Lesson 14
NYS COMMON CORE MATHEMATICS CURRICULUM
6•5
Lesson 14: Volume in the Real World
Student Outcomes

Students understand that volume is additive, and they apply volume formulas to determine the volume of
composite solid figures in real-world contexts.

Students apply volume formulas to find missing volumes and missing dimensions.
Lesson Notes
This lesson is a continuation of the three previous lessons, Lessons 11–13, in this module and Grade 5 Module 5 Topics A
and B.
Classwork
Example 1 (6 minutes)
Example 1
a.
𝟏
𝟐
𝟏
𝟖
The area of the base of a sandbox is 𝟗 𝐟𝐭 𝟐. The volume of the sandbox is 𝟕 𝐟𝐭 𝟑. Determine the height of
the sandbox.
MP.1
Students make sense of this problem on their own before discussing.

What information are we given in this problem?


How can we use the information to determine the height?


Note to Teacher:
We have been given the area of the base and the volume.
We know that the area of the base times the height gives the volume.
Since we already have the volume, we can do the opposite and divide to
get the height.
Notice that the number for the volume is less than the number for the area.
What does that tell us about the height?

If the product of the area of the base and the height is less than the area,
we know that the height must be less than 1.
Lesson 14:
Volume in the Real World
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This file derived from G6-M5-TE-1.3.0-10.2015
In these examples, it might be
easier for students to use
common denominators when
dividing and working with
dimensions. Students can use
the invert and multiply rule,
but it may cause more work
and make it harder to see the
relationships.
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Lesson 14
NYS COMMON CORE MATHEMATICS CURRICULUM

6•5
Calculate the height by solving a one step equation.
Volume = Area of the base × height

𝑉 = 𝑏ℎ
1
1
= (9 ) ℎ
8
2
57
19
= ( )ℎ
8
2
57 76
76
76
÷
= ( )ℎ ÷
8
8
8
8
57
=ℎ
76
3
=ℎ
4
7
MP.1
The height of the sandbox is

3
4
ft.
We could also calculate the height using the equation Height = Volume ÷ Area of the base. Solve using this
equation to determine if the height will be the same.

Height = Volume ÷ Area of the base
ℎ =𝑉÷𝑏
1
1
ℎ =7 ÷9
8
2
57 19
ℎ=
÷
8
2
57 76
ℎ=
÷
8
8
57
ℎ=
76
3
ℎ=
4
The height of the sandbox is
b.
3
4
ft.
The sandbox was filled with sand, but after the kids played, some of the sand spilled out. Now, the sand is at
𝟏
a height of 𝐟𝐭. Determine the volume of the sand in the sandbox after the children played in it.
𝟐

What new information have we been given in this problem?

This means that the sandbox is not totally filled. Therefore, the volume of sand used is not the same as
the volume of the sandbox.
Lesson 14:
Volume in the Real World
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Lesson 14
NYS COMMON CORE MATHEMATICS CURRICULUM

6•5
How will we determine the volume of the sand?

To determine the volume of the sand, I use the area of the base of the sandbox, but I use the height of
1
2
ft. instead of the height of the sandbox.
Volume = Area of the base × height
1 2 1
ft × ft.
2
2
19 2 1
Volume =
ft × ft.
2
2
19 3
Volume =
ft
4
3
Volume = 4 ft 3
4
Volume = 9
The volume of the sand is 4
3 3
ft .
4
Example 2 (6 minutes)
Example 2
A special-order sandbox has been created for children to use as an archeological digging area at the zoo. Determine the
volume of the sandbox.
𝟓𝐦
𝟐𝐦
𝟐
𝟏
𝐦
𝟓
𝟏
𝐦
𝟑
𝟑
𝟐 𝐦
𝟒
𝟒
𝟏
𝐦
𝟑
𝟏
𝟐 𝐦
𝟒

Describe this three-dimensional figure.

This figure looks like two rectangular prisms that have been placed together to form one large prism.

I could think of it as a piece on the left and a piece on the right.
MP.7
Lesson 14:
Volume in the Real World
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Lesson 14
NYS COMMON CORE MATHEMATICS CURRICULUM

MP.7

We can find the volume of each piece and then add the volumes together to get the volume of the
entire figure.
Does it matter which way we divide the shape when we calculate the volume?


Or, I could think of it as a piece in front and a piece behind.
How can we determine the volume of this figure?


6•5
Answers will vary.
At this point, you can divide the class in half and have each half determine the volume using one of the
described methods.

If the shape is divided into a figure on the left and a figure on the right, we would have the following:
Volume of prism on the left = 𝑙 𝑤 ℎ.
3
1
𝑉 =2 m×2m× m
4
5
11
1
𝑉=
m×2m× m
4
5
𝑉=

22 3
m
20
Volume of the prism on the right = 𝑙 𝑤 ℎ.
1
1
1
𝑉 =2 m×4 m× m
4
3
5
9
13
1
𝑉 = m×
m× m
4
3
5
117 3
𝑉=
m
60
39 3
𝑉=
m
20
Total volume = volume of left + volume of right
22 3 39 3
m +
m
20
20
61 3
1
Total volume =
m =3
m3
20
20
Total volume =
Lesson 14:
Volume in the Real World
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Lesson 14
NYS COMMON CORE MATHEMATICS CURRICULUM

6•5
If the shape is divided into a figure with a piece in front and piece behind, we have the following:
Volume of the back piece = 𝑙 𝑤 ℎ
1
𝑉 =5m×2m× m
5
𝑉 = 2 m3
Volume of the front piece = 𝑙 𝑤 ℎ
1
1
1
𝑉 =2 m×2 m× m
4
3
5
9
7
1
𝑉 = m× m× m
4
3
5
63 3
3
1 3
𝑉=
m =1
m3 = 1
m
60
60
20

Total volume = volume of back + volume of front
1
Total volume = 2 m3 + 1
m3
20
1 3
Total volume = 3
m
20
What do you notice about the volumes determined in each method?

The volume calculated with each method is the same. It does not matter how we break up the shape.
We still get the same volume.
Exercises (20 minutes)
Students work in pairs. When working with composite figures, have one student solve the problem using one method
and the other solve it another way so they can compare answers.
Exercises
2.
a.
The volume of the rectangular prism is
𝟑𝟔
𝟏𝟓
𝐲𝐝𝟑 . Determine the missing measurement using a one-step
equation.
𝑽 = 𝒃𝒉
𝟑𝟔
𝟒
= ( )𝒉
𝟏𝟓
𝟓
𝟑𝟔 𝟏𝟐
𝟏𝟐
𝟏𝟐
÷
= ( )𝒉 ÷
𝟏𝟓 𝟏𝟓
𝟏𝟓
𝟏𝟓
𝟑𝟔
=𝒉
𝟏𝟐
𝟑=𝒉
𝐀𝐫𝐞𝐚 =
𝟒
𝐲𝐝𝟐
𝟓
𝐡𝐞𝐢𝐠𝐡𝐭 = ?
The height is 𝟑 𝐲𝐝.
Lesson 14:
Volume in the Real World
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Lesson 14
NYS COMMON CORE MATHEMATICS CURRICULUM
b.
𝟒𝟓
The volume of the box is
𝟔
𝐦𝟑 . Determine the area of the base using a one-step equation.
𝑽 = 𝒃𝒉
𝟒𝟓
𝟗
= 𝒃( )
𝟔
𝟐
𝟒𝟓 𝟐𝟕
𝟐𝟕
𝟐𝟕
÷
= 𝒃( ) ÷
𝟔
𝟔
𝟔
𝟔
𝟒𝟓
=𝒃
𝟐𝟕
𝟓
=𝒃
𝟑
The area of the base is
3.
𝟓
𝟑
6•5
𝟗
𝐦
𝟐
𝐦𝟐 .
Marissa’s fish tank needs to be filled with more water.
a.
Determine how much water the tank can hold.
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐞𝐧𝐭𝐢𝐫𝐞 𝐭𝐚𝐧𝐤 = 𝒍 𝒘 𝒉
𝟑
𝟏
𝟑
𝑽 = ( 𝐦) ( 𝐦) ( 𝐦)
𝟒
𝟒
𝟓
𝟗
𝟑
𝑽=
𝐦
𝟖𝟎
b.
𝟑
𝐦
𝟓
𝟑
𝐦
𝟖
Determine how much water is already in the tank.
𝟑
𝐦
𝟒
𝟏
𝐦
𝟒
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐰𝐚𝐭𝐞𝐫 𝐢𝐧 𝐭𝐡𝐞 𝐭𝐚𝐧𝐤 = 𝒍 𝒘 𝒉
𝟑
𝟏
𝟑
𝑽 = ( 𝐦) ( 𝐦) ( 𝐦)
𝟒
𝟒
𝟖
𝟗
𝟑
𝑽=
𝐦
𝟏𝟐𝟖
c.
How much more water is needed to fill the tank?
Height of empty part of tank:
𝒉=
𝟑
𝟑
𝟐𝟒
𝟏𝟓
𝟗
𝐦− 𝐦=
𝐦−
𝐦=
𝐦
𝟓
𝟖
𝟒𝟎
𝟒𝟎
𝟒𝟎
𝐕𝐨𝐥𝐮𝐦𝐞 𝐧𝐞𝐞𝐝𝐞𝐝 𝐭𝐨 𝐟𝐢𝐥𝐥 = 𝒍 𝒘 𝒉
𝟑
𝟏
𝟗
𝑽 = ( 𝐦) ( 𝐦) (
𝐦)
𝟒
𝟒
𝟒𝟎
𝟐𝟕
𝑽=
𝐦𝟑
𝟔𝟒𝟎
Lesson 14:
Volume in the Real World
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Lesson 14
NYS COMMON CORE MATHEMATICS CURRICULUM
4.
6•5
Determine the volume of the composite figures.
a.
𝟑
𝟏𝟐 𝐦
𝟐
𝟗
𝟏
𝐦
𝟐
𝟏
𝐦
𝟒
𝟏
𝐦
𝟑
𝟐𝐦
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐥𝐞𝐟𝐭 𝐩𝐢𝐞𝐜𝐞 = 𝒍 𝒘 𝒉
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐛𝐚𝐜𝐤 𝐫𝐢𝐠𝐡𝐭 = 𝒍 𝒘 𝒉
𝟏
𝑽 = (𝟐 𝐦)(𝟏𝟐 𝐦) (𝟑 𝐦)
𝟒
𝟏𝟑
𝑽 = (𝟐 𝐦)(𝟏𝟐 𝐦) (
𝐦)
𝟒
𝟏
𝟏
𝟏
𝐦) (𝟐 𝐦) (𝟑 𝐦)
𝟐
𝟑
𝟒
𝟏𝟗
𝟕
𝟏𝟑
𝑽=(
𝐦) ( 𝐦) (
𝐦)
𝟐
𝟑
𝟒
𝟏,𝟕𝟐𝟗 𝟑
𝟏
𝑽=
𝐦 = 𝟕𝟐
𝐦𝟑
𝟐𝟒
𝟐𝟒
𝑽 = (𝟗
𝑽 = 𝟕𝟖 𝐦𝟑
𝐓𝐨𝐭𝐚𝐥 𝐯𝐨𝐥𝐮𝐦𝐞 = 𝟕𝟖 𝐦𝟑 + 𝟕𝟐
𝟏
b.
𝟏
𝐟𝐭.
𝟐
𝟏
𝟏
𝐦𝟑 = 𝟏𝟓𝟎
𝐦𝟑
𝟐𝟒
𝟐𝟒
𝟏
𝐟𝐭.
𝟐
𝟏
𝐟𝐭.
𝟐
𝟑
𝐟𝐭.
𝟒
𝟏
𝐟𝐭.
𝟒
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐥𝐨𝐧𝐠 𝐛𝐚𝐜𝐤 𝐩𝐢𝐞𝐜𝐞 = 𝒍 𝒘 𝒉
𝑽 = (𝟏
𝟏
𝟏
𝐟𝐭.
𝟒
𝟏
𝟏
𝟑
𝐟𝐭.) ( 𝐟𝐭.) ( 𝐟𝐭. )
𝟐
𝟐
𝟒
𝟑
𝟏
𝟑
𝑽 = ( 𝐟𝐭.) ( 𝐟𝐭.) ( 𝐟𝐭.)
𝟐
𝟐
𝟒
𝑽=
𝟗
𝐟𝐭 𝟑
𝟏𝟔
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐦𝐢𝐝𝐝𝐥𝐞 𝐩𝐢𝐞𝐜𝐞 = 𝒍 𝒘 𝒉
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐟𝐫𝐨𝐧𝐭 𝐩𝐢𝐞𝐜𝐞 = 𝒍 𝒘 𝒉
𝟑
𝟏
𝟑
𝑽 = ( 𝐟𝐭.) ( 𝐟𝐭.) ( 𝐟𝐭.)
𝟒
𝟐
𝟒
𝑽 = ( 𝐟𝐭.) ( 𝐟𝐭.) ( 𝐟𝐭.)
𝑽=
𝟗
𝐟𝐭 𝟑
𝟑𝟐
𝟏
𝟒
𝑽=
𝟏
𝟒
𝟑
𝟒
𝟑
𝐟𝐭 𝟑
𝟔𝟒
𝐓𝐨𝐭𝐚𝐥 𝐯𝐨𝐥𝐮𝐦𝐞 = 𝐬𝐮𝐦 𝐨𝐟 𝐭𝐡𝐞 𝐭𝐡𝐫𝐞𝐞 𝐯𝐨𝐥𝐮𝐦𝐞𝐬
𝐓𝐨𝐭𝐚𝐥 𝐯𝐨𝐥𝐮𝐦𝐞 =
𝟗
𝟗
𝟑
𝐟𝐭 𝟑 +
𝐟𝐭 𝟑 +
𝐟𝐭 𝟑
𝟏𝟔
𝟑𝟐
𝟔𝟒
𝐓𝐨𝐭𝐚𝐥 𝐯𝐨𝐥𝐮𝐦𝐞 =
𝟑𝟔 𝟑 𝟏𝟖 𝟑
𝟑
𝐟𝐭 +
𝐟𝐭 +
𝐟𝐭 𝟑
𝟔𝟒
𝟔𝟒
𝟔𝟒
𝐓𝐨𝐭𝐚𝐥 𝐯𝐨𝐥𝐮𝐦𝐞 =
𝟓𝟕 𝟑
𝐟𝐭
𝟔𝟒
Lesson 14:
Volume in the Real World
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NYS COMMON CORE MATHEMATICS CURRICULUM
Lesson 14
6•5
Another possible solution:
𝑽 = (𝟏
𝟏
𝟏
𝟑
𝟑
𝟏
𝟑
𝟏
𝟏
𝟑
𝐟𝐭.) ( 𝐟𝐭.) ( 𝐟𝐭.) + ( 𝐟𝐭.) ( 𝐟𝐭.) ( 𝐟𝐭.) + ( 𝐟𝐭.) ( 𝐟𝐭.) ( 𝐟𝐭.)
𝟐
𝟐
𝟒
𝟒
𝟐
𝟒
𝟒
𝟒
𝟒
𝑽=
𝟗
𝟗
𝟑
𝐟𝐭 𝟑 +
𝐟𝐭 𝟑 +
𝐟𝐭 𝟑
𝟏𝟔
𝟑𝟐
𝟔𝟒
𝑽=
𝟑𝟔 𝟑 𝟏𝟖 𝟑
𝟑
𝐟𝐭 +
𝐟𝐭 +
𝐟𝐭 𝟑
𝟔𝟒
𝟔𝟒
𝟔𝟒
𝑽=
𝟓𝟕 𝟑
𝐟𝐭
𝟔𝟒
Closing (5 minutes)
Students take time to share their solutions with the class. Discuss the differences between the types of problems and
how working with volume and the many formulas or methods for solving can help in determining how to get to a
solution.
Exit Ticket (8 minutes)
Lesson 14:
Volume in the Real World
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Lesson 14
NYS COMMON CORE MATHEMATICS CURRICULUM
Name
6•5
Date
Lesson 14: Volume in the Real World
Exit Ticket
1.
Determine the volume of the water that would be needed to fill the rest of the tank.
1
m
2
3
m
4
1
m
2
1
1 m
4
2.
Determine the volume of the composite figure.
5
ft.
8
1
ft.
4
1
ft.
6
1
ft.
3
1
ft.
4
Lesson 14:
Volume in the Real World
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Lesson 14
NYS COMMON CORE MATHEMATICS CURRICULUM
6•5
Exit Ticket Sample Solutions
1.
Determine the volume of the water that would be needed to fill the rest of the tank.
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐭𝐚𝐧𝐤 = 𝒍 𝒘 𝒉
𝟏
𝟏
𝟑
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐭𝐚𝐧𝐤 = (𝟏 𝐦) ( 𝐦) ( 𝐦)
𝟒
𝟐
𝟒
𝟏𝟓 𝟑
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐭𝐚𝐧𝐤 =
𝐦
𝟑𝟐
𝟏
𝐦
𝟐
𝟑
𝐦
𝟒
𝐑𝐞𝐦𝐚𝐢𝐧𝐢𝐧𝐠 𝐰𝐚𝐭𝐞𝐫 𝐧𝐞𝐞𝐝𝐞𝐝 =
2.
𝟏
𝐦
𝟐
𝟏
𝟏 𝐦
𝟒
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐰𝐚𝐭𝐞𝐫 = 𝒍 𝒘 𝒉
𝟏
𝟏
𝟏
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐰𝐚𝐭𝐞𝐫 = (𝟏 𝐦) ( 𝐦) ( 𝐦)
𝟒
𝟐
𝟐
𝟓
𝟏𝟎 𝟑
𝟑
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐰𝐚𝐭𝐞𝐫 =
𝐦 =
𝐦
𝟏𝟔
𝟑𝟐
𝟏𝟓 𝟑 𝟏𝟎 𝟑
𝟓
𝐦 −
𝐦 =
𝐦𝟑
𝟑𝟐
𝟑𝟐
𝟑𝟐
Determine the volume of the composite figure.
𝟓
𝐟𝐭.
𝟖
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐛𝐚𝐜𝐤 𝐩𝐢𝐞𝐜𝐞 = 𝒍 𝒘 𝒉
𝟓
𝟏
𝟏
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐛𝐚𝐜𝐤 𝐩𝐢𝐞𝐜𝐞 = ( 𝐟𝐭.) ( 𝐟𝐭.) ( 𝐟𝐭.)
𝟖
𝟑
𝟒
𝟓
𝟑
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐛𝐚𝐜𝐤 𝐩𝐢𝐞𝐜𝐞 =
𝐟𝐭
𝟗𝟔
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐟𝐫𝐨𝐧𝐭 𝐩𝐢𝐞𝐜𝐞 = 𝒍 𝒘 𝒉
𝟏
𝟏
𝟏
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐟𝐫𝐨𝐧𝐭 𝐩𝐢𝐞𝐜𝐞 = ( 𝐟𝐭.) ( 𝐟𝐭.) ( 𝐟𝐭.)
𝟒
𝟔
𝟒
𝟏
𝟑
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐟𝐫𝐨𝐧𝐭 𝐩𝐢𝐞𝐜𝐞 =
𝐟𝐭
𝟗𝟔
𝐓𝐨𝐭𝐚𝐥 𝐯𝐨𝐥𝐮𝐦𝐞 =
𝟏
𝐟𝐭.
𝟒
𝟏
𝐟𝐭.
𝟔
𝟏
𝐟𝐭.
𝟑
𝟏
𝐟𝐭.
𝟒
𝟓
𝟏
𝟔
𝟐
𝐟𝐭 𝟑 +
𝐟𝐭 𝟑 =
𝐟𝐭 𝟑 =
𝐟𝐭 𝟑
𝟗𝟔
𝟗𝟔
𝟗𝟔
𝟑𝟐
Problem Set Sample Solutions
1.
The volume of a rectangular prism is
𝟐𝟏
𝟏𝟐
𝟑
𝐟𝐭𝟑 , and the height of the prism is 𝐟𝐭. Determine the area of the base.
𝟒
𝑽 = 𝒃𝒉
𝟐𝟏
𝟑
= 𝒃( )
𝟏𝟐
𝟒
𝟐𝟏 𝟗
𝟗
𝟗
÷
= 𝒃( ) ÷
𝟏𝟐 𝟏𝟐
𝟏𝟐
𝟏𝟐
𝟐𝟏
=𝒃
𝟗
The area of the base is
Lesson 14:
𝟐𝟏
𝟗
𝐟𝐭𝟐 OR
𝟕
𝟑
𝐟𝐭𝟐 .
Volume in the Real World
This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from G6-M5-TE-1.3.0-10.2015
209
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 14
NYS COMMON CORE MATHEMATICS CURRICULUM
2.
The volume of a rectangular prism is
𝟏𝟎
𝟐𝟏
6•5
𝟐
𝐟𝐭𝟑 . The area of the base is 𝐟𝐭𝟐 . Determine the height of the rectangular
𝟑
prism.
𝐇𝐞𝐢𝐠𝐡𝐭 = 𝐕𝐨𝐥𝐮𝐦𝐞 ÷ 𝐀𝐫𝐞𝐚 𝐨𝐟 𝐭𝐡𝐞 𝐛𝐚𝐬𝐞
3.
𝐇𝐞𝐢𝐠𝐡𝐭 =
𝟏𝟎 𝟑 𝟐 𝟐
𝐟𝐭 ÷ 𝐟𝐭
𝟐𝟏
𝟑
𝐇𝐞𝐢𝐠𝐡𝐭 =
𝟏𝟎 𝟑 𝟏𝟒 𝟐
𝐟𝐭 ÷
𝐟𝐭
𝟐𝟏
𝟐𝟏
𝐇𝐞𝐢𝐠𝐡𝐭 =
𝟏𝟎
𝟓
𝐟𝐭. OR
𝐟𝐭.
𝟏𝟒
𝟕
Determine the volume of the space in the tank that still needs to be filled with water if the water is
𝟏
𝟑
𝐟𝐭. deep.
𝟐 𝐟𝐭.
𝟏
𝟓 𝐟𝐭.
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐭𝐚𝐧𝐤 = 𝒍 𝒘 𝒉
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐰𝐚𝐭𝐞𝐫 = 𝒍 𝒘 𝒉
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐭𝐚𝐧𝐤 = (𝟓 𝐟𝐭.) (𝟏
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐭𝐚𝐧𝐤 =
4.
𝟐
𝐟𝐭.
𝟑
𝟐
𝐟𝐭.) (𝟐 𝐟𝐭.)
𝟑
𝟓𝟎 𝟑
𝐟𝐭
𝟑
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐰𝐚𝐭𝐞𝐫 = (𝟓 𝐟𝐭.) (𝟏
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐰𝐚𝐭𝐞𝐫 =
𝐕𝐨𝐥𝐮𝐦𝐞 𝐭𝐨 𝐛𝐞 𝐟𝐢𝐥𝐥𝐞𝐝 =
𝟓𝟎 𝟑 𝟐𝟓 𝟑
𝐟𝐭 − 𝐟𝐭
𝟑
𝟗
𝐕𝐨𝐥𝐮𝐦𝐞 𝐭𝐨 𝐛𝐞 𝐟𝐢𝐥𝐥𝐞𝐝 =
𝟏𝟓𝟎 𝟑 𝟐𝟓 𝟑
𝐟𝐭 − 𝐟𝐭
𝟗
𝟗
𝐕𝐨𝐥𝐮𝐦𝐞 𝐭𝐨 𝐛𝐞 𝐟𝐢𝐥𝐥𝐞𝐝 =
𝟏𝟐𝟓 𝟑
𝐟𝐭
𝟗
𝟐
𝟏
𝐟𝐭.) ( 𝐟𝐭.)
𝟑
𝟑
𝟐𝟓 𝟑
𝐟𝐭
𝟗
Determine the volume of the composite figure.
𝟑
𝐦
𝟒
𝟏
𝐦
𝟖
𝟏
𝐦
𝟑
𝟏
𝐦
𝟑
𝟏
𝐦
𝟒
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐛𝐚𝐜𝐤 𝐩𝐢𝐞𝐜𝐞 = 𝒍 𝒘 𝒉
𝟑
𝟒
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐛𝐚𝐜𝐤 𝐩𝐢𝐞𝐜𝐞 = ( 𝐦) (
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐛𝐚𝐜𝐤 𝐩𝐢𝐞𝐜𝐞 =
𝐓𝐨𝐭𝐚𝐥 𝐯𝐨𝐥𝐮𝐦𝐞 =
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐟𝐫𝐨𝐧𝐭 𝐩𝐢𝐞𝐜𝐞 = 𝒍 𝒘 𝒉
𝟏
𝟏
𝐦) ( 𝐦)
𝟖
𝟑
𝟑
𝐦𝟑
𝟗𝟔
𝟏
𝟒
𝟏
𝟖
𝟏
𝟑
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐟𝐫𝐨𝐧𝐭 𝐩𝐢𝐞𝐜𝐞 = ( 𝐦) ( 𝐦) ( 𝐦)
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐟𝐫𝐨𝐧𝐭 𝐩𝐢𝐞𝐜𝐞 =
𝟏 𝟑
𝐦
𝟗𝟔
𝟏 𝟑
𝟑 𝟑
𝟏
𝟒
𝐦 + 𝐦𝟑 =
𝐦𝟑 OR
𝐦
𝟗𝟔
𝟗𝟔
𝟗𝟔
𝟐𝟒
Lesson 14:
Volume in the Real World
This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from G6-M5-TE-1.3.0-10.2015
210
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 14
NYS COMMON CORE MATHEMATICS CURRICULUM
5.
6•5
Determine the volume of the composite figure.
𝑽 = (𝟏 𝐢𝐧.) (𝟏
𝟏 𝐢𝐧.
𝟏
𝟏
𝟏
𝟏
𝐢𝐧.) (𝟏 𝐢𝐧.) + (𝟑 𝐢𝐧.) (𝟐 𝐢𝐧.) ( 𝐢𝐧.)
𝟐
𝟒
𝟐
𝟒
𝟏
𝟑
𝟓
𝟓
𝟏
𝑽 = (𝟏 𝐢𝐧.) ( 𝐢𝐧.) ( 𝐢𝐧.) + (𝟑 𝐢𝐧.) ( 𝐢𝐧.) ( 𝐢𝐧.)
𝟐
𝟒
𝟐
𝟒
𝑽=
𝑽=
𝟏
𝐢𝐧.
𝟐
𝟏
𝐢𝐧.
𝟒
𝟏
𝟏 𝐢𝐧.
𝟒
𝟏𝟓 𝟑 𝟏𝟓 𝟑
𝐢𝐧 +
𝐢𝐧
𝟖
𝟖
𝟐
𝟑𝟎 𝟑
𝟔
𝟑
𝐢𝐧 = 𝟑 𝐢𝐧𝟑 OR 𝟑 𝐢𝐧𝟑
𝟖
𝟖
𝟒
𝟏
𝐢𝐧.
𝟐
𝟑 𝐢𝐧.
6.
𝟏𝐦
𝟑
𝟏
𝐦
𝟐
𝟏
𝟏
𝐦
𝟒
𝟐𝐦
𝟕
a.
Write an equation to represent the volume of the composite figure.
𝑽 = (𝟑
b.
𝟏
𝐦
𝟒
𝟏
𝟏
𝟑
𝟏
𝐦 × 𝟐 𝐦 × 𝟏 𝐦) + (𝟑 𝐦 × 𝟐 𝐦 × 𝟐 𝐦)
𝟐
𝟒
𝟒
𝟒
Use your equation to calculate the volume of the composite figure.
𝟏
𝟏
𝟑
𝟏
𝐦 × 𝟐 𝐦 × 𝟏 𝐦) + (𝟑 𝐦 × 𝟐 𝐦 × 𝟐 𝐦)
𝟐
𝟒
𝟒
𝟒
𝟕
𝟐
𝟓
𝟏𝟓
𝟐
𝟗
𝑽 = ( 𝐦 × 𝐦 × 𝐦) + (
𝐦 × 𝐦 × 𝐦)
𝟐
𝟏
𝟒
𝟒
𝟏
𝟒
𝟕𝟎 𝟑 𝟐𝟕𝟎 𝟑
𝑽=
𝐦 +
𝐦
𝟖
𝟏𝟔
𝟕𝟎 𝟑 𝟏𝟑𝟓 𝟑
𝑽=
𝐦 +
𝐦
𝟖
𝟖
𝟐𝟎𝟓 𝟑
𝑽=
𝐦
𝟖
𝟓
𝑽 = 𝟐𝟓 𝐦𝟑
𝟖
𝑽 = (𝟑
Lesson 14:
Volume in the Real World
This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from G6-M5-TE-1.3.0-10.2015
211
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.