Lesson 26
NYS COMMON CORE MATHEMATICS CURRICULUM
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:
Volume of bottom prism:
𝐕𝐨𝐥𝐮𝐦𝐞𝐭𝐨𝐩 𝐩𝐫𝐢𝐬𝐦 = (𝟒 𝐦)(𝟓 𝐦)(𝟓 𝐦)
= 𝟏𝟎𝟎 𝐦
𝐕𝐨𝐥𝐮𝐦𝐞𝐛𝐨𝐭𝐭𝐨𝐦 𝐩𝐫𝐢𝐬𝐦 = (𝟏𝟎 𝐦)(𝟏𝟎 𝐦)(𝟑 𝐦)
𝟑
= 𝟑𝟎𝟎 𝐦𝟑
The volume of the object is 𝟏𝟎𝟎 𝐦𝟑 + 𝟑𝟎𝟎 𝐦𝟑 = 𝟒𝟎𝟎 𝐦𝟑.
There are different ways the volume of a composite figure may be calculated. If the figure is like the one shown 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:
Volume of Composite Three-Dimensional Objects
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Lesson 26
NYS 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.
𝐀𝐫𝐞𝐚 𝐨𝐟 𝐁𝐚𝐬𝐞𝐛𝐚𝐜𝐤 𝐩𝐫𝐢𝐬𝐦
𝐀𝐫𝐞𝐚 𝐨𝐟 𝐁𝐚𝐬𝐞𝐟𝐫𝐨𝐧𝐭 𝐩𝐫𝐢𝐬𝐦
= (𝟐 𝐢𝐧. )( 𝟏𝟎 𝐢𝐧. )
= (𝟔 𝐢𝐧. )(𝟐 𝐢𝐧. )
= 𝟐𝟎 𝐢𝐧.𝟐
= 𝟏𝟐 𝐢𝐧𝟐
𝐀𝐫𝐞𝐚 𝐨𝐟 𝐁𝐚𝐬𝐞 = 𝟐𝟎 𝐢𝐧𝟐 + 𝟏𝟐 𝐢𝐧𝟐 = 𝟑𝟐 𝐢𝐧𝟐
The volume of the object is (𝟗 𝐢𝐧. )(𝟑𝟐 𝐢𝐧𝟐 ) = 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.
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐨𝐛𝐣𝐞𝐜𝐭 = 𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐫𝐞𝐜𝐭𝐚𝐧𝐠𝐮𝐥𝐚𝐫 𝐩𝐫𝐢𝐬𝐦 + 𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐭𝐫𝐢𝐚𝐧𝐠𝐮𝐥𝐚𝐫 𝐩𝐫𝐢𝐬𝐦
Volume of rectangular prism:
Volume of triangular prism:
𝐕𝐨𝐥𝐮𝐦𝐞𝐫𝐞𝐜𝐭𝐚𝐧𝐠𝐮𝐥𝐚𝐫 𝐩𝐫𝐢𝐬𝐦 = 𝑩𝒉
𝟏
𝐕𝐨𝐥𝐮𝐦𝐞𝐭𝐫𝐢𝐚𝐧𝐠𝐮𝐥𝐚𝐫 𝐩𝐫𝐢𝐬𝐦 = 𝑩𝒉 = ( 𝒍𝒘) 𝒉
𝟐
𝟏
𝟏
𝟏
= ( ∙ 𝟑 𝐜𝐦 ∙ 𝟐 𝐜𝐦) ∙ 𝟏 𝐜𝐦
𝟐
𝟒
𝟐
𝟏
𝟑
=𝟓
𝐜𝐦
𝟏𝟔
= (𝒍𝒘)𝒉
𝟏
= (𝟑 𝐜𝐦 ∙ 𝟐 𝐜𝐦) ∙ 𝟏 𝐜𝐦
𝟐
= 𝟗 𝐜𝐦𝟑
MP.1
The volume of the object is 𝟗 𝐜𝐦𝟑 + 𝟓
𝟏
𝟏
𝐜𝐦𝟑 = 𝟏𝟒
𝐜𝐦𝟑 .
𝟏𝟔
𝟏𝟔
Strategy 2
The volume of a right prism is equal to the area of its base times its height. The base consists of a rectangle and a
triangle.
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐨𝐛𝐣𝐞𝐜𝐭 = 𝑩𝒉
𝑩 = 𝐀𝐫𝐞𝐚𝐫𝐞𝐜𝐭𝐚𝐧𝐠𝐥𝐞 + 𝐀𝐫𝐞𝐚𝐭𝐫𝐢𝐚𝐧𝐠𝐥𝐞
Lesson 26:
Volume of object:
Volume of Composite Three-Dimensional Objects
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Lesson 26
NYS COMMON CORE MATHEMATICS CURRICULUM
𝐀𝐫𝐞𝐚𝐫𝐞𝐜𝐭𝐚𝐧𝐠𝐥𝐞 = 𝟑 𝐜𝐦 ∙ 𝟐 𝐜𝐦 = 𝟔 𝐜𝐦𝟐
𝐕𝐨𝐥𝐮𝐦𝐞𝐨𝐛𝐣𝐞𝐜𝐭 = 𝑩𝒉
𝟑
𝟏
= (𝟗 𝐜𝐦𝟐 ) (𝟏 𝐜𝐦)
𝟖
𝟐
𝟏
𝟑
= 𝟏𝟒
𝐜𝐦
𝟏𝟔
𝟏
𝟏
𝟑
𝐀𝐫𝐞𝐚𝐭𝐫𝐢𝐚𝐧𝐠𝐥𝐞 = ∙ 𝟑 𝐜𝐦 ∙ 𝟐 𝐜𝐦 = 𝟑 𝐜𝐦𝟐
𝟐
𝟒
𝟖
𝟑
𝟑
𝑩 = 𝟔 𝐜𝐦𝟐 + 𝟑 𝐜𝐦𝟐 = 𝟗 𝐜𝐦𝟐
𝟖
𝟖
The volume of the object is 𝟏𝟒

1
2
1
4
1
2
1
2
1
4
1
2
(3 cm ∙ 2 cm + ∙ 3 cm ∙ 2 cm) (1 cm)
How do the numeric expressions represent the problem differently?


1
2
(3 cm ∙ 2 cm) ∙ 1 cm + ( ∙ 3 cm ∙ 2 cm) ∙ 1 cm
Write a numeric expression to represent the volume of the figure in Strategy 2.


𝟏
𝐜𝐦𝟑.
𝟏𝟔
Write a numeric expression to represent the volume of the figure in Strategy 1.


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.
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐨𝐛𝐣𝐞𝐜𝐭 = 𝐕𝐨𝐥𝐮𝐦𝐞𝐥𝐚𝐫𝐠𝐞 𝐩𝐫𝐢𝐬𝐦 − 𝐕𝐨𝐥𝐮𝐦𝐞𝐬𝐦𝐚𝐥𝐥 𝐩𝐫𝐢𝐬𝐦
Volume of large prism:
𝐕𝐨𝐥𝐮𝐦𝐞𝐥𝐚𝐫𝐠𝐞 𝐩𝐫𝐢𝐬𝐦
Volume of small prism:
𝟏
𝟏
= ( ∙ 𝟑 𝐜𝐦 ∙ 𝟒 𝐜𝐦) 𝟒 𝐜𝐦
𝟐
𝟐
= 𝟐𝟕 𝐜𝐦𝟑
𝟏 𝟏
𝟏
𝐕𝐨𝐥𝐮𝐦𝐞𝐬𝐦𝐚𝐥𝐥 𝐩𝐫𝐢𝐬𝐦 = ( ∙ 𝟏 𝐜𝐦 ∙ 𝟐 𝐜𝐦) 𝟒 𝐜𝐦
𝟐 𝟐
𝟐
𝟑
= 𝟔 𝐜𝐦𝟑
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.
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐨𝐛𝐣𝐞𝐜𝐭 = 𝑩𝒉
𝑩 = 𝐀𝐫𝐞𝐚𝐥𝐚𝐫𝐠𝐞 𝐭𝐫𝐢𝐚𝐧𝐠𝐥𝐞 − 𝐀𝐫𝐞𝐚𝐬𝐦𝐚𝐥𝐥 𝐭𝐫𝐢𝐚𝐧𝐠𝐥𝐞
Lesson 26:
Volume of object:
Volume of Composite Three-Dimensional Objects
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Lesson 26
NYS COMMON CORE MATHEMATICS CURRICULUM
𝟏
∙ 𝟑 𝐜𝐦 ∙ 𝟒 𝐜𝐦 = 𝟔 𝐜𝐦𝟐
𝟐
𝟏 𝟏
𝟏
𝐀𝐫𝐞𝐚𝐬𝐦𝐚𝐥𝐥 𝐭𝐫𝐢𝐚𝐧𝐠𝐥𝐞 = ∙ 𝟏 𝐜𝐦 ∙ 𝟐 𝐜𝐦 = 𝟏 𝐜𝐦𝟐
𝟐 𝟐
𝟐
𝟏
𝟏
𝑩 = 𝟔 𝐜𝐦𝟐 − 𝟏 𝐜𝐦𝟐 = 𝟒 𝐜𝐦𝟐
𝟐
𝟐
7•6
𝐕𝐨𝐥𝐮𝐦𝐞𝐨𝐛𝐣𝐞𝐜𝐭 = 𝑩𝒉
𝐀𝐫𝐞𝐚𝐥𝐚𝐫𝐠𝐞 𝐭𝐫𝐢𝐚𝐧𝐠𝐥𝐞 =
𝟏
𝟏
= (𝟒 𝐜𝐦𝟐 ∙ 𝟒 𝐜𝐦)
𝟐
𝟐
𝟏
𝟑
= 𝟐𝟎 𝐜𝐦
𝟒
𝟏
𝟒
The volume of the object is 𝟐𝟎 𝐜𝐦𝟑.

Write a numeric expression to represent the volume of the figure in Strategy 1.
1
2

1
2
1
2
1
2
4 cm
How do the numeric expressions represent the problem differently?


1
2
Write a numeric expression to represent the volume of the figure in Strategy 2.


1
2
( ⋅ 3 cm ⋅ 4 cm) 4 cm − ( ⋅ 1 cm ⋅ 2 cm) 4 cm

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 𝐟𝐭. ∙ 𝟐 𝐟𝐭. ∙ 𝟎. 𝟐𝟓 𝐟𝐭. ) = 𝟐. 𝟓 𝐟𝐭 𝟑
𝑽𝟐 = 𝟒(𝟐 𝐟𝐭. ∙ 𝟏. 𝟐𝟓 𝐟𝐭. ∙ 𝟎. 𝟐𝟓 𝐟𝐭. )= 𝟐. 𝟓 𝐟𝐭 𝟑
𝑽𝟏 + 𝑽𝟐 = 𝟐. 𝟓 𝐟𝐭 𝟑 + 𝟐. 𝟓 𝐟𝐭 𝟑 = 𝟓 𝐟𝐭 𝟑
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 𝐟𝐭. ⋅ 𝟐 𝐟𝐭. ⋅ 𝟏. 𝟕𝟓 𝐟𝐭. ) − (𝟐 𝐟𝐭. ⋅ 𝟏. 𝟓 𝐟𝐭. ⋅ 𝟏. 𝟐𝟓 𝐟𝐭. ) = 𝟖. 𝟕𝟓 𝐟𝐭 𝟑 − 𝟑. 𝟕𝟓 𝐟𝐭 𝟑 = 𝟓 𝐟𝐭 𝟑
Lesson 26:
Volume of Composite Three-Dimensional Objects
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NYS COMMON CORE MATHEMATICS CURRICULUM
Lesson 26:
Volume of Composite Three-Dimensional Objects
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This file derived from G7-M6-TE-1.3.0-10.2015
Lesson 26
7•6
294
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NYS COMMON CORE MATHEMATICS CURRICULUM
Lesson 26
7•6
Closing (2 minutes)
Lesson Summary
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 and 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:
Volume of Composite Three-Dimensional Objects
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This file derived from G7-M6-TE-1.3.0-10.2015
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Lesson 26
NYS 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.
Lesson 26:
Volume of Composite Three-Dimensional Objects
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Lesson 26
NYS COMMON CORE MATHEMATICS CURRICULUM
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 𝐦𝐦 ∙ 𝟏𝟑 𝐦𝐦) (𝟏𝟒 𝐦𝐦) = 𝟏, 𝟒𝟓𝟔 𝐦𝐦𝟑
Volume of the rectangular prism: (𝟔 𝐦𝐦 ∙ 𝟑 𝐦𝐦 ∙ 𝟏𝟒 𝐦𝐦) = 𝟐𝟓𝟐 𝐦𝐦𝟑
Volume of the composite prism: 𝟏, 𝟒𝟓𝟔 𝐦𝐦𝟑 − 𝟐𝟓𝟐 𝐦𝐦𝟑 = 𝟏, 𝟐𝟎𝟒 𝐦𝐦𝟑
The calculations above subtract the volume of the cutout 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 𝐢𝐧. ∙ 𝟏𝟐 𝐢𝐧. ∙ 𝟑 𝐢𝐧. )
𝑽 = 𝟒 𝐢𝐧.⋅ 𝟒 𝐢𝐧.⋅ 𝟖 𝐢𝐧.
= 𝟖𝟔𝟒 𝐢𝐧𝟑
= 𝟏𝟐𝟖 𝐢𝐧𝟑
The volume of the object is 𝟖𝟔𝟒 𝐢𝐧𝟑 + 𝟏𝟐𝟖 𝐢𝐧𝟑 = 𝟗𝟗𝟐 𝐢𝐧𝟑.
2.
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?
𝐕𝐨𝐥𝐮𝐦𝐞𝐨𝐛𝐣𝐞𝐜𝐭 = 𝐕𝐨𝐥𝐮𝐦𝐞𝐬𝐦𝐚𝐥𝐥 𝐜𝐮𝐛𝐞 + 𝐕𝐨𝐥𝐮𝐦𝐞𝐥𝐚𝐫𝐠𝐞 𝐜𝐮𝐛𝐞
𝟏
𝟐𝟕
𝐜𝐦𝟑 +
𝐜𝐦𝟑
𝟖
𝟖
𝟏
= 𝟑 𝐜𝐦𝟑
𝟐
𝟑
𝟏
𝐕𝐨𝐥𝐮𝐦𝐞𝐬𝐦𝐚𝐥𝐥 𝐜𝐮𝐛𝐞 = ( 𝐜𝐦)
𝟐
𝟏
= 𝐜𝐦𝟑
𝟖
𝑽=
𝟑
𝟑
𝐕𝐨𝐥𝐮𝐦𝐞𝐥𝐚𝐫𝐠𝐞 𝐜𝐮𝐛𝐞 = ( 𝐜𝐦)
𝟐
𝟐𝟕
=
𝐜𝐦𝟑
𝟖
The total volume of the object is 𝟑
Lesson 26:
𝟏
𝐜𝐦𝟑 .
𝟐
Volume of Composite Three-Dimensional Objects
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This file derived from G7-M6-TE-1.3.0-10.2015
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Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 26
NYS COMMON CORE MATHEMATICS CURRICULUM
3.
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.
6.57
7.2
Figure 1
a.
b.
Figure 2
Find the expression that represents the volume in each case; show that the volumes are equal.
Figure 1
Figure 2
𝟏
𝟐 ( ∙ 𝟏𝟒. 𝟔 𝐦𝐦 ∙ 𝟑. 𝟔 𝐦𝐦) ∙ 𝟗 𝐦𝐦
𝟐
𝟒𝟕𝟑. 𝟎𝟒 𝐦𝐦𝟑
((𝟖 𝐦𝐦 ∙ 𝟔. 𝟓𝟕 𝐦𝐦) ∙ 𝟗 𝐦𝐦)
𝟒𝟕𝟑. 𝟎𝟒 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.
4.
Find the volume of wood needed to construct the following side table composed of right rectangular prisms.
Volume of bottom
legs:
𝑽 = 𝟐(𝟖 𝐢𝐧. ∙ 𝟏 𝐢𝐧. ∙ 𝟎. 𝟕𝟓 𝐢𝐧. )
Volume of vertical
legs:
𝑽 = 𝟐(𝟏 𝐢𝐧. ∙ 𝟗. 𝟓 𝐢𝐧. ∙ 𝟎. 𝟕𝟓 𝐢𝐧. )
Volume of tabletop:
= 𝟏𝟐 𝐢𝐧𝟑
= 𝟏𝟒. 𝟐𝟓 𝐢𝐧𝟑
𝑽 = 𝟖 𝐢𝐧.⋅ 𝟔 𝐢𝐧.⋅ 𝟏. 𝟓 𝐢𝐧.
= 𝟕𝟐 𝐢𝐧𝟑
The volume of the table is
𝟏𝟐 𝐢𝐧𝟑 + 𝟏𝟒. 𝟐𝟓 𝐢𝐧𝟑 + 𝟕𝟐 𝐢𝐧𝟑 = 𝟗𝟖. 𝟐𝟓 𝐢𝐧𝟑 .
Lesson 26:
Volume of Composite Three-Dimensional Objects
This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from G7-M6-TE-1.3.0-10.2015
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Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 26
NYS COMMON CORE MATHEMATICS CURRICULUM
5.
7•6
A plastic die (singular for dice) for a game has an edge length of 𝟏. 𝟓 𝐜𝐦. Each face of the cube has the number of
cubic cutouts as its marker is supposed to indicate (i.e., the face marked 𝟑 has 𝟑 cutouts). What is the volume of the
die?
Number of cubic cutouts:
𝟏 + 𝟐 + 𝟑 + 𝟒 + 𝟓 + 𝟔 = 𝟐𝟏
Volume of cutout cubes:
Volume of large cube:
𝑽 = 𝟐𝟏(𝟐 𝐦𝐦)𝟑
𝟑
𝑽 = (𝟏. 𝟓 𝐜𝐦)𝟑
𝑽 = 𝟏𝟔𝟖 𝐦𝐦 = 𝟎. 𝟏𝟔𝟖 𝐜𝐦
𝟑
𝑽 = 𝟑. 𝟑𝟕𝟓 𝐜𝐦𝟑
The total volume of the die is
𝟑. 𝟑𝟕𝟓 𝐜𝐦𝟑 − 𝟎. 𝟏𝟔𝟖 𝐜𝐦𝟑 = 𝟑. 𝟐𝟎𝟕 𝐜𝐦𝟑.
6.
A wooden cube with an edge length of 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
(𝟔 𝐢𝐧. )𝟑 − 𝟔 𝐢𝐧𝟑 − 𝟓 𝐢𝐧𝟑 − 𝟓 𝐢𝐧𝟑 = 𝟐𝟎𝟎 𝐢𝐧𝟑.
7.
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%.
8.
A solid is created by putting together right rectangular prisms. If each of the side lengths is increased 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:
Volume of Composite Three-Dimensional Objects
This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from G7-M6-TE-1.3.0-10.2015
299
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.