Tag: Eureka Math Grade 6

Engage NY Eureka Math Grade 6 Module 5 Lesson 17 Answer Key

Eureka Math Grade 6 Module 5 Lesson 17 Opening Exercise Answer Key

Opening Exercise:

a. Write a numerical equation for the area of the figure below. Explain and identify different parts of the figure.

i.
Answer:
A = \(\frac{1}{2}\) (14 cm)(12 cm) = 84cm²
14 cm represents the bose of the figure because 5 cm + 9 cm = 14 cm, and 12 cm represents the altitude of the figure because it forms a right angle with the base.

ii. How would you write an equation that shows the area of a triangle with base b and height h?
Answer:
A = \(\frac{1}{2}\) bh

b. Write a numerical equation for the area of the figure below. Explain and identify different parts of the figure.

Answer:
A = (28 ft.)(18 ft) = 504 ft²
28 ft. represents the base of the rectangle, 18 ft. and 18 ft. represents the height of the rectangle.

ii. How would you write an equation that shows the area of a rectangle with base b and height h?
Answer:
A = bh.

Eureka Math Grade 6 Module 5 Lesson 17 Example Answer Key

Example 1:

Use the net to calculate the surface area of the figure. (Note: all measurements are in centimeters.)

Answer:
→ When you are calculating the area of a figure, what are you finding?
The area of a figure is the amount of space inside a two-dimensional figure.

→ The surface area is similar to the area, but the surface area is used to describe three-dimensional figures. What do you think is meant by the surface area of a solid?
The surface area of a three-dimensional figure is the area of each face added together.

→ What type of figure does the net create? How do you know?
It creates a rectangular prism because there are six rectangular faces.

If the boxes on the grid paper represent a 1 cm × 1 cm box, label the dimensions of the net.

→ The surface area of a figure is the sum of the areas of all faces. Calculate the area of each face, and record this value inside the corresponding rectangle.

→ In order to calculate the surface area, we have to find the sum of the areas we calculated since they represent the area of each face. There are two faces that have an area of 4 cm² and four faces that have an area of 2 cm². How can we use these areas to write a numerical expression to show how to calculate the surface area of the net?
The numerical expression to calculate the surface area of the net would be
(1 cm × 2 cm) + (1 cm × 2 cm) + (1 cm × 2 cm) + (1 cm × 2 cm) + (2 cm × 2 cm)+ (2 cm × 2 cm).

→ Write the expression more compactly, and explain what each part represents on the net.
4(1 cm × 2 cm) + 2(2 cm × 2 cm)
The expression means there are 4 rectangles that have dimensions 1 cm × 2 cm on the net and 2 rectangles that have dimensions 2 cm × 2 cm on the net.

→ What is the surface area of the net?
The surface area of the net is 16 cm².

Example 2:

Use the net to write an expression for surface area. (Note: all measurements are in square feet.)

Answer:
→ What type of figure does the net create? How do you know?
It creates a square pyramid because one face is a square and the other four faces are triangles.

→ If the boxes on the grid paper represent a 1 ft. × 1 ft. square, label the dimensions of the net.

→ How many faces does the rectangular pyramid have?
5

→ Knowing the figure has 5 faces, use the knowledge you gained in Example ito calculate the surface area of the rectangular pyramid.
Area of Base: 3 ft. × 3 ft. = 9 ft²
Area of Triangles: \(\frac{1}{2}\) × 3 ft. × 2 ft. = 3 ft²
Surface Area: 9 ft² + 3 ft² + 3 ft² + 3 ft² + 3 ft² = 21 ft²

Eureka Math Grade 6 Module 5 Lesson 17 Exercise Answer Key

Exercises:

Name the solid the net would create, and then write an expression for the surface area. Use the expression to determine the surface area. Assume that each box on the grid paper represents a 1 cm × 1 cm square. Explain how the expression represents the figure.

Exercise 1.

Answer:
Name of Shape: Rectangular Pyramid, but more specifically a Square Pyramid
Surface Area: 4 cm × 4 cm + 4(\(\frac{1}{2}\) × 4 cm × 3 cm)
Work: 16 cm² + 4(6 cm²) = 40 cm²
The surface area is 40 cm². The figure is made up of a square base that measures 4 cm × 4 cm and four triangles, each with a base of 4 cm and a height of 3 cm.

Exercise 2.

Answer:
Name of Shape: Rectangular Prism
Surface Area: 2(5 cm × 5 cm) + 4(5 cm × 2 cm)
Work: 2(25 cm² ) + 4(10 cm²) = 90 cm²
The surface area is 90 cm². The figure has 2 square faces, each of which measures 5 cm × 5 cm, and 4 rectangular faces, each of which measures 5 cm × 2 cm.

Exercise 3.

Answer:
Name of Shape: Rectangular Pyramid
SurfaceArea: 3 cm × 4 cm + 2(\(\frac{1}{2}\) × 4 cm × 4 cm)+ 2(\(\frac{1}{2}\) × 4 cm × 3 cm)
Work: 12 cm² + 2(8 cm²) + 2(6 cm²) = 40 cm²
The surface area is 40 cm². The figure has 1 rectangular base that measures 3 cm × 4 cm, 2 triangular faces, each with a bose of 4 cm and a height of 4 cm, and 2 other triangular faces, each with a base of 3 cm and a height of 4 cm.

Exercise 4.

Answer:
Name of Shape: Rectangular Prism
Surface Area: 2(6 cm × 5 cm) + 2(5 cm × 1 cm) + 2(6 cm × 1 cm)
Work: 2(30 cm²) + 2(5 cm²) + 2(6 cm²) = 82 cm²
The surface area is 82 cm². The figure has two 6 cm × 5 cm rectangular faces, two 5 cm × 1 cm rectangular faces, and two 6 cm × 1 cm rectangular faces.

Eureka Math Grade 6 Module 5 Lesson 17 Problem Set Answer Key

Name the shape, and write an expression for surface area. Calculate the surface area of the figure. Assume each box on the grid paper represents a 1 ft. × 1 ft. square.

Question 1.

Answer:
Name of Shape: Rectangular Prism
SurfaceArea: (2 ft. × 4 ft.) + (2 ft. × 4 ft.)+ (4 ft. × 7 ft.) + (4 ft. × 7 ft.) + (7 ft. × 2 ft.)+ (7 ft. × 2 ft.)
Work: 2(2 ft. × 4 ft.) + 2(4 ft. × 7 ft.) + 2(7 ft. × 2 ft.)
= 16 ft² + 56 ft² + 28 ft²
= 100 ft²

Question 2.

Answer:
Name of Shape: Rectangular Pyramid
SurfaceArea: (2 ft. × 5 ft.) + (\(\frac{1}{2}\) × 2ft. × 4ft.)+ (\(\frac{1}{2}\) × 2 ft. × 4ft.) + (\(\frac{1}{2}\) × 5 ft. × 4ft.) + \(\frac{1}{2}\) × 5ft. × 4 ft.)
Work: 2 ft. × 5 ft. + 2(\(\frac{1}{2}\) × 2 ft. × 4 ft.) + 2(\(\frac{1}{2}\) × 5 ft. × 4ft.)
= 10 ft² + 8 ft² + 20 ft² = 38 ft²

Explain the error in each problem below. Assume each box on the grid paper represents a 1 m × 1 m square.

Question 3.

Name of Shape: Rectangular Pyramid, but more specifically a Square Pyramid
Area of Base: 3m × 3m = 9m²
Area of Triangles: 3 m × 4m = 12 m²
SurfaceArea: 9m² + 12m² + 12m² + 12m² + 12m² = 57m²
Answer:
The error in the solution is the area of the triangles. In order to calculate the correct area of the triangles, you must use the correct formula A = \(\frac{1}{2}\)bh. Therefore, the area of each triangle would be 6 m² and not 12 m².

Question 4.

Name of Shape: Rectangular Prism or, more specifically. a Cube
Area of Faces: 3m × 3m = 9m²
Surface Area: 9 m² + 9 m² + 9 m² + 9 m² + 9 m² = 45m²
Answer:
The surface area is incorrect because the student did not find the sum of all 6 faces. The solution is shown above only calculates the sum of 5 faces. Therefore, the correct surface area should be 9 m² + 9 m² + 9 m² + 9 m² + 9 m² + 9m² = 54m² and not 45m².

Question 5.
Sofia and Ella are both writing expressions to calculate the surface area of a rectangular prism. However, they wrote different expressions.

a. Examine the expressions below, and determine if they represent the same value. Explain why or why not. Sofia’s Expression:
(3 cm × 4 cm) + (3 cm × 4 cm) + (3 cm × 5 cm) + (3 cm × 5 cm) + (4 cm × 5 cm) + (4 cm × 5 cm)

Ella’s Expression:
2(3 cm × 4 cm) + 2(3 cm × 5 cm) + 2(4 cm × 5 cm)
Answer:
Sofia’s and Ella’s expressions are the same, but Ella used the distributive property to make her expression more compact than Sofia’s.

b. What fact about the surface area of a rectangular prism does Ella’s expression show more clearly than Sofia’s?
Answer:
A rectangular prism is composed of three pairs of sides with identical areas.

Eureka Math Grade 6 Module 5 Lesson 17 Exit Ticket Answer Key

Question 1.
Name the shape, and then calculate the surface area of the figure. Assume each box on the grid paper represents a 1 in. × 1 in. square.

Answer:
Name of Shape: Rectangular Pyramid
Area of Base: 5 in. × 4 in. = 20 in²
Area of Triangles: \(\frac{1}{2}\) × 4 in. × 4 in. = 8 in², \(\frac{1}{2}\) × 5 in. × 4 in. = 10 in²
SurfaceArea: 20 in² +8 in² + 8in² + 10 in² + 10 in² = 56 in²

Eureka Math Grade 6 Module 5 Lesson 17 Addition and Subtraction Equations Answer Key

Addition and Subtraction Equations – Round 1:

Directions: Find the value of m in each equation.

Question 1.
m + 4 = 11
Answer:
m = 7

Question 2.
m + 2 = 5
Answer:
m = 3

Question 3.
m + 5 = 8
Answer:
m = 3

Question 4.
m – 7 = 10
Answer:
m = 17

Question 5.
m – 8 = 1
Answer:
m = 9

Question 6.
m – 4 = 2
Answer:
m = 6

Question 7.
m + 12 = 34
Answer:
m = 22

Question 8.
m + 25 = 45
Answer:
m = 20

Question 9.
m + 43 = 89
Answer:
m = 46

Question 10.
m – 20 = 31
Answer:
m = 51

Question 11.
m – 13 = 34
Answer:
m = 47

Question 12.
m – 45 = 68
Answer:
m = 113

Question 13.
m + 34 = 41
Answer:
m = 7

Question 14.
m + 29 = 52
Answer:
m = 23

Question 15.
m + 37 = 61
Answer:
m = 24

Question 16.
m – 43 = 63
Answer:
m = 106

Question 17.
m – 21 = 40
valuem = 61

Question 18.
m – 54 = 37
Answer:
m = 91

Question 19.
4 + m = 9
Answer:
m = 5

Question 20.
6 + m = 13
Answer:
m = 7

Question 21.
2 + m = 31
Answer:
m = 29

Question 22.
15 = m + 11
Answer:
m = 4

Question 23.
24 = m + 13
Answer:
m = 11

Question 24.
32 = m + 28
Answer:
m = 4

Question 25.
4 = m – 7
Answer:
m = 11

Question 26.
3 = m – 5
Answer:
m = 8

Question 27.
12 = m – 14
Answer:
m = 26

Question 28.
23.6 = m – 7.1
Answer:
m = 30.7

Question 29.
14.2 = m – 33.8
Answer:
m = 48

Question 30.
2.5 = m -41.8
Answer:
m = 44.3

Question 31.
64.9 = m + 23.4
Answer:
m = 41.5

Question 32.
72.2 = m + 38.7
Answer:
m = 33.5

Question 33.
1.81 = m – 15.13
Answer:
m = 16.94

Question 34.
24.68 = m – 56.82
Answer:
m = 81.5

Addition and Subtraction Equations – Round 2:

Directions: Find the value of m in each equation.

Question 1.
m + 2 = 7
Answer:
m = 5

Question 2.
m + 4 = 10
Answer:
m = 6

Question 3.
m + 8 = 15
Answer:
m = 7

Question 4.
m + 7 = 23
Answer:
m = 16

Question 5.
m + 12 = 16
Answer:
m = 4

Question 6.
m – 5 = 2
Answer:
m = 7

Question 7.
m – 3 = 8
Answer:
m = 11

Question 8.
m – 4 = 12
Answer:
m = 16

Question 9.
m – 14 = 45
Answer:
m = 59

Question 10.
m + 23 = 40
Answer:
m = 17

Question 11.
m + 13 = 31
Answer:
m = 18

Question 12.
m – 23 = 48
Answer:
m = 25

Question 13.
m + 38 = 52
Answer:
m = 14

Question 14.
m – 14 = 27
Answer:
m = 41

Question 15.
m – 23 = 35
Answer:
m = 58

Question 16.
m – 17 = 18
Answer:
m = 35

Question 17.
m – 64 = 1
Answer:
m = 65

Question 18.
6 = m + 3
Answer:
m = 3

Question 19.
12 = m + 7
Answer:
m = 5

Question 20.
24 = m + 16
Answer:
m = 8

Question 21.
13 = m + 9
Answer:
m = 4

Question 22.
32 = m – 3
Answer:
m = 35

Question 23.
22 = m – 12
Answer:
m = 34

Question 24.
34 = m – 10
Answer:
m = 44

Question 25.
48 = m + 29
Answer:
m = 19

Question 26.
21 = m + 17
Answer:
m = 4

Question 27.
52 = m + 37
Answer:
m = 15

Question 28.
\(\frac{6}{7}\) + m = \(\frac{4}{7}\)
Answer:
m = \(\frac{2}{7}\)

Question 29.
\(\frac{2}{3}\) = m – \(\frac{5}{3}\)
Answer:
m = \(\frac{7}{3}\)

Question 30.
\(\frac{1}{4}\) – m = \(\frac{8}{3}\)
Answer:
m = \(\frac{35}{12}\)

Question 31.
\(\frac{5}{6}\) = m – \(\frac{7}{12}\)
Answer:
m = \(\frac{17}{12}\)

Question 32.
\(\frac{7}{8}\) = m – \(\frac{5}{12}\)
Answer:
m = \(\frac{31}{24}\)

Question 33.
\(\frac{7}{6}\) + m = \(\frac{16}{3}\)
Answer:
m = \(\frac{25}{6}\)

Question 34.
\(\frac{1}{3}\) + m = \(\frac{13}{15}\)
Answer:
m = \(\frac{8}{15}\)

Engage NY Eureka Math 6th Grade Module 3 Lesson 19 Answer Key

Eureka Math Grade 6 Module 3 Lesson 19 Exercise Answer Key

Exploratory Challenge

The Length of a Line Segment is the Distance Between its End Points

Exercise 1.
Locate and label (4, 5) and (4, – 3). Draw the line segment between the end points given on the coordinate plane. How long is the line segment that you drew? Explain.

Answer:
The length of the line segment is also 8 units. I found that the distance between (4, – 3) and (4, 5) is 8 units. Because the end points are on opposite sides of zero, I added the absolute values of the second coordinates together, so the distance from end to end is 8 units.

Exercise 2.
Draw a horizontal line segment starting at (4, – 3) that has a length of 9 units. What are the possible coordinates of the other end point of the line segment? (There is more than one answer.)
Answer:
(- 5, – 3) or (13, – 3)

Which point did you choose to be the other end point of the horizontal line segment? Explain how and why you chose that point. Locate and label the point on the coordinate grid.
Answer:
The other end point of the horizontal line segment is (- 5, – 3). I chose this point because the other option, (13, – 3), is located off of the given coordinate grid.
Note: Students may choose the end point (13, – 3), but they must change the number scale of the x-axis to do so.

Exercise 3.
Extending Lengths of Line Segments to Sides of Geometric Figures
The two line segments that you have just drawn could be seen as two sides of a rectangle. Given this, the end points of the two line segments would be three of the vertices of this rectangle.
a. Find the coordinates of the fourth vertex of the rectangle. Explain how you find the coordinates of the fourth vertex using absolute value.
Answer:
The fourth vertex is (- 5, 5). The opposite sides of a rectangle are the same length, so the length of the vertical side starting at (- 5, – 3) has to be 8 units long. Also, the side from (- 5, – 3) to the remaining vertex is a vertical line, so the end points must have the same first coordinate. |- 3| = 3, and 8 – 3 = 5, so the remaining vertex must be five units above the x-axis.
Note: Students can use a similar argument using the length of the horizontal side starting at (4, 5), knowing it has to be 9 units long.

b. How does the fourth vertex that you found relate to each of the consecutive vertices in either direction?
Explain.
The fourth vertex has the same first coordinate as (- 5, – 3) because they are the end points of a vertical line segment. The fourth vertex has the same second coordinate as (4, 5) since they are the end points of a horizontal line segment.

c. Draw the remaining sides of the rectangle.
Answer:

Using Lengths of Sides of Geometric Figures to Solve Problems

Exercise 4.
Using the vertices that you have found and the lengths of the line segments between them, find the perimeter of the rectangle.
Answer:
8 + 9 + 8 + 9 = 34; the perimeter of the rectangle is 34 units.

Exercise 5.
Find the area of the rectangle.
Answer:
9 × 8 = 72; the area of the rectangle is 72 units².

Exercise 6.
Draw a diagonal line segment through the rectangle with opposite vertices for end points. What geometric figures are formed by this line segment? What are the areas of each of these figures? Explain.
Answer:
The diagonal line segment cuts the rectangle into two right triangles. The areas of the triangles are 36 units² each because the triangles each make up half of the rectangle, and half of 72 is 36.

Extension (If time allows): Line the edge of a piece of paper up to the diagonal in the rectangle. Mark the length of the diagonal on the edge of the paper. Align your marks horizontally or vertically on the grid, and estimate the length of the diagonal to the nearest integer. Use that estimation to now estimate the perimeter of the triangles.
Answer:
The length of the diagonal is approximately 12 units, and the perimeter of each triangle is approximately 29 units.

Exercise 7
Construct a rectangle on the coordinate plane that satisfies each of the criteria listed below. Identify the coordinate of each of its vertices.
→ Each of the vertices lies in a different quadrant.
→ Its sides are either vertical or horizontal.
→ The perimeter of the rectangle is 28 units.
Answers will vary. The example to the right shows a rectangle with side lengths 10 and 4 units. The coordinates of the rectangle’s vertices are (- 6,3), (4, 3), (4, – 1), and (- 6, – 1).

Using absolute value, show how the lengths of the sides of your rectangle provide a perimeter of 28 units.
Answer:
|- 6| = 6, |4| = 4, and 6 + 4 = 10, so the width of my rectangle is 10 units.
|3| = 3, |-1| = 1, and 3 + 1 = 4, so the height of my rectangle is 4 units.
10 + 4 + 10 + 4 = 28, so the perimeter of my rectangle is 28 units.

Eureka Math Grade 6 Module 3 Lesson 19 Problem Set Answer Key

Question 1.
One end point of a line segment is (-3, -6). The length of the line segment is 7 units. Find four points that could serve as the other end point of the given line segment.
Answer:
(- 10, – 6); (4, – 6); (- 3, 1); (- 3, – 13)

Question 2.
Two of the vertices of a rectangle are (1, – 6) and (- 8, – 6). If the rectangle has a perimeter of 26 units, what are the coordinates of its other two vertices?
Answer:
(1, – 2) and (- 8, – 2), or (1, – 10) and (- 8, – 10)

Question 3.
A rectangle has a perimeter of 28 units, an area of 48 square units, and sides that are either horizontal or vertical. If one vertex is the point (- 5, – 7) and the origin is In the interior of the rectangle, find the vertex of the rectangle that is opposite (- 5, – 7).
Answer:
(1, 1)

Eureka Math Grade 6 Module 3 Lesson 19 Exit Ticket Answer Key

Question 1.
The coordinates of one end point of a line segment are (- 2, – 7). The line segment is 12 units long. Give three
possible coordinates of the line segment’s other end point.
Answer:
(10, – 7); (- 14, – 7); (- 2, 5); (- 2, – 19)

Question 2.
Graph a rectangle with an area of 12 units² such that its vertices lie in at least two of the four quadrants in the coordinate plane. State the lengths of each of the sides, and use absolute value to show how you determined the lengths of the sides.
Answer:
Answers will vary. The rectangle can have side lengths of 6 and 2 or 3 and 4. A sample is provided on the grid on the right. 6 × 2 = 12

Eureka Math Grade 6 Module 3 Lesson 19 Opening Exercise Answer Key

Question 1.
In the coordinate plane, find the distance between the points using absolute value.

Answer:
The distance between the points is 8 units. The points have the same first coordinates and, therefore, lie on the same vertical line. |- 3| = 3, and |5| = 5, and the numbers lie on opposite sides of 0, so their absolute values are added together; 3 + 5 = 8. We can check our answer by just counting the number of units between the two points.

Engage NY Eureka Math 6th Grade Module 4 Lesson 1 Answer Key

Eureka Math Grade 6 Module 4 Lesson 1 Exercise Answer Key

Exercise 1.
Predict what will happen when a tape diagram has a large number of squares, some squares are removed, and then the same amount of squares are added back on.
Answer:
Possible answer: When some squares are removed from a tape diagram, and then the same amount of squares are added back on, the tape diagram will end up with the same amount of squares that it started with.

Exercise 2.
Build a tape diagram with 10 squares.
a. Remove six squares. Write an expression to represent the tape diagram.
Answer:
10 – 6

b. Add six squares onto the tape diagram. Alter the original expression to represent the current tape diagram.
Answer:
10 – 6 + 6

c. Evaluate the expression.
Answer:
10

Exercise 3.
Write an equation, using variables, to represent the identities we demonstrated with tape diagrams.
Answer:
Possible answer: w – x + x = w

Exercise 4.
Using your knowledge of identities, fill In each of the blanks.
a. 4 + 5 – ______ = 4
Answer:
5

b. 25 – _____ + 10 = 25
Answer:
10

C. ______ + 16 – 16 = 45
Answer:
45

d. 56 – 20 + 20 = ______
Answer:
56

Exercise 5.
Using your knowledge of identities, fill In each of the blanks.
a. a + b – ______ = a
Answer:
b

b. c – d + d = ______
Answer:
C

c. e + _______ – f = e
Answer:
f

d. ________ – h + h = g
Answer:
g

Eureka Math Grade 6 Module 4 Lesson 1 Problem Set Answer Key

Question 1.
Fill in each blank.
a. ____ +15 – 15 = 21
21

b. 450 – 230 + 230 = ____
Answer:
450

c. 1289 – ______ + 856 = 1289
Answer:
856

Question 2.
Why are the equations w – x + x = w and w + x – x = w called identities?
Answer:
Possible answer: These equations are called identities because the variables can be replaced with any numbers, and after completing the operations, I returned to the original value.

Eureka Math Grade 6 Module 4 Lesson 1 Exit Ticket Answer Key

Question 1.
Draw a series of tape diagrams to represent the following number sentences.
a. 3 + 5 – 5 = 3
Answer:

b. 8 – 2 + 2 = 8
Answer:

Question 2.
Fill in each blank.
a. 65+ _____ – 15 = 65
Answer:
15

b. ______ + g – g = k
Answer:
k

c. a + b – _______ = a
Answer:
b

d. 367 – 93 + 93 = _________
Answer:
367

Eureka Math Grade 6 Module 4 Lesson 1 Opening Exercise Answer Key

a. Draw a tape diagram to represent the following expression: 5 + 4.
Answer:

b. Write an expression for each tape diagram.
i.

Answer:

ii.

Answer:

Eureka Math Grade 6 Module 4 Lesson 1 Multiplication of Decimals Answer Key

Progression of Exercises

Question 1.
0.5 × 0.5 =
Answer:
0.25

Question 2.
0.6 × 0.6 =
Answer:
0.36

Question 3.
0.7 × 0.7 =
Answer:
0.49

Question 4.
0.5 × 0.6 =
Answer:
0.3

Question 5.
1.5 × 1.5 =
Answer:
2.25

Question 6.
2.5 × 2.5 =
Answer:
6.25

Question 7.
0.25 × 0.25 =
Answer:
0. 0625

Question 8.
0.1 × 0.1 =
Answer:
0.01

Question 9.
0.1 × 123.4 =
Answer:
12.34

Question 10.
0.01 × 123.4 =
Answer:
1.234

Engage NY Eureka Math Grade 6 Module 5 Lesson 16 Answer Key

Eureka Math Grade 6 Module 5 Lesson 16 Exercise Answer Key

Opening Exercise:

Question 1.
Sketch the faces in the area below. Label the dimensions.

Answer:
Display this graphic using a document camera or other device.

Eureka Math Grade 6 Module 5 Lesson 16 Exploratory Challenge Answer Key

Exploratory Challenge 1: Rectangular Prisms

a. Use the measurements from the solid figures to cut and arrange the faces into a net. (Note: All measurements are in centimeters.)

Answer:
One possible configuration of rectangles is shown here:

b. A juice box measures 4 inches high, 3 inches long, and 2 inches wide. Cut and arrange all 6 faces into a net. (Note: All measurements are in inches.)

Answer:
One possible configuration of faces is shown here:

c. Challenge: Write a numerical expression for the total area of the net for part (b). Explain each term in your expression.
Answer:
Possible answer: 2(2 in. × 3 in.) + 2(2 in. × 4 in.) + 2(3 in. × 4 in.). There are two sides that have dimensions 2 in. by 3 in., two sides that are 2 in. by 4 in., and two sides that are 3 in. by 4 in.

Exploratory Challenge 2: Triangular Prisms

a. Use the measurements from the triangular prism to cut and arrange the faces Into a net. (Note: All measurements are in inches.)

Answer:
One possible configuration of rectangles and triangles is shown here:

Exploratory Challenge 3: Pyramids

Pyramids are named for the shape of the base.

a. Use the measurements from this square pyramid to cut and arrange the faces into a net. Test your net to be sure it folds into a square pyramid.

Answer:
One possible configuration of square and triangles is shown below. (Note: all measurements are in centimeters.)

b. A triangular pyramid that has equilateral triangles for faces is called a tetrahedron. Use the measurements from this tetrahedron to cut and arrange the faces into a net.
All edges are 4 in. in length.

Answer:
One possible configuration of triangles is shown below. (Note: All measurements are in inches.)

Eureka Math Grade 6 Module 5 Lesson 16 Problem Set Answer Key

Question 1.
Sketch and label the net of the following solid figures, and label the edge lengths.

a. A cereal box that measures 13 inches high, 7 inches long, and 2 inches wide
Answer:
One possible configuration of faces is shown below. (Note: all measurements are in inches.)

b. A cubic gift box that measures 8 cm on each edge.
Answer:
One possible configuration of faces is shown here:

All edges are 8 cm in length.

c. challenge: Write a numerical expression for the total area of the net in part (b). Tell what each of the terms in your expression means.
Answer:
6(8 cm × 8 cm) or
(8 cm × 8 cm) + (8 cm × 8 cm) + (8 cm × 8 cm) + (8 cm × 8 cm) + (8 cm × 8 cm) + (8 cm × 8 cm). There are 6 faces in the cube, and each has dimensions 8 cm by 8 cm.

Question 2.
This tent is shaped like a triangular prism. It has equilateral bases that measure 5 feet on each side. The tent is 8 feet long. Sketch the net of the tent, and label the edge lengths.

Answer:
Possible net:

Question 3.
The base of a table ¡s shaped like a square pyramid. The pyramid has equilateral faces that measure 25 inches on each side. The base is 25 inches long. Sketch the net of the table base, and label the edge lengths.
Answer:
Possible net:

Question 4.
The roof of a shed is in the shape of a triangular prism. It has equilateral bases that measure 3 feet on each side. The length of the roof is 10 feet. Sketch the net of the roof, and label the edge lengths.
Answer:
Possible net:

Eureka Math Grade 6 Module 5 Lesson 16 Exit Ticket Answer Key

Question 1.
Sketch and label a net of this pizza box. It has a square top that measures 16 inches on a side, and the height is 2 inches. Treat the box as a prism, without counting the interior flaps that a pizza box usually has.

Answer:
One possible configuration of faces is shown below. (Note: All measurements are in inches.)

Engage NY Eureka Math Grade 6 Module 5 Lesson 15 Answer Key

Eureka Math Grade 6 Module 5 Lesson 15 Exercise Answer Key

Exercise: Cube

Exercise 1.
Nets are two-dimensional figures that can be folded into three-dimensional solids. Some of the drawings below are nets of a cube. Others are not cubed nets; they can be folded, but not into a cube.

a. Experiment with the larger cut-out patterns provided. Shade in each of the figures above that can fold into a cube.
Answer:

b. Write the letters of the figures that can be folded into a cube.
Answer:
A, B, C, E, G, I, L, M, O, P, and T.

c. Write the letters of the figures that cannot be folded into a cube.
Answer:
D, F, H, J, K, N, Q, R, and S.

Eureka Math Grade 6 Module 5 Lesson 15 Problem Set Answer Key

Question 1.
Match the following nets to the picture of its solid. Then, write the name of the solid.

Answer:

Question 2.
Sketch a net that can fold into a cube.
Answer:
Sketches will vary but should match one of the shaded ones from earlier in the lesson. Here are the 11 possible nets for a cube.

Question 3.
Below are the nets for a variety of prisms and pyramids. Classify the solids as prisms or pyramids, and identify the shape of the base(s). Then, write the name of the solid.

Answer:

Eureka Math Grade 6 Module 5 Lesson 15 Exit Ticket Answer Key

Question 1.
What is a net? Describe it in your own words.
Answer:
Answers will vary but should capture the essence of the definition used in this lesson. A net is a two-dimensional figure that can be folded to create a three-dimensional solid.

Question 2.
Which of the following can fold to make a cube? Explain how you know.

Answer:
Evidence for claims will vary.

Engage NY Eureka Math Grade 6 Module 5 Lesson 14 Answer Key

Eureka Math Grade 6 Module 5 Lesson 14 Example Answer Key

Example 1:

a. The area of the base of a sandbox is 9\(\frac{1}{2}\) ft². The volume of the sandbox is 7\(\frac{1}{8}\) ft³. Determine the height of the sandbox.

Answer:
→ What information are we given in this problem?
We have been given the area of the base and the volume.

→ How can we use the information to determine the height?
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.

→ Calculate the height by solving a one-step equation.
Volume = Area of the base × height
V = bh

The height of the sandbox is \(\frac{3}{4}\) 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

The height of the sandbox is \(\frac{3}{4}\) ft.

b. The sandbox was filled with sand, but after the kids played, some of the sand spilled out. Now, the sand is at a height of \(\frac{1}{2}\) ft. Determine the volume of the sand in the sandbox after the children played in it.
Answer:

→ 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.

→ 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 \(\frac{1}{2}\)ft. instead of the height of the sandbox.
Volume = Area of the base × height
Volume = 9\(\frac{1}{2}\) ft² × \(\frac{1}{2}\) ft.
Volume = \(\frac{19}{2}\) ft² × \(\frac{1}{2}\) ft.
Volume = \(\frac{19}{4}\) ft³
Volume = 4\(\frac{3}{4}\) ft³
The volume of the sand is 4\(\frac{3}{4}\) ft³.

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.

Answer:

→ 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.

→ Or, I could think of it as a piece in front and a piece behind.

→ How can we determine the volume of this figure?
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?
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.
V = 2\(\frac{3}{4}\) m × 2 m × \(\frac{1}{5}\) m
V = \(\frac{11}{4}\) m × 2 m × \(\frac{1}{5}\) m
V = \(\frac{22}{20}\) m³

→ Volume of the prism on the right = l w h.
V = 2\(\frac{1}{4}\) m × 4\(\frac{1}{3}\) m × \(\frac{1}{5}\) m

V = \(\frac{9}{4}\) m × \(\frac{13}{3}\) m × \(\frac{1}{5}\) m

V = \(\frac{117}{60}\) m

V = \(\frac{39}{20}\) m³

Total volume = volume of left + volume of right

Total volume = \(\frac{22}{20}\) m³ + \(\frac{39}{20}\) m³

Total volume = \(\frac{61}{20}\) m³ = 3\(\frac{1}{20}\) m³

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
V = 5 m × 2 m × \(\frac{1}{5}\) m
V = 2m³

Volume of the front piece = l w h
V = 2\(\frac{1}{4}\) m × 2\(\frac{1}{3}\) m × \(\frac{1}{5}\) m
V = \(\frac{9}{4}\) m × \(\frac{7}{3}\) m × \(\frac{1}{5}\) m

V = \(\frac{63}{60}\) m3 = 1\(\frac{3}{60}\) m = 1\(\frac{1}{20}\) m³
Total volume = volume of back + volume of front
Total volume = 2 m³ + 1\(\frac{1}{20}\) m³
Total volume = 3\(\frac{1}{20}\) m³

→ 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.

Eureka Math Grade 6 Module 5 Lesson 14 Exercise Answer Key

Exercise 1.

a. The volume of the rectangular prism is \(\frac{36}{15}\) yd³. Determine the missing measurement using a one-step equation.

Answer:
V = bh
\(\frac{36}{15}\) = \(\frac{4}{5}\) h

\(\frac{36}{15}\) ÷ \(\frac{12}{15}\) = \(\frac{12}{15}\)h ÷ \(\frac{12}{15}\)

\(\frac{36}{12}\) = h
3 = h
The height is 3 yd.

b. The volume of the box is \(\frac{45}{6}\) m³. Determine the area of the base using a one-step equation.

Answer:
V = bh
\(\frac{45}{6}\) = b(\(\frac{2}{9}\))
\(\frac{45}{6}\) ÷ \(\frac{27}{6}\) = b(\(\frac{27}{6}\)) ÷ \(\frac{7}{6}\)
\(\frac{45}{27}\) = b
\(\frac{5}{3}\) = b
The area of the base is \(\frac{45}{6}\) m².

Exercise 2.
Marissa’s fish tank needs to be filled with more water.

a. Determine how much water the tank can hold.
Answer:
Volume of entire tank = l w h
V = (\(\frac{3}{4}\) m) (\(\frac{1}{4}\) m) (\(\frac{3}{5}\) m)
V = \(\frac{9}{80}\) m³.

b. Determine how much water is already in the tank.
Answer:
Volume of water in the tank = l w h
V = (\(\frac{3}{4}\) m) (\(\frac{1}{4}\) m) (\(\frac{3}{8}\) m)
V = \(\frac{9}{128}\) m³.

c. How much more water is needed to fill the tank?
Answer:
Height of empty part of tank:
h = \(\frac{3}{5}\) m – \(\frac{3}{8}\) m
= \(\frac{24}{40}\) m – \(\frac{15}{40}\) m
= \(\frac{9}{40}\) m

Volume needed to fill = l w h
V = (\(\frac{3}{4}\) m) (\(\frac{1}{4}\) m) (\(\frac{9}{40}\) m)
V = \(\frac{27}{640}\) m³

Exercise 3.
Determine the volume of composite figures.

a.
Answer:

b.
Answer:

Another possible solution:

Eureka Math Grade 6 Module 5 Lesson 14 Problem Set Answer Key

Question 1.
The volume of a rectangular prism is \(\frac{21}{12}\) ft³, the height of the prism is \(\frac{3}{4}\) ft. Determine the area of the base.
Answer:
V = bh
\(\frac{21}{12}\) = b (\(\frac{3}{4}\))
\(\frac{21}{9}\) = b
The area of the base is \(\frac{21}{9}\) ft² OR \(\frac{7}{3}\) ft².

Question 2.
The volume of a rectangular prism is \(\frac{10}{21}\) ft³. The area of the base is \(\frac{2}{3}\) ft2. Determine the height of the rectangular prism.
Answer:
Height = Volume ÷ Area of the base
Helght = \(\frac{10}{21}\) ft³ ÷ \(\frac{2}{3}\) ft²
Height = \(\frac{10}{21}\) ft³ ÷ \(\frac{14}{21}\) ft²
Height = \(\frac{10}{14}\) ft. OR \(\frac{5}{7}\) ft.

Question 3.
Determine the volume of the space in the tank that still needs to be tilled with water if the water is ft. deep.

Answer:
Volume of tank = l w h
Volume of tank = (5 ft.) (1\(\frac{2}{3}\) ft.) (2 ft.)
Volume of tank = \(\frac{50}{3}\) ft³
Volume to be filled = \(\frac{50}{3}\) ft³ – \(\frac{25}{9}\) ft³
Volume to be filled = \(\frac{150}{9}\) ft³ – \(\frac{25}{9}\) ft³
Volume to be filled = \(\frac{125}{9}\) ft³

Volume of water = l w h
VoIumeofwater = (5 ft.) (1\(\frac{2}{3}\) ft.) (\(\frac{1}{3}\) ft.)
Volume of water = \(\frac{25}{9}\) ft³

Question 4.
Determine the volume of composite figure.

Answer:
Volume of back piece = l w h
Volume of back piece = (\(\frac{3}{4}\) m) (\(\frac{1}{8}\) m) (\(\frac{1}{3}\) m)

Volume of back piece = \(\frac{3}{96}\) m³

Volume of front piece = l w h

Volume of front piece = (\(\frac{1}{4}\) m) (\(\frac{1}{8}\) m) (\(\frac{1}{3}\) m)
Volume of front piece = \(\frac{1}{96}\) m³
Total volume = \(\frac{3}{96}\) m³ + \(\frac{1}{96}\) m³ = \(\frac{4}{96}\) m³ OR \(\frac{1}{24}\)m³

Question 5.
Determine the volume of the composite figure.

Answer:
V = (1 in.) (1\(\frac{1}{2}\) in.) (1\(\frac{1}{4}\) in.) + (3 in.) (2\(\frac{1}{2}\) in.) (\(\frac{1}{4}\) in.)

V = (1 in.) (\(\frac{3}{2}\) in.) (\(\frac{5}{4}\) in.) + (3 in.) (\(\frac{5}{2}\) in.) (\(\frac{1}{4}\) in.)
V = \(\frac{15}{8}\) in³ + \(\frac{15}{8}\) in³
V = \(\frac{30}{8}\) in³ = 3\(\frac{6}{8}\) in³ OR 3\(\frac{3}{4}\) in³

Question 6.

a. Write an equation to represent the volume of the composite figure.
Answer:
V = (3\(\frac{1}{2}\) m × 2 m × 1\(\frac{1}{4}\) m) + (3\(\frac{3}{4}\) m × 2 m × 2\(\frac{1}{4}\) m)

b. Use your equation to calculate the volume of the composite figure.
Answer:

Eureka Math Grade 6 Module 5 Lesson 14 Exit Ticket Answer Key

Question 1.
Determine the volume of the water that would be needed to fill the rest of the tank.

Answer:
Volume of tank = l w h
Volume of tank = (1\(\frac{1}{4}\) m) (\(\frac{1}{2}\) m) (\(\frac{3}{4}\) m)
Volume of tank = \(\frac{15}{32}\) m³

Volume of water = l w h
Volume of water = (1\(\frac{1}{4}\) m) (\(\frac{1}{2}\) m) (\(\frac{1}{2}\) m)
Volume of waler = \(\frac{5}{16}\) m³ = \(\frac{10}{32}\) m
Remaining water needed = \(\frac{15}{32}\) m³ – \(\frac{10}{32}\) m³ = \(\frac{5}{32}\)m³

Question 2.
Determine the volume of the composite figure.
Answer:
Volume of back piece = l w h
Volume of back piece = (\(\frac{5}{8}\) ft.) (\(\frac{1}{3}\) ft.) (\(\frac{1}{4}\) ft.)
Volume of back piece = \(\frac{5}{96}\) ft³

Volume of front piece = l w h
Volume of front piece = (\(\frac{1}{4}\) ft.) (\(\frac{1}{6}\) ft.) (\(\frac{1}{4}\) ft.)
Volume of front piece = \(\frac{1}{96}\) ft³
Total volume = \(\frac{5}{96}\) ft³ + \(\frac{1}{96}\) ft³ = \(\frac{6}{96}\) ft³ = \(\frac{2}{32}\) ft³

Engage NY Eureka Math Grade 6 Module 5 Lesson 13 Answer Key

Eureka Math Grade 6 Module 5 Lesson 13 Example Answer Key

Example 1:
Determine the volume of a cube with side lengths of 2\(\frac{1}{4}\) cm.
Answer:
V = l w h
V = (2\(\frac{1}{4}\) cm) (2\(\frac{1}{4}\) cm) (2\(\frac{1}{4}\) cm)
V = \(\frac{9}{4}\) cm × \(\frac{9}{4}\) cm × \(\frac{9}{4}\) cm
V = \(\frac{729}{64}\) cm³

Example 2:
Determine the volume of a rectangular prism with a base area of \(\frac{7}{12}\) ft² and a height of \(\frac{1}{3}\) ft.
Answer:
V = Area of the base × height
V = (\(\frac{7}{12}\) ft²) (\(\frac{1}{3}\) ft.)
V = \(\frac{7}{36}\) ft³

Eureka Math Grade 6 Module 5 Lesson 13 Exercise Answer Key

Exercise 1.
Use the rectangular prism to answer the next set of questions.

a. Determine the volume of the prism.
Answer:
V = Area of the base × height
V = (\(\frac{13}{2}\) ft²) (\(\frac{5}{3}\) ft.)
V = \(\frac{65}{6}\) ft³

b. Determine the volume of the prism if the height of the prism is doubled.
Answer:
Height × 2 = (\(\frac{5}{3}\) ft. × 2) = \(\frac{10}{3}\) ft.
V = Area of base × height
V=(\(\frac{13}{2}\) ft²)(\(\frac{10}{3}\) ft.)
V = \(\frac{130}{6}\) ft³ or 21\(\frac{2}{3}\) ft³

c. Compare the volume of the rectangular prism in part (a) with the volume of the prism in part (b). What do you notice?
Answer:
When the height of the rectangular prism is doubled, the volume is also doubled.

d. Complete and use the table below to determine the relationship between the height and volume.

Answer:

What happened to the volume when the height was tripled?
Answer:
The volume tripled.

What happened to the volume when the height was quadrupled?
Answer:
The volume quadrupled.

What conclusions can you make when the base area stays constant and only the height changes?
Answer:
Answers will vary but should include the idea of a proportional relationship. Each time the height is multiplied by a number, the original volume is multiplied by the same amount.

Exercise 2.

a. If B represents the area of the base and h represents the height, write an expression that represents the volume.
Answer:
Bh

b. If we double the height, write an expression for the new height.
Answer:
2h

c. Write an expression that represents the volume with the doubled height.
Answer:
B2h

d. Write an equivalent expression using the commutative and associative properties to show the volume Is
twice the original volume.
Answer:
2(Bh)

Exercise 3.
Use the cube to answer the following questions.

a. Determine the volume of the cube.
Answer:
V= l w h
V = (3m) (3m) (3m)
V = 27 m³

b. Determine the volume of a cube whose side lengths are half as long as the side lengths of the original cube.
Answer:
V = l w h
V = (\(\frac{3}{2}\) m) (\(\frac{3}{2}\) m) (\(\frac{3}{2}\) m)
V = \(\frac{27}{8}\) m³

c. Determine the volume if the side lengths are one-fourth as long as the original cube’s side lengths.
Answer:
V = l w h
V = (\(\frac{3}{4}\) m) (\(\frac{3}{4}\) m) (\(\frac{3}{4}\) m)
V = \(\frac{27}{64}\) m³

d. Determine the volume if the side lengths are one-sixth as long as the original cube’s side lengths.
Answer:
V = lwh
V = (\(\frac{3}{6}\) m) (\(\frac{3}{6}\) m) (\(\frac{3}{6}\) m)
V = \(\frac{27}{216}\)m³
OR
V = \(\frac{1}{8}\) m³

e. Explain the relationship between the side lengths and the volumes of the cubes.
Answer:
If each of the sides are changed by the same fractional amount, \(\frac{1}{a^{\prime}}\) of the original, then the volume of the new figure will be \(\left(\frac{1}{a}\right)^{3}\) of the original volume. For example, if the sides are \(\frac{1}{2}\) as long, then the volume will be \(\left(\frac{1}{2}\right)^{3}\) = \(\frac{1}{8}\) as much.

Exercise 4.
Check to see if the relationship you found in Exercise 3 is the same for rectangular prisms.

a. Determine the volume of the rectangular prism.
Answer:
V = l w h
V = (9 ft.) (2 ft.) (3 ft.)
V = 54 ft³

b. Determine the volume if all of the sides are half as long as the original lengths.
Answer:
V = l w h
V = (\(\frac{9}{2}\) ft.) (\(\frac{2}{2}\) ft.) (\(\frac{3}{2}\) ft.)
V = \(\frac{9}{2}\) ft³
OR
V = \(\frac{27}{4}\) ft³

c. Determine the volume if all of the sides are one-third as long as the original lengths.
Answer:
V = l w h
V = (\(\frac{9}{3}\) ft.) (\(\frac{2}{3}\) ft.) (\(\frac{3}{3}\) ft.)
V = \(\frac{54}{27}\) ft³
OR
V = 2 ft³

d. Is the relationship between the side lengths and the volume the same as the one that occurred in Exercise 3? Explain your answer.
Answer:
Yes, the relationship that was found in the problem with the cubes still holds true with this rectangular prism. When I found the volume of o prism with side lengths that were one-third the original, the volume was
(\(\frac{1}{3}\))³ = \(\frac{1}{27}\) the original.

Exercise 5.

a. If e represents a side length of the cube, create an expression that shows the volume of the cube.
Answer:
e³

b. If we divide the side lengths by three, create an expression for the new side length.
Answer:
\(\frac{1}{3}\) e or \(\frac{\mathrm{e}}{3}\)

c. Write an expression that represents the volume of the cube with one-third the side length.
Answer:
(\(\frac{1}{3}\) e)³ or (\(\frac{\mathrm{e}}{3}\))³

d. Write an equivalent expression to show that the volume is of the original volume.
Answer:

Eureka Math Grade 6 Module 5 Lesson 13 Problem Set Answer Key

Question 1.
Determine the volume of the rectangular prism.

Answer:
V = Area of the base × height
V = (\(\frac{30}{7}\) cm²) (\(\frac{1}{3}\) cm)
V = \(\frac{30}{21}\) cm³
OR
V = \(\frac{10}{7}\) cm³

Question 2.
Determine the volume of the rectangular prism in Problem 1 if the height is quadrupled (multiplied by four). Then, determine the relationship between the volumes in Problem 1 and this prism.
Answer:
V = Area of base × height
V = (\(\frac{30}{7}\) cm²) (\(\frac{4}{3}\) cm)
V = \(\frac{120}{21}\) cm³
OR
V = \(\frac{40}{7}\) cm³
When the height was quadrupled, the volume was also quadrupled.

Question 3.
The area of the base of a rectangular prism can be represented by B, and the height is represented by h.

a. Write an equation that represents the volume of the prism.
Answer:
V = Bh

b. If the area of the base is doubled, write an equation that represents the volume of the prism.
Answer:
V = 2Bh

c. If the height of the prism is doubled, write an equation that represents the volume of the prism.
Answer:
V = B2h = 2Bh

d. Compare the volume in parts (b) and (c). What do you notice about the volumes?
Answer:
The expressions in part (b) and part (c) are equal to each other.

e. Write an expression for the volume of the prism if both the height and the area of the base are doubled.
Answer:
V = 2B2h = 4Bh

Question 4.
Determine the volume of a cube with a side length of 5 In.
Answer:
V = l w h
V = (5\(\frac{1}{3}\) in.) (5\(\frac{1}{3}\) in.) (5\(\frac{1}{3}\) in.)
V = \(\frac{16}{3}\) in. × \(\frac{16}{3}\) in. × \(\frac{16}{3}\) in.
V = \(\frac{4,096}{27}\) in³.

Question 5.
Use the information in Problem 4 to answer the following:

a. Determine the volume of the cube in Problem 4 if all of the side lengths are cut in half.
Answer:
V = l w h
V = (2\(\frac{2}{3}\) in.)(2\(\frac{2}{3}\) in.)(2\(\frac{2}{3}\) in.)
V = \(\frac{8}{3}\) in. × \(\frac{8}{3}\) in. × \(\frac{8}{3}\) in.
V = \(\frac{512}{27}\) in³

b. How could you determine the volume of the cube with the side lengths cut in half using the volume in Problem 4?
Answer:
Because each side is hallas long, I know that the volume is \(\frac{1}{8}\) the volume of the cube in Problem 4. This is because the length, the width, and the height were all cut in half.
\(\frac{1}{2}\)l × \(\frac{1}{2}\)w × \(\frac{1}{2}\)h = \(\frac{1}{8}\)lwh
\(\frac{1}{8}\) × \(\frac{4,096}{27}\) in³ = \(\frac{512}{27}\) in³

Question 6.
Use the rectangular prism to answer the following questions.

a. Complete the table.

Answer:

b. How did the volume change when the length was one-third as long?
Answer:
415 one-third of 12. Therefore, when the length is one-third as long, the volume is also one-third as much.

c. How did the volume change when the length was tripled?
Answer:
36 is three times as much as 12. Therefore, when the length is three times as long, the volume is also three times as much.

d. What conclusion can you make about the relationship between the volume and the length?
Answer:
When the length changes but the width and height stay the same, the change in the volume is proportional to the change in the length.

Question 7.
The sum of the volumes of two rectangular prisms, Box A and Box B, are 14.325 cm3. Box A has a volume of 5.61 cm3.

a. Let B represent the volume of Box B in cubic centimeters. Write an equation that could be used to determine the volume of Box B.
Answer:
14.325 cm³ = 5.61 cm³ + B

b. Solve the equation to determine the volume of Box B.
Answer:
B = 8.715 cm³

c. If the area of the base of Box B is 1.5 cm2, write an equation that could he used to determine the height of
Box B. Let h represent the height of Box B in centimeters.
Answer:
8.715 cm³ = (1.5 cm²)h

d. Solve the equation to determine the height of Box B.
Answer:
h = 5.81cm

Eureka Math Grade 6 Module 5 Lesson 13 Exit Ticket Answer Key

Question 1.
A new company wants to mail out samples of its hair products. The company has a sample box that is a rectangular prism with a rectangular base with an area of 23 -in. The height of the prism is 1in. Determine the volume of the sample box.
Answer:
V = Area of base × height
V = (23\(\frac{1}{3}\) in²) (1\(\frac{1}{4}\) in)
V = \(\frac{70}{3}\) in² × \(\frac{5}{4}\) in.
V = \(\frac{350}{12}\) in³
OR
V = \(\frac{175}{6}\) in²

Question 2.
A different sample box has a height that is twice as long as the original box described in Problem 1. What is the volume of this sample box? How does the volume of this sample box compare to the volume of the sample box in Problem 1?
Answer:
V = Area of base × height
V = (23\(\frac{1}{3}\) in²)(2\(\frac{1}{2}\) in.)
V = (\(\frac{70}{3}\) in²) (\(\frac{5}{2}\) in.)
V = \(\frac{350}{6}\) in³
OR
V = \(\frac{175}{3}\) in³
By doubling the height, we have also doubled the volume.

Eureka Math Grade 6 Module 5 Lesson 13 Multiplication and Division Equations with Fractions Answer Key

Multiplication and Division Equations with Fractions:

Progression of Exercises:

Question 1.
5y = 35
Answer:
y=7

Question 2.
3m = 135
Answer:
m = 45

Question 3.
12k = 156
Answer:
k = 13

Question 4.
\(\frac{f}{3}\) = 24
Answer:
f = 72

Question 5.
\(\frac{x}{7}\) = 42
Answer:
x = 294

Question 6.
\(\frac{c}{13}\) = 18
Answer:
c = 234

Question 7.
\(\frac{2}{3}\) g = 6
Answer:
g = 9

Question 8.
\(\frac{3}{5}\) k = 9
Answer:
k = 15

Question 9.
\(\frac{3}{4}\) y = 10
Answer:
y = \(\frac{40}{3}\) = 13 \(\frac{1}{3}\)

Question 10.
\(\frac{5}{8}\) j = 9
Answer:
j = \(\frac{72}{5}\) = 14 \(\frac{2}{5}\)

Question 11.
\(\frac{3}{7}\) h = 13
Answer:
h = \(\frac{91}{3}\) = 30\(\frac{1}{3}\)

Question 12.
\(\frac{m}{4}\) = \(\frac{3}{5}\)
Answer:
m = \(\frac{12}{5}\) = 2 \(\frac{2}{5}\)

Question 13.
\(\frac{f}{3}\) = \(\frac{2}{7}\)
Answer:
f = \(\frac{6}{7}\)

Question 14.
\(\frac{2}{5}\) p = \(\frac{3}{7}\)
Answer:
p = \(\frac{15}{14}\) = 1\(\frac{1}{14}\)

Question 15.
\(\frac{3}{4}\) k = \(\frac{5}{8}\)
Answer:
k = \(\frac{20}{24}\) = \(\frac{5}{6}\)

Engage NY Eureka Math Grade 6 Module 5 Lesson 12 Answer Key

Eureka Math Grade 6 Module 5 Lesson 12 Example Answer Key

Example 1

a. Write a numerical expression for the volume of each of the rectangular prisms above.
Answer:
(15 in,) (1\(\frac{1}{2}\) in.) (3 in.)
(15 in.) (1 \(\frac{1}{2}\) in.) (6 in.)
(15 in.) (1\(\frac{1}{2}\) in.) (9 in.)

b. What do all of these expressions have in common? What do they represent?
Answer:
All of the expressions have (15 in.) (1\(\frac{1}{2}\) in.). This is the area of the base.

c. Rewrite the numerical expressions to show what they have in common.
Answer:
(22\(\frac{1}{2}\) in²)(3 in.) (22\(\frac{1}{2}\) in²) (6 in.) (22\(\frac{1}{2}\) in²) (9 in.)

d. If we know volume for a rectangular prism as length times width times height, what is another formula for volume that we could use based on these examples?
Answer:
We could use (area of the base) (height), or area of the base times height.

e. What is the area of the base for all of the rectangular prisms?
Answer:
(15 in.) (1\(\frac{1}{2}\) in.) = 22 \(\frac{1}{2}\) in²

f. Determine the volume of each rectangular prism using either method.
Answer:
(15 in.)(1\(\frac{1}{2}\) in.)(3 in.) = 67\(\frac{1}{2}\) in³ or (22\(\frac{1}{4}\) in²) (3 in.) = 67\(\frac{1}{2}\) in³

(15 in.)(1\(\frac{1}{2}\) in.)(6 in.) = 135 in³ or (22\(\frac{1}{2}\) in²)(6 in.)= 135 in³

(15 in.)(1\(\frac{1}{2}\) in.)(9 in.) = 202\(\frac{1}{2}\) in³ or (22\(\frac{1}{2}\) in²)(9 in.) = 204\(\frac{1}{2}\) in³

g. How do the volumes of the first and second rectangular prisms compare? The volumes of the first and third?
Answer:
135 in³ = 67 in³ × 2
202\(\frac{1}{2}\) in³ = 67\(\frac{1}{2}\) in³ × 3

The volume of the second prism is twice that of the first because the height is doubled. The volume of the third prism is three times as much as the first because the height is triple the first prism’s height.

Example 2:
The base of a rectangular prism has an area of 3\(\frac{1}{4}\) in². The height of the prism is 2\(\frac{1}{2}\) in. Determine the volume of the rectangular prism.
Answer:
V = Area of base × height
V = (3\(\frac{1}{4}\) in²) (2\(\frac{1}{2}\) in.)
V = (\(\frac{13}{4}\) in²) (\(\frac{5}{2}\)in.)
V = \(\frac{65}{8}\) in³

Extension:

Question 1.
A company is creating a rectangular prism that must have a volume of 6 ft³. The company also knows that the area of the base must be 2\(\frac{1}{2}\) ft². How can you use what you learned today about volume to determine the height of the rectangular prism?
Answer:
I know that the volume can be calculated by multiplying the area of the base times the height. So, if I needed the height instead, I would do the opposite. I would divide the volume by the area of the base to determine the height.
V = Area of base × height
6ft³ = (2\(\frac{1}{2}\) ft²)h
6 ft³ ÷ 2\(\frac{1}{2}\) ft² = h
2\(\frac{2}{5}\) ft. = h

Eureka Math Grade 6 Module 5 Lesson 12 Exercise Answer Key

Eureka Math Grade 6 Module 5 Lesson 12 Problem Set Answer Key

Question 1.
Determine the volume of the rectangular prism.

Answer:
V = l w h
V = (1\(\frac{1}{2}\) m) (\(\frac{1}{2}\) m) (\(\frac{7}{8}\) m)
V = \(\frac{21}{32}\) m³

Question 2.
The area of the base of a rectangular prism is 4ft², and the height is 2 ft. Determine the volume of the rectangular prism.
Answer:

Question 3.
The length of a rectangular prism is 3\(\frac{1}{2}\) times as long as the width. The height is \(\frac{1}{4}\) of the width. The width is 3 cm. Determine the volume.
Answer:
Width = 3cm
Length = 3 cm × 3\(\frac{1}{2}\) = \(\frac{21}{2}\) cm
Height = 3 cm × \(\frac{1}{4}\) = \(\frac{3}{4}\) cm
V = l w h
V = (\(\frac{21}{2}\) cm) (3 cm) (\(\frac{3}{4}\) cm)
V = \(\frac{189}{8}\) cm³

Question 4.

a. Write numerical expressions to represent the volume in two different ways, and explain what each reveals.
Answer:
(10\(\frac{1}{2}\) in.) (1\(\frac{2}{3}\) in.) (6 in.) represents the product of three edge lengths. (\(\frac{35}{2}\) in²) (6 in.) represents the product of the base area times the height. Answers will vary.

b. Determine the volume of the rectangular prism.
Answer:
(10\(\frac{1}{2}\) in.)(1\(\frac{2}{3}\) in.)(6 in.)= 105 in³ or
(\(\frac{35}{2}\) in²) (6 in.)= 105 in³

Question 5.
An aquarium in the shape of a rectangular prism has the following dimensions: length = 50 cm, width = 25\(\frac{1}{2}\) cm, and height = 30 cm.

a. Write numerical expressions to represent the volume in two different ways, and explain what each reveals.
Answer:
(50 cm) (25\(\frac{1}{2}\) cm) (30\(\frac{1}{2}\) cm) represents the product of the three edge lengths. (1,275 cm²) (30\(\frac{1}{2}\) cm) represents the base area times the height.

b. Determine the volume of the rectangular prism.
Answer:
(1.275 cm²) (30\(\frac{1}{2}\) cm) = 38,887\(\frac{1}{2}\) cm³

Question 6.
The area of the base in this rectangular prism is fixed at 36 cm2. As the height of the rectangular prism changes, the volume will also change as a result.

a. Complete the table of values to determine the various heights and volumes.

Answer:

b. Write an equation to represent the relationship in the table. Be sure to define the variables used In the equation.
Answer:
Let X be the height of the rectangular prism in centimeters.
Let y be the volume of the rectangular prism in cubic centimeters.
36x = y

c. What is the unit rate for this proportional relationship? What does it mean in this situation?
Answer:
The unit rate is 36.
For every centimeter of height, the volume increases by 36 cm³ because the area of the base is 36 cm². In order to determine the volume, multiply the height by 36.

Question 7.
The volume of a rectangular prism is 16.328 cm³. The height is 3. 14 cm.

a. Let B represent the area of the base of the rectangular prism. Write an equation that relates the volume, the area of the base, and the height.
Answer:
16.328 = 3.14B

b. Solve the equation for B.
Answer:
\(\frac{16.328}{3.14}=\frac{3.14 B}{3.14}\)
B = 5.2
The area of the base is 5.2 cm².

Eureka Math Grade 6 Module 5 Lesson 12 Exit Ticket Answer Key

Question 1.
Determine the volume of the rectangular prism in two different ways.

Answer:
V = l. w. h
V = (\(\frac{3}{4}\) ft.) (\(\frac{3}{8}\) ft.) (\(\frac{3}{4}\) ft.)
V = \(\frac{27}{128}\) ft³

V = Area of base. height
V = (\(\frac{9}{32}\) ft²) . (\(\frac{3}{4}\) ft².)
V = \(\frac{27}{128}\) ft³

Question 2.
The area of the base of a rectangular prism is 12 cm2, and the height is 3 cm. Determine the volume of the rectangular prism.
Answer:
V = Area of base. height
V = (12 cm²) (3\(\frac{1}{3}\) cm)
V = \(\frac{120}{3}\) cm³
V = 40 cm³