Eureka Math

Engage NY Eureka Math 8th Grade Module 7 Lesson 23 Answer Key

Eureka Math Grade 8 Module 7 Lesson 23 Exercise Answer Key

Mathematical Modeling Exercise
A ladder of length L ft. leaning against a wall is sliding down. The ladder starts off being flush with (right up against) the wall. The top of the ladder slides down the vertical wall at a constant speed of v ft. per second. Let the ladder in the position L₁ slide down to position L₂ after 1 second, as shown below.

Will the bottom of the ladder move at a constant rate away from point O?
Answer:
→ Identify what each of the symbols in the diagram represents.
O represents the corner where the floor and the wall intersect.
L₁ represents the position of the ladder after it has slid down the wall.
L₂ represents the position of the ladder after it has slid one second after position L₁ down the wall.
A represents the starting position of the top of the ladder.
A’ represents the position of the top of the ladder after it has slid down the wall for one second.
v represents the distance that the ladder slid down the wall in one second.
B represents the starting position of the bottom of the ladder.
B’ represents the position of the bottom of the ladder after it has slid for one second.
h represents the distance the ladder has moved along the ground after sliding down the wall in one second.

→ The distance from point A to point A’ is v ft. Explain why.
Since the ladder is sliding down the wall at a constant rate of v ft. per second, then after 1 second, the ladder moves v feet. Since we are given that the time it took for the ladder to go from position L₁ to L₂ is one second, then we know the distance between those points must be v feet.

→ The bottom of the ladder then slides on the floor to the left so that in 1 second it moves from B to B’ as shown. Therefore, the average speed of the bottom of the ladder is h ft. per second in this 1-second interval. Will the bottom of the ladder move at a constant rate away from point O? In other words, if the ladder moves at a constant rate, will the distance it has moved (shown as D’ in the image below) coincide with point E where |EO| is the length of the ladder?

Answer:
Provide time for students to discuss the answer to the question in pairs or small groups, and then have students share their reasoning. This question is the essential question of the lesson. The answer to this question is the purpose of the entire investigation.
Consider prompting their thinking with the following questions:

→ Think about the distance the base of the ladder is away from the wall as a function of time t. What do you think the beginning placement of the ladder should be for time t = 0?
→ Flush vertical against the wall
→ How far away from the wall is the base of the ladder then at time t = 0?
→ It is right up against the wall.
→ If we let d represent the distance of the base of the ladder from the wall, we know that at t = 0, d = 0.
→ Can the ladder continue sliding forever? Is there a time at which it must stop sliding?
Yes, it must stop sliding when it lies flat on the floor.
Can we compute at what time that will occur?
Allow students time to struggle with this. They may or may not be able to develop the equation below.
t = \(\frac{L}{v}\)
→ So our question is, is d a linear function of time? That is, does the value of d change by constant amounts over constant time intervals?
Students may suggest modeling the experiment using a note card against a vertical book and observing how the distance changes as the card slides down the book. Students may compare extreme cases where the base of the ladder moves a greater distance in the first second of sliding down the wall as opposed to the last second.

Students should recognize that this situation cannot be described by a linear function. Specifically, if the top of the ladder was v feet from the floor as shown below, it would reach O in one second (because the ladder slides down the wall at a constant rate of v per second). Then after 1 second, the ladder will be flat on the floor, and the foot of the ladder would be at the point where |EO| = L, or the length of the ladder. The discussion points below may be useful if students were unable to determine that the motion is non-linear.

→ If the rate of change could be described by a linear function, then the point D would move to D’ after 1 second, where |D’ D| = h ft . (where h is defined as the length the ladder moved from D to D’ in one second). But this is impossible.

→ Recall that the length of the ladder, L, is |EO|. When the ladder is flat on the floor, then at most, the foot of the ladder will be at point E from point O. If the rate of change of the ladder were linear, then the foot of the ladder would be at D’ because the linear rate of change would move the ladder a distance of h feet every 1 second. From the picture (on the previous page) you can see that D’≠E. Therefore, it is impossible that the rate of change of the ladder could be described by a linear function.

→ Intuitively, if you think about when the top of the ladder, C, is close to the floor (point O), a change in the height of C would produce very little change in the horizontal position of the bottom of the ladder, D.

→ Consider the three right triangles shown below. If we let the length of the ladder be 8 ft., then we can see that a constant change of 1 ft. in the vertical distance produces very little change in the horizontal distance. Specifically, the change from 3 ft. to 2 ft. produces a horizontal change of approximately 0.3 ft., and the change from 2 ft. to 1 ft. produces a horizontal change of approximately 0.2 ft. A change from 1 ft. to 0 ft., meaning that the ladder is flat on the floor, would produce a horizontal change of just 0.1 ft. (the difference between the length of the ladder, 8 ft. and 7.9 ft.)

Consider the three right triangles shown below, specifically the change in the length of the base as the height decreases in increments of 1 ft.

Answer:
→ In particular, when the ladder is flat on the floor so that C = O, then the bottom cannot be further left than the point E because |EO| = L, the length of the ladder. Therefore, the ladder will never reach point D’, and the function that describes the movement of the ladder cannot be linear.

→ We want to show that our intuitive sense of the movement of the ladder is accurate. Our goal is to derive a formula, d, for the function of the distance of the bottom of the ladder from O over time t. Because the top of the ladder slides down the wall at a constant rate of v ft . per second, the top of the ladder is now at point A, which is vt ft. below the vertical height of L feet, and the bottom of the ladder is at point B, as shown below. We want to determine |BO|, which by definition is the formula for the function, d.

Answer:
→ Explain the expression vt. What does it represent?
The expression vt represents the distance the ladder has slid down the wall after t seconds. Since v is the rate at which the ladder slides down the wall, then vt is the distance it slides after t seconds.

→ How can we determine |BO|?
The shape formed by the ladder, wall, and floor is a right triangle, so we can use the Pythagorean theorem to find |BO|.

→ What is the length of |AO|?
|AO| is the length of the ladder L minus the distance the ladder slides down the wall after t seconds (i.e., vt). Therefore, |AO| = L – vt.

→ What is the length of the hypotenuse of the right triangle?
The length of the hypotenuse is the length of the ladder, L.
Use the Pythagorean theorem to write an expression that gives |BO| (i.e., d).
Provide students time to work in pairs to write the expression for |BO|. Give guidance as necessary.
By the Pythagorean theorem,
(L-vt)²+d² = L²
d² = L²-(L-vt)²
\(\sqrt{d^{2}}\) = \(\sqrt{L^{2}-(L-v t)^{2}}\)
d = \(\sqrt{L^{2}-(L-v t)^{2}}\)
Pause after deriving the equation d = \(\sqrt{L^{2}-(L-v t)^{2}}\). Ask students to explain what the equation represents. Students should recognize that the equation gives the distance the foot of the ladder is from the wall, which is |BO|.

→ Let’s return to an earlier question: Will the ladder continue to slide forever?
The goal is for students to conclude that when d = L, the ladder will no longer slide. That is, when
\(\sqrt{L^{2}-(L-v t)^{2}}\) = L, the ladder is finished sliding. If students need convincing, consider the following situation with concrete numbers.
What would happen if t were very large? Suppose the constant rate, v, of the ladder falling down the wall is
2 feet per second, the length of the ladder, L, is 10 feet, and the time t is 100 seconds—what is d equal to?
The value of d is
d = \(\sqrt{L^{2}-(L-v t)^{2}}\)
= \(\sqrt{10^{2}-(10-2(100))^{2}}\)
= \(\sqrt{100-(190)^{2}}\)
= \(\sqrt{100-36,100}\)
= \(\sqrt{-36,000}\)

If the value of t were very large, then the formula would make no sense because the length of |BO| would be equal to the square of a negative number.
For this reason, we can only consider values of t so that the top of the ladder is still above the floor. Symbolically, vt≤L, where vt is the expression that describes the distance the ladder has moved at a specific rate v for a specific time t. We need that distance to be less than or equal to the length of the ladder.

→ What happens when t = \(\frac{L}{v}\)? Substitute \(\frac{L}{v}\) for t in our formula.
Substituting \(\frac{L}{v}\) for t,
d = \(\sqrt{L^{2}-(L-v t)^{2}}\)
= \(\sqrt{L^{2}-\left(L-v\left(\frac{L}{v}\right)\right)^{2}}\)
= \(\sqrt{L^{2}-(L-L)^{2}}\)
= \(\sqrt{L^{2}-0^{2}}\)
= \(\sqrt{L^{2}}\)
= L.
When t = \(\frac{L}{v}\), the top of the ladder will be at the point O, and the ladder will be flat on the floor because d represents the length of |BO|. If that length is equal to L, then the ladder must be on the floor.

→ Back to our original concern: What kind of function describes the rate of change of the movement of the bottom of the ladder on the floor? It should be clear that by the equation d = \(\sqrt{L^{2}-(L-v t)^{2}}\), which represents |BO| for any time t, that the motion (rate of change) is not one of constant speed. Nevertheless, thanks to the concept of a function, we can make predictions about the location of the ladder for any value of t as long as t≤\([latex]\frac{L}{v}\)[/latex]

→ We will use some concrete numbers to compute the average rate of change over different time intervals. Suppose the ladder is 15 feet long, L = 15, and the top of the ladder is sliding down the wall at a constant speed of 1 ft. per second, v = 1. Then, the horizontal distance of the bottom of the ladder from the wall (|BO|) is given by the formula
d = \(\sqrt{15^{2}-(15-1 t)^{2}}\)
= \(\sqrt{225-(15-t)^{2}}\)
Determine the outputs the function would give for the specific inputs. Use a calculator to approximate the lengths. Round to the hundredths place.

Answer:

Make at least three observations about what you notice from the data in the table. Justify your observations mathematically with evidence from the table.
→ Sample observations given below.
The average rate of change between 0 and 1 second is 5.39 feet per second.
\(\frac{5.39-0}{1-0}\) = 5.39
The average rate of change between 3 and 4 seconds is 1.2 feet per second.
\(\frac{10.2-9}{4-3}\) = 1.2
The average rate of change between 7 and 8 seconds is 0.58 feet per second.
\(\frac{13.27-12.69}{8-7}\) = 0.58
The average rate of change between 14 and 15 seconds is 0.03 feet per second.
\(\frac{15-14.97}{15-14}\) = 0.03

→ Now that we have computed the average rate of change over different time intervals, we can make two conclusions: (1) The motion at the bottom of the ladder is not linear, and (2) there is a decrease in the average speeds; that is, the rate of change of the position of the ladder is slowing down as observed in the four 1-second intervals we computed. These conclusions are also supported by the graph of the situation shown on the next page. The data points do not form a line; therefore, the rate of change with respect to the position of the bottom of the ladder is not linear.

Eureka Math Grade 8 Module 7 Lesson 23 Problem Set Answer Key

Question 1.
Suppose the ladder is 10 feet long, and the top of the ladder is sliding down the wall at a rate of 0.8 ft. per second. Compute the average rate of change in the position of the bottom of the ladder over the intervals of time from 0 to 0.5 seconds, 3 to 3.5 seconds, 7 to 7.5 seconds, 9.5 to 10 seconds, and 12 to 12.5 seconds. How do you interpret these numbers?

Answer:

The average rate of change between 0 and 0.5 seconds is 5.6 feet per second.
\(\frac{2.8-0}{0.5-0}\) = \(\frac{2.8}{0.5}\) = 5.6
The average rate of change between 3 and 3.5 seconds is 0.88 feet per second.
\(\frac{6.94-6.5}{3.5-3}\) = \(\frac{0.44}{0.5}\) = 0.88
The average rate of change between 7 and 7.5 seconds is 0.38 feet per second.
\(\frac{9.17-8.98}{7.5-5}\) = \(\frac{0.19}{0.5}\) = 0.38
The average rate of change between 9.5 and 10 seconds is 0.18 feet per second.
\(\frac{9.8-9.71}{10-9.5}\) = \(\frac{0.09}{0.5}\) = 0.18
The average rate of change between 12 and 12.5 seconds is 0.02 feet per second.
\(\frac{10-9.99}{12.5-12}\) = \(\frac{0.01}{0.5}\) = 0.02
The average speeds show that the rate of change in the position of the bottom of the ladder is not linear. Furthermore, it shows that the rate of change in the position at the bottom of the ladder is quick at first, 5.6 feet per second in the first half second of motion, and then slows down to 0.02 feet per second in the half-second interval from 12 to 12.5 seconds.

Question 2.
Will any length of ladder, L, and any constant speed of sliding of the top of the ladder, v ft. per second, ever produce a constant rate of change in the position of the bottom of the ladder? Explain.
Answer:
No, the rate of change in the position at the bottom of the ladder will never be constant. We showed that if the rate were constant, there would be movement in the last second of the ladder sliding down that wall that would place the ladder in an impossible location. That is, if the rate of change were constant, then the bottom of the ladder would be in a location that exceeds the length of the ladder. Also, we determined that the distance that the bottom of the ladder is from the wall over any time period can be found using the formula d = \(\sqrt{L^{2}-(L-v t)^{2}}\), which is a non-linear equation. Since graphs of functions are equal to the graph of a certain equation, the graph of the function represented by the formula d = \(\sqrt{L^{2}-(L-v t)^{2}}\) is not a line, and the rate of change in position at the bottom of the ladder is not constant.

Eureka Math Grade 8 Module 7 Lesson 23 Exit Ticket Answer Key

Question 1.
Suppose a book is 5.5 inches long and leaning on a shelf. The top of the book is sliding down the shelf at a rate of 0.5 in. per second. Complete the table below. Then, compute the average rate of change in the position of the bottom of the book over the intervals of time from 0 to 1 second and 10 to 11 seconds. How do you interpret these numbers?

Answer:

The average rate of change between 0 and 1 second is 2.29 inches per second.
\(\frac{2.29-0}{1-0}\) = \(\frac{2.29}{1}\) = 2.29.
The average rate of change between 10 and 11 seconds is 0.02 inches per second.
\(\frac{5.5-5.48}{11-10}\) = \(\frac{0.02}{1}\) = 0.02.
The average speeds show that the rate of change of the position of the bottom of the book is not linear. Furthermore, it shows that the rate of change of the bottom of the book is quick at first, 2.29 inches per second in the first second of motion, and then slows down to 0.02 inches per second in the one second interval from 10 to 11 seconds.

Engage NY Eureka Math 8th Grade Module 7 End of Module Assessment Answer Key

Eureka Math Grade 8 Module 7 End of Module Assessment Task Answer Key

When using a calculator to complete the assessment, use the π key and the full display of the calculator for computations.
Question 1.
a. Is a triangle with side lengths of 7 cm, 24 cm, and 25 cm a right triangle? Explain.
Answer:
7² + 24² = 25²
49 + 576 = 625
625 = 625
Yes. The lengths 7,24,25 satisfy the Pythagorean theorem, therefore, it is a right triangle.

b. Is a triangle with side lengths of 4 mm, 11 mm, and 15 mm a right triangle? Explain.
Answer:
4² + 11² = 15²
16 + 121 = 225
137 ≠ 225
No. The lengths 4, 11, 15 do not satisfy the Pythagorean theorem, therefore, it is not a right triangle.

c. The area of the right triangle shown below is 30 ft². The segment XY has a length of 5 ft. Find the length of the hypotenuse.

Answer:

\(\frac{1}{2}\) = 30
5h = 60
h = 12
5² + 12² = x²
25 + 144 = x²
169 = x²
\(\sqrt{169}\) = x
13 = x
The length of the hypotenuse is 13ft.

d. Two paths from school to the store are shown below: One uses Riverside Drive, and another uses Cypress and Central Avenues. Which path is shorter? By about how much? Explain how you know.

Answer:
Let c be the hypotenuse in miles.
7² + 9² = c²
49 + 81 = c²
130 = c²
\(\sqrt{130}\) = \(\sqrt{c^{2}}\)
\(\sqrt{130}\) = c
11.4 ≈ c
The path along Riverside Drive is shorter, about 11.4miles, compared to the path along Cypress and Control Avenues, 16 miles. The difference is about 4.6 miles. The Pythagorean theorem allowed me to calculate the distance along Riverside Drive because the three roads from a right triangle.

e. What is the distance between points A and B?

Answer:

Let c be the distance between points A & B.
2² + 5² = c²
4 + 25 = c²
\(\sqrt{29}\) = \(\sqrt{c^{2}}\)
\(\sqrt{29}\) = c
5.4 ≈ c
The distance between points A & B is about 5.4 units.

f. Do the segments connecting the coordinates (-1, 6), (4, 2), and (7, 6) form a right triangle? Show work that leads to your answer.

Answer:

3² + 4² = 5²
9 + 16 = 25
25 = 25
4² + 5² = c²
16 + 25 = c²
\(\sqrt{41}\) = \(\sqrt{c^{2}}\)
\(\sqrt{41}\) = c
6 < \(\sqrt{41}\) < 7, So side 8 units is the longest.
5² + (\(\sqrt{41}\))² = 8 ²
25 + 41 = 64
66 ≠ 64
No. The segment connecting (-1, 6), (4, 2), and (7, 6) do not form a right triangle because their lengths do not satisfy the Pythagorean theorem.

g. Using an example, illustrate and explain the Pythagorean theorem.
Answer:
Given a right triangle ABC, the sides a, b, c (Where c is the hypotenuse) satisfy a² + b² = c²

a = 3, b = 4, c = 5
3² + 4² = 5²
9 + 16 = 25
25 = 25

h. Using a different example than in part (g), illustrate and explain the converse of the Pythagorean theorem.
Answer:
Given a triangle ABC with side lengths a, b, c (where c is the hypotenuse) that satisfies a² + b² = c², then triangle ABC is a right triangle.

6² + 8² = 10²
36 + 64 = 100
100 = 100
Therefore, triangle ABC is a right triangle because it satisfies the converse of the Pythagorean theorem.

i. Explain a proof of the Pythagorean theorem and its converse.
Answer:
See rubric to locate proofs of the theorem and its converse within the module.

Question 2.
Dorothy wants to purchase a container that will hold the most sugar. Assuming each of the containers below can be completely filled with sugar, write a note recommending a container, including justification for your choice.

Note: The figures are not drawn to scale.

Answer:
Cylinder:
V = 6² π (15)
= 540π

\(\frac{1}{2}\) Sphere:
V = \(\frac{1}{2}\) (\(\frac{4}{3}\)) π 6³
= \(\frac{2}{3}\) (216) π
= 144π

Total Volume:
540π + 144π = 684π

Cylinder:
V = 6² π (14)
= 504π

Cone:
V = \(\frac{1}{3}\)π (6²) 8
= 96π

Total Volume:
504π + 96π = 600π
Dorothy,
You should choose the container with the half sphere on top because it has a greater volume than the container with the cone on top. The containers have volumes of 684π cm³ and 600π cm³, respectively. Since 684 π is greater than 600π, then the container with the half sphere will held more sugar compared to the container with the cone on top.

Question 3.
a. Determine the volume of the cone shown below. Give an answer in terms of π and an approximate answer rounded to the tenths place.

Answer:

8² + h² = 15²
64 + h² = 225
h² = 161
\(\sqrt{h^{2}}\) = \(\sqrt{161}\)
h = \(\sqrt{161}\)
V = \(\frac{1}{3}\) π(64) (\(\sqrt{161}\))
= \(\frac{64}{3}\) \(\sqrt{161}\) π
≈ 850.4
The volume of the cone is exactly \(\frac{64}{3}\) \(\sqrt{161}\) π mm³ and approximately 850.4 mm³

b. The distance between the two points on the surface of the sphere shown below is 10 inches. Determine the volume of the sphere. Give an answer in terms of π and an approximate answer rounded to a whole number.

Answer:

x² + x² = 10²
2x² = 100
x² = 50
\(\sqrt{x^{2}}\) = \(\sqrt{50}\)
x = \(\sqrt{50}\)
V = \(\frac{4}{3}\)π (\(\sqrt{50}\))³
≈ 1481
The volume of the sphere is exactly \(\frac{4}{3}\)π (\(\sqrt{50}\))³ in³ and approximately 1481 in³.

c. A sphere has a volume of 457 \(\frac{1}{3}\) π in³. What is the radius of the sphere?
Answer:

The radius of the sphere is 7 in.

Engage NY Eureka Math 8th Grade Module 7 Mid Module Assessment Answer Key

Eureka Math Grade 8 Module 7 Mid Module Assessment Task Answer Key

Question 1.
a. What is the decimal expansion of the number \(\frac{35}{7}\)? Is the number \(\frac{35}{7}\) rational or irrational? Explain.
Answer:
\(\frac{35}{7}\) = 5.000….
The number \(\frac{35}{7}\) is a rational number. Rational numbers have decimal expansions that repeat. In this case, the decimal that repeats is a zero.

b. What is the decimal expansion of the number \(\frac{4}{33}\)? Is the number \(\frac{4}{33}\) rational or irrational? Explain.
Answer:
\(\frac{4}{33}\) = 0.1212……… = \(0 . \overline{12}\)

The number \(\frac{4}{33}\) is a rational number. Rational numbers have decimal expansions that repeat. The digits 12 repeat in the decimal expansion of \(\frac{4}{33}\). So, \(\frac{4}{33}\) is rational.

Question 2.
a. Write \(0 . \overline{345}\) as a fraction.
Answer:
Let x be \(0 . \overline{345}\)
10³x = 10³(\(0 . \overline{345}\))
1000 x = 1000(\(0 . \overline{345}\))
1000x = \(345 . \overline{345}\)
1000x = 345 + x
1000x – x = 345 + x – x
999x = 345
x = \(\frac{345}{999}\) = \(\frac{115}{333}\)

b. Write \(2.8 \overline{40}\) as a fraction.
Answer:
Let x be \(2.8 \overline{40}\)
10x = \(2.8 \overline{40}\)
10x = 28 + \(0. \overline{40}\)
10x = \(\frac{(28)(99)+40}{99}\)
x = \(\frac{2812}{99}\)(\(\frac{1}{10}\))
x = \(\frac{2812}{990}\) = \(\frac{1406}{495}\)

Let y be \(0. \overline{40}\)
10²y = 10² (\(0. \overline{40}\))
100y = \(40. \overline{40}\)
100y = 40 + y
100y – y = 40 + y – y
99y = 40
y = \(\frac{40}{99}\)
\(2.8 \overline{40}\) = \(\frac{2812}{990}\) = \(\frac{1406}{495}\)

c. Brandon stated that 0.66 and \(\frac{2}{3}\) are equivalent. Do you agree? Explain why or why not.
Answer:
No, I do not agree with brandon. The decimal 0.66 is equal to the fraction \(\frac{66}{100}\) = \(\frac{33}{50}\), not \(\frac{2}{3}\). Also, the number \(\frac{2}{3}\) is equal to the infinit decimal \(0 . \overline{6}\). The number 0.66 is a finite decimal. Therefore, 0.66 and \(\frac{2}{3}\) are not equivlent.

d. Between which two positive integers does \(\sqrt{33}\) lie?
Answer:
The number \(\sqrt{33}\) is between 5 and 6 because
5² < (\(\sqrt{33}\))² < 6²

e. For what integer x is \(\sqrt{x}\) closest to 5.25? Explain.
Answer:
(5.25)² = 27.5625
Since \(\sqrt{x}\) is the square root of x, then x² will give me the integer that belongs in the square root.
(5.25)² = 27.5625, which is closest to the integer 28.

Question 3.
Identify each of the following numbers as rational or irrational. If the number is irrational, explain how you know.
a. \(\sqrt{29}\)
Answer:
Irrational because 29 is not a perfect square and \(\sqrt{29}\) has an infinite decimal expansion that does not repeat.

b. \(5 . \overline{39}\)
Answer:
Rational

c. \(\frac{12}{4}\)
Answer:
Rational

d. \(\sqrt{36}\)
Answer:
Rational

e. \(\sqrt{5}\)
Answer:
Irrational because 5 is not a perfect square and \(\sqrt{5}\) has an infinite decimal expansion that does not repeat.

f. \(\sqrt [ 3 ]{ 27 }\)
Answer:
Rational

g. π=3.141592…
Answer:
Irrational because pi has a decimal expansion that does not repeat.

h. Order the numbers in parts (a)–(g) from least to greatest, and place them on a number line.
Answer:
\(\sqrt{29}\) : 5² < (\(\sqrt{29}\))² < 6², 5.3² < (\(\sqrt{29}\))² < 5.4², 5.38² < (\(\sqrt{29}\))² < 5.39²

Question 4.
Circle the greater number in each of the pairs (a)–(e) below.
a. Which is greater, 8 or \(\sqrt{60}\) ?
Answer:
8

b. Which is greater, 4 or \(\sqrt{26}\) ?
Answer:
\(\sqrt{26}\)

c. Which is greater, \(\sqrt [ 3 ]{ 64 }\) or \(\sqrt{16}\) ?
Answer:
The numbers are equal: \(\sqrt [ 3 ]{ 64 }\) = 4, \(\sqrt{16}\) = 4

d. Which is greater, \(\sqrt [ 3 ]{ 125 }\)5 or \(\sqrt{30}\) ?
Answer:
\(\sqrt{30}\)

e. Which is greater, -7 or –\(\sqrt{42}\) ?
Answer:
–\(\sqrt{42}\)

f. Put the numbers 9, \(\sqrt{52}\) , and \(\sqrt [ 3 ]{ 216 }\) in order from least to greatest. Explain how you know which order to put them in.
Answer:
\(\sqrt{52}\) is between 7 and 8
\(\sqrt [ 3 ]{ 216 }\) = 6
In order from least to greatest:
\(\sqrt [ 3 ]{ 216 }\), \(\sqrt{52}\), 9

Question 5.

Answer:

a. Between which two labeled points on the number line would \(\sqrt{5}\) be located?
Answer:
The number \(\sqrt{5}\) is between 2.2 and 2.3.

b. Explain how you know where to place \(\sqrt{5}\) on the number line.
Answer:
I knew that \(\sqrt{5}\) was between 2 and 3 but closer to 2. So next, I checked intervals of tenths beginning with 2.0 to 2.1. The interval that \(\sqrt{5}\) fit between was 2.2 and 2.3 because 2.2² < (\(\sqrt{5}\))² < 2.3², 4.84 < 5 < 5.29.

c. How could you improve the accuracy of your estimate?
Answer:
To improve the estimate, I would have to continue with the method of rational approximation to determine which interval of hundreths \(\sqrt{5}\) fits between. Once I knew the interval of hundreths, I would check the interval of thousandths, and so on.
\(\sqrt{5}\) : 2² < (\(\sqrt{5}\))² < 3², 2.2² < (\(\sqrt{5}\))² < 2.3² 4 < 5 < 9
4.84 < 5 < 5.29

Question 6.
Determine the positive solution for each of the following equations.
a. 121 = x²
Answer:
\(\sqrt{121}\) = \(\sqrt{x^{2}}\)
11 = x
11² = 121
121 = 121

b. x³=1000
Answer:
\(\sqrt[3]{x^{3}}\) = \(\sqrt [ 3 ]{ 1000 }\)
x = 10
10³ = 1000
1000 = 1000

c. 17 + x² = 42
Answer:
17 – 17 + x² = 42 – 17
x² = 25
\(\sqrt{x^{2}}\) = \(\sqrt{25}\)
x = 5
17 + 5² = 42
17 + 25 = 42
42 = 42

d. x³ + 3x – 9 = x – 1 + 2x
Answer:
x³ + 3x – 3x – 9 = x – 1 + 2x – 3x
x³ – 9 = -1
x³ – 9 + 9 = -1 + 9
x³ = 8
\(\sqrt[3]{x^{3}}\) = \(\sqrt [ 3 ]{ 8 }\)
x = 2
2³ + (3)(2) – 9 = 2 – 1 + (2)(2)
8 + 6 – 9 = 2 – 1 + 4
14 – 9 = 1 + 4
5 = 5

e. The cube shown has a volume of 216 cm³.
i. Write an equation that could be used to determine the length, l, of one side.

Answer:
V = l³
216 = l³

ii. Solve the equation, and explain how you solved it.
Answer:
216 = l³
\(\sqrt [ 3 ]{ 216 }\) = \(\sqrt[3]{l^{3}}\)
6 = l
The length of one side is 6 cm.
To solve the equation, I had to take the cube root of both sides of the equation. The cube root of l³, \(\sqrt[3]{l^{3}}\), is l. The cube root of 216, \(\sqrt [ 3 ]{ 216 }\), is 6 because 6³ = 216.
Therefore, the length of one side of the cube is 6 cm.

Engage NY Eureka Math 8th Grade Module 4 Lesson 31 Answer Key

Eureka Math Grade 8 Module 4 Lesson 31 Exercise Answer Key

Exercises

Exercise 1.
Identify two Pythagorean triples using the known triple 3, 4, 5 (other than 6, 8, 10).
Answer:
Answers will vary. Accept any triple that is a whole number multiple of 3, 4, 5.

Exercise 2.
Identify two Pythagorean triples using the known triple 5, 12, 13.
Answer:
Answers will vary. Accept any triple that is a whole number multiple of 5, 12, 13.

Exercise 3.
Identify two triples using either 3, 4, 5 or 5, 12, 13.
Answer:
Answers will vary.

Use the system to find Pythagorean triples for the given values of s and t. Recall that the solution in the form of (\(\frac{c}{b}\), \(\frac{a}{b}\)) is the triple a, b, c.
Exercise 4.
s = 4, t = 5
Answer:

x + y + x – y = \(\frac{5}{4}\) + \(\frac{4}{5}\)
2x = \(\frac{5}{4}\) + \(\frac{4}{5}\)
2x = \(\frac{41}{20}\)
x = \(\frac{41}{40}\)

\(\frac{41}{40}\) + y = \(\frac{5}{4}\)
y = \(\frac{5}{4}\) – \(\frac{41}{40}\)
y = \(\frac{9}{40}\)

Then the solution is (\(\frac{41}{40}\), \(\frac{9}{40}\)), and the triple is 9, 40, 41.

Exercise 5.
s = 7, t = 10
Answer:

x + y + x – y = \(\frac{10}{7}\) + \(\frac{7}{10}\)
2x = \(\frac{149}{70}\)
x = \(\frac{149}{140}\)

\(\frac{149}{140}\) + y = \(\frac{10}{7}\)
y = \(\frac{10}{7}\) – \(\frac{149}{140}\)
y = \(\frac{51}{140}\)
Then the solution is (\(\frac{149}{140}\), \(\frac{51}{140}\)), and the triple is 51, 140, 149.

Exercise 6.
s = 1, t = 4
Answer:

x + y + x – y = 4 + \(\frac{1}{4}\)
2x = \(\frac{17}{4}\)
x = \(\frac{17}{8}\)

\(\frac{17}{8}\) + y = \(\frac{4}{1}\)
y = 4 – \(\frac{17}{8}\)
y = \(\frac{15}{8}\)
Then the solution is (\(\frac{17}{8}\), \(\frac{15}{8}\)), and the triple is 15, 8, 17.

Exercise 7.
Use a calculator to verify that you found a Pythagorean triple in each of the Exercises 4–6. Show your work below.
Answer:
For the triple 9, 40, 41:
9² + 40² = 41²
81 + 1600 = 1681
1681 = 1681

For the triple 51, 140, 149:
51² + 140² = 149²
2601 + 19600 = 22201
22201 = 22201

For the triple 15, 8, 17:
15² + 8² = 17²
225 + 64 = 289
289 = 289

Eureka Math Grade 8 Module 4 Lesson 31 Problem Set Answer Key

Question 1.
Explain in terms of similar triangles why it is that when you multiply the known Pythagorean triple 3, 4, 5 by 12, it generates a Pythagorean triple.
Answer:
The triangle with lengths 3, 4, 5 is similar to the triangle with lengths 36, 48, 60. They are both right triangles whose corresponding side lengths are equal to the same constant.
\(\frac{36}{3}\) = \(\frac{48}{4}\) = \(\frac{60}{5}\) = 12
Therefore, the triangles are similar, and we can say that there is a dilation from some center with scale factor r = 12 that makes the triangles congruent.

Question 2.
Identify three Pythagorean triples using the known triple 8, 15, 17.
Answer:
Answers will vary. Accept any triple that is a whole number multiple of 8, 15, 17.

Question 3.
Identify three triples (numbers that satisfy a² + b² = c², but a, b, c are not whole numbers) using the triple 8, 15, 17.
Answer:
Answers will vary. Accept any triple that is not a set of whole numbers.

Use the system to find Pythagorean triples for the given values of s and t. Recall that the solution, in the form of (c/b, a/b), is the triple, a, b, c.
Question 4.
s = 2, t = 9
Answer:

x + y + x – y = \(\frac{9}{2}\) + \(\frac{2}{9}\)
2x = \(\frac{85}{18}\)
x = \(\frac{85}{36}\)

\(\frac{85}{36}\) + y = \(\frac{9}{2}\)
y = \(\frac{9}{2}\) – \(\frac{85}{36}\)
y = \(\frac{77}{36}\)
Then the solution is (\(\frac{85}{36}\), \(\frac{77}{36}\)), and the triple is 77, 36, 85.

Question 5.
s = 6, t = 7
Answer:

x + y + x – y = \(\frac{7}{6}\) + 6/7
2x = \(\frac{85}{42}\)
x = \(\frac{85}{84}\)

\(\frac{85}{84}\) + y = \(\frac{7}{6}\)
y = \(\frac{7}{6}\) – \(\frac{85}{84}\)
y = \(\frac{13}{84}\)
Then the solution is (\(\frac{85}{84}\), \(\frac{13}{84}\)), and the triple is 13, 84, 85.

Question 6.
s = 3, t = 4
Answer:

x + y + x – y = \(\frac{4}{3}\) + \(\frac{3}{4}\)
2x = \(\frac{25}{12}\)
x = \(\frac{25}{24}\)

\(\frac{25}{24}\) + y = \(\frac{4}{3}\)
y = \(\frac{4}{3}\) – \(\frac{25}{24}\)
y = \(\frac{7}{24}\)
Then the solution is (\(\frac{25}{24}\), \(\frac{7}{24}\)), and the triple is 7, 24, 25.

Question 7.
Use a calculator to verify that you found a Pythagorean triple in each of the Problems 4–6. Show your work.
Answer:
For the triple 77, 36, 85:
77² + 36² = 85²
5929 + 1296 = 7225
7225 = 7225

For the triple 13, 84, 85:
13² + 84² = 85²
169 + 7056 = 7225
7225 = 7225

For the triple 7, 24, 25:
7² + 24² = 25²
49 + 576 = 625
625 = 625

Eureka Math Grade 8 Module 4 Lesson 31 Exit Ticket Answer Key

Use a calculator to complete Problems 1–3.
Question 1.
Is 7, 20, 21 a Pythagorean triple? Is 1, \(\frac{15}{8}\), \(\frac{17}{8}\) a Pythagorean triple? Explain.
Answer:
The set of numbers 7, 20, 21 is not a Pythagorean triple because 7² + 20² ≠ 21².
The set of numbers 1, \(\frac{15}{8}\), \(\frac{17}{8}\) is not a Pythagorean triple because the numbers \(\frac{15}{8}\) and \(\frac{17}{8}\) are not whole numbers.
But they are a triple because 1² + (\(\frac{15}{8}\))² = (\(\frac{17}{8}\))².

Question 2.
Identify two Pythagorean triples using the known triple 9, 40, 41.
Answer:
Answers will vary. Accept any triple that is a whole number multiple of 9, 40, 41.

Question 3.
Use the system to find Pythagorean triples for the given values of s = 2 and t = 3. Recall that the solution in the form of (\(\frac{c}{b}\), \(\frac{a}{b}\)) is the triple a, b, c. Verify your results.
Answer:

x + y + x – y = \(\frac{3}{2}\) + \(\frac{2}{3}\)
2x = \(\frac{13}{6}\)
x = \(\frac{13}{12}\)

\(\frac{13}{12}\) + y = \(\frac{3}{2}\)
y = \(\frac{3}{2}\) – \(\frac{13}{12}\)
y = \(\frac{5}{12}\)
Then the solution is (\(\frac{13}{12}\), \(\frac{5}{12}\)), and the triple is 5, 12, 13.
5² + 12² = 13²
25 + 144 = 169
169 = 169

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

Engage NY Eureka Math Grade 6 Module 5 Lesson 11 Answer Key

Eureka Math Grade 6 Module 5 Lesson 11 Example Answer Key

Example 1.
A box with the same dimensions as the prism in the Opening Exercise is used to ship miniature dice whose side lengths have been cut in half. The dice are \(\frac{1}{2}\) in. × \(\frac{1}{2}\) in. × \(\frac{1}{2}\) in. cubes. How many dice of this size can fit in the box?

Answer:

→ How many cubes could we fit across the length? The width? The height?
Two cubes would fit across a 1-inch length. So, I would need to double the lengths to get the number of cubes. Twenty cubes will fit across the 10-inch length, 8 cubes will fit across the 4 -inch width and 12 cubes will fit across the 6-inch height.

→ How can you use this information to determine the number of \(\frac{1}{2}\) in. × \(\frac{1}{2}\) in. × \(\frac{1}{2}\) in. cubes it takes to fill the box?
I can multiply the number of cubes in length, width, and height.
20 × 8 × 12 = 1,920, so 1,920 of the smaller cubes will fill the box.

→ How many of these smaller cubes can fit into the 1 in. × 1 in. × 1 in. cube?
Two confits across the length, two across the width, and two for the height. 2 × 2 × 2 = 8. Eight smaller cubes confit in the larger cube.

→ How does the number of cubes in this example compare to the number of cubes that would be needed in the
Opening Exercise?
\(\frac{\text { new }}{\text { old }}=\frac{1,920}{240}=\frac{8}{1}\)
If I fill the same box with cubes that are half the length, I need 8 times as many.

→ How is the volume of the box related to the number of cubes that will fit in it?
The volume of the box is of the number of cubes that will fit in it.

→ What is the volume of 1 cube?
V= \(\frac{1}{2}\) in. × \(\frac{1}{2}\) in. × \(\frac{1}{2}\) in.

→ What is the product of the number of cubes and the volume of the cubes? What does this product represent?
1920 × \(\frac{1}{8}\) = 240
The product represents the volume of the original box.

Example 2.
A \(\frac{1}{5}\) in. cube is used to fill the prism. How many in. cubes does it take to fill the prism? What is the volume of the prism? How is the number of cubes related to the volume?

Answer:

→ How would you determine, or find, the number of cubes that fill the prism?
One method would be to determine the number of cubes that will fit across the length, width, and height. Then, I would multiply. 6 will fit across the length, 4 across the width, and 15 across the height. 6 × 4 × 15 = 360, so 360 cubes will fill the prism.

→ How is the number of cubes and the volume related?
The volume is equal to the number of cubes times the volume of one cube.
The volume of one cube is \(\frac{1}{4}\) in. × \(\frac{1}{4}\) in. × \(\frac{1}{4}\) in. = \(\frac{1}{64}\) in³.

360 cubes × \(\frac{1}{64}\) in³ = \(\frac{360}{64}\) in³ = \(\frac{540}{64}\) in³ =5 \(\frac{5}{8}\)in³

→ What other method can be used to determine the volume?
V = l w h
V = (1 \(\frac{1}{2}\) in.) (1 in.) (3 \(\frac{3}{4}\) in.)
V = \(\frac{3}{2}\) in. × \(\frac{1}{1}\) in. × \(\frac{15}{4}\) in.
V = \(\frac{45}{8}\) in³ = 5 \(\frac{5}{8}\) in³

→ Would any other size cubes fit perfectly inside the prism with no space left over?
We would not be able to use cubes with side lengths of \(\frac{1}{2}\) in., \(\frac{1}{3}\) in., or \(\frac{2}{3}\) in. because there would be spaces left over. However, we could use a cube with a side length of \(\frac{1}{8}\) in. without having spaces left over.

Eureka Math Grade 6 Module 5 Lesson 11 Exercise Answer Key

Opening Exercise:

Exercise 1.
Which prism holds more 1 in. × 1 in. × 1 in. cubes? How many more cubes does the prism hold?

Answer:
Students discuss their solutions with a partner.

→ How many 1 in. × 1 in. × 1 in. cubes can fit across the bottom of the first rectangular prism?
Answer:
40 cubes can fit across the bottom.

→ How did you determine this number?
Answer:
Answers will vary. I determined how many cubes could fill the bottom layer of the prism and then decided how many layers were needed.

Students who are English language learners may need a model of what “layers” means in this context.

→ How many layers of 1 in. × 1 in. × 1 in. cubes can fit inside the rectangular prism?
There are 6 inches in the height; therefore, 6 layers of cubes can fit inside.

→ How many 1 in. × 1 in. × 1 in. cubes can fit across the bottom of the second rectangular prism?
40 cubes can fit across the bottom.

→ How many layers do you need?
I need 12 layers because the prism is 12 in. tall.

→ Which rectangular prism can hold more cubes?
The second rectangular prism can holds more cubes.

→ How did you determine this?
Both rectangular prisms hold the same number of cubes in one layer, but the second rectangular prism has more layers.

→ How many more layers does the second rectangular prism hold?
It holds 6 more layers.

→ How many more cubes does the second rectangular prism hold?
The second rectangular prism has 6 more layers than the first, with 40 cubes in each layer.
6 × 40 = 240, so the second rectangular prism holds 240 more cubes than the first.

> What other ways can you determine the volume of a rectangular prism?
We can also use the formula V = l. w . h.

Exercises:

Exercise 1.
Use the prism to answer the following questions.

a. Calculate the volume.
Answer:

b. If you have to fill the prism with cubes whose side lengths are less than 1 cm, what size would be best?
Answer:
The best choice would be a cube with side lengths of \(\frac{1}{3}\) cm.

c. How many of the cubes would fit in the prism?
Answer:
16 × 2 × 4 = 128, so 128 cubes will fit in the prism.

d. Use the relationship between the number of cubes and the volume to prove that your volume calculation Is correct.
Answer:
The volume of one cube would be \(\frac{1}{3}\) cm × \(\frac{1}{3}\)cm × \(\frac{1}{3}\) cm = \(\frac{1}{27}\) cm³

Since there are 128 cubes, the volume would be 128 × \(\frac{1}{27}\) cm³ = \(\frac{128}{27}\) cm³ or 4\(\frac{20}{27}\) cm³.

Exercise 2.
Calculate the volume of the following rectangular prisms.

a.
Answer:

b.
Answer:

Exercise 3.
A toy company is packaging its toys to be shipped. Each small toy is placed inside a cube-shaped box with side lengths of \(\frac{1}{2}\) in. These smaller boxes are then placed into a larger box with dimensions of 12 in. × 4\(\frac{1}{2}\) in. × 3\(\frac{1}{2}\) in.

a. What is the greatest number of small toy boxes that can be packed into the larger box for shipping?
Answer:
24 × 9 × 7 = 1,512, so 1,512 toys can be packed into the larger box.

b. Use the number of small toy boxes that can be shipped in the larger box to help determine the volume of the shipping box.
Answer:
One small box would have a volume of \(\frac{1}{2}\) in. × \(\frac{1}{2}\) in. × \(\frac{1}{2}\) in. = \(\frac{1}{8}\) in³.
Now, I multiply the number of cubes by the volume of the cube.
1,512 × \(\frac{1}{8}\) in³ = \(\frac{1512}{8}\) in³ = 189 in³

Exercise 4.
A rectangular prism with a volume of 8 cubic units is filled with cubes twice: once with cubes with side lengths of \(\frac{1}{2}\) unit and once with cubes with side lengths of \(\frac{1}{3}\) unit.

a. How many more of the cubes with \(\frac{1}{3}\) unit side lengths than cubes with \(\frac{1}{2}\) unit side lengths are needed to fill the prism?
Answer:
There are 8 cubes with \(\frac{1}{2}\) unit side lengths in 1 cubic unit because the volume of one cube is \(\frac{1}{8}\) cubic unit. Since we have 8 cubic units, we would have 64 total cubes with \(\frac{1}{2}\) unit side lengths because 8 × 8 = 64.

There are 27 cubes with \(\frac{1}{3}\) unit side lengths in 1 cubic unit because the volume of one cube is \(\frac{1}{27}\) cubic units. Since we have 8 cubic units, we would have 216 total cubes with \(\frac{1}{3}\) unit side lengths because 8 × 27 = 216. 216 – 64 = 152, so 152 more cubes with \(\frac{1}{3}\) unit side lengths are needed to fill the prism.

b. Why does it take more cubes with \(\frac{1}{3}\) unit side lengths to fill the prism than it does with cubes with \(\frac{1}{2}\) unit side lengths?
Answer:
\(\frac{1}{3}\) < \(\frac{1}{2}\). The side length is shorter for the cube with a \(\frac{1}{3}\) unit side length, so it takes more to fill the rectangular prism.

Exercise 5.
Calculate the volume of the rectangular prism. Show two different methods for determining the volume.

Answer:
Method 1:

Method 2:
Fill the rectangular prism with cubes that are \(\frac{1}{4}\) m × \(\frac{1}{4}\) m × \(\frac{1}{4}\) m.
The volume of each cube is \(\frac{1}{64}\) m³.
We would have 6 cubes across the length, 3 cubes across the width, and 18 cubes across the height.
6 × 3 × 18 = 324, so the rectangular prism could be filled with 324 cubes with \(\frac{1}{4}\) m side lengths.
324 × \(\frac{1}{64}\) m³ = 5 \(\frac{1}{16}\) m³.

Eureka Math Grade 6 Module 5 Lesson 11 Problem Set Answer Key

Question 1.
Answer the following questions using this rectangular prism:

a. What is the volume of the prism?
Answer:
V = l w h
V = (9 in.) (1 \(\frac{1}{3}\) in.) (4 \(\frac{2}{3}\)in.)
V = ( \(\frac{9}{1}\) in.) (\(\frac{4}{3}\) in.) (\(\frac{14}{3}\) in.)
V = \(\frac{504}{9}\) in³
V = 56 in³.

b. Linda fills the rectangular prism with cubes that have side lengths of \(\frac{1}{3}\) in. How many cubes does she need to fill the rectangular prism?
Answer:
She needs 27 across by 4 wide and 14 high.
Number of cubes = 27 × 4 × 14 = 1,512.
Linda needs 512 cubes with \(\frac{1}{3}\) in. side lengths to fill the rectangular prism.

c. How is the number of cubes related to the volume?
Answer:
56 × 27 = 1,512
The number of cubes needed is 27 times larger than the volume.

d. Why is the number of cubes needed different from the volume?
Answer:
Because the cubes are not each 1 in., the volume is different from the number of cubes. However, I could multiply the number of cubes by the volume of one cube and still get the original volume.

e. Should Linda try to fill this rectangular prism with cubes that are \(\frac{1}{2}\) in. long on each side? Why or why not?
Answer:
Because some of the lengths are \(\frac{1}{3}\) in. and some are \(\frac{2}{3}\) in., it would not be possible to use side lengths of \(\frac{1}{2}\) in. to fill the prism.

Question 2.
Calculate the volume of the following prisms.

a.
Answer:
V = l w h
V = (24 cm) (2\(\frac{2}{3}\) cm) (4\(\frac{1}{2}\) cm)
V = (24cm) (\(\frac{8}{3}\) cm) (\(\frac{9}{2}\) cm)
V = \(\frac{1728}{6}\) cm³
V = 288 cm³

b.
Answer:

Question 3.
A rectangular prism with a volume of 12 cubic units is filled with cubes twice: once with cubes with \(\frac{1}{2}\) unit side lengths and once with cubes with \(\frac{1}{3}\)-unit side lengths.

a. How many more of the cubes with \(\frac{1}{3}\)-unit side lengths than cubes with \(\frac{1}{2}\)-unit side lengths are needed to fill the prism?
Answer:
There are 8 cubes with \(\frac{1}{2}\)-unit side lengths in 1 cubic unit because the volume of one cube is \(\frac{1}{8}\) cubic unit. Since we have 12 cubic units, we would have 96 total cubes with \(\frac{1}{2}\)-unit side lengths because 12 × 8 = 96.

There are 27 cubes with \(\frac{1}{3}\)-unit side lengths in 1 cubic unit because the volume of one cube is \(\frac{1}{27}\) cubic unit. Since we have 12 cubic units, we would have 324 total cubes with \(\frac{1}{3}\)-unit side lengths because 12 × 27 = 324.

324 – 96 = 228, so there are 228 more cubes with \(\frac{1}{3}\)-unit side lengths needed than there are cubes with \(\frac{1}{2}\)-unit side lengths needed.

b. Finally, the prism is filled with cubes whose side lengths are \(\frac{1}{4}\) unit. How many \(\frac{1}{4}\) unit cubes would it take to fill the prism?
Answer:
There are 64 cubes with \(\frac{1}{4}\)-unit side lengths in 1 cubic unit because the volume of one cube is \(\frac{1}{64}\) cubic unit.
Since there are 12 cubic units, we would have 768 total cubes with side lengths of \(\frac{1}{4}\) unit because 12 × 64 = 768.

Question 4.
A toy company is packaging its toys to be shipped. Each toy is placed inside a cube-shaped box with side lengths of 3\(\frac{1}{2}\) in. These smaller boxes are then packed into a larger box with dimensions of 14 in. × 7 in. × 3\(\frac{1}{2}\) in.

a. What is the greatest number of toy boxes that can be packed into the larger box for shipping?
Answer:
4 × 2 × 1 = 8, so 8 toy boxes can be packed into the larger box for shipping.

b. Use the number of toy boxes that can be shipped in the large box to determine the volume of the shipping box.
Answer:
One small box would haveavolume of 3\(\frac{1}{2}\) in. × 3\(\frac{1}{2}\) in. × 3\(\frac{1}{2}\) in. = 42\(\frac{7}{8}\) in³.
Now, I will multiply the number of cubes by the volume of the cube. 8 × 42\(\frac{7}{8}\) in³ = 343 in³

Question 5.
A rectangular prism has a volume of 34.224 cubic meters. The height of the box is 3. 1 meters, and the length is 2.4 meters.

a. Write an equation that relates the volume to the length, width, and height. Let w represent the width, in meters.
Ans;
34.224 = (3.1) (2. 4)w

b. Solve the equation.
Answer:
34.224 = 7.44 w
w = 4.6
The width is 4.6 m.

Eureka Math Grade 6 Module 5 Lesson 11 Exit Ticket Answer Key

Question 1.
Calculate the volume of the rectangular prism using two different methods. Label your solutions Method 1 and Method 2.

Answer:
Method 1:
V = l w h
V = (1\(\frac{3}{8}\) cm) (\(\frac{5}{8}\) cm) (2\(\frac{1}{4}\) cm)

V = \(\frac{11}{8}\) cm × \(\frac{5}{8}\) cm × \(\frac{9}{4}\) cm

V = \(\frac{495}{256}\) cm³

Method 2:
Fill shape with \(\frac{1}{8}\) cm cubes.
11 × 5 × 18 = 990, so 990 cubes could be used to fill the prism.
Each cube has a volume of \(\frac{1}{8}\) cm × \(\frac{1}{8}\) cm × \(\frac{1}{8}\) cm = \(\frac{1}{512}\) cm³

V = 990 × \(\frac{1}{512}\) cm³ = \(\frac{990}{512}\) cm³ = \(\frac{495}{256}\) cm³

Eureka Math Grade 6 Module 5 Lesson 11 Multiplication of Fractions II Answer Key

Multiplication of Fractions II – Round 1:

Directions: Determine the product of the fractions and simplify.

Question 1.
\(\frac{1}{2}\) × \(\frac{5}{8}\)
Answer:
\(\frac{5}{16}\)

Question 2.
\(\frac{3}{4}\) × \(\frac{3}{5}\)
Answer:
\(\frac{9}{20}\)

Question 3.
\(\frac{1}{4}\) × \(\frac{7}{8}\)
Answer:
\(\frac{7}{32}\)

Question 4.
\(\frac{3}{9}\) × \(\frac{2}{5}\)
Answer:
\(\frac{6}{45}\) = \(\frac{2}{15}\)

Question 5.
\(\frac{5}{8}\) × \(\frac{3}{7}\)
Answer:
\(\frac{15}{56}\)

Question 6.
\(\frac{3}{7}\) × \(\frac{4}{9}\)
Answer:
\(\frac{12}{63}\) = \(\frac{4}{21}\)

Question 7.
\(\frac{2}{5}\) × \(\frac{3}{8}\)
Answer:
\(\frac{6}{40}\) = \(\frac{3}{20}\)

Question 8.
\(\frac{4}{9}\) × \(\frac{5}{9}\)
Answer:
\(\frac{20}{81}\)

Question 9.
\(\frac{2}{3}\) × \(\frac{5}{7}\)
Answer:
\(\frac{10}{21}\)

Question 10.
\(\frac{2}{7}\) × \(\frac{3}{10}\)
Answer:
\(\frac{6}{70}\) = \(\frac{3}{35}\)

Question 11.
\(\frac{3}{4}\) × \(\frac{9}{10}\)
Answer:
\(\frac{27}{40}\)

Question 12.
\(\frac{3}{5}\) × \(\frac{2}{9}\)
Answer:
\(\frac{6}{45}\) = \(\frac{2}{15}\)

Question 13.
\(\frac{2}{10}\) × \(\frac{5}{6}\)
Answer:
\(\frac{10}{60}\) = \(\frac{1}{6}\)

Question 14.
\(\frac{5}{8}\) × \(\frac{7}{10}\)
Answer:
\(\frac{35}{80}\) = \(\frac{7}{16}\)

Question 15.
\(\frac{3}{5}\) × \(\frac{7}{9}\)
Answer:
\(\frac{21}{45}\) = \(\frac{7}{15}\)

Question 16.
\(\frac{2}{9}\) × \(\frac{3}{8}\)
Answer:
\(\frac{6}{72}\) = \(\frac{1}{12}\)

Question 17.
\(\frac{3}{8}\) × \(\frac{8}{9}\)
Answer:
\(\frac{24}{72}\) = \(\frac{1}{3}\)

Question 18.
\(\frac{3}{4}\) × \(\frac{7}{9}\)
Answer:
\(\frac{21}{36}\) = \(\frac{7}{12}\)

Question 19.
\(\frac{3}{5}\) × \(\frac{10}{13}\)
Answer:
\(\frac{30}{65}\) = \(\frac{6}{13}\)

Question 20.
1 \(\frac{2}{7}\) × \(\frac{7}{8}\)
Answer:
\(\frac{63}{56}\) = 1 \(\frac{1}{8}\)

Question 21.
3 \(\frac{1}{2}\) × 3 \(\frac{5}{6}\)
Answer:
\(\frac{161}{12}\) = 13 \(\frac{5}{12}\)

Question 22.
\(\frac{1}{4}\) × \(\frac{1}{4}\)
Answer:
\(\frac{1}{4}\)

Question 23.
1 \(\frac{7}{8}\) × 5 \(\frac{1}{5}\)
Answer:
\(\frac{390}{40}\) = 9 \(\frac{3}{4}\)

Question 24.
7 \(\frac{2}{5}\) × 2 \(\frac{3}{8}\)
Answer:
\(\frac{703}{40}\) = 17 \(\frac{23}{40}\)

Question 25.
4 \(\frac{2}{3}\) × 2 \(\frac{3}{10}\)
Answer:
\(\frac{322}{30}\) = 10 \(\frac{11}{15}\)

Question 26.
3 \(\frac{3}{5}\) × 6 \(\frac{1}{4}\)
Answer:
\(\frac{450}{20}\) = 22 \(\frac{1}{2}\)

Question 27.
2 \(\frac{7}{9}\) × 5 \(\frac{1}{3}\)
Answer:
\(\frac{400}{27}\) = 14 \(\frac{22}{27}\)

Question 28.
4 \(\frac{3}{8}\) × 3 \(\frac{1}{5}\)
Answer:
\(\frac{560}{40}\) = 4

Question 29.
3 \(\frac{1}{3}\) × 5 \(\frac{2}{5}\)
Answer:
\(\frac{270}{15}\) = 18 \(\frac{2}{3}\)

Question 30.
2 \(\frac{2}{3}\) × 7
Answer:
\(\frac{56}{3}\) = 18 \(\frac{2}{3}\)

Multiplication of Fractions II – Round 2:

Directions: Determine the product of the fractions and simplify.

Question 1.
\(\frac{2}{3}\) × \(\frac{5}{7}\)
Answer:
\(\frac{10}{21}\)

Question 2.
\(\frac{1}{4}\) × \(\frac{3}{5}\)
Answer:
\(\frac{3}{20}\)

Question 3.
\(\frac{2}{3}\) × \(\frac{2}{5}\)
Answer:
\(\frac{4}{15}\)

Question 4.
\(\frac{5}{9}\) × \(\frac{5}{8}\)
Answer:
\(\frac{25}{72}\)

Question 5.
\(\frac{5}{8}\) × \(\frac{3}{7}\)
Answer:
\(\frac{15}{56}\)

Question 6.
\(\frac{3}{4}\) × \(\frac{7}{8}\)
Answer:
\(\frac{21}{32}\)

Question 7.
\(\frac{2}{5}\) × \(\frac{3}{8}\)
Answer:
\(\frac{6}{40}\) = \(\frac{3}{20}\)

Question 8.
\(\frac{3}{4}\) × \(\frac{3}{4}\)
Answer:
\(\frac{9}{16}\)

Question 9.
\(\frac{7}{8}\) × \(\frac{3}{10}\)
Answer:
\(\frac{21}{80}\)

Question 10.
\(\frac{4}{9}\) × \(\frac{1}{2}\)
Answer:
\(\frac{4}{18}\) = \(\frac{2}{9}\)

Question 11.
\(\frac{6}{11}\) × \(\frac{3}{8}\)
Answer:
\(\frac{18}{88}\) = \(\frac{9}{44}\)

Question 12.
\(\frac{5}{6}\) × \(\frac{9}{10}\)
Answer:
\(\frac{45}{60}\) = \(\frac{3}{4}\)

Question 13.
\(\frac{3}{4}\) × \(\frac{2}{9}\)
Answer:
\(\frac{6}{36}\) = \(\frac{1}{6}\)

Question 14.
\(\frac{4}{11}\) × \(\frac{5}{8}\)
Answer:
\(\frac{20}{88}\) = \(\frac{5}{22}\)

Question 15.
\(\frac{2}{3}\) × \(\frac{9}{10}\)
Answer:
\(\frac{18}{30}\) = \(\frac{3}{5}\)

Question 16.
\(\frac{3}{11}\) × \(\frac{2}{9}\)
Answer:
\(\frac{6}{99}\) = \(\frac{2}{33}\)

Question 17.
\(\frac{3}{5}\) × \(\frac{10}{21}\)
Answer:
\(\frac{30}{105}\) = \(\frac{2}{7}\)

Question 18.
\(\frac{4}{9}\) × \(\frac{3}{10}\)
Answer:
\(\frac{12}{90}\) = \(\frac{2}{15}\)

Question 19.
\(\frac{3}{8}\) × \(\frac{4}{5}\)
Answer:
\(\frac{12}{40}\) = \(\frac{3}{10}\)

Question 20.
\(\frac{6}{11}\) × \(\frac{2}{15}\)
Answer:
\(\frac{12}{165}\) = \(\frac{4}{55}\)

Question 21.
1 \(\frac{2}{3}\) × \(\frac{3}{5}\)
Answer:
\(\frac{15}{15}\) = 1

Question 22.
2 \(\frac{1}{6}\) × \(\frac{3}{4}\)
Answer:
\(\frac{39}{24}\) = \(\frac{13}{8}\) = 1 \(\frac{5}{8}\)

Question 23.
1 \(\frac{2}{5}\) × 3 \(\frac{2}{3}\)
Answer:
\(\frac{77}{15}\) = 5 \(\frac{2}{15}\)

Question 24.
4 \(\frac{2}{3}\) × 1 \(\frac{1}{4}\)
Answer:
\(\frac{70}{12}\) = 5 \(\frac{10}{12}\) = 5 \(\frac{5}{6}\)

Question 25.
3 \(\frac{1}{2}\) × 2 \(\frac{4}{5}\)
Answer:
\(\frac{98}{10}\) = 9 \(\frac{8}{10}\) = 9 \(\frac{4}{5}\)

Question 26.
3 × 5 \(\frac{3}{4}\)
Answer:
\(\frac{69}{4}\) = 17 \(\frac{1}{4}\)

Question 27.
1 \(\frac{2}{3}\) × 3 \(\frac{1}{4}\)
Answer:
\(\frac{65}{12}\) = 5 \(\frac{5}{12}\)

Question 28.
2 \(\frac{3}{5}\) × 3
Answer:
\(\frac{39}{5}\) = 7 \(\frac{4}{5}\)

Question 29.
1 \(\frac{5}{7}\) × 3 \(\frac{1}{2}\)
Answer:
\(\frac{84}{14}\) = 6

Question 30.
3 \(\frac{1}{3}\) × 1 \(\frac{9}{10}\)
Answer:
\(\frac{190}{30}\) = 6 \(\frac{10}{30}\) = 6 \(\frac{1}{3}\)

Engage NY Eureka Math Grade 6 Module 5 Lesson 10 Answer Key

Eureka Math Grade 6 Module 5 Lesson 10 Exercise Answer Key

Opening Exercise:

Question a.
Find the area and perimeter of this rectangle:

Answer:
A = bh = 9 cm × 5 cm = 45 cm²
P = 9 cm + 9 cm + 5 cm + 5 cm = 28 cm

Question b.
Find the width of this rectangle. The area is 1.2 m², and the length is 1.5 m.

Answer:
A = l × w
1.2 m² = 1.5 m × w
\(\frac{1.2 m^{2}}{1.5 m}\) = \(\frac{1.5 \mathrm{~m} \times \mathrm{w}}{1.5 \mathrm{~m}}\)
0.8 m = w

Eureka Math Grade 6 Module 5 Lesson 10 Example Answer Key

Example: Student Desks or Tables

Question 1.
Measure the dimensions of the top of your desk.
Answer:

Question 2.
How do you find the area of the top of your desk?
Answer:

Question 3.
How do you find the perimeter?
Answer:

Question 4.
Record these on your paper in the appropriate column.
Answer:

Exploratory Challenge:

Question 1.
Estimate and predict the area and perimeter of each object. Then measure each object, and calculate both the area and perimeter of each.

Answer:

Optional Challenge:

Answer:

Eureka Math Grade 6 Module 5 Lesson 10 Problem Set Answer Key

Question 1.
How is the length of the side of a square related to its area and perimeter? The diagram below shows the first four squares stacked on top of each other with their upper left-hand corners lined up. The length of one side of the smallest square is 1 foot.

a. Complete this chart calculating area and perimeter for each square.

Answer:

b. In a square, which numerical value is greater, the area or the perimeter?
Answer:
It depends. For side length less than 4 feet, perimeter is greater; however, for side length greater than 4 feet, area is greater.

c. When is the numerical value of a square’s area (in square units) equal to Its perimeter (in units)?
Answer:
When the side length is exactly 4 feet.

d. Why is this true?
Answer:
n² = 4n is only true when n = 4.

Question 2.
This drawing shows a school pool. The walkway around the pool needs special non skid strips installed but only at the edge of the pool and the outer edges of the walkway.

a. Find the length of nonskid strips that is needed for the job.
Answer:
50 m+ 50 m + 15 m + 15 m + 90 m + 90 m + 25 m + 25 m = 360 m

b. The nonskid strips are sold only in rolls of 50 m. How many rolls need to be purchased for the job?
Answer:
360 m ÷ 50 \(\frac{\mathrm{m}}{\text { roll }}\) = 7.2 rolls
Therefore, 8 rolls need to be purchased.

Question 3.
A homeowner called in a painter to paint the walls and ceiling of one bedroom. His bedroom is 18 ft. long, 12 ft. wide, and 8 ft. high. The room has two doors, each 3 ft. by 7 ft., and three windows each 3 ft. by 5 ft. The doors and windows will not be painted. A gallon of paint can cover 300 ft². A hired painter claims he needs a minimum of 4 gallons. Show that his estimate is too high.
Answer:
Area of 2 long walls:        2(18 ft. × 8 ft.) = 288 ft²
Area of 2 short walls:       2(12 ft. × 8 ft.) = 192 ft²
Area of ceiling:                18 ft. × 12 ft. = 216 ft²
Area of2 doors:                2(3 ft. × 7 ft.) 42 ft²
Area of 3 windows:          3(3 ft. × 5 ft.) = 45 ft²
Area to be painted:          (288 ft² + 192 ft² + 216 ft²) – (42 ft² + 45 ft²2) = 609 ft²
Gallons of point needed:  609 + 300 = 2.03

The painter will need a little more than 2 gallons. The pointer’s estimate for how much paint Is necessary was too high.

Question 4.
Theresa won a gardening contest and was awarded a roll of deer-proof fencing. The fencing is 36 feet long. She and her husband, John, discuss how to best use the fencing to make a rectangular garden. They agree that they should only use whole numbers of feet for the length and width of the garden.

a. What are all of the possible dimensions of the garden?
Answer:

b. Which plan yields the maximum area for the garden? Which plan yields the minimum area?
Answer:

The 9 ft. by 9 ft. garden would have the maximum area (81 ft²), while the 17 ft. by 1 ft. garden would have only 17 ft² of garden space.

Question 5.
Write and then solve the equation to find the missing value below.

Answer:
A = l × w
182 m² = 1.4m × w
\(\frac{1.82 \mathrm{~m}^{2}}{1.4 \mathrm{~m}}\) = w
1.3 m = w

Question 6.
Challenge: This is a drawing of the flag of the Republic of the Congo. The area of this flag is 3\(\frac{3}{4}\) ft².

a. Using the area formula, tell how you would determine the value of the base. This figure is not drawn to scale.
Answer:
A = bh
A ÷ h = b
3\(\frac{3}{4}\) ft² ÷ 1\(\frac{1}{2}\)ft. = b
2\(\frac{1}{2}\) ft. = b

b. Using what you found in part (a), determine the missing value of the base.
Ans:
2\(\frac{1}{2}\) ft. = 1\(\frac{1}{2}\) ft. + x
1 ft. = x

Eureka Math Grade 6 Module 5 Lesson 10 Exit Ticket Answer Key

Question 1.
The local school is building a new playground. This plan shows the part of the playground that needs to be framed with wood for the swing set. The unit of measure is feet. Determine the number of feet of wood needed to frame the area.

Answer:
Perimeter: 10 ft. + 6 ft. + 6 ft. + 3 ft. + 3 ft. + 8 ft. + 8 ft. + 4 ft. = 48 ft.

Question 2.
The school wants to fill the area enclosed with wood with mulch for safety. Determine the number of square feet that needs to be covered by the mulch.
Answer:

Area of Left Rectangle = bh = (6 ft. x 10 ft.) = 60 ft²
Area of Right Rectangle = bh = (8 ft. X 4 ft.) = 32 ft²
Total Area = 60 ft² + 32 ft² = 92 ft²

Engage NY Eureka Math Grade 6 Module 5 Lesson 9 Answer Key

Eureka Math Grade 6 Module 5 Lesson 9 Example Answer Key

Example 1
Jasjeet has made a scale drawing of a vegetable garden she plans to make in her backyard. She needs to determine the perimeter and area to know how much fencing and dirt to purchase. Determine both the perimeter and area.

Answer:

AB = 4 units BC = 7 units CD = 4 units
DE = 6 units EF = 8 units AF = 13 units

Perimeter = 4 units + 7 units + 4 units + 6 units + 8 units + 13 units
Perimeter = 42 units

The area is determined by making a horizontal cut from (1, 1) to point C.
Area of Top
A = lw
A = (4 units)(7 units)
A = 28 units²

Area of Bottom
A = lw
A = (8 units) (6 units)
A = 48 units²

Total Area = 28 units² + 48 units²
Total Area = 76 units²

Example 2:

Calculate the area of the polygon using two different methods. Write two expressions to represent the two methods, and compare the structure of the expressions.

Answer:
Answers will vary. The following are two possible methods. However, students could also break the shape into two
triangles and a rectangle or use another correct method.

Method one:

Method Two:

Area of Triangle 1 and 4
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (4 units)(3 units)
A = \(\frac{1}{2}\) (12 units²)
A = 6 units2
Since there are 2, we have a total area of 12 units².

Area of Triangle 2 and 3
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (8 units)(3 units)
A = \(\frac{1}{2}\) (24 units²)
A = 12 units2
Since there are 2, we have a total area of 24 units².

A = lw
A = (8 units) (6 units)
A = 48 units²

A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (2 units)(3 units)
A = 3 units²

There are 4 triangles of equivalent base and height.
4(3 units2) = 12 units²

Total Area = 48 units² – 12 units²
Total area = 36 units²

Total Area = 12 units² + 24 units² = 36 units²
Total area = 36 units²

Expressions:
2(\(\frac{1}{2}\)(4)(3)) + 2(\(\frac{1}{2}\)(8)(3)) or (8)(6) – 4(\(\frac{1}{2}\)(2)(3))

The first expression shows terms being added together because I separated the hexagon into smaller pieces and had to add their areas back together.
The second expression shows terms being subtracted because I made a larger outside shape and then had to take away the extra pieces.

Eureka Math Grade 6 Module 5 Lesson 9 Exercise Answer Key

Question 1.
Determine the area of the following shapes.

a.
Answer:

Area of rectangle = lw
A = (10 units) (9 units)
A = 90 units ²

Area of Triangle
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (3 units) (3 units)
A = 4.5 units ²

4 triangles with equivalent base and height
4 (4.5 units ²) = 18 units²

Area = 90 units² – 18 units²
Area = 72 units²

Please note the students may also choose to solve by decomposing. Here is another option.

b.
Answer:

Area of Triangle 1
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (8 units)(3 units)
A = (4 units)(3 units)
A = 12 units²

Area of Triangles 2 and 4
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (5 units)(3 units)
A = \(\frac{1}{2}\) (15 units²)
A = 7.5 units²

Since triangles 2 and 4 have the same base and height measurements, the combined area is 15 units.

Area of Rectangle 3
A = bh
A = (5 units)(2 units)
A = 10 units²

Total Area = 12 units² + 15 units² + 10 units²
Total Area = 37 units²

Another correct solution might start with the following diagram:

Question 2.
Determine the area and perimeter of the following shapes.

a.
Answer:

Area of Large Square

A = s²
A = (10 units)²
A = 100units²

Removed Piece
A = bh
A = (6 units)(4 units)
A = 24 units²

Area = 100 units² – 24 units²
Area = 76 units²

Perimeter = 10 units + 6 units + 6 units + 4 units + 4 units + 10 units
Perimeter = 40 units

Other correct solutions might start with the following diagrams:

b.
Answer:

Area:

Horizontal Area
A = bh
A = (15 units) (6 units)
A = 90 units²

Vertical Area
A = bh
A = (10 units) (4 units)
A = 40 units²

Total Area = 90 units² + 40 units²
Total Area = 130 units²

Perimeter = 15 units + 6 units + 4 units + 10 units + 4 units + 10 units + 7 units + 6 units
Perimeter = 62 units

Other correct solutions might start with the following diagrams:

Eureka Math Grade 6 Module 5 Lesson 9 Problem Set Answer Key

Question 1.
Determine the area of polygon.

Answer:

Area of Triange 1
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (5 units)(8 units)
A = \(\frac{1}{2}\) (40 units²)
A = 20 units²

Area of Triangle 2
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (12 units)(8 units)
A = \(\frac{1}{2}\) (96 units²)
A = 48 units²

Area of Tnangle 3
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (12 units)( 5 units)
A = \(\frac{1}{2}\) (60 units²)
A = 30 units²

Total Area = 20 units² + 48 units² + 30 units²
Total Area = 98 units²

Question 2.
Determine the area and perimeter of the polygon.

Answer:

Area:

Horizontal Rectangle
A = bh
A = (13 units) (5 units)
A = 65 units²

Vertical Rectangle
A = bh
A = (5 units) (9 units)
A = 45 units²

Square
A = S²
A = (2 units)²
A = 4 units²

Total Area = 65 units² + 45 units² +4 units²
Total Area = 114 units²

Perimeter:

Perimeter = 2 units + 2 units + 7 units + 13 units + 14 units + 5 units + 9 units + 6 units
Perimeter = 58 units

Question 3.
Determine the area of the polygon. Then, write an expression that could be used to determine the area.

Answer:

Area of Rectangle on Left
A = lw
A = (8 units) (7 units)
A = 56 units²

Area of Rectangle on Right
A = lw
A = (5 units) (8 units)
A = 40 units²

Area of Triangle on Top
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (5 units)(5 units)
A = 12.5 units²

Total Area = 56 units² + 40 units² + 12.5 units² = 108.5 units²
Expression: (8)(7) + (5)(8) + \(\frac{1}{2}\) (5)(5)

Question 4.
If the length of each square was worth 2 instead of 1, how would the area in Problem 3 change? How would your expression change to represent this area?
Answer:
If each length is twice as long, when they are multiplied, 2l × 2w = 41w. Therefore, the area will be four times larger when the side lengths are doubled.

I could multiply my entire expression by 4 to make it 4 times as big. 4 [(8)(7) + (5)(8) + \(\frac{1}{2}\) (5)(5)]

Question 5.
Determine the area of the polygon. Then, write an expression that represents the area.

Answer:

Area of Outside Rectangle
A = lw
A = (9 units) (16 units)
A = 144 units²

Area of Rectangle on Left
A = lw
A = (4 units) (8 units)
A = 32 units²

Area of Rectangle on Right
A = lw
A = (4 units) (3 units)
A = 12 units²

Total Area = 144 units² – 32 units² – 12 units²
Total Area = 100 units²
Expression: (9)(16) – (4)(8) – (4)(3)

Question 6.
Describe another method you could use to find the area of the polygon in Problem 5. Then, state how the expression for the area would be different than the expression you wrote.
Answer:
I could have broken up the large shape into many smaller rectangles. Then I would need to add all the areas of these rectangles together to determine the total area.

My expression showed subtraction because I created o rectangle that was larger than the original polygon, and then I had to subtract the extra oreos. If I break the shape into pieces, I would need to add the terms together instead of subtracting them to get the total area.

Question 7.
Write one of the letters from your name using rectangles on the coordinate plane. Then, determine the area and perimeter. (For help see Exercise 2(b). This irregular polygon looks sort of like a T.)
Answer:

Answers will vary.

Eureka Math Grade 6 Module 5 Lesson 9 Exit Ticket Answer Key

Question 1.
Determine the area and perimeter of the figure below. Note that each square unit is 1 unit in length.

Answer:

Area:

Area of Large Rectangle
A = bh
A = (11 units)(13 units)
A = 143 units²

Area of Small Square
A = s²
A = (4units)²
A = 16 units²

Area of Irregular Shape
A = 143 units² – 16 units²
A = 127 units²

Perimeter = 13 units + 4 units + 4 units + 4 units + 4 units + 3 units + 13 units + 11 units
Perimeter = 56 units

Other correct solutions might start with the following diagrams:

Eureka Math Grade 6 Module 5 Lesson 9 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
Answer:
m = 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 1:

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: 25
m =

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

Question 14.
m – 14 = 27
Answer:
m = 13

Question 15.
m – 23 = 35
Answer:
m = 12

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 21.
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: 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 17 Answer Key

Eureka Math Grade 6 Module 3 Lesson 17 Example Answer Key

Example 1.
Locate and label the points {(3, 2), (8, 4), (- 3, 8), (- 2, – 9), (0, 6), (- 1, – 2), (10, – 2)) on the grid above.
Answer:

Example 2.
Drawing the Coordinate Plane Using an Increased Number Scale for One Axis
Draw a coordinate plane on the grid below, and then locate and label the following points:
{(- 4, 20), (- 3, 35), (1, – 35), (6, 10), (9, – 40)}.
Answer:

Example 3.
Drawing the Coordinate Plane Using a Decreased Number Scale for One Axis
Draw a coordinate plane on the grid below, and then locate and label the following points:
{(0. 1, 4), (0. 5, 7), (- 0.7, – 5), (- 0.4, 3), (0.8, 1))}.
Answer:

Example 4.
Drawing the Coordinate Plane Using a Different Number Scale for Both Axes
Determine a scale for the x-axis that will allow all x-coordinates to be shown on your grid.
Answer:
The grid is 16 units wide, and the x-coordinates range from – 14 to 14. If I let each grid line represent 2 units, then the x-axis will range from – 16 to 16.

Determine a scale for the y-axis that will allow all y-coordinates to be shown on your grid.
Answer:
The grid is 16 units high, and the y-coordinates range from – 4 to 3. 5. I could let each grid line represent one unit, but if I let each grid line represent \(\frac{1}{2}\) of a unit, the points will be easier to graph.

Draw and label the coordinate plane, and then locate and label the set of points.
{(- 14, 2), (- 4, – 0. 5), (6,- 3. 5), (14, 2. 5), (0, 3.5), (- 8, – 4)}
Answer:

Eureka Math Grade 6 Module 3 Lesson 17 Problem Set Answer Key

Question 1.
Label the coordinate plane, and then locate and label the set of points below.
{(0.3, 0.9), (- 0.1, 0.7), (- 0.5, – 0.1), (- 0.9, 0.3), (0, – 0.4)}
Answer:

Question 2.
Label the coordinate plane, and then locate and label the set of points below.
{(90, 9), (- 110, – 11), (40, 4), (- 60, – 6), (- 80, – 8)}
Answer:

Extension:

Question 3.
Describe the pattern you see in the coordinates In Problem 2 and the pattern you see in the points. Are these patterns consistent for other points too?
Answer:
The x-coordinate for each of the given points is 10 times its y-coordinate. When I graphed the points, they appear to make a straight line. I checked other ordered pairs with the same pattern, such as (- 100, – 10), (20, 2), and even (0, 0), and it appears that these points are also on that line.

Eureka Math Grade 6 Module 3 Lesson 17 Exit Ticket Answer Key

Question 1.
Determine an appropriate scale for the set of points given below. Draw and label the coordinate plane, and then locate and label the set of points.
{(10, 0. 2), (- 25, 0.8), (0, – 0.4), (20, 1), (- 5, – 0. 8)}
Answer:
The x-coordinates range from – 25 to 20. The grid is 10 units wide. If I let each grid line represent 5 units, then the x-axis will range from – 25 to 25.

The y-coordinates range from – 0.8 to 1. The grid is 10 units high. If I let each grid line represent two-tenths of a unit, then the y-axis will range from – 1 to 1.

Eureka Math Grade 6 Module 3 Lesson 17 Opening Exercise Answer Key

Question 1.
Draw all necessary components of the coordinate plane on the blank 20 × 20 grid provided below, placing the origin at the center of the grid and letting each grid line represent 1 unit.

Answer:

Engage NY Eureka Math 6th Grade Module 3 Lesson 16 Answer Key

Eureka Math Grade 6 Module 3 Lesson 16 Example Answer Key

Example 1.
Extending Opposite Numbers to the Coordinate Plane
Extending Opposite Numbers to the Coordinates of Points on the Coordinate Plane Locate and label your points on the coordinate plane to the right. For each given pair of points in the table below, record your observations and conjectures in the appropriate cell. Pay attention to the absolute values of the coordinates and where the points lie in reference to each axis.



Answer:

Examples 2 – 3: Navigating the Coordinate Plane

Answer:

Eureka Math Grade 6 Module 3 Lesson 16 Exercise Answer Key

Exercises

In each column, write the coordinates of the points that are related to the given point by the criteria listed in the first column of the table. Point S(5, 3) has been reflected over the x- and y-axes for you as a guide, and its images are shown on the coordinate plane. Use the coordinate grid to help you locate each point and its corresponding coordinates.


Answer:

Exercise 1.
When the coordinates of two points are (x, y) and (- x, y), what line of symmetry do the points share? Explain.
Answer:
They share the y-axis because the y-coordinates are the same and the x-coordinates are opposites, which means the points will be the same distance from the y-axis but on opposite sides.

Exercise 2.
When the coordinates of two points are (x, y) and (x, – y). what line of symmetry do the points share? Explain.
Answer:
They share the x-axis because the x-coordinates are the same and the y-coordinates are opposites, which means the points will be the same distance from the x-axis but on opposite sides.

Eureka Math Grade 6 Module 3 Lesson 16 Problem Set Answer Key

Question 1.
Locate a point In Quadrant IV of the coordinate plane. Label the point A, and write its ordered pair next to it.
Answer:
Answers will vary; Quadrant IV (5, – 3)

a. Reflect point A over an axis so that its image is in Quadrant III. Label the image B, and write its ordered pair next to it. Which axis did you reflect over? What is the only difference in the ordered pairs of points A and B?
Answer:
B(- 5, – 3); reflected over the y-axis
The ordered pairs differ only by the sign of their x-coordinates: A(5, – 3) and B(- 5, – 3).

b. Reflect point B over an axis so that its image is in Quadrant II. Label the image C, and write its ordered pair next to it. Which axis did you reflect over? What is the only difference in the ordered pairs of points B and C? How does the ordered pair of point C relate to the ordered pair of point A?
Answer:
C(- 5, 3); reflected over the x-axis
The ordered pairs differ only by the signs of their y-coordinates: B(- 5, – 3) and C(- 5, 3).
The ordered pair for point C differs from the ordered pair for point A by the signs of both coordinates:
A(5, – 3) and C(- 5, 3).

c. Reflect point C over an axis so that its image is in Quadrant I. Label the image D, and write its ordered pair next to it. Which axis did you reflect over? How does the ordered pair for point D compare to the ordered pair for point C? How does the ordered pair for point D compare to points A and B?
Answer:
D(5, 3); reflected over the y-axis again
Point D differs from point C by only the sign of its x-coordinate: D(5, 3) and C(- 5, 3).
Point D differs from point B by the signs of both coordinates: D(5, 3) and B(- 5, – 3).
Point D differs from point A by only the sign of the y-coordinate: D(5, 3) and A(5, – 3).

Question 2.
Bobbie listened to her teacher’s directions and navigated from the point (- 1,0) to (5, – 3). She knows that she has the correct answer, but she forgot part of the teacher’s directions. Her teacher’s directions included the following:
“Move 7 units down, reflect about the   ?   -axis, move up 4 units, and then move right 4 units.”
Help Bobbie determine the missing axis in the directions, and explain your answer.
Answer:
The missing line is a reflection over the y-axis. The first line would move the location to (- 1, – 7). A reflection over the y-axis would move the location to (1, – 7) in Quadrant IV, which is 4 units left and 4 units down from the end point (5, – 3).

Eureka Math Grade 6 Module 3 Lesson 16 Exit Ticket Answer Key

Question 1.
How are the ordered pairs (4, 9) and (4, – 9) similar, and how are they different? Are the two points related by a reflection over an axis in the coordinate plane? If so, indicate which axis is the line of symmetry between the points. If they are not related by a reflection over an axis in the coordinate plane, explain how you know.
Answer:
The x-coordinates are the same, but the y-coordinates are opposites, meaning they are the same distance from zero on the x-axis and the same distance but on opposite sides of zero on the y-axis. Reflecting about the x-axis interchanges these two points.

Question 2.
Given the point (- 5, 2), write the coordinates of a point that is related by a reflection over the x- or y-axis. Specify which axis is the line of symmetry.
Answer:
Using the x-axis as a line of symmetry, (-3, -2); using the y-axis as a line of symmetry, (5,2)

Eureka Math Grade 6 Module 3 Lesson 16 Opening Exercise Answer Key

Question 1.
Give an example of two opposite numbers, and describe where the numbers lie on the number line. How are opposite numbers similar, and how are they different?
Answer:
Answers may vary. 2 and – 2 are opposites because they are both 2 units from zero on a number line but in opposite directions. Opposites are similar because they have the same absolute value, but they are different because opposites are on opposite sides of zero.

Engage NY Eureka Math 6th Grade Module 3 Lesson 15 Answer Key

Eureka Math Grade 6 Module 3 Lesson 15 Example Answer Key

Example 1. Extending the Axes Beyond Zero
The point below represents zero on the number line. Draw a number line to the right starting at zero. Then, follow directions as provided by the teacher.

Answer:
→ Use a straightedge to extend the x-axis to the left of zero to represent the real number line horizontally, and complete the number line using the same scale as on the right side of zero.
→ Describe the y-axis. What types of numbers should it include?
The y-axis is a vertical number line that includes numbers on both sides of zero (above and below), and so it includes both positive and negative numbers.
→ Use a straightedge to draw a vertical number line above zero.

Provide students with time to draw.
→ Extend the y-axis below zero to represent the real number line vertically, and complete the number line using
the same scale as above zero.

Example 2: Components of the Coordinate Plane
All points on the coordinate plane are described with reference to the origin. What is the origin, and what are its coordinates?
Answer:
The origin is the point where the x- and y-axes intersect. The coordinates of the origin are (0, 0).

To describe locations of points in the coordinate plane, we use _________________________ of numbers. Order is important, so on the coordinate plane, we use the form ( ). The first coordinate represents the point’s location from zero on the ________-axis, and the second coordinate represents the point’s location from zero on the ________-axis.
Answer:
To describe locations of points in the coordinate plane, we use   ordered   pairs of numbers. Order is important, so on the coordinate plane, we use the form (   x, y   ). The first coordinate represents the point’s location from zero on the    x   -axis, and the second coordinate represents the point’s location from zero on the   y   -axis.

Eureka Math Grade 6 Module 3 Lesson 15 Exercise Answer Key

Exercise 1.
Use the coordinate plane below to answer parts (a) – (c).
a. Graph at least five points on the x-axis, and label their coordinates.
Answer:
Points will vary.

b. What do the coordinates of your points have in common?
Answer:
Each point has a y-coordinate of 0.

c. What must be true about any point that lies on the x-axis? Explain.
Answer:
If a point lies on the x-axis, its y-coordinate must be 0 because the point is located 0 units above or below the x-axis. The x-axis intersects the y-axis at 0.

Answer:

Exercise 2.
Use the coordinate plane to answer parts (a) – (c).
a. Graph at least five points on the y-axis, and label their coordinates.
Answer:
Points will wary.

b. What do the coordinates of your points have in common?
Answer:
Each point has an x-coordinate of 0.

c. What must be true about any point that lies on the y-axis? Explain.
Answer:
If a point lies on the y-axis, its x-coordinate must be 0 because the point is located 0 units left or right of the y-axis. The y-axis intersects 0 on the x-axis.

Exercise 3.
If the origin is the only point with 0 for both coordinates, what must be true about the origin?
Answer:
The origin is the only point that is on both the x-axis and the y-axis.

Exercise 4.
Locate and label each point described by the ordered pairs below. Indicate which of the quadrants the points lie in.
a. (7, 2)
Answer:
Quadrant I

b. (3, – 4)
Answer:
Quadrant IV

c. (1, – 5)
Answer:
Quadrant IV

d. (- 3, 8)
Answer:
Quadrant II

e. (- 2, – 1)
Answer:
Quadrant III

Answer:

Exercise 5.
Write the coordinates of at least one other point in each of the four quadrants.
a. Quadrant I
Answer:
Answers will vary, but both numbers must be positive.

b. Quadrant II
Answer:
Answers will vary, but the x-coordinate must be negative, and the y-coordinate must be positive.

c. Quadrant III
Answer:
Answers will vary, but both numbers must be negative.

d. Quadrant IV
Answer:
Answers will vary, but the x-coordinate must be positive, and the y-coordinate must be negative.

Exercise 6.
Do you see any similarities in the points within each quadrant? Explain your reasoning.
Answer:
The ordered pairs describing the points in Quadrant I contain both positive values. The ordered pairs describing the points in Quadrant III contain both negative values. The first coordinates of the ordered pairs describing the points in Quadrant II are negative values, but their second coordinates are positive values. The first coordinates of the ordered pairs describing the points in Quadrant IV are positive values, but their second coordinates are negative values.

Eureka Math Grade 6 Module 3 Lesson 15 Problem Set Answer Key

Question 1.
Name the quadrant in which each of the points lies. If the point does not lie in a quadrant, specify which axis the point lies on.
a. (- 2, 5)
Quadrant II

b. (8, – 4)
Quadrant IV

c. (- 1, – 8)
Quadrant Ill

d. (9. 2, 7)
Quadrant I

e. (0, – 4)
None; the point is not in a quadrant because it lies on the y-axis.

Question 2.
Jackie claims that points with the same x- and y-coordinates must lie in Quadrant I or Quadrant Ill. Do you agree or disagree? Explain your answer.
Answer:
Disagree; most points with the same x- and y-coordinates lie in Quadrant I or Quadrant III, but the origin (o, 0) is on the x- and y-axes, not in any quadrant.

Question 3.
Locate and label each set of points on the coordinate plane. Describe similarities of the ordered pairs in each set, and describe the points on the plane.
a. {(- 2, 5), (- 2, 2), (- 2, 7), (- 2, – 3), (- 2, – 0. 8))
Answer:
The ordered pairs all have x-coordinates of – 2, and the points lie along a vertical line above and below – 2 on the x-axis.

b. {(- 9, 9), (- 4, 4), (- 2, 2), (1, – 1), (3, – 3), (0, 0)}
Answer:
The ordered pairs each have opposite values for their x- and y-coordinates. The points in the plane line up diagonally through Quadrant II, the origin, and Quadrant IV.

c. {(- 7, – 8), (5, – 8), (0, – 8), (10, – 8), (- 3, – 8)}
Answer:
The ordered pairs all have y-coordinates of – 8, and the points lie along a horizontal line to the left and right of – 8 on the y-axis.

Answer:

Question 4.
Locate and label at least five points on the coordinate plane that have an x-coordinate of 6.
a. What is true of the y-coordinates below the x-axis?
Answer:
The y-coordinates are all negative values.

b. What is true of the y-coordinates above the x-axis?
Answer:
The y-coordinates are all positive values.

c. What must be true of the y-coordinates on the x-axis?
Answer:
The y-coordinates on the x-axis must be 0.

Answer:

Eureka Math Grade 6 Module 3 Lesson 15 Exit Ticket Answer Key

Question 1.
Label the second quadrant on the coordinate plane, and then answer the following questions:
a. Write the coordinates of one point that lies in the second quadrant of the coordinate plane.
Answer:
Answers will vary.

b. What must be true about the coordinates of any point that lies in the second quadrant?
Answer:
The x-coordinate must be a negative value, and the y-coordinate must be a positive value.

Answer:

Question 2.
Label the third quadrant on the coordinate plane, and then answer the following questions:
a. Write the coordinates of one point that lies in the third quadrant of the coordinate plane.
Answer:
Answers will vary.

b. What must be true about the coordinates of any point that lies in the third quadrant?
Answer:
The x- and y-coordinates of any point in the third quadrant must both be negative values.

Question 3.
An ordered pair has coordinates that have the same sign. In which quadrant(s) could the point lie? Explain.
Answer:
The point would have to be located either in Quadrant I where both coordinates are positive values or in Quadrant III where both coordinates are negative values.

Question 4.
Another ordered pair has coordinates that are opposites. In which quadrant(s) could the point lie? Explain.
Answer:
The point would have to be located in either Quadrant II or Quadrant IV because those are the two quadrants where the coordinates have opposite signs. The point could also be located at the origin (0, 0) since zero is its own opposite.