𝒕𝒂𝒏𝒈𝒍𝒆 + 𝑨𝒓𝒆𝒂𝒕𝒓𝒊𝒂𝒏𝒈𝒍𝒆
𝑨𝒓𝒆𝒂𝒓𝒆𝒄𝒕𝒂𝒏𝒈𝒍𝒆 = 𝟑 cm ∙ 𝟐 cm
= 𝟔 cm2
𝟏
𝟏
∙ 𝟑 cm ∙ 𝟐 cm
𝟐
𝟒
𝟑
= 𝟑 cm2
𝟖
𝑨𝒓𝒆𝒂(𝒕𝒓𝒊𝒂𝒏𝒈𝒍𝒆) =
𝑩 = 𝟔 cm2 + 𝟑
Volume of object = 𝑩𝒉
Volume of object:
𝑽𝒐𝒍𝒖𝒎𝒆𝒐𝒃𝒋𝒆𝒄𝒕 = 𝑩𝒉
𝟑
𝟏
= (𝟗 𝒄𝒎𝟐 ) (𝟏 𝒄𝒎)
𝟖
𝟐
𝟏
= 𝟏𝟒
𝒄𝒎𝟑
𝟏𝟔
𝟑
𝟑
cm2 = 𝟗 cm2
𝟖
𝟖
The volume of the object is (𝟗 + 𝟓
Lesson 26:
Date:
𝟏
𝟏
) cm3 = 𝟏𝟒
cm3
𝟏𝟔
𝟏𝟔
Volume of Composite Three-Dimensional Objects
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Lesson 26
COMMON CORE MATHEMATICS CURRICULUM

Write a numeric expression to represent the volume of the figure in Strategy 1.


1
1
1
1
𝑐𝑚 + ( ∙ 3 𝑐𝑚 ∙ 2 𝑐𝑚) ∙ 1 𝑐𝑚
2
2
4
2
(3 𝑐𝑚 ∙ 2 𝑐𝑚 +
1
1
1
∙ 3 𝑐𝑚 ∙ 2 𝑐𝑚) (1 𝑐𝑚)
2
4
2
How do the numeric expressions represent the problem differently?


(3 𝑐𝑚 ∙ 2 𝑐𝑚) ∙ 1
Write a numeric expression to represent the volume of the figure in Strategy 2.


7•6
The first expression is appropriate to use when individual volumes of the decomposed figure are being
added together, whereas the second expression is used when the area of the base of the composite
figure is found and then multiplied by the height to determine the volume.
What property allows us to show that these representations are equivalent?

The distributive property.
Example 2 (10 minutes)
Example 2
Find the volume of the right prism shown in the diagram whose
base is the region between two right triangles. Use two different
strategies.
Strategy #1
The volume of the right prism is equal to the difference of the
volumes of the two triangular prisms.
𝑽𝒐𝒍𝒖𝒎𝒆𝒍𝒂𝒓𝒈𝒆 𝒑𝒓𝒊𝒔𝒎:
𝑽𝒐𝒍𝒖𝒎𝒆𝒍𝒂𝒓𝒈𝒆 𝒑𝒓𝒊𝒔𝒎
Volume of object = Volumelarge prism − Volumesmall prism
𝑽𝒐𝒍𝒖𝒎𝒆𝒔𝒎𝒂𝒍𝒍 𝒑𝒓𝒊𝒔𝒎:
𝟏
𝟏
𝟏 𝟏
𝟏
= ( ∙ 𝟑 𝒄𝒎 ∙ 𝟒 𝒄𝒎) 𝟒 𝒄𝒎
𝑽𝒐𝒍𝒖𝒎𝒆𝒔𝒎𝒂𝒍𝒍 𝒑𝒓𝒊𝒔𝒎 = ( ∙ 𝟏 𝒄𝒎 ∙ 𝟐 𝒄𝒎) 𝟒 𝒄𝒎
𝟐
𝟐
𝟐 𝟐
𝟐
𝟑
= 𝟐𝟕 𝒄𝒎𝟑
𝟑
= 𝟔 𝒄𝒎
𝟒
𝟏
𝟒
The volume of the object is 𝟐𝟎 cm3.
Strategy #2
The volume of a right prism is equal to the area of its base times its height. The base is the region between two right
triangles.
𝑩 = 𝑨𝒓𝒆𝒂𝒍𝒂𝒓𝒈𝒆 𝒕𝒓𝒊𝒂𝒏𝒈𝒍𝒆 − 𝑨𝒓𝒆𝒂𝒔𝒎𝒂𝒍𝒍 𝒕𝒓𝒊𝒂𝒏𝒈𝒍𝒆
𝟏
𝑨𝒓𝒆𝒂𝒍𝒂𝒓𝒈𝒆 𝒕𝒓𝒊𝒂𝒏𝒈𝒍𝒆 = ∙ 𝟑 𝒄𝒎 ∙ 𝟒 𝒄𝒎
𝟐
= 𝟔 𝒄𝒎𝟐
𝟏 𝟏
∙ 𝟏 𝒄𝒎 ∙ 𝟐 𝒄𝒎
𝟐 𝟐
𝟏
= 𝟏 𝒄𝒎𝟑
𝟐
𝑨𝒓𝒆𝒂𝒔𝒎𝒂𝒍𝒍 𝒕𝒓𝒊𝒂𝒏𝒈𝒍𝒆 =
Volume of object = 𝑩𝒉
Volume of object:
𝑽𝒐𝒍𝒖𝒎𝒆𝒐𝒃𝒋𝒆𝒄𝒕 = 𝑩𝒉
𝟏
𝟏
= (𝟒 𝒄𝒎𝟐 ∙ 𝟒 𝒄𝒎)
𝟐
𝟐
𝟏
= 𝟐𝟎 𝒄𝒎𝟑
𝟒
𝟏
𝟏
𝑩 = 𝟔 𝒄𝒎𝟐 − 𝟏 𝒄𝒎𝟐 = 𝟒 𝒄𝒎𝟐
𝟐
𝟐
𝟏
𝟒
The volume of the object is 𝟐𝟎 cm3.
Lesson 26:
Date:
Volume of Composite Three-Dimensional Objects
2/10/16
279
Lesson 26
COMMON CORE MATHEMATICS CURRICULUM

Write a numeric expression to represent the volume of the figure in Strategy 1.
1
2


( ∙ 3 𝑐𝑚 ∙ 4 𝑐𝑚) 4
1
2
1
2
[( ∙ 3 𝑐𝑚 ∙ 4 𝑐𝑚) − ( ∙ 1
1
1
𝑐𝑚 ∙ 2 𝑐𝑚)] 4 𝑐𝑚
2
2
How do the numeric expressions represent the problem differently?


1
1
1
1
𝑐𝑚 − ( ∙ 1 𝑐𝑚 ∙ 2 𝑐𝑚) 4 𝑐𝑚
2
2
2
2
Write a numeric expression to represent the volume of the figure in Strategy 2.


7•6
The first expression is appropriate to use when the volume of the smaller prism is being subtracted
away from the volume of the larger prism, whereas the second expression is used when the area of the
base of the composite figure is found and then multiplied by the height to determine the volume.
What property allows us to show that these representations are equivalent?

The distributive property.
Example 3 (10 minutes)
Example 3
A box with a length of 𝟐 ft., a width of 𝟏. 𝟓 ft., and a height of 𝟏. 𝟐𝟓 ft. contains fragile electronic equipment that is
packed inside a larger box with three inches of styrofoam cushioning material on each side (above, below, left side, right
side, front, and back).
a.
Give the dimensions of the larger box.
Length 𝟐. 𝟓 ft., width 𝟐 ft., and height 𝟏. 𝟕𝟓 ft.
b.
Design styrofoam right rectangular prisms that could be placed around the box to provide the cushioning; i.e.,
give the dimensions and how many of each size are needed.
Possible answer: Two pieces with dimensions 𝟐. 𝟓 ft. × 𝟐 ft. × 𝟑 in. and four pieces with dimensions
𝟐 ft. × 𝟏. 𝟐𝟓 ft. × 𝟑 in.
c.
Find the volume of the styrofoam cushioning material by adding the volumes of the right rectangular prisms
in the previous question.
𝑽𝟏 = 𝟐(𝟐. 𝟓 ∙ 𝟐 ∙ 𝟎. 𝟐𝟓) ft3 = 𝟐. 𝟓 ft3
𝑽𝟐 = 𝟒(𝟐 ∙ 𝟏. 𝟐𝟓 ∙ 𝟎. 𝟐𝟓) ft3 = 𝟐. 𝟓 ft3
𝑽𝟏 + 𝑽𝟐 = (𝟐. 𝟓 + 𝟐. 𝟓) ft3 = 𝟓 ft3
d.
Find the volume of the styrofoam cushioning material by computing the difference between the volume of
the larger box and the volume of the smaller box.
(𝟐. 𝟓 ∙ 𝟐 ∙ 𝟏. 𝟕𝟓 − 𝟐 ∙ 𝟏. 𝟓 ∙ 𝟏. 𝟐𝟓) ft3 = (𝟖. 𝟕𝟓 − 𝟑. 𝟕𝟓) ft3 = 𝟓 ft3
Closing (2 minutes)

To find the volume of a three-dimensional composite object, two or more distinct volumes must be added
together (if they are joined together) or subtracted from each other (if one is a missing section of the other).

There are two strategies to find the volume of a prism: Find the area of the base, then multiply times the
prism’s height; decompose the prism into two or more smaller prisms of the same height and add the volumes
of those smaller prisms.
Exit Ticket (5 minutes)
Lesson 26:
Date:
Volume of Composite Three-Dimensional Objects
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Lesson 26
COMMON CORE MATHEMATICS CURRICULUM
Name
7•6
Date
Lesson 26: Volume of Composite Three-Dimensional Objects
Exit Ticket
A triangular prism has a rectangular prism cut out of it from
one base to the opposite base, as shown in the figure.
Determine the volume of the figure, provided all dimensions
are in millimeters.
Is there any other way to determine the volume of the
figure? If so, please explain.
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Lesson 26
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7•6
Exit Ticket Sample Solutions
A triangular prism has a rectangular prism cut out of it from one
base to the opposite base, as shown in the figure. Determine the
volume of the figure, provided all dimensions are in millimeters.
Is there any other way to determine the volume of the figure? If
so, please explain.
Possible response:
𝟏
𝟐
Volume of the triangular prism: ( ∙ 𝟏𝟔 ∙ 𝟏𝟑) (𝟏𝟒) mm3 = 𝟏, 𝟒𝟓𝟔
mm3
Volume of the rectangular prism: (𝟔 ∙ 𝟑 ∙ 𝟏𝟒) mm3 = 𝟐𝟓𝟐 mm3
Volume of composite prism: (𝟏, 𝟒𝟓𝟔 − 𝟐𝟓𝟐) mm3 = 𝟏, 𝟐𝟎𝟒 mm3
The calculations above subtract the volume of the cut out prism from the volume of the main prism. Another strategy
would be to find the area of the base of the figure, which is the area of the triangle less the area of the rectangle, and
then multiply by the height to find the volume of the prism.
Problem Set Sample Solutions
1.
Find the volume of the three-dimensional object composed of right
rectangular prisms.
Volumeobject = Volumetop and bottom prisms + Volumemiddle prism
Volume of top and bottom prisms:
Volume of middle prism:
𝑽 = 𝟐(𝟏𝟐 ∙ 𝟏𝟐 ∙ 𝟑) in3
= 𝟖𝟔𝟒 in3
𝑽 = (𝟒 ∙ 𝟒 ∙ 𝟖) in3
= 𝟏𝟐𝟖 in3
The volume of the object is (𝟖𝟔𝟒 + 𝟏𝟐𝟖) in3 = 𝟗𝟗𝟐 in3
A smaller cube is stacked on top of a larger cube. An edge of the smaller cube measures
𝟏
𝟐
cm in length, while the
larger cube has an edge length three times as long. What is the total volume of the object?
Volume of object = Volumesmall cube + Volumelarge cube
𝟏 𝟐𝟕
) cm3
𝟖
𝟖
𝟏
= 𝟑 cm3
𝟐
𝟏 𝟑
𝟐
𝟏
= cm3
𝟖
𝑽=( +
𝑽𝒐𝒍𝒖𝒎𝒆𝒔𝒎𝒂𝒍𝒍 𝒄𝒖𝒃𝒆 = ( ) cm3
𝟑 𝟑
𝟐
𝟐𝟕
=
cm3
𝟖
𝑽𝒐𝒍𝒖𝒎𝒆𝒍𝒂𝒓𝒈𝒆 𝒄𝒖𝒃𝒆 = ( ) cm3
𝟏
𝟐
The total volume of the object is 𝟑 cm3.
Lesson 26:
Date:
Volume of Composite Three-Dimensional Objects
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Lesson 26
COMMON CORE MATHEMATICS CURRICULUM
7•6
Two students are finding the volume of a prism with a rhombus base but are provided different information
regarding the prism. One student receives Figure 1, while the other receives Figure 2.
Figure 1
a.
Figure 2
Find the expression that represents the volume in each case; show that the volumes are equal.
Figure 1
𝟏
𝑽𝒐𝒍𝒖𝒎𝒆 = 𝟐 ( ∙ 𝟏𝟒. 𝟔 ∙ 𝟑. 𝟐𝟔 ) ∙ 𝟗 mm3
𝟐
= 𝟒𝟐𝟖. 𝟒 mm3
b.
Figure 2
𝑽𝒐𝒍𝒖𝒎𝒆 = ((𝟖 ∙ 𝟓. 𝟗𝟓) ∙ 𝟗) mm3
= 𝟒𝟐𝟖. 𝟒 mm3
How does each calculation differ in the context of how the prism is viewed?
In Figure 1, the prism is treated as two triangular prisms joined together. The volume of each triangular prism
is found and then doubled, whereas in Figure 2, the prism has a base in the shape of a rhombus, and the
volume is found by calculating the area of the rhomboid base and then multiplying by the height.
Find the volume of wood needed to construct the following side table composed of right rectangular prisms.
Volume of bottom legs:
𝑽 = 𝟐(𝟖 ∙ 𝟏 ∙ 𝟎. 𝟕𝟓) in3
= 𝟏𝟐 in3
Volume of vertical legs:
𝑽 = 𝟐(𝟏 ∙ 𝟗. 𝟓 ∙ 𝟎. 𝟕𝟓) in3
= 𝟏𝟒. 𝟐𝟓 in3
Volume of table top:
𝑽 = (𝟖 ∙ 𝟔 ∙ 𝟏. 𝟓) in3
= 𝟕𝟐 in3
The volume of the table is (𝟏𝟐 + 𝟏𝟒. 𝟐𝟓 + 𝟕𝟐) in3 = 𝟗𝟖. 𝟐𝟓 in3.
A plastic die (singular for dice) of a game has an edge length of 𝟏. 𝟓 cm. Each face of the cube has the number of
cubic cut outs as its marker is supposed to indicate (i.e., the face marked 𝟑 has 𝟑 cut outs). What is the volume of
the die?
Number of cubic cut outs:
𝟏 + 𝟐 + 𝟑 + 𝟒 + 𝟓 + 𝟔 = 𝟐𝟏
Volume of cut out cubes:
𝟑
mm3
𝑽 = 𝟐𝟏(𝟐)
𝑽 = 𝟏𝟔𝟖 mm3 = 𝟎. 𝟏𝟔𝟖 cm3
Volume of large cube:
𝑽 = (𝟏. 𝟓)𝟑
𝑽 = 𝟑. 𝟑𝟕𝟓 cm3
The total volume of the die is (𝟑. 𝟑𝟕𝟓 − 𝟎. 𝟏𝟔𝟖) cm3 = 𝟑. 𝟐𝟎𝟕 cm3.
Lesson 26:
Date:
Volume of Composite Three-Dimensional Objects
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COMMON CORE MATHEMATICS CURRICULUM
Lesson 26
7•6
A wooden cube with edge length 𝟔 inches has square holes (holes in the
shape of right rectangular prisms) cut through the centers of each of the
three sides as shown in the figure. Find the volume of the resulting
solid if the square for the holes has an edge length of 𝟏 inch.
Think of making the square holes between opposite sides by cutting
three times: The first cut removes 𝟔 in3, and the second and third cuts
each remove 𝟓 in3. The resulting solid has a volume of (𝟔𝟑 − 𝟔 − 𝟓 −
𝟓) in3 = 𝟐𝟎𝟎 in3.
A right rectangular prism has each of its dimensions (length, width, and height) increased by 𝟓𝟎%. By what percent
is its volume increased?
𝑽 = (𝒍 ∙ 𝒘 ∙ 𝒉)
𝑽′ = (𝟏. 𝟓𝒍 ∙ 𝟏. 𝟓𝒘 ∙ 𝟏. 𝟓𝒉)
𝑽′ = 𝟑. 𝟑𝟕𝟓𝒍𝒘𝒉
The larger volume is 𝟑𝟑𝟕. 𝟓% of the smaller volume. The volume has increased by 𝟐𝟑𝟕. 𝟓%.
A solid is created by putting together right rectangular prisms. If each of the side lengths is increase by 𝟒𝟎%, by
what percent is the volume increased?
If each of the side lengths is increased by 𝟒𝟎%, then the volume of each right rectangular prism is multiplied by
𝟏. 𝟒𝟑 = 𝟐. 𝟕𝟒𝟒. Since this is true for each right rectangular prism, the volume of the larger solid, 𝑽′, can be found by
multiplying the volume of the smaller solid, 𝑽, by 𝟐. 𝟕𝟒𝟒 = 𝟐𝟕𝟒. 𝟒%; i.e. 𝑽′ = 𝟐. 𝟕𝟒𝟒𝑽. This is an increase of
𝟏𝟕𝟒. 𝟒%.
Lesson 26:
Date:
Volume of Composite Three-Dimensional Objects
2/10/16
284