## Engage NY Eureka Math 7th Grade Module 3 Lesson 26 Answer Key

### Eureka Math Grade 7 Module 3 Lesson 26 Example Answer Key

Example 1.

The insulated box shown is made from a large cube with a hollow inside that is a right rectangular prism with a square base. The figure on the right is what the box looks like from above.

a. Calculate the volume of the outer box.

Answer:

24 cm × 24 cm × 24 cm = 13,824 cm^{3}

b. Calculate the volume of the inner prism.

Answer:

18 cm × 18 cm × 21 \(\frac{1}{4}\) cm = 6,885 cm^{3}

c. Describe in words how you would find the volume of the insulation.

Answer:

Find the volume of the outer cube and the inner right rectangular prism, and then subtract the two volumes.

d. Calculate the volume of the insulation in cubic centimeters.

Answer:

13,824 cm^{3} – 6,885 cm^{3} = 6,939cm^{3}

e. Calculate the amount of water the box can hold in liters.

Answer:

6939 cm^{3} = 6939 mL = \(\frac{(6939 \mathrm{~mL})}{1000 \frac{\mathrm{mL}}{\mathrm{L}}}\) = 6.939 L

### Eureka Math Grade 7 Module 3 Lesson 26 Exercise Answer Key

Opening Exercise

Explain to your partner how you would calculate the area of the shaded region. Then, calculate the area.

Answer:

Find the area of the outer rectangle, and subtract the area of the inner rectangle.

6 cm × 3 cm – 5 cm × 2 cm = 8 cm^{2}

Exercise 1: Brick Planter Design

You have been asked by your school to design a brick planter that will be used by classes to plant flowers. The planter will be built in the shape of a right rectangular prism with no bottom so water and roots can access the ground beneath. The exterior dimensions are to be 12 ft. × 9 ft. × 2 \(\frac{1}{2}\) ft. The bricks used to construct the planter are 6 in. long, 3 \(\frac{1}{2}\) in. wide, and 2 in. high.

a. What are the interior dimensions of the planter if the thickness of the planter’s walls is equal to the length of the bricks?

Answer:

6 in = \(\frac{1}{2}\) ft.

Interior length:

12 ft. – \(\frac{1}{2}\) ft. – \(\frac{1}{2}\) ft. = 11 ft.

Interior width:

9 ft. – \(\frac{1}{2}\) ft. – \(\frac{1}{2}\) ft. = 8 ft.

Interior dimensions:

11 ft. × 8 ft. × 2 \(\frac{1}{2}\) ft.

b. What is the volume of the bricks that form the planter?

Answer:

Solution 1

Subtract the volume of the smaller interior prism V_S from the volume of the large exterior prism V_L.

V_{Brick} = V_{L} – V_{S}

V_{Brick} = (12 ft. × 9 ft. × 2 \(\frac{1}{2}\) ft.) – (11 ft. × 8 ft. × 2 \(\frac{1}{2}\) ft.)

V_{Brick} = 270 ft^{3} – 220 ft^{3}

V_{Brick} = 50 ft^{3}

Solution 2

The volume of the brick is equal to the area of the base times the height.

B = \(\frac{1}{2}\) ft. × (8 \(\frac{1}{2}\) ft. + 11 \(\frac{1}{2}\) ft. + 8 \(\frac{1}{2}\) ft. + 11 \(\frac{1}{2}\) ft.)

B = \(\frac{1}{2}\) ft. × (40 ft.) = 20 ft^{2}

V = Bh

V = (20 ft^{2} )(2 \(\frac{1}{2}\) ft.) = 50 ft^{3}

c. If you are going to fill the planter \(\frac{3}{4}\) full of soil, how much soil will you need to purchase, and what will be the height of the soil?

Answer:

The height of the soil will be \(\frac{3}{4}\) of 2 \(\frac{1}{2}\)feet.

\(\frac{3}{4}\) (\(\frac{5}{2}\) ft.) = \(\frac{15}{8}\) ft.; The height of the soil will be \(\frac{15}{8}\) ft. (or 1 \(\frac{7}{8}\) ft.).

The volume of the soil in the planter:

V = (11 ft × 8 ft. × \(\frac{15}{8}\) ft.)

V = (11 ft. × 15 ft^{2} ) = 165 ft^{3}

d. How many bricks are needed to construct the planter?

Answer:

P = 2(8 \(\frac{1}{2}\)ft.) + 2(11 \(\frac{1}{2}\)ft.)

P = 17 ft. + 23 ft. = 40 ft.

3 \(\frac{1}{2}\) in. = \(\frac{7}{24}\) ft.

We can then divide the perimeter by the width of each brick in order to determine the number of bricks needed for each layer of the planter.

40 ÷ \(\frac{7}{24}\) = \(\frac{960}{7}\)

40 × \(\frac{24}{7}\) = \(\frac{960}{7}\) ≈ 137.1

Each layer of the planter requires approximately 137.1 bricks.

2 in. = \(\frac{1}{6}\) ft.

The height of the planter, 2 \(\frac{1}{2}\) ft., is equal to the product of the number of layers of brick, n, and the height of each brick, \(\frac{1}{6}\) ft.

2\(\frac{1}{2}\) = l(\(\frac{1}{6}\))

6(2 \(\frac{1}{2}\)) = l

15 = l

There are 15 layers of bricks in the planter.

The total number of bricks, b, is equal to the product of the number of bricks in each layer (\(\frac{960}{7}\)) and the number of layers (15).

b = \(\frac{960}{7}\))(15)

b = \(\frac{14400}{7}\) ≈ 2057.1

It is not reasonable to purchase 0.1 brick; we must round up to the next whole brick, which is 2,058 bricks. Therefore, 2,058 bricks are needed to construct the planter.

e. Each brick used in this project costs $0.82 and weighs 4.5 lb. The supply company charges a delivery fee of $15 per whole ton (2000 lb) over 4000 lb How much will your school pay for the bricks (including delivery) to construct the planter?

Answer:

If the school purchases 2058 bricks, the total weight of the bricks for the planter,

2058(4.5 lb) = 9261 lb

The number of whole tons over 4,000 pounds,

9261 – 4000 = 5261

Since 1 ton = 2000 lb., there are 2 whole tons (4000 lb.) in 5,261 lb.

Total cost = cost of bricks + cost of delivery

Total cost = 0.82(2058) + 2(15)

Total cost = 1687.56 + 30 = 1717.56

The cost for bricks and delivery will be $1,717.56.

f. A cubic foot of topsoil weighs between 75 and 100 lb. How much will the soil in the planter weigh?

Answer:

The volume of the soil in the planter is 165 ft^{3}.

Minimum weight:

Minimum weight = 75 lb(165)

Minimum weight = 12375 lb.

Maximum weight:

Maximum weight = 100 lb(165)

Maximum weight = 16500 lb.

The soil in the planter will weigh between 12,375 lb. and 16,500 lb.

g. If the topsoil costs $0.88 per cubic foot, calculate the total cost of materials that will be used to construct the planter.

Answer:

The total cost of the top soil:

Cost = 0.88(165) = 145.2; The cost of the top soil will be $145.20.

The total cost of materials for the brick planter project:

Cost = (cost of bricks) + (cost of soil)

Cost = $1,717.56 + $145.20

Cost = $1,862.76

The total cost of materials for the brick planter project will be $1,862.76.

Exercise 2: Design a Feeder

You did such a good job designing the planter that a local farmer has asked you to design a feeder for the animals on his farm. Your feeder must be able to contain at least 100,000 cubic centimeters, but not more than 200,000 cubic centimeters of grain when it is full. The feeder is to be built of stainless steel and must be in the shape of a right prism but not a right rectangular prism. Sketch your design below including dimensions. Calculate the volume of grain that it can hold and the amount of metal needed to construct the feeder.

The farmer needs a cost estimate. Calculate the cost of constructing the feeder if \(\frac{1}{2}\) cm thick stainless steel sells for $93.25 per square meter.

Answer:

Answers will vary. Below is an example using a right trapezoidal prism.

This feeder design consists of an open – top container in the shape of a right trapezoidal prism. The trapezoidal sides of the feeder will allow animals easier access to feed at its bottom. The dimensions of the feeder are shown in the diagram.

B = \(\frac{1}{2}\)(b_{1} + b_{2} )h

B = \(\frac{1}{2}\) (100 cm + 80 cm)∙30 cm

B = \(\frac{1}{2}\) (180 cm)∙30 cm

B = 90 cm∙30 cm

B = 2700 cm^{2}

V = Bh

V = (2,700 cm^{2} )(60 cm)

V = 162,000 cm^{3}

The volume of the solid prism is 162,000 cm^{3}, so the volume that the feeder can contain is slightly less, depending on the thickness of the metal used.

The exterior surface area of the feeder tells us the area of metal required to build the feeder.

SA = (LA – A_{top}) + 2B

SA = 60 cm∙(40 cm + 80 cm + 40 cm) + 2(2,700 cm^{2} )

SA = 60 cm(160 cm) + 5,400 cm^{2}

SA = 9,600 cm^{2} + 5,400 cm^{2}

SA = 15,000 cm^{2}

The feeder will require 15,000 cm^{2} of metal.

1 m^{2} = 10,000 cm^{2}, so 15,000 cm^{2} = 1.5 m^{2}

Cost = 93.25(1.5) = 139.875

Since this is a measure of money, the cost must be rounded to the nearest cent, which is $139.88.

### Eureka Math Grade 7 Module 3 Lesson 26 Problem Set Answer Key

Question 1.

A child’s toy is constructed by cutting a right triangular prism out of a right rectangular prism.

a. Calculate the volume of the rectangular prism.

Answer:

10 cm × 10 cm × 12 \(\frac{1}{2}\) cm = 1250 cm^{3}

b. Calculate the volume of the triangular prism.

Answer:

\(\frac{1}{2}\) (5 cm × 2 \(\frac{1}{2}\) cm) × 12 \(\frac{1}{2}\) cm = 78 \(\frac{1}{8}\) cm^{3}

c. Calculate the volume of the material remaining in the rectangular prism.

Answer:

1250 cm^{3} – 78 \(\frac{1}{8}\) cm^{3} = 1171 \(\frac{7}{8}\) cm^{3}

d.

What is the largest number of triangular prisms that can be cut from the rectangular prism?

Answer:

\(\frac{1250 \mathrm{~cm}^{3}}{78 \frac{1}{8} \mathrm{~cm}^{3}}\) = 16

e. What is the surface area of the triangular prism (assume there is no top or bottom)?

Answer:

5.6 cm × 12 \(\frac{1}{2}\) cm + 2 \(\frac{1}{2}\) cm × 12 \(\frac{1}{2}\) cm + 5 cm × 12 \(\frac{1}{2}\) cm = 163 \(\frac{3}{4}\) cm^{2}

Question 2.

A landscape designer is constructing a flower bed in the shape of a right trapezoidal prism. He needs to run three identical square prisms through the bed for drainage.

a. What is the volume of the bed without the drainage pipes?

Answer:

\(\frac{1}{2}\) (14 ft. + 12 ft.) × 3 ft. × 16 ft. = 624 ft^{3}

b. What is the total volume of the three drainage pipes?

Answer:

3(\(\frac{1}{4}\) ft^{2} × 16 ft.) = 12 ft^{3}

c. What is the volume of soil if the planter is filled to 3/4 of its total capacity with the pipes in place?

Answer:

\(\frac{3}{4}\) (624 ft^{3} ) – 12 ft^{3} = 456 ft^{3}

d. What is the height of the soil? If necessary, round to the nearest tenth.

Answer:

\(\frac{456 \mathrm{ft}^{3}}{\frac{1}{2}(14 \mathrm{ft} + 12 \mathrm{ft}) \times 16 \mathrm{ft}}\)≈ 2.2 ft.

e. If the bed is made of 8 ft. × 4 ft. pieces of plywood, how many pieces of plywood will the landscape designer need to construct the bed without the drainage pipes?

Answer:

2(3 \(\frac{1}{4}\) ft. × 16 ft.) + 12 ft. × 16 ft. + 2(\(\frac{1}{2}\) (12 ft. + 14 ft.) × 3 ft.) = 374 ft^{2}

374 ft^{2} ÷ \(\frac{(8 \mathrm{ft} \times 4 \mathrm{ft})}{\text { piece of plywood }}\) = 11.7, or 12 pieces of plywood

f. If the plywood needed to construct the bed costs $35 per 8 ft. × 4 ft. piece, the drainage pipes cost $125 each, and the soil costs $1.25/cubic foot, how much does it cost to construct and fill the bed?

Answer:

\(\frac{\$ 35}{\text { piece of plywood }}\)(12 pieces of plywood) + \(\frac{\$ 125}{\text { pipe }}\) (3 pipes) + \(\frac{\$ 1.25}{f t^{3} \text { soil }}\)(456 ft^{3} soil) = $1,365.00

### Eureka Math Grade 7 Module 3 Lesson 26 Exit Ticket Answer Key

Lawrence is designing a cooling tank that is a square prism. A pipe in the shape of a smaller 2 ft × 2 ft square prism passes through the center of the tank as shown in the diagram, through which a coolant will flow.

a. What is the volume of the tank including the cooling pipe?

Answer:

7 ft. × 3 ft. × 3 ft. = 63 ft^{3}

b. What is the volume of coolant that fits inside the cooling pipe?

Answer:

2 ft. × 2 ft. × 7 ft. = 28 ft^{3}

c. What is the volume of the shell (the tank not including the cooling pipe)?

Answer:

63 ft^{3} – 28 ft^{3} = 35 ft^{3}

d. Find the surface area of the cooling pipe.

Answer:

2 ft. × 7 ft. × 4 = 56 ft^{2}