Eureka Math

Engage NY Eureka Math 6th Grade Module 3 Lesson 14 Answer Key

Eureka Math Grade 6 Module 3 Lesson 14 Example Answer Key

Example 1: The Order in Ordered Pairs
The first number of an ordered pair is called the ___________ .
Answer:
first coordinate.

The second number of an ordered pair is called the ___________ .
Answer:
second coordinate.

Example 2.
Using Ordered Pairs to Name Locations
Describe how the ordered pair is being used in your scenario. Indicate what defines the first coordinate and what defines the second coordinate in your scenario.
Answer:
Ordered pairs are like a set of directions; they indicate where to go in one direction and then indicate where to go in the second direction.
→ Scenario 1: The seats in a college football stadium are arranged into 210 sections, with 144 seats in each
section. Your ticket to the game indicates the location of your seat using the ordered pair of numbers
(123,37). Describe the meaning of each number in the ordered pair and how you would use them to find your seat.

→ Scenario 2: Airline pilots use measurements of longitude and latitude to determine their location and to find airports around the world. Longitude is measured as 0 – 180° east or 0 – 180° west of a line stretching from the North Pole to the South Pole through Greenwich, England, called the prime meridian. Latitude is measured as 0 – 90° north or 0 – 90° south of the earth’s equator. A pilot has the ordered pair (90° west, 30° north). What does each number in the ordered pair describe? How would the pilot locate the airport on a map? Would there be any confusion if a pilot were given the ordered pair (90°, 30°)? Explain.

→ Scenario 3: Each room in a school building is named by an ordered pair of numbers that indicates the number of the floor on which the room lies, followed by the sequential number of the room on the floor from the main staircase. A new student at the school is trying to get to science class in room 4 – 13. Describe to the student what each number means and how she should use the number to find her classroom. Suppose there are classrooms below the main floor. How might these rooms be described?

Eureka Math Grade 6 Module 3 Lesson 14 Exercise Answer Key

The first coordinates of the ordered pairs represent the numbers on the line labeled X, and the second coordinates represent the numbers on the line labeled y.

Exercise 1.
Name the letter from the grid below that corresponds with each ordered pair of numbers below.

a. (1, 4)
Answer:
Point F

b. (0, 5)
Answer:
Point A

c. (4, 1)
Answer:
Point B

d. (8.5, 8)
Answer:
Point L

e. (5, – 2)
Answer:
Point G

f. (5, 4.2)
Answer:
Point H

g. (2,- 1)
Answer:
Point C

h. (0, 9)
Answer:
Point E

Exercise 2.
List the ordered pair of numbers that corresponds with each letter from the grid below.

a. Point M
Answer:
(5, 7)

b. Point S
Answer:
(- 2, 3)

c. Point N
Answer:
(6, 0)

d. Point T
Answer:
(- 3, 2)

e. Point P
Answer:
(0, 6)

f. Point U
Answer:
(7, 5)

g. Point Q
Answer:
(2, 3)

h. Point V
Answer:
(- 1, 6)

I. Point R
Answer:
(0, 3)

Eureka Math Grade 6 Module 3 Lesson 14 Problem Set Answer Key

Question 1.
Use the set of ordered pairs below to answer each question.
{(4, 20), (8, 4), (2, 3), (15, 3), (6, 15), (6, 30), (1, 5), (6, 18), (0, 3)}

a. Write the ordered pair(s) whose first and second coordinate have a greatest common factor of 3.
Answer:
(15, 3) and (6, 15)

b. Write the ordered pair(s) whose first coordinate is a factor of its second coordinate.
Answer:
(4, 20), (6, 30), (1, 5), and (6, 18)

c. Write the ordered pair(s) whose second coordinate is a prime number.
Answer:
(2, 3), (15, 3), (1, 5), and (0, 3)

Question 2.
Write ordered pairs that represent the location of points A, B, C, and D, where the first coordinate represents the horizontal direction, and the second coordinate represents the vertical direction.

Answer:
A (4, 1); B (1, – 3); C (6, 0); D (1, 4)

Extension:

Question 3.
Write ordered pairs of integers that satisfy the criteria in each part below. Remember that the origin is the point whose coordinates are (0, 0). When possible, give ordered pairs such that (i) both coordinates are positive, (ii) both coordinates are negative, and (iii) the coordinates have opposite signs in either order.

a. These points’ vertical distance from the origin Is twice their horizontal distance.
Answer:
Answers will vary; examples are (5, 10), (- 2, 4), (- 5, – 10), (2, – 4).

b. These points’ horizontal distance from the origin Is two units more than the vertical distance.
Answer:
Answers will vary; examples are (3, 1), (- 3, 1), (- 3,- 1), (3, – 1).

c. These points’ horizontal and vertical distances from the origin are equal, but only one coordinate is positive.
Answer:
Answers will vary; examples are (3, – 3), (- 8,8).

Eureka Math Grade 6 Module 3 Lesson 14 Exit Ticket Answer Key

Question 1.
On the map below, the fire department and the hospital have one matching coordinate. Determine the proper order of the ordered pairs In the map, and write the correct ordered pairs for the locations of the fire department and hospital. Indicate which of their coordinates are the same.
Answer:
The order of the numbers is (x, y); fire department: (6, 7) and hospital: (10, 7); they have the same second coordinate.

Answer:

Question 2.
On the map above, locate and label the location of each description below:
a. The local bank has the same first coordinate as the fire department, but its second coordinate is half of the fire department’s second coordinate. What ordered pair describes the location of the bank? Locate and label the bank on the map using point B.
Answer:
(6, 3. 5); see the map image for the correct location of point B.

b. The Village Police Department has the same second coordinate as the bank, but its first coordinate is – 2. What ordered pair describes the location of the Village Police Department? Locate and label the Village Police Department on the map using point P.
Answer:
(- 2, 3.5); see the map image for the correct location of point P.

Engage NY Eureka Math 6th Grade Module 3 Lesson 13 Answer Key

Eureka Math Grade 6 Module 3 Lesson 13 Example Answer Key

Example 1. Ordering Numbers In the Real World
A $25 credit and a $25 charge appear similar, yet they are very different.
Describe what is similar about the two transactions.
Answer:
The transactions look similar because they are described using the same number. Both transactions have the same magnitude (or absolute value) and, therefore, result in a change of $25 to an account balance.

How do the two transactions differ?
Answer:
The credit would cause an increase to an account balance and, therefore, should be represented by 25, while the charge would instead decrease an account balance and should be represented by – 25. The two transactions represent changes that are opposites.

Example 2: Using Absolute Value to Solve Real-World Problems
The captain of a fishing vessel is standing on the deck at 23 feet above sea level. He holds a rope tied to his fishing net that is below him underwater at a depth of 38 feet.

Draw a diagram using a number line, and then use absolute value to compare the lengths of rope in and out of the water.
Answer:

The captain is above the water, and the fishing net is below the water’s surface. Using the water level as reference point zero, I can draw the diagram using a vertical number line. The captain is located at 23, and the fishing net is located at – 38.

|23| = 23 and |- 38| = 38,so there is more rope underwater than above.

38 – 23 = 15
The length of rope below the water’s surface is 15 feet longer than the rope above water.

Example 3: Making Sense of Absolute Value and Statements of Inequality
A recent television commercial asked viewers, “Do you have over $10,000 in credit card debt?”

What types of numbers are associated with the word debt, and why? Write a number that represents the value from the television commercial.
Answer:
Negative numbers; debt describes money that is owed; – 10,000

Give one example of “over $10,000 in credit card debt.” Then, write a rational number that represents your example.
Answer:
Answers will vary, but the number should have a value of less than – 10,000. Credit card debt of $11,000; – 11,000

How do the debts compare, and how do the rational numbers that describe them compare? Explain.
Answer:
The example $11, 000 is greater than $10, 000 from the commercial; however, the rational numbers that represent these debt values have the opposite order because they are negative numbers. – 11,000 < – 10, 000. The absolute values of negative numbers have the opposite order of the negative values themselves.

Eureka Math Grade 6 Module 3 Lesson 13 Exercise Answer Key

Exercise 1.
Scientists are studying temperatures and weather patterns in the Northern Hemisphere. They recorded
temperatures (in degrees Celsius) in the table below as reported in emails from various participants. Represent each reported temperature using a rational number. Order the rational numbers from least to greatest. Explain why the rational numbers that you chose appropriately represent the given temperatures.

Answer:

– 13 < – 8 < – 6 < – 5 < – 4 < 0 < 2 < 12
The words “below zero” refer to negative numbers because they are located below zero on a vertical number line.

Exercise 2.
Jami’s bank account statement shows the transactions below. Represent each transaction as a rational number describing how it changes Jami’s account balance. Then, order the rational numbers from greatest to least. Explain why the rational numbers that you chose appropriately reflect the given transactions.

Answer:

5.5 > 4.08 > – 1.5 > – 3 > – 3.95 > – 12.2 > – 20
The words “debit,” “charge,” and “withdrawal” all describe transactions in which money is taken out of Jami’s account, decreasing its balance. These transactions are represented by negative numbers. The words “credit” and “deposit” describe transactions that will put money into Jami’s account, increasing its balance. These transactions are represented by positive numbers.

Exercise 3.
During the summer, Madison monitors the water level in her parents’ swimming pool to make sure it is not too far above or below normal. The table below shows the numbers she recorded In July and August to represent how the water levels compare to normal. Order the rational numbers from least to greatest. Explain why the rational numbers that you chose appropriately reflect the given water levels.

Answer:

\(-1 \frac{1}{4}<-\frac{3}{4}<-\frac{1}{2}<-\frac{3}{8}<\frac{1}{8}<\frac{1}{4}<\frac{1}{2}\)
The measurements are taken in reference to normal level, which is considered to be 0. The words “above normal” refer to the positive numbers located above zero on a vertical number line, and the words “below normal” refer to the negative numbers located below zero on a vertical number line.

Exercise 4.
Changes in the weather can be predicted by changes in the barometric pressure. Over several weeks, Stephanie recorded changes in barometric pressure seen on her barometer to compare to local weather forecasts. Her observations are recorded In the table below. Use rational numbers to record the indicated changes In the pressure in the second row of the table. Order the rational numbers from least to greatest. Explain why the rational numbers that you chose appropriately represent the given pressure changes.

Answer:

Eureka Math Grade 6 Module 3 Lesson 13 Problem Set Answer Key

Question 1.
Negative air pressure created by an air pump makes a vacuum cleaner able to collect air and dirt into a bag or other container. Below are several readings from a pressure gauge. Write rational numbers to represent each of the readings, and then order the rational numbers from least to greatest.

Answer:

– 13 < – 7.8 < – 6.3 < – 1.9 < 2 < 7.8 < 25

Question 2.
The fuel gauge in Nic’s car says that he has 26 miles to go until his tank is empty. He passed a fuel station 19 miles ago, and a sign says there is a town only 8 miles ahead. If he takes a chance and drives ahead to the town and there isn’t a fuel station there, does he have enough fuel to go back to the last station? Include a diagram along a number line, and use absolute value to find your answer.
Answer:
No, he does not have enough fuel to drive to the town and then back to the fuel station. He needs 8 miles’ worth of fuel to get to the town, which lowers his limit to 18 miles. The total distance between the fuel station and the town is 27 miles; |8| + |- 19| = 8 + 19 = 27. Nic would be9miles short on fuel. It would be safer to go back to the fuel station without going to the town first.

Eureka Math Grade 6 Module 3 Lesson 13 Exit Ticket Answer Key

Question 1.
Loni and Daryl call each other from different sides of Watertown. Their locations are shown on the number line below using miles. Use absolute value to explain who is a farther distance (in miles) from Watertown. How much closer is one than the other?

Answer:

Loni’s location is – 6, and |- 6| = 6 because – 6 is 6 units from 0 on the number line. Daryl’s location is 10, and |10| = 10 because 10 is 10 units from 0 on the number line. We know that 10 > 6, so Daryl is farther from Watertown than Loni.
10 – 6 = 4; Loni is 4 miles closer to Watertown than Daryl.

Question 2.
Claude recently read that no one has ever scuba dived more than 330 meters below sea level. Describe what this means in terms of elevation using sea level as a reference point.
Answer:
330 meters below sea level is an elevation of – 330 feet. “More than 330 meters below sea level” means that no diver has ever had more than 330 meters between himself and sea level when he was below the water’s surface while scuba diving.

Eureka Math Grade 6 Module 3 Lesson 13 Opening Exercise Answer Key

Question 1.
A radio disc jockey reports that the temperature outside his studio has changed 10 degrees since he came on the air this morning. Discuss with your group what listeners can conclude from this report.
Answer:
The report is not specific enough to be conclusive because 10 degrees of change could mean an increase or a decrease in temperature. A listener might assume the report says an increase in temperature; however, the word “changed” is not specific enough to conclude a positive or negative change.

Engage NY Eureka Math 6th Grade Module 3 Lesson 12 Answer Key

Eureka Math Grade 6 Module 3 Lesson 12 Example Answer Key

Example 1: Comparing Order of Integers to the Order of Their Absolute Values
Write an Inequality statement relating the ordered integers from the Opening Exercise. Below each integer, write its absolute value.
Answer:

Circle the absolute values that are in increasing numerical order and their corresponding integers. Describe the circled values.
Answer:

The circled integers are all positive values except zero. The positive integers and their absolute values have the same order.

Rewrite the integers that are not circled in the space below. How do these integers differ from the ones you circled?
Answer:
– 12, – 9, – 5, – 2, – 1
They are all negative integers.

Rewrite the negative integers in ascending order and their absolute values in ascending order below them.
Answer:

Describe how the order of the absolute values compares to the order of the negative integers.
Answer:
The orders of the negative integers and their corresponding absolute values are opposite.

Example 2: The Order of Negative Integers and Their Absolute Values
Draw arrows starting at the dashed line (zero) to represent each of the integers shown on the number line below. The arrows that correspond with 1 and 2 have been modeled for you.

Answer:

As you approach zero from the left on the number line, the integers ___________, but the absolute values of those integers ___________. This means that the order of negative integers is ___________ the order of their absolute values.
Answer:
As you approach zero from the left on the number line, the integers   increase   , but the absolute values of those integers    decrease  . This means that the order of negative integers is   opposite   the order of their absolute values.

Eureka Math Grade 6 Module 3 Lesson 12 Exercise Answer Key

Complete the steps below to order these numbers:

{2.1, – 4\(\frac{1}{2}\), – 6. 0.25, – 1.5, 0, 3.9, – 6.3, – 4, 2\(\frac{3}{4}\), 3.99, – 9\(\frac{1}{4}\)}

a. Separate the set of numbers into positive rational numbers, negative rational numbers, and zero in the top cells below (order does not matter).
Answer:

b. Write the absolute values of the rational numbers (order does not matter) in the bottom cells below.

Answer:

c. Order each subset of absolute values from least to greatest.

Answer:

d. Order each subset of rational numbers from least to greatest.

Answer:

e. Order the whole given set of rational numbers from least to greatest.
Answer:
– 9\(\frac{1}{4,}\), – 6.3, – 6, – 4\(\frac{1}{2}\), – 4, – 1.5, 0, 0.25, 2.1, 2\(\frac{3}{4}\), 3.9, 3.99

Exercise 2.
a. Find a set of four integers such that their order and the order of their absolute values are the same.
Answer:
Answers will vary. An example follows: 4, 6, 8, 10

b. Find a set of four integers such that their order and the order of their absolute values are opposite.
Answer:
Answers will vary. An example follows: – 10, – 8, – 6, – 4

c. Find a set of four non-integer rational numbers such that their order and the order of their absolute values are the same.
Answer:
Answers will vary. An example follows: 2\(\frac{1}{2}\), 3\(\frac{1}{2}\), 4\(\frac{1}{2}\), 5\(\frac{1}{2}\)

d. Find a set of four non-integer rational numbers such that their order and the order of their absolute values are opposite.
Answer:
Answers will vary. An example follows: – 5\(\frac{1}{2}\), – 4\(\frac{1}{2}\), – 3\(\frac{1}{2}\), – 2\(\frac{1}{2}\)

e. Order all of your numbers from parts (a) – (d) in the space below. This means you should be ordering 16 numbers from least to greatest.
Answer:
Answers will vary. An example follows:
– 10, – 8, – 6, – 5\(\frac{1}{2}\), – 4\(\frac{1}{2}\), – 4, – 3\(\frac{1}{2}\), – 2\(\frac{1}{2}\), 2\(\frac{1}{2}\), 3\(\frac{1}{2}\), 4, 4\(\frac{1}{2}\), 5\(\frac{1}{2}\), 6, 8, 10

Eureka Math Grade 6 Module 3 Lesson 12 Problem Set Answer Key

Question 1.
Micah and Joel each have a set of five rational numbers. Although their sets are not the same, their sets of numbers have absolute values that are the same. Show an example of what Micah and Joel could have for numbers. Give the sets in order and the absolute values in order.
Answer:
Examples may vary, If Micah had 1, 2, 3, 4, 5, then his order of absolute values would be the same: 1, 2, 3, 4, 5. If Joel had the numbers – 5, – 4, – 3, -2, -1, then his order of absolute values would also be 1, 2, 3, 4, 5.

Enrichment Extension: Show an example where Micah and Joel both have positive and negative numbers.
Answer:
If Micah had the numbers: – 5, – 3, – 1, 2, 4, his order of absolute values would be 1, 2, 3, 4, 5. If Joel hod the numbers – 4, – 2, 1, 3, 5, then the order of his absolute values would also be 1, 2, 3, 4, 5.

Question 2.
For each pair of rational numbers below, place each number In the Venn diagram based on how it compares to the other.
a. – 4, – 8
b. 4,8
C. 7,- 3
d. – 9,2
e. 6,1
f. – 5,5
g. – 2,0
Answer:

Eureka Math Grade 6 Module 3 Lesson 12 Exit Ticket Answer Key

Question 1.
Bethany writes a set of rational numbers in increasing order. Her teacher asks her to write the absolute values of these numbers in Increasing order. When her teacher checks Bethany’s work, she Is pleased to see that Bethany has not changed the order of her numbers. Why is this?
Answer:
All of Bethany’s rational numbers are positive or 0. The positive rational numbers have the same order as their absolute values. If any of Bethany’s rational numbers are negative, then the order would be different.

Question 2.
Mason was ordering the following rational numbers In math class: – 3. 3, – 15, – 8\(\frac{8}{9}\).
a. Order the numbers from least to greatest.
Answer:
– 15, – 8\(\frac{8}{9}\), – 3.3

b. List the order of their absolute values from least to greatest.
Answer:
3.3, 8\(\frac{8}{9}\), 15

c. Explain why the orderings in parts (a) and (b) are different.
Answer:
Since these are all negative numbers, when I ordered them from least to greatest, the one farthest away from zero (farthest to the left on the number line) came first. This number is – 15. Absolute value is the numbers’ distance from zero, and so the number farthest away from zero has the greatest absolute value, so 15 will be greatest in the list of absolute values, and so on.

Eureka Math Grade 6 Module 3 Lesson 12 Opening Exercise Answer Key

Record your integer values in order from least to greatest in the space below.
Answer:
Sample answer: – 12, – 9, – 5, – 2, – 1, 0, 2, 5, 7, 8

Engage NY Eureka Math 6th Grade Module 3 Lesson 11 Answer Key

Eureka Math Grade 6 Module 3 Lesson 11 Example Answer Key

Example 1. The Absolute Value of a Number
The absolute value often is written as |10|. On the number line, count the number of units from 10 to 0. How many units is 10 from 0?
Answer:
|10| = 10

Answer:

What other number has an absolute value of 10? Why?
Answer:
|- 10| = 10 because – 10 is 10 units from zero and – 10 and 10 are opposites.

The   absolute   value of a number is the distance between the number and zero on the number line.

Example 2. Using Absolute Value to Find Magnitude
Mrs. Owens received a call from her bank because she had a checkbook balance of – $45. What was the magnitude of the amount overdrawn?
Answer:
|- 45| = 45
Mrs. Owens overdrew her checking account by $45.

The   magnitude    of a measurement is the absolute value of its measure.

Eureka Math Grade 6 Module 3 Lesson 11 Exercise Answer Key

Exercise 1 – 3
Complete the following chart.

Answer:

For each scenario below, use absolute value to determine the magnitude of each quantity.

Exercise 4.
Maria was sick with the flu, and her weight change as a result of it is represented by – 4 pounds. How much weight did Maria lose?
Answer:
|-4| = 4 Maria lost 4 pounds.

Exercise 5.
Jeffrey owes his friend $5. How much is Jeffrey’s debt?
Answer:
|- 5| = 5 Jeffrey has a $5 debt.

Exercise 6.
The elevation of Niagara Falls, which is located between Lake Erie and Lake Ontario, is 326 feet. How far is this above sea level?
Answer:
|326| = 326 It is 326 feet above sea level.

Exercise 7.
How far below zero is – 16 degrees Celsius?
Answer:
|- 16| = 16 – 16°C is 16 degrees below zero.

Exercise 8.
Frank received a monthly statement for his college savings account. It listed a deposit of $100 as + 100. 00. It listed a withdrawal of $25 as – 25.00. The statement showed an overall ending balance of $835. 50. How much money did Frank add to his account that month? How much did he take out? What is the total amount Frank has saved for college?
Answer:
|100| = 100 Frank added $100 to his account.
|- 25| = 25 Frank took $25 out of his account.
|835. 50| = 835. 50 The total amount of Frank’s savings for college is $835. 50.

Exercise 9.
Meg is playing a card game with her friend, lona. The cards have positive and negative numbers printed on them. Meg exclaims: “The absolute value of the number on my card equals 8.” What is the number on Meg’s card?
Answer:
|- 8| = 8 or |8| = 8
Meg either has 8 or – 8 on her card.

Exercise 10.
List a positive and negative number whose absolute value is greater than 3. Justify your answer using the number line.
Answer:
Answers may vary. |-4| = 4 and |7| = 7; 4 > 3 and 7 > 3. On a number line, the distance from zero to – 4 is 4 units. So, the absolute value of – 4 is 4. The number 4 is to the right of 3 on the number line, so 4 is greater than 3. The distance from zero to 7 on a number line is 7 units, so the absolute value of 7 is 7. Since 7 is to the right of 3 on the number line, 7 is greater than 3.

Exercise 11.
Which of the following situations can be represented by the absolute value of 10? Check all that apply.
_________ The temperature is 10 degrees below zero. Express this as an integer.
_________ Determine the size of Harold’s debt if he owes $10.
_________ Determine how far – 10 is from zero on a number line.
_________ 10 degrees is how many degrees above zero?
Answer:
The temperature is 10 degrees below zero. Express this as an integer.
  X    Determine the size of Harold’s debt if he owes $10.
  X    Determine how far – 10 is from zero on a number line.
  X    10 degrees is how many degrees above zero?

Exercise 12.
Julia used absolute value to find the distance between 0 and 6 on a number line. She then wrote a similar
statement to represent the distance between 0 and – 6. Below is her work. Is it correct? Explain.
Answer:
|6| = 6 and |- 6| = – 6
No. The distance is 6 units whether you go from 0 to 6 or 0 to – 6. So, the absolute value 0f – 6 should also be 6, but Julia said it was – 6.

Exercise 13.
Use absolute value to represent the amount, in dollars, of a $238. 25 profit.
Answer:
|1238. 25| = 238.25

Exercise 14.
Judy lost 15 pounds. Use absolute value to represent the number of pounds Judy lost.
Answer:
|- 15| = 15

Exercise 15.
In math class, Carl and Angela are debating about Integers and absolute value. Carl said two integers can have the same absolute value, and Angela said one integer can have two absolute values. Who is right? Defend your answer.
Answer:
Carl is right. An integer and Its opposite are the same distance from zero. So, they have the same absolute values because absolute value is the distance between the number and zero.

Exercise 16.
Jamie told his math teacher: “Give me any absolute value, and I can tell you two numbers that have that absolute value.” Is Jamie correct? For any given absolute value, will there always be two numbers that have that absolute value?
Answer:
No, Jamie is not correct because zero is its own opposite. Only one number has an absolute value oJO, and that would be O.

Exercise 17.
Use a number line to show why a number and its opposite have the same absolute value.
Answer:
A number and its opposite are the same distance from zero but on opposite sides. An example is 5 and – 5. These numbers are both 5 units from zero. Their distance is the same, so they have the same absolute value, 5.

Exercise 18.
A bank teller assisted two customers with transactions. One customer made a $25 withdrawal from a savings account. The other customer made a $15 deposit. Use absolute value to show the size of each transaction. Which transaction involved more money?
Answer:
|- 25| = 25 and |15| = 15. The $25 withdrawal involved more money.

Exercise 19.
Which is farther from zero: – 7\(\frac{3}{4}\) or 7\(\frac{1}{2}\)? Use absolute value to defend your answer.
Answer:
The number that is farther from 0 is – 7\(\frac{3}{4}\). This is because = |-7\(\frac{3}{4}\)| and |7\(\frac{1}{2}\)| = 7\(\frac{1}{2}\). Absolute value is a number’s distance from zero. I compared the absolute value of each number to determine which was farther from zero. The absolute value of – 7\(\frac{3}{4}\) is 7\(\frac{3}{4}\). The absolute value of 7 is 7\(\frac{1}{2}\). We know that 7\(\frac{3}{4}\) is greater than 7\(\frac{1}{2}\). Therefore, – 7\(\frac{3}{4}\) is farther from zero than 7\(\frac{1}{2}\).
Therefore, – 7\(\frac{3}{4}\) is farther from zero than 7\(\frac{1}{2}\).

Eureka Math Grade 6 Module 3 Lesson 11 Problem Set Answer Key

For each of the following two quantities in Problems 1 – 4, which has the greater magnitude? (Use absolute value to defend your answers.)

Question 1.
33 dollars and – 52 dollars
Answer:
|- 52| = 52 |33| = 33                                       52 > 33, so – 52 dollars has the greater magnitude.

Question 2.
– 14 feet and 23 feet
Answer:
|- 14| = 14 |23| = 23                                      14 < 23, so 23 feet has the greater magnitude.

Question 3.
– 24.6 pounds and – 24.58 pounds
Answer:
|- 24.6| = 24.6                                |- 24. 58| = 24.58
24.6 > 24.58, so – 24.6 pounds has the greater magnitude.

Question 4.
– 11\(\frac{1}{4}\) degrees and 11 degrees
Answer:
|-11\(\frac{1}{4}\)| = 11\(\frac{1}{4}\)                              |11| = 11
11\(\frac{1}{4}\) > 11, so – 11\(\frac{1}{4}\) degrees has the greater magnitude.

For Problems 5-7, answer true or false. If false, explain why.

Question 5.
The absolute value of a negative number will always be a positive number.
Answer:
True

Question 6.
The absolute value of any number will always be a positive number.
Answer:
False. Zero is the exception since the absolute value of zero is zero, and zero is not positive.

Question 7.
Positive numbers will always have a higher absolute value than negative numbers.
Answer:
False. A number and its opposite have the same absolute value.

Question 8.
Write a word problem whose solution is |20| = 20.
Answer:
Answers will vary. Kelli flew a kite 20 feet above the ground. Determine the distance between the kite and the ground.

Question 9.
Write a word problem whose solution is |- 70| = 70.
Answer:
Answers will vary. Paul dug a hole in his yard 70 inches deep to prepare for an in-ground swimming pool. Determine the distance between the ground and the bottom of the hole that Paul dug.

Question 10.
Look at the bank account transactions listed below, and determine which has the greatest impact on the account balance. Explain.
a. A withdrawal of $60
b. A deposit of $55
c. A withdrawal of $58.50
Answer:
|- 60| = 60                         |55| = 55                           |- 58.50| = 58.50
60 > 58. 50 > 55, so a withdrawal of $60 has the greatest impact on the account balance.

Eureka Math Grade 6 Module 3 Lesson 11 Exit Ticket Answer Key

Jessie and his family drove up to a picnic area on a mountain. In the morning, they followed a trail that led to the mountain summit, which was 2,000 feet above the picnic area. They then returned to the picnic area for lunch. After lunch, they hiked on a trail that led to the mountain overlook, which was 3, 500 feet below the picnic area.

a. Locate and label the elevation of the mountain summit and mountain overlook on a vertical number line. The picnic area represents zero. Write a rational number to represent each location.
Answer:
Picnic area:   0 
Mountain summit:   2,000  
Mountain overlook:   3,500  

b. Use absolute value to represent the distance on the number line of each location from the picnic area.
Answer:
Distance from the picnic area to the mountain summit    |2,000| =   2,000   
Distance from the picnic area to the mountain overlook:    |- 3,500| =   3,500   

c. What is the distance between the elevations of the summit and overlook? Use absolute value and your number line from part (a) to explain your answer.
Answer:
Summit to picnic area and picnic area to overlook: 2,000 + 3,500 = 5,500
There are 2,000 units from zero to 2,000 on the number line.
There are 3,500 units from zero to – 3, 500 on the number line.
Altogether, that equals 5, 500 units, which represents the distance on the number line between the two elevations. Therefore, the difference in elevations is 5,500 feet.

Eureka Math Grade 6 Module 3 Lesson 11 Opening Exercise Answer Key

Answer:
After two minutes:
→ What are some examples you found (pairs of numbers that are the same distance from zero)?
\(\frac{1}{2}\) and \(\frac{1}{2}\), 8.01 and – 8.01, – 7 and 7.
→ What is the relationship between each pair of numbers?
They are opposites.
→ How does each pair of numbers relate to zero?
Both numbers in each pair are the same distance from zero.

Engage NY Eureka Math 6th Grade Module 3 Lesson 10 Answer Key

Eureka Math Grade 6 Module 3 Lesson 10 Example Answer Key

Example 1.
Writing Inequality Statements Involving Rational Numbers
Write one inequality statement to show the relationship among the following shoe sizes: 10\(\frac{1}{2}\), 8, and 9.
a. From least to greatest:
Answer:
8 < 9 < 10\(\frac{1}{2}\)

b. From greatest to least:
Answer:
10\(\frac{1}{2}\) > 9 > 8

Example 2.
Interpreting Data and Writing Inequality Statements
Mary is comparing the rainfall totals for May, June, and July. The data Is reflected in the table below. Fill in the blanks below to create inequality statements that compare the changes in Total Rainfall for each month (the right-most column of the table).

Write one inequality to order the Changes in Total Rainfall:
From least to greatest:   – 1.4 < 0.3 < 0.5   
From greatest to least:   0.5 >0.3 > -1.4     

In this case, does the greatest number indicate the greatest change in rainfall? Explain.
Answer:
No. In this situation, the greatest change is for the month of May since the average total rainfall went down from last year by 1.4 inches, but the greatest number in the inequality statement is 0. 5.

Eureka Math Grade 6 Module 3 Lesson 10 Exercise Answer Key

Exercise 1.
Graph your answer from the Opening Exercise part (a) on the number line below.
Answer:

Exercise 2.
Also, graph the points associated with 4 and 5 on the number line.
Answer:

Exercise 3.
Explain in words how the location of the three numbers on the number line supports the inequality statements you wrote in the Opening Exercise parts (b) and (c).
Answer:
The numbers are ordered from least to greatest when I look at the number line from left to right. So, 4 is less than 4. 75, and 4.75 is less than 5.

Exercise 4.
Write one inequality statement that shows the relationship among all three numbers.

Answer:
4 < 4.75 < 5

Exercise 5.
Mark’s favorite football team lost yards on two back-to-back plays. They lost 3 yards on the first play. They lost 1 yard on the second play. Write an inequality statement using integers to compare the forward progress made on each play.
Answer:
– 3 < – 1

Exercise 6.
Sierra had to pay the school for two textbooks that she lost. One textbook cost $55, and the other cost $75. Her mother wrote two separate checks, one for each expense. Write two integers that represent the change to her mother’s checking account balance. Then, write an inequality statement that shows the relationship between these two numbers.
Answer:
– 55 and – 75; – 55 > – 75

Exercise 7.
Jason ordered the numbers – 70, – 18, and – 18. 5 from least to greatest by writing the following statement: – 18 < – 18.5 < – 70. Is this a true statement? Explain.
Answer:
No, it is not a true statement because 18 < 18. 5 < 70, so the opposites of these numbers are in the opposite order. The order should be – 70 < – 18.5 < 18.

Exercise 8.
Write a real-world situation that is represented by the following inequality: – 19 < 40. Explain the position of the numbers on a number line.
Answer:
The coldest temperature in January was – 19 degrees Fahrenheit, and the warmest temperature was 40 degrees Fahrenheit. Since the point associated with 40 is above zero on a vertical number line and – 19 is below zero, we know that 40 is greater than – 19. This means that 40 degrees Fahrenheit is warmer than – 19 degrees Fahrenheit.

Exercise 9.
A Closer Look at the Sprint
Look at the following two examples from the Sprint.

a. Fill in the numbers In the correct order.
– 1< – <\(\frac{1}{4}\) < 0 and 0 > – \(\frac{1}{4}\) > – 1

b. Explain how the position of the numbers on the number line supports the inequality statements you created.
Answer:
– 1 is the farthest left on the number line, so it is the least value. 0 is farthest right, so it is the greatest value, and – \(\frac{1}{4}\) is in between.

c. Create a new pair of greater than and less than inequality statements using three other rational numbers.
Answer:
Answers will vary. 8 > 0. 5 > – 1.8 and – 1.8 < 0.5 < 8

Eureka Math Grade 6 Module 3 Lesson 10 Problem Set Answer Key

For each of the relationships described below, write an Inequality that relates the rational numbers.

Question 1.
Seven feet below sea level Is farther below sea level than 4\(\frac{1}{2}\) feet below sea level.
Answer:
– 7 < – 4\(\frac{1}{2}\)

Question 2.
Sixteen degrees Celsius is warmer than zero degrees Celsius.
Answer:
16 > 0

Question 3.
Three and one-half yards of fabric Is less than five and one-half yards of fabric.
Answer:
3\(\frac{1}{2}\) < 5\(\frac{1}{2}\)

Question 4.
A loss of $500 in the stock market is worse than a gain of $200 in the stock market.
Answer:
– 500 < 200

Question 5.
A test score of 64 is worse than a test score of 65, and a test score of 65 is worse than a test score of 67\(\frac{1}{2}\)
Answer:
64 < 65 < 67\(\frac{1}{2}\)

Question 6.
In December, the total snowfall was 13. 2 inches, which is more than the total snowfall in October and November, which was 3.7 inches and 6. 15 inches, respectively.
Answer:
13.2 > 6.15 > 3.7

For each of the following, use the information given by the inequality to describe the relative position of the numbers on a horizontal number line.

Question 7.
– 0.2 < – 0.1
Answer:
– 0.2 is to the left of – 0.1, or – 0.1 is to the right of – 0.2.

Question 8.
8\(\frac{1}{4}\) > -8\(\frac{1}{4}\)
Answer:
8\(\frac{1}{4}\) is to the right of – 8\(\frac{1}{4}\) or – 8\(\frac{1}{4}\) is to the left of 8\(\frac{1}{4}\).

Question 9.
– 2 < 0 < 5
Answer:
– 2 is to the left of zero and zero is to the left of 5, or 5 is to the right of zero and zero is to the right of – 2.

Question 10.
– 99 > – 100
Answer:
– 99 is to the right of – 100, or – 100 is to the left of – 99.

Question 11.
– 7.6 <- 7\(\frac{1}{2}\) – 7
Answer:
– 7.6 is to the left of – 7\(\frac{1}{2}\) and – 7\(\frac{1}{2}\) is to the left of – 7, or – 7 is to the right of – 7\(\frac{1}{2}\) and – 7\(\frac{1}{2}\) is to the right of – 7.6.

Fill in the blanks with numbers that correctly complete each of the statements.

Question 12.
Three integers between -4 and 0
Answer:
– 3 < – 2 < – 1

Question 13.
Three rational numbers between 16 and 15
Answer:
15.3 < 15.6 < 15.7
Other answers are possible.

Question 14.
Three rational numbers between -1 and -2
Answer:
– 1.9 < – 1.55 < – 1.02
Other answers are possible.

Question 15.
Three integers between 2 and – 2
Answer:
– 1 < 0 < 1

Eureka Math Grade 6 Module 3 Lesson 10 Exit Ticket Answer Key

Question 1.
Kendra collected data for her science project. She surveyed people asking them how many hours they sleep during a typical night. The chart below shows how each person’s response compares to 8 hours (which is the answer she expected most people to say).

a. Plot and label each of the numbers in the right-most column of the table above on the number line below.

Answer:

b. List the numbers from least to greatest
Answer:
– 1.0, – \(\frac{1}{4}\), 0, 0.5, 1.5

c. Using your answer from part (b) and inequality symbols, write one statement that shows the relationship among all the numbers.
Answer:
– 1.0 < –\(\frac{1}{4}\) < 0 < 0.5 < 1.5 or 0.5 > 0.5 > 0 > –\(\frac{1}{4}\) > -1.0

Eureka Math Grade 6 Module 3 Lesson 10 Opening Exercise Answer Key

“The amount of money I have in my pocket is less than $5 but greater than $4.”
a. One possible value for the amount of money in my pocket is ___________ .
Answer:
$4.75

b. Write an inequality statement comparing the possible value of the money in my pocket to $4.
Answer:
4.00 < 4.75

c. Write an inequality statement comparing the possible value of the money in my pocket to $5.
Answer:
4.75 < 5.00

Eureka Math Grade 6 Module 3 Lesson 10 Inequality Statements Answer Key

Rational Numbers: Inequality Statements – Round 1
Directions: Work in numerical order to answer Problems 1 – 33. Arrange each set of numbers in order according to the inequality symbols.

Question 1.

Answer:

Question 2

Answer:

Question 3.

Answer:

Question 4.

Answer:

Question 5.

Answer:

Question 6.

Answer:

Question 7.

Answer:

Question 8.

Answer:

Question 9.

Answer:

Question 10.

Answer:

Question 11.

Answer:

Question 12.

Answer:

Question 13.

Answer:

Question 14

Answer:

Question 15.

Answer:

Question 16.

Answer:

Question 17.

Answer:

Question 18.

Answer:

Question 19.

Answer:

Question 20.

Answer:

Question 21.

Answer:

Question 22.

Answer:

Question 23.

Answer:

Question 24.

Answer:

Question 25.

Answer:

Question 26.

Answer:

Question 27.

Answer:

Question 28.

Answer:

Question 29.

Answer:

Question 30.

Answer:

Question 31.

Answer:

Question 32.

Answer:

Question 33.

Answer:

Rational Numbers: Inequality Statements – Round 2
Directions: Work in numerical order to answer Problems 1 – 33. Arrange each set of numbers in order according to the inequality symbols.

Answer:

Engage NY Eureka Math 8th Grade Module 3 Lesson 9 Answer Key

Eureka Math Grade 8 Module 3 Lesson 9 Exploratory Challenge Answer Key

Exploratory Challenge 1.
The goal is to show that if △ABC is similar to △A’ B’ C’, then △A’ B’ C’ is similar to △ABC. Symbolically,
if △ABC~△A’B’C’, then △A’ B’ C’~△ABC.

a. First, determine whether or not △ABC is in fact similar to △A’ B’ C’. (If it isn’t, then no further work needs to be done.) Use a protractor to verify that the corresponding angles are congruent and that the ratios of the corresponding sides are equal to some scale factor.
Answer:

The corresponding angles are congruent: ∠A≅∠A’, ∠B≅∠B’, and ∠C≅∠C’, therefore
|∠A|=|∠A’ |=49°, |∠B|=|∠B’ |=99°, and |∠C|=|∠C’ |=32°.
The ratios of the corresponding sides are equal: \(\frac{4}{8}\)=\(\frac{3}{6}\)=\(\frac{2}{4}\)=r.

b. Describe the sequence of dilation followed by a congruence that proves △ABC~△A’ B’ C’.
Answer:

To map △ABC onto △A’ B’ C’, dilate △ABC from center O by scale factor r=\(\frac{1}{2}\), noted in the diagram above by the red triangle. Then, translate the red triangle up two units and five units to the right. Next, rotate the red triangle 180 degrees around point A’ until AC coincides with A’C’.

c. Describe the sequence of dilation followed by a congruence that proves △A’B’C’~△ABC.
Answer:
Note that in the diagram below, both axes have been compressed.

To map △A’B’C’ onto △ABC, dilate △A’B’C’ from center O by scale factor r=2, noted by the blue triangle in the diagram. Then, translate the blue triangle ten units to the left and two units down. Next, rotate the blue triangle 180 degrees around point A until side A’C’ coincides with side AC.

d. Is it true that △ABC~△A’ B’ C’ and △A’ B’ C’~△ABC? Why do you think this is so?
Answer:
Yes, it is true that △ABC~△A’ B’ C’ and △A’ B’ C’~△ABC. I think it is true because when we say figures are similar, it means that they are the same figure, just a different size because one has been dilated by a scale factor. For that reason, if one figure, like △ABC, is similar to another, like △A’ B’ C’, it must mean that △A’ B’ C’~△ABC. However, the sequence you would use to map one of the figures onto the other is different.

Exploratory Challenge 2.
The goal is to show that if △ABC is similar to △A’B’C’ and △A’B’C’ is similar to △A”B”C”, then △ABC is similar to
△A”B”C”. Symbolically, if △ABC~△A’ B’ C’ and △A’B’C’~△A”B”C”, then △ABC~△A”B”C”.

a. Describe the similarity that proves △ABC~△A’B’C’.
Answer:

To map △ABC onto △A’B’C’, we need to first determine the scale factor that makes △ABC the same size as △A’ B’ C’. Then, \(\frac{3}{1}\)=\(\frac{6.3}{2.1}\)=\(\frac{9}{3}\)=r. Dilate △ABC from center O by scale factor r=3, shown in red in the diagram. Then, translate the red triangle 5 units up.

b. Describe the similarity that proves △A’B’C’~△A”B”C”.
Answer:

To map △A’B’C’ onto △A”B”C”, we need to first determine the scale factor that makes △A’B’C’ the same size as △A”B”C”. Then, \(\frac{4.2}{6.3}\)=\(\frac{6}{9}\)=\(\frac{2}{3}\)=r. Dilate △A’B’C’ from center O by scale factor r=\(\frac{2}{3}\), shown in blue in the diagram. Then, translate the blue triangle 3.5 units down and 5 units to the right. Next, rotate the blue triangle 90 degrees clockwise around point A” until the blue triangle coincides with
△A”B”C”.

c. Verify that, in fact, △ABC~△A”B”C” by checking corresponding angles and corresponding side lengths. Then, describe the sequence that would prove the similarity △ABC~△A” B” C”.
Answer:

The corresponding angles are congruent: ∠A≅∠A”, ∠B≅∠B”, and ∠C≅∠C”; therefore, |∠A|=|∠A” |=18°, |∠B|=|∠B” |=117°, and |∠C|=|∠C” |=45°. The ratio of the corresponding sides is equal:
\(\frac{4.2}{2.1}\)=\(\frac{6}{3}\)=\(\frac{2}{1}\)=r. Dilate △ABC from center O by scale factor r=2, shown as the pink triangle in the diagram. Then, translate the pink triangle 5 units to the right. Finally, rotate the pink triangle 90 degrees clockwise around point A” until the pink triangle coincides with △A”B”C”.

d. Is it true that if △ABC~△A’ B’ C’ and △A’ B’ C’~△A”B”C”, then △ABC~△A”B”C”? Why do you think this is so?
Answer:
Yes, it is true that if △ABC~△A’ B’ C’ and △A’ B’ C’~△A” B” C”, then △ABC~△A” B” C”. Again, because these figures are similar, it means that they have equal angles and are made different sizes based on a specific scale factor. Since dilations map angles to angles of the same degree, it makes sense that all three figures would have the “same shape.” Also, using the idea that similarity is a symmetric relation, the statement that △ABC~△A’ B’ C’ implies that △A’ B’ C’~△ABC.
Since we know that △A’ B’ C’~△A”B”C”, it is reasonable to conclude that △ABC~△A”B”C”.

Eureka Math Grade 8 Module 3 Lesson 9 Problem Set Answer Key

Question 1.
Would a dilation alone be enough to show that similarity is symmetric? That is, would a dilation alone prove that if △ABC~ △A’B’C’, then △A’ B’ C’~ △ABC? Consider the two examples below.
a. Given △ABC~ △A’ B’ C’, is a dilation enough to show that △A’ B’ C’~ △ABC? Explain.

Answer:
For these two triangles, a dilation alone is enough to show that if △ABC~△A’B’C’, then △A’ B’ C’~△ABC. The reason that dilation alone is enough is because both of the triangles have been dilated from the same center. Therefore, to map one onto the other, all that would be required is a dilation.

b. Given △ABC~△A’ B’ C’, is a dilation enough to show that △A’ B’ C’~△ABC? Explain.

Answer:
For these two triangles, a dilation alone is not enough to show that if △ABC~ △A’ B’ C’, then △A’ B’ C’~
△ABC. The reason is that a dilation would just make them the same size. It would not show that you could map one of the triangles onto the other. To do that, you would need a sequence of basic rigid motions to demonstrate the congruence.

c. In general, is dilation enough to prove that similarity is a symmetric relation? Explain.
Answer:
No, in general, a dilation alone does not prove that similarity is a symmetric relation. In some cases, like part (a), it would be enough, but because we are talking about general cases, we must consider figures that require a sequence of basic rigid motions to map one onto the other. Therefore, in general, to show that there is a symmetric relationship, we must use what we know about similar figures, a dilation followed by a congruence, as opposed to dilation alone.

Question 2.
Would a dilation alone be enough to show that similarity is transitive? That is, would a dilation alone prove that if
△ABC~△A’B’C’, and △A’ B’ C’~△A”B”C”, then △ABC~△A”B”C”? Consider the two examples below.
a. Given △ABC~△A’ B’ C’ and △A’ B’ C’~△A”B”C”, is a dilation enough to show that △ABC~△A”B”C”? Explain.

Answer:
Yes, in this case, we could dilate by different scale factors to show that all three triangles are similar to each other.

b. Given △ABC~△A’ B’ C’ and △A’ B’ C’~△A”B”C”, is a dilation enough to show that △ABC~△A”B”C”? Explain.
Answer:

In this case, it would take more than just a dilation to show that all three triangles are similar to one another. Specifically, it would take a dilation followed by a congruence to prove the similarity among the three.

c. In general, is dilation enough to prove that similarity is a transitive relation? Explain.
Answer:
In some cases, it might be enough, but the general case requires the use of dilation and a congruence. Therefore, to prove that similarity is a transitive relation, you must use both a dilation and a congruence.

Question 3.
In the diagram below, △ABC~△A’ B’ C’ and △A’ B’ C’~△A”B”C”. Is △ABC~△A”B”C”? If so, describe the dilation followed by the congruence that demonstrates the similarity.

Answer:
Yes, △ABC~△A”B”C” because similarity is transitive. Since r|AB|=|A”B”|, then r×4=2, which means
r=\(\frac{1}{2}\). Then, a dilation from the origin by scale factor r=\(\frac{1}{2}\) makes △ABC the same size as △A”B”C”. Translate the dilated image of △ABC 6 \(\frac{1}{2}\) units to the left and then reflect across line A”B”. The sequence of the dilation and the congruence map △ABC onto △A” B” C”, demonstrating the similarity.

Eureka Math Grade 8 Module 3 Lesson 9 Exit Ticket Answer Key

Use the diagram below to answer Problems 1 and 2.

Question 1.
Which two triangles, if any, have similarity that is symmetric?
Answer:
△S~△R and △R~△S
△S~△T and △T~△S
△T~△R and △R~△T

Question 2.
Which three triangles, if any, have similarity that is transitive?
Answer:
One possible solution: Since △S~△R and △R~△T, then △S~△T.
Note that △U and △V are not similar to each other or any other triangles. Therefore, they should not be in any solution.

Engage NY Eureka Math 8th Grade Module 3 Lesson 7 Answer Key

Eureka Math Grade 8 Module 3 Lesson 7 Exercise Answer Key

Use the diagram below to prove the theorem: Dilations preserve the measures of angles.

Let there be a dilation from center O with scale factor r. Given ∠PQR, show that since P’=Dilation(P),
Q’=Dilation(Q), and R’=Dilation(R), then |∠PQR|=|∠P’Q’R’|. That is, show that the image of the angle after a dilation has the same measure, in degrees, as the original.

Answer:
Using FTS, we know that the line containing \(\overrightarrow{Q^{\prime} P^{\prime}}\) is parallel to the line containing \(\overrightarrow{Q P}\) and that the line containing \(\overrightarrow{Q R}\) is parallel to the line containing \(\overrightarrow{Q R}\). We also know that there exists just one line through a given point, parallel to a given line. Therefore, we know that must intersect at a point. We know this because there is already a line that goes through point Q’ that is parallel to , and that line is . Since cannot be parallel to , it must intersect it. We let the intersection of and be named point B. Alternate interior angles of parallel lines cut by a transversal are equal in measure. Parallel lines QR and Q’R’ are cut by transversal \(\overline{\boldsymbol{Q}^{\prime} \boldsymbol{B}}\). Therefore, the alternate interior angles P’Q’R’ and Q’BQ are equal in measure. Parallel lines and are cut by transversal \(\overline{Q B} .\) Therefore, the alternate interior angles PQR and Q’BQ are equal in measure. Since the measures of ∠P’Q’R’ and ∠PQR are equal to the measure of ∠QB’Q, then the measure of ∠PQR is equal to the measure of ∠P’ Q’ R’.

Eureka Math Grade 8 Module 3 Lesson 7 Problem Set Answer Key

Question 1.
A dilation from center O by scale factor r of a line maps to what? Verify your claim on the coordinate plane.
Answer:
The dilation of a line maps to a line.
Sample student work is shown below.

Question 2.
A dilation from center O by scale factor r of a segment maps to what? Verify your claim on the coordinate plane.
Answer:
The dilation of a segment maps to a segment.
Sample student work is shown below.

Question 3.
A dilation from center O by scale factor r of a ray maps to what? Verify your claim on the coordinate plane.
Answer:
The dilation of a ray maps to a ray.
Sample student work is shown below.

Question 4.
Challenge Problem:
Prove the theorem: A dilation maps lines to lines.
Let there be a dilation from center O with scale factor r so that P’=Dilation(P) and Q’=Dilation(Q). Show that line PQ maps to line P’Q’ (i.e., that dilations map lines to lines). Draw a diagram, and then write your informal proof of the theorem. (Hint: This proof is a lot like the proof for segments. This time, let U be a point on line PQ that is not between points P and Q.)
Answer:
Sample student drawing and response are below:

Let U be a point on line PQ. By the definition of dilation, we also know that U’=Dilation(U). We need to show that U’ is a point on line P’Q’. If we can, then we have proven that a dilation maps lines to lines.
By the definition of dilation and FTS, we know that \(\frac{\left|O P^{\prime}\right|}{|O P|}\) =\(\frac{\left|O Q^{\prime}\right|}{|O Q|}\) and that \(\overleftrightarrow{P Q}\) is parallel to . Similarly, we know that and that line QU is parallel to line Q’U’. Since U is a point on line PQ, then we also know that line PQ is parallel to line Q’U’. But we already know that is parallel to . Since there can only be one line that passes through Q’ that is parallel to line PQ, then line P’Q’ and line Q’U’ must coincide. That places the dilation of point U, U’, on the line P’Q’, which proves that dilations map lines to lines.

Eureka Math Grade 8 Module 3 Lesson 7 Exit Ticket Answer Key

Dilate ∠ABC with center O and scale factor r=2. Label the dilated angle, ∠A’B’C’.

Answer:

Question 1.
If ∠ABC=72°, then what is the measure of ∠A’ B’ C’?
Answer:
Since dilations preserve angles, then ∠A’B’C’=72°.

Question 2.
If the length of segment AB is 2 cm, what is the length of segment A’B’?
Answer:
The length of segment A’B’ is 4 cm.

Question 3.
Which segments, if any, are parallel?
Answer:
Since dilations map segments to parallel segments, then \(\overline{A B}\) || \(\overline{A^{\prime} B^{\prime}}\), and \(\overline{\boldsymbol{B C}}\) || \(\overline{\boldsymbol{B}^{\prime} \boldsymbol{C}^{\prime}}\).

Engage NY Eureka Math 8th Grade Module 3 Lesson 8 Answer Key

Eureka Math Grade 8 Module 3 Lesson 8 Example Answer Key

Example 1.
In the picture below, we have triangle ABC that has been dilated from center O by a scale factor of r=\(\frac{1}{2}\). It is noted by A’B’C’. We also have triangle A”B”C”, which is congruent to triangle A’B’C’ (i.e., △A’B’C’≅△A”B”C”).

Describe the sequence that would map triangle A”B”C” onto triangle ABC.

→ Based on the definition of similarity, how could we show that triangle A”B”C” is similar to triangle ABC?
→ To show that △A” B” C”~△ABC, we need to describe a dilation followed by a congruence.
→ We want to describe a sequence that would map triangle A”B”C” onto triangle ABC. There is no clear way to do this, so let’s begin with something simpler: How can we map triangle A’B’C’ onto triangle ABC? That is, what is the precise dilation that would make triangle A’B’C’ the same size as triangle ABC?
→ A dilation from center O with scale factor r=2
→ Remember, our goal was to describe how to map triangle A”B”C” onto triangle ABC. What precise dilation would make triangle A”B”C” the same size as triangle ABC?
→ A dilation from center O with scale factor r=2 would make triangle A”B”C” the same size as triangle ABC.
→ (Show the picture below with the dilated triangle A”B”C” noted by A”’B”’C”’.) Now that we know how to make triangle A”B”C” the same size as triangle ABC, what rigid motion(s) should we use to actually map triangle A”B”C” onto triangle ABC? Have we done anything like this before?

→ Problem 2 of the Problem Set from Lesson 2 was like this. That is, we had two figures dilated by the same scale factor in different locations on the plane. To get one to map to the other, we just translated along a vector.
→ Now that we have an idea of what needs to be done, let’s describe the translation in terms of coordinates. How many units and in which direction do we need to translate so that triangle A”’B”’C”’ maps to triangle ABC?
→ We need to translate triangle A”’B”’C”’ 20 units to the left and 2 units down.
→ Let’s use precise language to describe how to map triangle A”B”C” onto triangle ABC. We need information about the dilation and the translation.
→ The sequence that would map triangle A”B”C” onto triangle ABC is as follows: Dilate triangle A”B”C” from center O by scale factor r=2. Then, translate the dilated triangle 20 units to the left and 2 units down.
→ Since we were able to map triangle A”B”C” onto triangle ABC with a dilation followed by a congruence, we can write that triangle A”B”C” is similar to triangle ABC, in notation, △A” B” C”~△ABC.

Example 2.
In the picture below, we have triangle DEF that has been dilated from center O, by scale factor r=3. It is noted by D’ E’ F’. We also have triangle D” E” F”, which is congruent to triangle D’ E’ F’ (i.e., △D’ E’ F’≅
△D”E”F”).

→ We want to describe a sequence that would map triangle D”E”F” onto triangle DEF. This is similar to what we did in the last example. Can someone summarize the work we did in the last example?
→ First, we figured out what scale factor r would make the triangles the same size. Then, we used a sequence of translations to map the magnified figure onto the original triangle.
→ What is the difference between this problem and the last?
→ This time, the scale factor is greater than one, so we need to shrink triangle D”E”F” to the size of triangle DEF. Also, it appears as if a translation alone does not map one triangle onto another.
→ Now, since we want to dilate triangle D”E”F” to the size of triangle DEF, we need to know what scale factor r to use. Since triangle D”E”F” is congruent to D’E’F’, then we can use those triangles to determine the scale factor needed. We need a scale factor so that |OF|=r|OF’|. What scale factor do you think we should use, and why?
→ We need a scale factor r=\(\frac{1}{3}\) because we want |OF|=r|OF’|.
→ What precise dilation would make triangle D”E”F” the same size as triangle DEF?
→ A dilation from center O with scale factor r=\(\frac{1}{3}\) would make triangle D”E”F” the same size as triangle DEF.
→ (Show the picture below with the dilated triangle D”E”F” noted by D”’E”’F”’.) Now we should use what we know about rigid motions to map the dilated version of triangle D”E”F” onto triangle DEF. What should we do first?

→ We should translate triangle D”’E”’F”’ 2 units to the right.
→ (Show the picture below, the translated triangle noted in red.) What should we do next (refer to the translated triangle as the red triangle)?

→ Next, we should reflect the red triangle across the x-axis to map the red triangle onto triangle DEF.
→ Use precise language to describe how to map triangle D”E”F” onto triangle DEF.
→ The sequence that would map triangle D”E”F” onto triangle DEF is as follows: Dilate triangle D”E”F” from center O by scale factor r=\(\frac{1}{3}\) . Then, translate the dilated image of triangle D”E”F”, noted by D”’ E”’ F”’, two units to the right. Finally, reflect across the x-axis to map the red triangle onto triangle DEF.
→ Since we were able to map triangle D”E”F” onto triangle DEF with a dilation followed by a congruence, we can write that triangle D”E”F” is similar to triangle DEF. (In notation: △D”E”F”~△DEF)

Example 3.
In the diagram below, △ABC ~△A’B’C’. Describe a sequence of a dilation followed by a congruence that would prove these figures to be similar.

→ Let’s begin with the scale factor. We know that r|AB|=|A’B’|. What scale factor r makes △ABC the same size as △A’B’C’?
→ We know that r⋅2=1; therefore, r=\(\frac{1}{2}\) makes △ABC the same size as △A’ B’ C’.
→ If we apply a dilation from the origin of scale factor r=\(\frac{1}{2}\), then the triangles are the same size (as shown and noted by triangle A”B”C”). What sequence of rigid motions would map the dilated image of △ABC onto △A’B’C’?

→ We could translate the dilated image of △ABC, △A”B”C”, 3 units to the right and 4 units down and then reflect the triangle across line A’B’.
→ The sequence that would map △ABC onto △A’B’C’ to prove the figures similar is a dilation from the origin by scale factor r=\(\frac{1}{2}\), followed by the translation of the dilated version of △ABC 3 units to the right and 4 units down, followed by the reflection across line A’B’.

Example 4.
In the diagram below, we have two similar figures. Using the notation, we have △ABC ~△DEF. We want to describe a sequence of the dilation followed by a congruence that would prove these figures to be similar.

→ First, we need to describe the dilation that would make the triangles the same size. What information do we have to help us describe the dilation?
→ Since we know the length of side \(\overline{A C}\) and side \(\overline{D F}\), we can determine the scale factor.
→ Can we use any two sides to calculate the scale factor? Assume, for instance, that we know that side \(\overline{A C}\) is
18 units in length and side \(\overline{E F}\) is 2 units in length. Could we find the scale factor using those two sides, \(\overline{A C}\) and \(\overline{E F}\)? Why or why not?
→ No, we need more information about corresponding sides. Sides \(\overline{A C}\) and \(\overline{D F}\) are the longest sides of each triangle (they are also opposite the obtuse angle in the triangle). Side \(\overline{A C}\) does not correspond to side \(\overline{E F}\). If we knew the length of side \(\overline{B C}\), we could use sides \(\overline{B C}\) and \(\overline{E F}\).
→ Now that we know that we can find the scale factor if we have information about corresponding sides, how would we calculate the scale factor if we were mapping △ABC onto △DEF?
→ |DF|=r|AC|, so 6=r⋅18, and r=\(\frac{1}{3}\).
→ If we were mapping △DEF onto △ABC, what would the scale factor be?
→ |AC|=r|DF|, so 18=r∙6, and r=3.
→ What is the precise dilation that would map △ABC onto △DEF?
→ Dilate △ABC from center O, by scale factor r=\(\frac{1}{3}\).
→ (Show the picture below with the dilated triangle noted as △A’B’C’.) Now we have to describe the congruence. Work with a partner to determine the sequence of rigid motions that would map △ABC onto △DEF.

→ Translate the dilated version of △ABC 7 units to the right and 2 units down. Then, rotate d degrees around point E so that segment B’C’ maps onto segment EF. Finally, reflect across line EF.
Note that “d degrees” refers to a rotation by an appropriate number of degrees to exhibit similarity. Students may choose to describe this number of degrees in other ways.
→ The sequence of a dilation followed by a congruence that proves △ABC ~△DEF is as follows: Dilate △ABC from center O by scale factor r=\(\frac{1}{3}\). Translate the dilated version of △ABC 7 units to the right and 2 units down. Next, rotate around point E by d degrees so that segment B’C’ maps onto segment EF, and then reflect the triangle across line EF.

Example 5.
→ Knowing that a sequence of a dilation followed by a congruence defines similarity also helps determine if two figures are in fact similar. For example, would a dilation map triangle ABC onto triangle DEF (i.e., is △ABC ~△DEF)?

→ No. By FTS, we expect the corresponding side lengths to be in proportion and equal to the scale factor. When we compare side \(\overline{A C}\) to side \(\overline{D F}\) and \(\overline{B C}\) to \(\overline{E F}\), we get \(\frac{18}{6}\)≠\(\frac{15}{4}\).
→ Therefore, the triangles are not similar because a dilation does not map one to the other.

Example 6.
→ Again, knowing that a dilation followed by a congruence defines similarity also helps determine if two figures are in fact similar. For example, would a dilation map Figure A onto Figure A’ (i.e., is Figure A ~ Figure A’)?

→ No. Even though we could say that the corresponding sides are in proportion, there exists no single rigid motion or sequence of rigid motions that would map a four-sided figure to a three-sided figure. Therefore, the figures do not fulfill the congruence part of the definition for similarity, and Figure A is not similar to Figure A’.

Eureka Math Grade 8 Module 3 Lesson 8 Exercise Answer Key

Exercises
Allow students to work in pairs to describe sequences that map one figure onto another.

Exercise 1.
Triangle ABC was dilated from center O by scale factor r=\(\frac{1}{2}\). The dilated triangle is noted by A’B’C’. Another triangle A”B”C” is congruent to triangle A’B’C’ (i.e., △A”B”C”≅△A’B’C’). Describe a dilation followed by the basic rigid motion that would map triangle A”B”C” onto triangle ABC.

Answer:
Triangle A”B”C” needs to be dilated from center O, by scale factor r=2 to bring it to the same size as triangle ABC. This produces a triangle noted by A”’B”’C”’. Next, triangle A”’B”’C”’ needs to be translated 4 units up and 12 units left. The dilation followed by the translation maps triangle A”B”C” onto triangle ABC.

Exercise 2.
Describe a sequence that would show △ABC ~△A’ B’ C’.

Answer:
Since r|AB|=|A’ B’ |, then r⋅2=6, and r=3. A dilation from the origin by scale factor r=3 makes △ABC the same size as △A’B’C’. Then, a translation of the dilated image of △ABC ten units right and five units down, followed by a rotation of 90 degrees around point C’, maps △ABC onto △A’ B’ C’, proving the triangles to be similar.

Exercise 3.
Are the two triangles shown below similar? If so, describe a sequence that would prove △ABC ~△A’B’C’. If not, state how you know they are not similar.

Answer:
Yes, △ABC ~△A’B’C’. The corresponding sides are in proportion and equal to the scale factor:
\(\frac{10}{15}\)=\(\frac{4}{6}\)=\(\frac{12}{18}\)=\(\frac{2}{3}\)=r
To map triangle ABC onto triangle A’B’C’, dilate triangle ABC from center O, by scale factor r=\(\frac{2}{3}\). Then, translate triangle ABC along vector \(\overrightarrow{A A^{\prime}}\). Next, rotate triangle ABC d degrees around point A.

Exercise 4.
Are the two triangles shown below similar? If so, describe the sequence that would prove △ABC ~△A’B’C’. If not, state how you know they are not similar.

Answer:
Yes, triangle △ABC ~△A’B’C’. The corresponding sides are in proportion and equal to the scale factor:
\(\frac{4}{3}\)=\(\frac{8}{6}\)=\(\frac{4}{3}\)=\(1.3 \overline{3}\); \(\frac{10.67}{8}\)=1.33375; therefore, r=1.33 which is approximately equal to \(\frac{4}{3}\)
To map triangle ABC onto triangle A’B’C’, dilate triangle ABC from center O, by scale factor r=\(\frac{4}{3}\). Then, translate triangle ABC along vector \(\overrightarrow{A A^{\prime}}\). Next, rotate triangle ABC 180 degrees around point A’.

Eureka Math Grade 8 Module 3 Lesson 8 Problem Set Answer Key

Students practice dilating a curved figure and describing a sequence of a dilation followed by a congruence that maps one figure onto another.

Question 1.
In the picture below, we have triangle DEF that has been dilated from center O by scale factor r=4. It is noted by D’E’F’. We also have triangle D”E”F”, which is congruent to triangle D’E’F’ (i.e., △D’E’F’≅△D”E”F”). Describe the sequence of a dilation, followed by a congruence (of one or more rigid motions), that would map triangle D”E”F” onto triangle DEF.

Answer:

First, we must dilate triangle D”E”F” by scale factor r=\(\frac{1}{4}\) to shrink it to the size of triangle DEF. Next, we must translate the dilated triangle, noted by D”’E”’F”’, one unit up and two units to the right. This sequence of the dilation followed by the translation would map triangle D”E”F” onto triangle DEF.

Question 2.
Triangle ABC was dilated from center O by scale factor r=\(\frac{1}{2}\). The dilated triangle is noted by A’B’C’. Another triangle A”B”C” is congruent to triangle A’B’C’ (i.e., △A”B”C”≅△A’B’C’). Describe the dilation followed by the basic rigid motions that would map triangle A”B”C” onto triangle ABC.

Answer:

Triangle A”B”C” needs to be dilated from center O by scale factor r=2 to bring it to the same size as triangle ABC. This produces a triangle noted by A”’B”’C”’. Next, triangle A”’B”’C”’ needs to be translated 18 units to the right and two units down, producing the triangle shown in red. Next, rotate the red triangle d degrees around point B so that one of the segments of the red triangle coincides completely with segment BC. Then, reflect the red triangle across line BC. The dilation, followed by the congruence described, maps triangle A”B”C” onto triangle ABC.

Question 3.
Are the two figures shown below similar? If so, describe a sequence that would prove the similarity. If not, state how you know they are not similar.

Answer:
No, these figures are not similar. There is no single rigid motion, or sequence of rigid motions, that would map Figure A onto Figure B.

Question 4.
Triangle ABC is similar to triangle A’B’C’ (i.e., △ABC ~△A’B’C’). Prove the similarity by describing a sequence that would map triangle A’B’C’ onto triangle ABC.

Answer:
The scale factor that would magnify triangle A’B’C’ to the size of triangle ABC is r=3. The sequence that would prove the similarity of the triangles is a dilation from center O by a scale factor of r=3, followed by a translation along vector \(\overrightarrow{A^{\prime} A}\), and finally, a reflection across line AC.

Question 5.
Are the two figures shown below similar? If so, describe a sequence that would prove △ABC ~△A’B’C’. If not, state how you know they are not similar.

Answer:
Yes, the triangles are similar. The scale factor that triangle ABC has been dilated is r=\(\frac{1}{5}\). The sequence that proves the triangles are similar is as follows: dilate triangle A’B’C’ from center O by scale factor r=5; then, translate triangle A’B’C’ along vector \(\overrightarrow{C^{\prime} C}\); next, rotate triangle A’B’C’ d degrees around point C; and finally, reflect triangle A’B’C’ across line AC.

Question 6.
Describe a sequence that would show △ABC ~△A’ B’ C’.

Answer:
Since r|AB|=|A’ B’|, then r∙3=1 and r = \(\frac{1}{3}\). A dilation from the origin by scale factor r\(\frac{1}{3}\) makes △ABC the same size as △A’B’C’. Then, a translation of the dilated image of △ABC four units down and one unit to the right, followed by a reflection across line A’ B’, maps △ABC onto △A’ B’ C’, proving the triangles to be similar.

Eureka Math Grade 8 Module 3 Lesson 8 Exit Ticket Answer Key

In the picture below, we have triangle DEF that has been dilated from center O by scale factor r=\(\frac{1}{2}\). The dilated triangle is noted by D’E’F’. We also have a triangle D”EF, which is congruent to triangle DEF (i.e., △DEF≅△D”EF). Describe the sequence of a dilation, followed by a congruence (of one or more rigid motions), that would map triangle D’E’F’ onto triangle D”EF.

Answer:
Triangle D’E’F’ needs to be dilated from center O by scale factor r=2 to bring it to the same size as triangle DEF. This produces the triangle noted by DEF. Next, triangle DEF needs to be reflected across line EF. The dilation followed by the reflection maps triangle D’E’F’ onto triangle D”EF.

Engage NY Eureka Math 8th Grade Module 3 Lesson 6 Answer Key

Eureka Math Grade 8 Module 3 Lesson 6 Example Answer Key

Example 1.
Students learn the multiplicative effect of scale factor on a point. Note that this effect holds when the center of dilation is the origin. In this lesson, the center of any dilation used is always assumed to be (0,0).
Show the diagram below, and ask students to look at and write or share a claim about the effect that dilation has on the coordinates of dilated points.
→ The graph below represents a dilation from center (0,0) by scale factor r=2.

Show students the second diagram below so they can check if their claims were correct. Give students time to verify the claims that they made about the above graph with the one below. Then, have them share their claims with the class. Use the discussion that follows to crystallize what students observed.

→ The graph below represents a dilation from center (0,0) by scale factor r=4.

→ In Lesson 5, we found the location of a dilated point by using the knowledge of dilation and scale factor, as well as the lines of the coordinate plane to ensure equal angles, to find the coordinates of the dilated point. For example, we were given the point A(5,2) and told the scale factor of dilation was r=2. Remember that the center of this dilation is (0,0). We created the following picture and determined the location of A’to be (10,4).

→ We can use this information and the observations we made at the beginning of class to develop a shortcut for finding the coordinates of dilated points when the center of dilation is the origin.
→ Notice that the horizontal distance from the y-axis to point A was multiplied by a scale factor of 2. That is, the x-coordinate of point A was multiplied by a scale factor of 2. Similarly, the vertical distance from the x-axis to point A was multiplied by a scale factor of 2.
→ Here are the coordinates of point A(5,2) and the dilated point A'(10,4). Since the scale factor was 2, we can more easily see what happened to the coordinates of A after the dilation if we write the coordinates of A’as (2∙5,2∙2), that is, the scale factor of 2 multiplied by each of the coordinates of A to get A’.
→ The reasoning goes back to our understanding of dilation. The length r|OB|=|OB’|, by the definition of dilation, and the length r|AB|=|A’B’|; therefore,
r=\(\frac{\left|O B^{\prime}\right|}{|O B|}\) =\(\frac{\left|A^{\prime} B^{\prime}\right|}{|A B|}\),
where the length of the segment OB’is the x-coordinate of the dilated point (i.e., 10), and the length of the segment A’B’is the y-coordinate of the dilated point (i.e., 4).
In other words, based on what we know about the lengths of dilated segments, when the center of dilation is the origin, we can determine the coordinates of a dilated point by multiplying each of the coordinates in the original point by the scale factor.

Example 2.
Students learn the multiplicative effect of scale factor on a point.
→ Let’s look at another example from Lesson 5. We were given the point A(7,6) and asked to find the location of the dilated point A’when r=\(\frac{11}{7}\). Our work on this problem led us to coordinates of approximately (11,9.4) for point A’. Verify that we would get the same result if we multiply each of the coordinates of point A by the scale factor.
A'(\(\frac{11}{7}\)∙7,\(\frac{11}{7}\)∙6)
\(\frac{11}{7}\)∙7=11
and
\(\frac{11}{7}\)∙6=\(\frac{66}{7}\)≈9.4
Therefore, multiplying each coordinate by the scale factor produced the desired result.

Example 3.
→ The coordinates in other quadrants of the graph are affected in the same manner as we have just seen. Based on what we have learned so far, given point A(-2,3), predict the location of A’when A is dilated from a center at the origin, (0,0), by scale factor r=3.
Provide students time to predict, justify, and possibly verify, in pairs, that A'(3∙(-2),3∙3)=(-6,9). Verify the fact on the coordinate plane, or have students share their verifications with the class.

→ As before, mark a point B on the x-axis. Then, |OB’|=3|OB|. Where is point B’located?
→ Since the length of |OB|=2, then |OB’|=3∙2=6. But we are looking at a distance to the left of zero; therefore, the location of B’is (-6,0).

→ Now that we know where B’is, we can easily find the location of A’. It is on the ray \(\overrightarrow{O A}\), but at what location?
→ The location of A'(-6,9), as desired

Example 4.
Students learn the multiplicative effect of scale factor on a two-dimensional figure.
→ Now that we know the multiplicative relationship between a point and its dilated location (i.e., if point P(p₁,p₂) is dilated from the origin by scale factor r, then P'(rp₁,rp₂ )), we can quickly find the coordinates of any point, including those that comprise a two-dimensional figure, under a dilation of any scale factor.
→ For example, triangle ABC has coordinates A(2,3), B(-3,4), and C(5,7). The triangle is being dilated from the origin with scale factor r=4. What are the coordinates of triangle A’B’C’?
→ First, find the coordinates of A’.
→ A'(4⋅2,4⋅3)=A'(8,12)
→ Next, locate the coordinates of B’.
→ B'(4⋅(-3),4⋅4)=B'(-12,16)
→ Finally, locate the coordinates of C’.
→ C'(4∙5,4∙7)=C'(20,28)
→ Therefore, the vertices of triangle A’B’C’have coordinates of (8,12), (-12,16), and (20,28), respectively.

Example 5.
→ Students learn the multiplicative effect of scale factor on a two-dimensional figure.
→ Parallelogram ABCD has coordinates of (-2,4), (4,4), (2,-1), and (-4,-1), respectively. Find the coordinates of parallelogram A’B’C’D’after a dilation from the origin with a scale factor r=\(\frac{1}{2}\).
→ A'(\(\frac{1}{2}\)∙(-2),\(\frac{1}{2}\)∙4)=A'(-1,2)
→ B'(\(\frac{1}{2}\)∙4,\(\frac{1}{2}\)∙4)=B'(2,2)
→ C'(\(\frac{1}{2}\)∙2,\(\frac{1}{2}\)∙(-1))=C'(1,-\(\frac{1}{2}\))
→ D'(\(\frac{1}{2}\)∙(-4),\(\frac{1}{2}\)∙(-1))=D'(-2,-\(\frac{1}{2}\))
Therefore, the vertices of parallelogram A’B’C’D’have coordinates of (-1,2), (2,2), (1,-\(\frac{1}{2}\)), and (-2,-\(\frac{1}{2}\)), respectively.

Eureka Math Grade 8 Module 3 Lesson 6 Exercise Answer Key

Exercises 1–5.
Point A(7,9) is dilated from the origin by scale factor r=6. What are the coordinates of point A’?
Answer:
A'(6∙7,6∙9)=A'(42,54)

Exercise 2.
Point B(-8,5) is dilated from the origin by scale factor r=\(\frac{1}{2}\). What are the coordinates of point B’?
Answer:
B'(\(\frac{1}{2}\)∙(-8),\(\frac{1}{2}\)∙5)=B'(-4,\(\frac{5}{2}\))

Exercise 3.
Point C(6,-2) is dilated from the origin by scale factor r=\(\frac{3}{4}\). What are the coordinates of point C’?
Answer:
C'(\(\frac{3}{4}\)∙6,\(\frac{3}{4}\)∙(-2))=C'(\(\frac{9}{2}\),-\(\frac{3}{2}\))

Exercise 4.
Point D(0,11) is dilated from the origin by scale factor r=4. What are the coordinates of point D’?
Answer:
D'(4∙0,4∙11)=D'(0,44)

Exercise 5.
Point E(-2,-5) is dilated from the origin by scale factor r=\(\frac{3}{2}\). What are the coordinates of point E’?
Answer:
E'(\(\frac{3}{2}\)∙(-2),\(\frac{3}{2}\)∙(-5))=E'(-3,-\(\frac{15}{2}\))

Exercises 6–8.

Exercise 6.
The coordinates of triangle ABC are shown on the coordinate plane below. The triangle is dilated from the origin by scale factor r=12. Identify the coordinates of the dilated triangle A’B’C’.

Point A(-2,2), so A'(12∙(-2),12∙2)=A'(-24,24).
Point B(-3,-3), so B'(12∙(-3),12∙(-3))=B'(-36,-36).
Point C(6,1), so C'(12∙6,12∙1)=C'(72,12).
The coordinates of the vertices of triangle A’B’C’are (-24,24), (-36,-36), and (72,12), respectively.

Exercise 7.
Figure DEFG is shown on the coordinate plane below. The figure is dilated from the origin by scale factor r=\(\frac{2}{3}\). Identify the coordinates of the dilated figure D’E’F’G’, and then draw and label figure D’E’F’G’ on the coordinate plane.

Answer:
Point D(-6,3), so D'(\(\frac{2}{3}\)∙(-6),\(\frac{2}{3}\)∙3)=D'(-4,2).
Point E(-4,-3), so E'(\(\frac{2}{3}\)∙(-4),\(\frac{2}{3}\)∙(-3))=E'(-\(\frac{8}{3}\),-2).
Point F(5,-2), so F'(\(\frac{2}{3}\)∙5,\(\frac{2}{3}\)∙(-2))=F'(\(\frac{10}{3}\),-\(\frac{4}{3}\) ). –
Point G(-3,3), so G'(\(\frac{2}{3}\)∙(-3),\(\frac{2}{3}\)∙3)=G'(-2,2).
The coordinates of the vertices of figure D’E’F’G’are (-4,2), (-\(\frac{8}{3}\),-2), (\(\frac{10}{3}\),-\(\frac{4}{3}\)), and (-2,2), respectively.

Exercise 8.
The triangle ABC has coordinates A(3,2), B(12,3), and C(9,12). Draw and label triangle ABC on the coordinate plane. The triangle is dilated from the origin by scale factor r=\(\frac{1}{3}\). Identify the coordinates of the dilated triangle

A’B’C’, and then draw and label triangle A’B’C’on the \(\frac{1}{3}\) coordinate plane.
Answer:
Point A(3,2), then A'(\(\frac{1}{3}\)∙3,\(\frac{1}{3}\)∙2)=A'(1,\(\frac{2}{3}\)).
Point B(12,3), so B'(\(\frac{1}{3}\)∙12,\(\frac{1}{3}\)∙3)=B'(4,1).
Point C(9,12), so C'(\(\frac{1}{3}\)∙9,\(\frac{1}{3}\)∙12)=C'(3,4).
The coordinates of triangle A’B’C’are (1,\(\frac{2}{3}\)), (4,1), and (3,4), respectively.

Eureka Math Grade 8 Module 3 Lesson 6 Problem Set Answer Key

Students practice finding the coordinates of dilated points of two-dimensional figures.

Question 1.
Triangle ABC is shown on the coordinate plane below. The triangle is dilated from the origin by scale factor r=4. Identify the coordinates of the dilated triangle A’B’C’.

Answer:
Point A(-10,6), so A'(4∙(-10),4∙6)=A'(-40,24).
Point B(-11,2), so B'(4∙(-11),4∙2)=B'(-44,8).
Point C(-4,4), so C'(4∙(-4),4∙4)=C'(-16,16).
The coordinates of the vertices of triangle A’B’C’are (-40,24), (-44,8), and (-16,16), respectively.

Question 2.
Triangle ABC is shown on the coordinate plane below. The triangle is dilated from the origin by scale factor r=\(\frac{5}{4}\). Identify the coordinates of the dilated triangle A’B’C’.

Answer:
Point A(-14,-8), so A'(\(\frac{5}{4}\)∙(-14),\(\frac{5}{4}\)∙(-8))=A'(-\(\frac{35}{2}\),-10).
Point B(-12,-1), so B'(\(\frac{5}{4}\)∙(-12),\(\frac{5}{4}\)∙(-1))=B'(-15,-\(\frac{5}{4}\)).
Point C(-4,-1), so C'(\(\frac{5}{4}\)∙(-4),\(\frac{5}{4}\)∙(-1))=B'(-5,-\(\frac{5}{4}\)).
The coordinates of the vertices of triangle A’B’C’are (-\(\frac{35}{2}\),-10), (-15,-\(\frac{5}{4}\)), and (-5,-\(\frac{5}{4}\)), respectively.

Question 3.
The triangle ABC has coordinates A(6,1), B(12,4), and C(-6,2). The triangle is dilated from the origin by a scale factor r=\(\frac{1}{2}\). Identify the coordinates of the dilated triangle A’B’C’.
Answer:
Point A(6,1), so A'(\(\frac{1}{2}\)∙6,\(\frac{1}{2}\)∙1)=A'(3,\(\frac{1}{2}\)).
Point B(12,4), so B'(\(\frac{1}{2}\)∙12,\(\frac{1}{2}\)∙4)=B'(6,2).
Point C(-6,2), so C'(\(\frac{1}{2}\)∙(-6),\(\frac{1}{2}\)∙2)=C'(-3,1).
The coordinates of the vertices of triangle A’B’C’are (3,\(\frac{1}{2}\)), (6,2), and (-3,1), respectively.

Question 4.
Figure DEFG is shown on the coordinate plane below. The figure is dilated from the origin by scale factor r=\(\frac{3}{2}\). Identify the coordinates of the dilated figure D’E’F’G’, and then draw and label figure D’E’F’G’ on the coordinate plane.

Answer:
Point D(-3,1), so D'(\(\frac{3}{2}\)∙(-3),\(\frac{3}{2}\)∙1)=D'(-\(\frac{9}{2}\),\(\frac{3}{2}\)).
Point E(-1,-1), so E'(\(\frac{3}{2}\)∙(-1),\(\frac{3}{2}\)∙(-1))=E'(-\(\frac{3}{2}\),-\(\frac{3}{2}\)).
Point F(6,1), so F'(\(\frac{3}{2}\)∙6,\(\frac{3}{2}\)∙1)=F'(9,\(\frac{3}{2}\)).
Point G(0,5), so G'(\(\frac{3}{2}\)∙0,\(\frac{3}{2}\)∙5)=G'(0,\(\frac{15}{2}\)).
The coordinates of the vertices of figure D’E’F’G’are (-\(\frac{9}{2}\),\(\frac{3}{2}\)), (-\(\frac{3}{2}\),-\(\frac{3}{2}\)), (9,\(\frac{3}{2}\)), and (0,\(\frac{15}{2}\)), respectively.

Question 5.
Figure DEFG has coordinates D(1,1), E(7,3), F(5,-4), and G(-1,-4). The figure is dilated from the origin by scale factor r=7. Identify the coordinates of the dilated figure D’E’F’G’.
Answer:
Point D(1,1), so D'(7∙1,7∙1)=D'(7,7).
Point E(7,3), so E'(7∙7,7∙3)=E'(49,21).
Point F(5,-4), so F'(7∙5,7∙(-4))=F'(35,-28).
Point G(-1,-4), so G'(7∙(-1),7∙(-4))=G'(-7,-28).
The coordinates of the vertices of figure D’E’F’G’are (7,7), (49,21),(35,-28), and (-7,-28), respectively.

Eureka Math Grade 8 Module 3 Lesson 6 Exit Ticket Answer Key

Question 1.
The point A(7,4) is dilated from the origin by a scale factor r=3. What are the coordinates of point A’?
Answer:
Since point A(7,4), then A'(3∙7,3∙4)=A'(21,12).

Question 2.
The triangle ABC, shown on the coordinate plane below, is dilated from the origin by scale factor r=\(\frac{1}{2}\). What is the location of triangle A’B’C’? Draw and label it on the coordinate plane.

Answer:

Point A(3,4), so A'(\(\frac{1}{2}\)∙3,\(\frac{1}{2}\)∙4)=A'(\(\frac{3}{2}\),2).
Point B(-7,2), so B'(\(\frac{1}{2}\)∙(-7),\(\frac{1}{2}\)∙2)=B'(-\(\frac{7}{2}\),1).
Point C(2,2), so C'(\(\frac{1}{2}\)∙2,\(\frac{1}{2}\)∙2)=C'(1,1).
The coordinates of the vertices of triangle A’B’C’are (\(\frac{3}{2}\),2), (-\(\frac{7}{2}\),1), and (1,1), respectively.

Engage NY Eureka Math 8th Grade Module 3 Lesson 5 Answer Key

Eureka Math Grade 8 Module 3 Lesson 5 Exercise Answer Key

Exercise 1.
In the diagram below, points P and Q have been dilated from center O by scale factor r. \(\overline{P Q}\) || \(\overline{\boldsymbol{P}^{\prime} \boldsymbol{Q}^{\prime}}\) |PQ|=5 cm, and |P’Q’|=10 cm.

b. Determine the scale factor r.
Answer:
According to FTS, |P’Q’|=r|PQ|. Therefore, 10=r∙5, so r=2.

b. Locate the center O of dilation. Measure the segments to verify that |OP’|=r|OP| and |OQ’|=r|OQ|. Show your work below.
Answer:
Center O and measurements are shown above.
|OP’|=r|OP|
6=2∙3
6=6
|OQ’|=r|OQ|
8=2∙4
8=8

Exercise 2.
In the diagram below, you are given center O and ray \(\overrightarrow{0 A}\). Point A is dilated by a scale factor r=4. Use what you know about FTS to find the location of point A’.

Answer:
Point A’ must be located at (12,12).

Exercise 3.
In the diagram below, you are given center O and ray \(\overrightarrow{0 A}\). Point A is dilated by a scale factor r=\(\frac{5}{12}\). Use what you know about FTS to find the location of point A’.

Answer:
The x-coordinate of A’ is 5. The y-coordinate is equal to the length of segment A’B’. Since |A’B’|=r|AB|, then |A’B’|=\(\frac{5}{12}\)∙8=\(\frac{40}{12}\)≈3.3. The location of A’ is (5,3.3).

Eureka Math Grade 8 Module 3 Lesson 5 Exit Ticket Answer Key

In the diagram below, you are given center O and ray \(\overrightarrow{O A}\). Point A is dilated by a scale factor r= \(\frac{6}{4}\). Use what you know about FTS to find the location of point A’.

Answer:

The y-coordinate of A’ is 6. The x-coordinate is equal to the length of segment A’B’. Since |A’B’|=r|AB|, then
|A’B’|=\(\frac{6}{4}\)∙3=\(\frac{18}{4}\)=4.5. The location of A’ is (4.5,6).

Eureka Math Grade 8 Module 3 Lesson 5 Problem Set Answer Key

Students practice using the first consequences of FTS in terms of dilated points and their locations on the coordinate plane.

Question 1.
Dilate point A, located at (3,4) from center O, by a scale factor r=\(\frac{5}{3}\).

What is the precise location of point A’?
Answer:
The y-coordinate of point A’ is the length of segment A’B’. Since |A’B’|=r|AB|, then |A’B’|=\(\frac{5}{3}\)∙4=\(\frac{20}{3}\). The location of point A’ is
(5,\(\frac{20}{3}\)), or approximately (5,6.7).

Question 2.
Dilate point A, located at (9,7) from center O, by a scale factor r=\(\frac{4}{9}\). Then, dilate point B, located at (9,5) from center O, by a scale factor of r=\(\frac{4}{9}\). What are the coordinates of points A’ and B’? Explain.

Answer:
The y-coordinate of point A’ is the length of A’C’. Since |A’C’|=r|AC|, then |A’C’|=\(\frac{4}{9}\)∙7=\(\frac{28}{9}\). The location of point A’ is (4, \(\frac{28}{9}\)), or approximately (4,3.1). The y-coordinate of point B’ is the length of B’C’. Since |B’C’|=r|BC|, then |B’C’|=\(\frac{4}{9}\)∙5=\(\frac{20}{9}\). The location of point B’ is (4, \(\frac{20}{9}\)), or approximately (4,2.2).

Question 3.
Explain how you used the fundamental theorem of similarity in Problems 1 and 2.
Answer:
Using what I knew about scale factor, I was able to determine the placement of points A’ and B’, but I did not know the actual coordinates. So, one of the ways that FTS was used was actually in terms of the converse of FTS. I had to make sure I had parallel lines. Since the lines of the coordinate plane guarantee parallel lines, I knew that |A’C’|=r|AC|. Then, since I knew the length of segment AC and the scale factor, I could find the precise location of A’. The precise location of B’ was found in a similar way but using |B’C’|=r|BC|.

Engage NY Eureka Math 8th Grade Module 3 Lesson 4 Answer Key

Eureka Math Grade 8 Module 3 Lesson 4 Exercise Answer Key

Exercise
In the diagram below, points R and S have been dilated from center O by a scale factor of r=3.

a. If |OR|=2.3 cm, what is |OR’|?
Answer:
|OR’|=3(2.3 cm)=6.9 cm

b. If |OS|=3.5 cm, what is |OS’|?
Answer:
|OS’|=3(3.5 cm)=10.5 cm

c. Connect the point R to the point S and the point R’ to the point S’. What do you know about the lines that contain segments RS and R’ S’?
Answer:
The lines containing the segments RS and R’S’ are parallel.

d. What is the relationship between the length of segment RS and the length of segment R’ S’?
Answer:
The length of segment R’ S’ is equal to the length of segment RS, times the scale factor of 3 (i.e., |R’ S’|=3|RS|).

e. Identify pairs of angles that are equal in measure. How do you know they are equal?
Answer:
|∠ORS|=|∠OR’ S’| and |∠OSR|=|∠OS’ R’| They are equal because they are corresponding angles of parallel lines cut by a transversal.

Eureka Math Grade 8 Module 3 Lesson 4 Problem Set Answer Key

Students verify that the fundamental theorem of similarity holds true when the scale factor r is 0<r<1.

Question 1.
Use a piece of notebook paper to verify the fundamental theorem of similarity for a scale factor r that is
0<r<1.
✓ Mark a point O on the first line of notebook paper.
✓ Mark the point P on a line several lines down from the center O. Draw a ray, \(\overrightarrow{O P}\). Mark the point P’ on the ray and on a line of the notebook paper closer to O than you placed point P. This ensures that you have a scale factor that is 0<r<1. Write your scale factor at the top of the notebook paper.
✓ Draw another ray, \(\overrightarrow{O Q}\), and mark the points Q and Q’ according to your scale factor.
✓ Connect points P and Q. Then, connect points P’ and Q’.
✓ Place a point, A, on the line containing segment PQ between points P and Q. Draw ray \(\overrightarrow{O A}\). Mark point A’ at the intersection of the line containing segment P’Q’ and ray \(\overrightarrow{O A}\).
Answer:
Sample student work is shown in the picture below:

a. Are the lines containing segments PQ and P’Q’ parallel lines? How do you know?
Answer:
Yes, the lines containing segments PQ and P’Q’ are parallel. The notebook lines are parallel, and these lines fall on the notebook lines.

b. Which, if any, of the following pairs of angles are equal in measure? Explain.
i. ∠OPQ and ∠OP’Q’
ii. ∠OAQ and ∠OA’Q’
iii. ∠OAP and ∠OA’P’
iv. ∠OQP and ∠OQ’P’
Answer:
All four pairs of angles are equal in measure because each pair of angles are corresponding angles of parallel lines cut by a transversal. In each case, the parallel lines are line PQ and line P’ Q’, and the transversal is the respective ray.

c. Which, if any, of the following statements are true? Show your work to verify or dispute each statement.
i. |OP’|=r|OP|
ii. |OQ’|=r|OQ|
iii. |P’A’|=r|PA|
iv. |A’Q’|=r|AQ|
Answer:
All four of the statements are true. Verify that students have shown that the length of the dilated segment was equal to the scale factor multiplied by the original segment length.

d. Do you believe that the fundamental theorem of similarity (FTS) is true even when the scale factor is 0<r<1? Explain.
Answer:
Yes, because I just experimentally verified the properties of FTS for when the scale factor is 0<r<1.

Question 2.
Caleb sketched the following diagram on graph paper. He dilated points B and C from center O.
Answer:

a. What is the scale factor r? Show your work.
Answer:
|OB’|=r|OB|
2=r∙6
\(\frac{2}{6}\)=r
\(\frac{1}{3}\)=r

b. Verify the scale factor with a different set of segments.
Answer:
|B’ C’|=r|BC|
3=r∙9
\(\frac{3}{9}\)=r
\(\frac{1}{3}\)=r

c. Which segments are parallel? How do you know?
Answer:
Segment BC and B’C’ are parallel. They lie on the lines of the graph paper, which are parallel.

d. Which angles are equal in measure? How do you know?
Answer:
|∠OB’ C’|=|∠OBC|, and |∠OC’ B’|=|∠OCB| because they are corresponding angles of parallel lines cut by a transversal.

Question 3.
Points B and C were dilated from center O.

a. What is the scale factor r? Show your work.
Answer:
|OC’|=r|OC|
6=r∙3
\(\frac{6}{3}\)=r
2=r

b. If |OB|=5, what is |OB’|?
Answer:
|OB’|=r|OB|
|OB’|=2∙5
|OB’|=10

c. How does the perimeter of triangle OBC compare to the perimeter of triangle OB’C’?
Answer:
The perimeter of triangle OBC is 12 units, and the perimeter of triangle OB’C’ is 24 units.

d. Did the perimeter of triangle OB’C’=r×(perimeter of triangle OBC)? Explain.
Answer:
Yes, the perimeter of triangle OB’C’ was twice the perimeter of triangle OBC, which makes sense because the dilation increased the length of each segment by a scale factor of 2. That means that each side of triangle OB’C’ was twice as long as each side of triangle OBC.

Eureka Math Grade 8 Module 3 Lesson 4 Exit Ticket Answer Key

Steven sketched the following diagram on graph paper. He dilated points B and C from point O. Answer the following questions based on his drawing.

Question 1.
What is the scale factor r? Show your work.
Answer:

|OB’|=r|OB|
7=r∙3
\(\frac{7}{3}\)=r

Question 2.
Verify the scale factor with a different set of segments.
Answer:
|B’ C’|=r|BC|
7=r∙3
\(\frac{7}{3}\)=r

Question 3.
Which segments are parallel? How do you know?
Answer:
Segments BC and B’C’ are parallel since they lie on the grid lines of the paper, which are parallel.

Question 4.
Are ∠OBC and ∠OB’C’ right angles? How do you know?
Answer:
The grid lines on graph paper are perpendicular, and since perpendicular lines form right angles, ∠OBC and ∠OB’C’ are right angles.

Engage NY Eureka Math 8th Grade Module 3 Lesson 3 Answer Key

Eureka Math Grade 8 Module 3 Lesson 3 Example Answer Key

Example 1.
Dilate circle A from center O at the origin by scale factor r=3.

→ Are three points enough? Let’s try.
(Show a picture of three dilated points.) If we connect these three dilated points, what image do we get?
→ With just three points, the image looks like a triangle.

→ What if we dilate a fourth point? Is that enough? Let’s try.
→ (Show the picture of four dilated points.) If we connect these four dilated points, what image do we get?
→ With four points, the image looks like a quadrilateral.

→ What if we dilate five, six, or ten points? What do you think?
→ The more points that are dilated, the more the image looks like a circle.
→ (Show the picture with many dilated points.)

→ Notice that the shape of the dilated image is now unmistakably a circle. Dilations map circles to circles, so it is important that when we dilate a circle we choose our points carefully.
→ Would we have an image that looked like a circle if all of the points we considered were located on just one part of the circle? For example, what if all of our points were located on just the top half of the circle? Would the dilated points produce an image of the circle?

→ Or consider the image when we select points just on the lower half of the circle:

→ Consider the image when the points are focused on just the sides of the circle:

→ The images are not good enough to truly show that the original figure was a circle.
→ How should we select points to dilate when we have a curved figure?
→ We should select points on all parts of the curve, not just those points focused in one area.
→ The number of points to dilate that is enough is as many as are needed to produce a dilated image that looks like the original. For curved figures, like this circle, the more points you dilate the better. The location of the points you choose to dilate is also important. The points selected to dilate should be spread out evenly on the curve.

Example 2.
In the picture below, we have triangle ABC, that has been dilated from center O, by a scale factor of r=\(\frac{1}{3}\). It is noted by A’B’C’.

Ask students what can be done to map this new triangle, triangle A’B’C’, back to the original. Tell them to be as specific as possible. Students should write their conjectures or share with a partner.
→ Let’s use the definition of dilation and some side lengths to help us figure out how to map triangle A’B’C’ back onto triangle ABC. How are the lengths |OA’ | and |OA| related?
→ We know by the definition of dilation that |OA’|=r|OA|.
We know that r=\(\frac{1}{3}\). Let’s say that the length of segment OA is 6 units (we can pick any number, but 6 makes it easy for us to compute). What is the length of segment OA’?
→ Since |OA’|=\(\frac{1}{3}\) |OA|, and we are saying that the length of segment OA is 6, then |OA’|=\(\frac{1}{3}\)∙=2, and |OA’|=2 units.
→ Now, since we want to dilate triangle A’B’C’ to the size of triangle ABC, we need to know what scale factor r is required so that |OA|=r|OA’|. What scale factor should we use, and why?
→ We need a scale factor r=3 because we want |OA|=r|OA’|. Using the lengths from before, we have 6=r⋅2. Therefore, r=3.
→ Now that we know the scale factor, what precise dilation would map triangle A’B’C’ onto triangle ABC?
→ A dilation from center O with scale factor r=3

Example 3.
In the picture below, we have triangle DEF that has been dilated from center O by a scale factor of r=4. It is
noted by D’E’F’.

→ Based on the example we just did, make a conjecture about how we could map this new triangle D’E’F’ back onto the original triangle.
Let students attempt to prove their conjectures on their own or with a partner. If necessary, use the scaffolding questions that follow.
→ What is the difference between this problem and the last?
→ This time the scale factor is greater than one, so we need to shrink triangle D’E’F’ to the size of triangle DEF.
→ We know that r=4. Let’s say that the length of segment OF is 3 units. What is the length of segment OF’?
→ Since |OF’|=r|OF| and we are saying that the length of segment OF is 3, then |OF’|=4⋅3=12, and |OF’|=12 units.
→ Now, since we want to dilate triangle D’E’F’ to the size of triangle DEF, we need to know what scale factor r is required so that |OF|=r|OF’|. What scale factor should we use, and why?
→ We need a scale factor r=\(\frac{1}{4}\) because we want |OF|=r|OF’|. Using the lengths from before, we have 3=r⋅12. Therefore, r=\(\frac{1}{4}\).
→ What precise dilation would make triangle D’E’F’ the same size as triangle DEF?
→ A dilation from center O with scale factor r=\(\frac{1}{4}\) would make triangle D’E’F’ the same size as triangle DEF.

Eureka Math Grade 8 Module 3 Lesson 3 Exercise Answer Key

Exercises 1–2.

Exercise 1.
Dilate ellipse E, from center O at the origin of the graph, with scale factor r=2. Use as many points as necessary to develop the dilated image of ellipse E.
Answer:
The dilated image of E is shown in red below. Verify that students have dilated enough points evenly placed to get an image that resembles an ellipse.

Exercise 2.
What shape was the dilated image?
Answer:
The dilated image was an ellipse. Dilations map ellipses to ellipses.

Exercise 3.
Triangle ABC has been dilated from center O by a scale factor of r=\(\frac{1}{4}\) denoted by triangle A’B’C’. Using a centimeter ruler, verify that it would take a scale factor of r=4 from center O to map triangle A’B’C’ onto triangle ABC.

Answer:
Sample measurements provided. Note that, due to print variations, the measurements may be slightly different.
Verify that students have measured the lengths of segments from center O to each of the dilated points. Then, verify that students have multiplied each of the lengths by 4 to see that it really is the length of the segments from center O to the original points.

Eureka Math Grade 8 Module 3 Lesson 3 Problem Set Answer Key

Students practice dilating a curved figure and stating the scale factor that would bring a dilated figure back to its original size.

Question 1.
Dilate the figure from center O by a scale factor r=2. Make sure to use enough points to make a good image of the original figure.
Answer:
Sample student work is shown below. Verify that students used enough points to produce an image similar to the original.

Question 2.
Describe the process for selecting points when dilating a curved figure.
Answer:
When dilating a curved figure, you have to make sure to use a lot of points to produce a decent image of the original figure. You also have to make sure that the points you choose are not all concentrated in just one part of the figure.

Question 3.
A figure was dilated from center O by a scale factor of r=5. What scale factor would shrink the dilated figure back to the original size?
Answer:
A scale factor of r=\(\frac{1}{5}\) would bring the dilated figure back to its original size.

Question 4.
A figure has been dilated from center O by a scale factor of r=\(\frac{7}{6}\). What scale factor would shrink the dilated figure back to the original size?
A scale factor of r=\(\frac{6}{7}\) would bring the dilated figure back to its original size.

Question 5.
A figure has been dilated from center O by a scale factor of r=\(\frac{3}{10}\). What scale factor would magnify the dilated figure back to the original size?
Answer:
A scale factor of r=\(\frac{10}{3}\) would bring the dilated figure back to its original size.

Eureka Math Grade 8 Module 3 Lesson 3 Exit Ticket Answer Key

Question 1.
Dilate circle A from center O by a scale factor r=\(\frac{1}{2}\). Make sure to use enough points to make a good image of the original figure.

Answer:

Student work is shown below. Verify that students used enough points to produce an image similar to the original.

Question 2.
What scale factor would magnify the dilated circle back to the original size of circle A? How do you know?
Answer:
A scale factor of r=2 would bring the dilated circle back to the size of circle A. Since the circle was dilated by a scale factor of r=\(\frac{1}{2}\), then to bring it back to its original size, you must dilate by a scale factor that is the reciprocal of \(\frac{1}{2}\), which is 2.

Engage NY Eureka Math 8th Grade Module 3 Lesson 2 Answer Key

Eureka Math Grade 8 Module 3 Lesson 2 Example Answer Key

Example 1.
Examples 1–3 demonstrate that dilations map lines to lines and how to use a compass to dilate.
→ This example shows that a dilation maps a line to a line. This means that the image of a line, after undergoing a dilation, is also a line. Given line L, we dilate with a scale factor r=2 from center O. Before we begin, exactly how many lines can be drawn through two points?
→ Only one line can be drawn through two points.
To dilate the line, we choose two points on L (points P and Q) to dilate. When we connect the dilated images of P and Q (P’ and Q’), we have the dilated image of line L, L’.

→ First, let’s select a center O off the line L and two points P and Q on line L.

Examples 1–2:
Dilations Map Lines to Lines

→ Second, we draw rays from center O through each of the points P and Q. We want to make sure that the points O, P, and P’ (the dilated P) lie on the same line (i.e., are collinear). That is what keeps the dilated image “in proportion.” Think back to the last lesson where we saw how the size of the picture changed when pulling the corners compared to the sides. Pulling from the corners kept the picture “in proportion.” The way we achieve this in diagrams is by drawing rays and making sure that the center, the point, and the dilated point are all on the same line.

→ Next, we use our compass to measure the distance from O to P. Do this by putting the point of the compass on point O, and adjust the radius of the compass to draw an arc through point P. Once you have the compass set, move the point of the compass to P, and make a mark along the ray OP (without changing the radius of the compass) to mark P’. Recall that the dilated point P’ is the distance 2|OP| (i.e., |OP’ |=2|OP|). The compass helps us to find the location of P’ so that it is exactly twice the length of segment OP. Use your ruler to prove this to students.

→ Next, we repeat this process to locate Q’.

→ Finally, connect points P’ and Q’ to draw line L’.

→ Return to your conjecture from before, or look at our class list. Which conjectures were accurate? How do you know?
→ Answers may vary depending on conjectures made by the class. Students should identify that the conjecture of a line mapping to a line under a dilation is correct.
→ What do you think would happen if we selected a different location for the center or for the points P and Q?
→ Points O, P, and Q are arbitrary points. That means that they could have been anywhere on the plane. For that reason, the results would be the same; that is, the dilation would still produce a line, and the line would be parallel to the original.
→ Look at the drawing again, and imagine using our transparency to translate the segment OP along vector \(\overrightarrow{O Q}\) to segment PP’ and to translate the segment OQ along vector \(\overrightarrow{O Q}\) to segment QQ’. With that information, can you say anything more about lines L and L’?
→ Since P and Q are arbitrary points on line L, and translations map lines to parallel lines when the vector is not parallel to or part of the original line, we can say that L is parallel to L’.
→ How would the work we did change if the scale factor were r=3 instead of r=2?
→ We would have to find a point P’ so that it is 3 times the length of segment OP instead of twice the length of segment OP. Same for the point Q’

Example 2.
→ Do you think line L would still be a line under a dilation with scale factor r=3? Would the dilated line, L’, still be parallel to L? (Allow time for students to talk to their partners and make predictions.)
→ Yes, it is still a line, and it would still be parallel to line L. The scale factor being three instead of two simply means that we would perform the translation of the points P’ and Q’ more than once, but the result would be the same.
→ Here is what would happen with scale factor r=3.

Example 3.
Dilations Map Lines to Lines

→ What would happen if the center O were on line L? (Allow time for students to talk to their partners and make predictions.)
→ If the center O were on line L, and if we pick points P and Q on L, then the dilations of points P and Q, P’ and Q’, would also be on L. That means that line L and its dilated image, line L’, would coincide.

→ What we have shown with these three examples is that a line, after a dilation, is still a line. Mathematicians like to say that dilations map lines to lines.

Example 4.
Example 4 demonstrates that dilations map rays to rays. It also demonstrates how to use a ruler to dilate with scale factor r=\(\frac{1}{2}\). Similar to Example 1, before this example, discuss the conjectures students developed about rays. Also, consider getting students started and then asking them to finish with a partner.
→ This example shows that a dilation maps a ray to a ray. Given ray \(\overrightarrow{A B}\), we dilate with a scale factor r=\(\frac{1}{2}\) from center O.
→ To dilate the ray, we choose a center O off of the ray. Like before, we draw rays from center O through points A and B.

→ Since our scale factor is r=\(\frac{1}{2}\), we need to use a ruler to measure the length of segments OA and OB. When you get into high school Geometry, you learn how to use a compass to handle scale factors that are greater than zero but less than one, like our r=\(\frac{1}{2}\). For now, we use a ruler.

→ Since our scale factor is r=\(\frac{1}{2}\), we know that the dilated segment, OA’, must be equal to \(\frac{1}{2}\) the length of segment OA (i.e., |OA’ |=\(\frac{1}{2}\) |OA|). Then, |OA’ | must be \(\frac{1}{2}\)∙5; therefore, |OA’ |=2.5. What must the |OB’ | be?
→ We know that |OB’ |=\(\frac{1}{2}\) |OB|; therefore, |OB’ |=\(\frac{1}{2}\)⋅8=4, and |OB’ |=4.
→ Now that we know the lengths of segments OA’ and OB’, we use a ruler to mark off those points on their respective rays.

→ Finally, we connect point A’ through point B’. With your partner, evaluate your conjecture. What happened to our ray after the dilation?
→ When we connect point A’ through point B’, then we have the ray \(\overrightarrow{A^{\prime} B^{\prime}}\).

→ What do you think would have happened if we selected our center O as a point on the ray \(\overrightarrow{A B}\)?
→ If our center O were on the ray, then the ray \(\overrightarrow{A B}\) would coincide with its dilated image \(\overrightarrow{A^{\prime} B^{\prime}},\), which is similar to what we saw when the line L was dilated from a center O on it.

Eureka Math Grade 8 Module 3 Lesson 2 Exercise Answer Key

Given center O and triangle ABC, dilate the triangle from center O with a scale factor r=3.

a. Note that the triangle ABC is made up of segments AB, BC, and CA. Were the dilated images of these segments still segments?
Answer:
Yes, when dilated, the segments were still segments.

b. Measure the length of the segments AB and A’ B’. What do you notice? (Think about the definition of dilation.)
Answer:
The segment A’ B’ was three times the length of segment AB. This fits with the definition of dilation, that is, |A’ B’ |=r|AB|.

c. Verify the claim you made in part (b) by measuring and comparing the lengths of segments BC and B’ C’ and segments CA and C’ A’. What does this mean in terms of the segments formed between dilated points?
Answer:
This means that dilations affect segments in the same way they do points. Specifically, the lengths of segments are dilated according to the scale factor.

d. Measure ∠ABC and ∠A’B’C’. What do you notice?
Answer:
The angles are equal in measure.

e. Verify the claim you made in part (d) by measuring and comparing the following sets of angles:
(1) ∠BCA and ∠B’ C’ A’ and
(2) ∠CAB and ∠C’ A’ B’. What does that mean in terms of dilations with respect to angles and their degrees?
Answer:
It means that dilations map angles to angles, and the dilation preserves the measures of the angles.

Eureka Math Grade 8 Module 3 Lesson 2 Exit Ticket Answer Key

Question 1.
Given center O and quadrilateral ABCD, using a compass and ruler, dilate the figure from center O by a scale factor of r=2. Label the dilated quadrilateral A’B’C’D’.

Answer:
Sample student work is shown below. Verify that students have magnified the image ABCD.

Question 2.
Describe what you learned today about what happens to lines, segments, rays, and angles after a dilation.
Answer:
We learned that a dilation maps a line to a line, a segment to a segment, a ray to a ray, and an angle to an angle. Further, the length of the dilated line segment is exactly r (the scale factor) times the length of the original segment. Also, the measure of a dilated angle remains unchanged compared to the original angle.

Eureka Math Grade 8 Module 3 Lesson 2 Problem Set Answer Key

Students practice dilating figures with different scale factors.

Question 1.
Use a ruler to dilate the following figure from center O, with scale factor r=\(\frac{1}{2}\).

Answer:
The dilated figure is shown in red below. Verify that students have dilated according to the scale factor r=\(\frac{1}{2}\).

Question 2.
Use a compass to dilate the figure ABCDE from center O, with scale factor r=2.

Answer:
The figure in red below shows the dilated image of ABCDE.

a. Dilate the same figure, ABCDE, from a new center, O’, with scale factor r=2. Use double primes (A”B”C”D”E”) to distinguish this image from the original.
Answer:

The figure in blue below shows the dilated figure A”B”C”D”E”.

b. What rigid motion, or sequence of rigid motions, would map A”B”C”D”E” to A’B’C’D’E’?
Answer:

A translation along vector \(\overrightarrow{\boldsymbol{A}^{\prime \prime} \boldsymbol{A}^{\prime}}\) (or any vector that connects a point of A”B”C”D”E” and its corresponding point of A’B’C’D’E’) would map the figure A”B”C”D”E” to A’B’C’D’E’.
The image below (with rays removed for clarity) shows the vector \(\overrightarrow{\boldsymbol{A}^{\prime \prime} \boldsymbol{A}^{\prime}}\).

Question 3.
Given center O and triangle ABC, dilate the figure from center O by a scale factor of r=\(\frac{1}{4}\). Label the dilated triangle A’B’C’.

Answer:

Question 4.
A line segment AB undergoes a dilation. Based on today’s lesson, what is the image of the segment?
Answer:
The segment dilates as a segment.

Question 5.
∠GHI measures 78°. After a dilation, what is the measure of ∠G’H’I’? How do you know?
Answer:
The measure of ∠G’H’I’ is 78°. Dilations preserve angle measure, so it remains the same size as ∠GHI.

Engage NY Eureka Math 8th Grade Module 3 Lesson 1 Answer Key

Eureka Math Grade 8 Module 3 Lesson 1 Exploratory Challenge Answer Key

Exploratory Challenge.
Two geometric figures are said to be similar if they have the same shape but not necessarily the same size. Using that informal definition, are the following pairs of figures similar to one another? Explain.
Pair A:

Answer:
Yes, these figures appear to be similar. They are the same shape, but one is larger than the other, or one is smaller than the other.

Pair B:

Answer:
No, these figures do not appear to be similar. One looks like a square and the other like a rectangle.

Pair C:

Answer:
These figures appear to be exactly the same, which means they are congruent.

Pair D:

Answer:
Yes, these figures appear to be similar. They are both circles, but they are different sizes.

Pair E:

Answer:
Yes, these figures appear to be similar. They are the same shape, but they are different in size.

Pair F:

Answer:
Yes, these figures appear to be similar. The faces look the same, but they are just different in size.

Pair G:

Answer:
They do not look to be similar, but I’m not sure. They are both happy faces, but one is squished compared to the other.

P air H:

Answer:
No, these two figures do not look to be similar. Each is curved but shaped differently.

Eureka Math Grade 8 Module 3 Lesson 1 Exercise Answer Key

Question 1.
Given |OP|=5 in.
a. If segment OP is dilated by a scale factor r=4, what is the length of segment OP’?
Answer:
|OP |=20 in. because the scale factor multiplied by the length of the original segment is 20; that is,
4 ∙ 5=20.

b. If segment OP is dilated by a scale factor r=\(\frac{1}{2}\), what is the length of segment OP’?
Answer:
|OP’ |=2.5 in. because the scale factor multiplied by the length of the original segment is 2.5; that is,
(\(\frac{1}{2}\))∙5=2.5.

Use the diagram below to answer Exercises 2–6. Let there be a dilation from center O. Then, Dilation(P)=P’ and Dilation(Q)=Q’. In the diagram below, |OP|=3 cm and |OQ|=4 cm, as shown.

Question 2.
If the scale factor is r=3, what is the length of segment OP’?
Answer:
The length of the segment OP’ is 9 cm.

Question 3.
Use the definition of dilation to show that your answer to Exercise 2 is correct.
Answer:
|OP’ |=r |OP|; therefore, |OP’|=3∙3 cm=9 cm.

Question 4.
If the scale factor is r=3, what is the length of segment OQ’?
Answer:
The length of the segment OQ’ is 12 cm.

Question 5.
Use the definition of dilation to show that your answer to Exercise 4 is correct.
Answer:
|OQ’ |=r|OQ|; therefore, |OQ’ |=3∙4 cm=12 cm.

Question 6.
If you know that |OP|=3, |OP’ |=9, how could you use that information to determine the scale factor?
Answer:
Since we know |OP’|=r|OP|, we can solve for r: \(\frac{\left|o P^{\prime}\right|}{|O P|}\) =r, which is \(\frac{9}{3}\)=r or 3=r.

Eureka Math Grade 8 Module 3 Lesson 1 Exit Ticket Answer Key

Question 1.
Why do we need a better definition for similarity than “same shape, not the same size”?
Answer:
We need a better definition that includes dilation and a scale factor because some figures may look to be similar (e.g., the smiley faces), but we cannot know for sure unless we can check the proportionality. Other figures (e.g., the parabolas) may not look similar but are. We need a definition so that we are not just guessing if they are similar by looking at them.

Question 2.
Use the diagram below. Let there be a dilation from center O with scale factor 3. Then, Dilation(P)=P’. In the diagram below, |OP|=5 cm. What is |OP’ |? Show your work.

Answer:
Since |OP’|=r|OP|, then
|OP’|=3∙5 cm,
|OP’|=15 cm.

Question 3.
Use the diagram below. Let there be a dilation from center O. Then, Dilation(P)=P’. In the diagram below, |OP|=18 cm and |OP’|=9 cm. What is the scale factor r? Show your work.

Answer:
Since |OP’ |=r|OP|, then
9 cm=r∙18 cm,
\(\frac{1}{2}\)=r.

Eureka Math Grade 8 Module 3 Lesson 1 Problem Set Answer Key

Have students practice using the definition of dilation and finding lengths according to a scale factor.

Question 1.
Let there be a dilation from center O. Then, Dilation(P)=P’ and Dilation(Q)=Q’. Examine the drawing below. What can you determine about the scale factor of the dilation?
Answer:

The scale factor must be greater than one, r>1, because the dilated points are farther from the center than the original points.

Question 2.
Let there be a dilation from center O. Then, Dilation(P)=P’, and Dilation(Q)=Q’. Examine the drawing below. What can you determine about the scale factor of the dilation?

Answer:
The scale factor must be greater than zero but less than one, 0<r<1, because the dilated points are closer to the center than the original points.

Question 3.
Let there be a dilation from center O with a scale factor r=4. Then, Dilation(P)=P’ and Dilation(Q)=Q’. |OP|=3.2 cm, and |OQ|=2.7 cm, as shown. Use the drawing below to answer parts (a) and (b). The drawing is not to scale.

a. Use the definition of dilation to determine |OP’ |.
Answer:
|OP’ |=r|OP|; therefore, |OP’ |=4 (3.2 cm)=12.8 cm.

b. Use the definition of dilation to determine |OQ’ |.
Answer:
|OQ’ |=r|OQ|; therefore, |OQ’ |=4 (2.7 cm)=10.8 cm.

Question 4.
Let there be a dilation from center O with a scale factor r. Then, Dilation(A)=A’, Dilation(B)=B’, and Dilation(C)=C’. |OA|=3, |OB|=15, |OC|=6, and |OB’ |=5, as shown. Use the drawing below to answer parts (a)–(c).

a. Using the definition of dilation with lengths OB and OB’, determine the scale factor of the dilation.
Answer:
|OB’ |=r|OB|, which means 5=r×15; therefore, r=\(\frac{1}{3}\).

b. Use the definition of dilation to determine |OA’|.
Answer:
|OA’ |=\(\frac{1}{3}\) |OA|; therefore, |OA’ |=\(\frac{1}{3}\)∙3=1, and |OA’|=1.

c. Use the definition of dilation to determine |OC’|.
Answer:
|OC’ |=\(\frac{1}{3}\) |OC|; therefore, |OC’|=\(\frac{1}{3}\)∙6=2, and |OC’|=2.

Engage NY Eureka Math 8th Grade Module 2 Mid Module Assessment Answer Key

Eureka Math Grade 8 Module 2 Mid Module Assessment Task Answer Key

Question 1.
Translate △XYZ along \(\overrightarrow{A B}\). Label the image of the triangle with X’, Y’, and Z’.
Answer:

b. Reflect △XYZ across the line of reflection, l. Label the image of the triangle with X’, Y’, and Z’.
Answer:

c. Rotate △XYZ around the point (1,0) clockwise 90°. Label the image of the triangle with X’, Y’, and Z’.
Answer:

Question 2.
Use the picture below to answer the questions.
Answer:

a. Can Figure A be mapped onto Figure B using only translation? Explain. Use drawings as needed in your explanation.
Answer:
No, if i translate along vector \(\overrightarrow{A B}\) i can get the longer point of figure A to map onto the lower left point of figure B(one pair of corresponding points). But no other points of the figures coincide.)

b. Can Figure A be mapped onto Figure B using only reflection? Explain. Use drawings as needed in your explanation. Use the graphs below to answer parts (a) and (b).
Answer:
No, when i connect a point of figure A to its image on figure B, the line of reflection should bisect the segment. When i connect midpoints of \(\overline{x x^{\prime}}\) & \(\overline{y y^{\prime}}\) i get a possible line of reflection, but when i check, figure A does not map onto figure B.

Question 3.
Reflect △XYZ over the horizontal line (parallel to the x-axis) through point (0,1). Label the reflected image with X’Y’Z’.

a. Reflect △XYZ over the horizontal line (parallel to the x-axis) through point (0,1). Label the reflected image with X’Y’Z’.
Answer:

b. One triangle in the diagram below can be mapped onto the other using two reflections. Identify the lines of reflection that would map one onto the other. Can you map one triangle onto the other using just one basic rigid motion? If so, explain.
Answer:

A reflection across the x-axis maps △ABC to △A’B’C’ and a reflection across the y-axis maps △A’B’C’ to △A”B”C”.
Since AB || A”B”, BC = B”C”, AC = A”C”, then A 180° rotation about the origin will map △ABC to △A”B”C”.