Engage NY Eureka Math 3rd Grade Module 7 Lesson 30 Answer Key
Eureka Math Grade 3 Module 7 Lesson 30 Pattern Sheet Answer Key
Multiply.
multiply by 9 (6–10)
Answer:
Explanation:
9 × 5 = 45
9 × 6 = 54
9 × 7 = 63
9 × 8 = 72
9 × 9 = 81
9 × 10 = 90.
Eureka Math Grade 3 Module 7 Lesson 30 Problem Set Answer Key
Use this form to critique your classmate’s problemsolving work.
Classmate: 

Problem Number:  
Strategies My Classmate Used:  
Things My Classmate Did Well:  
Suggestions for Improvement:  
Strategies I Would Like to Try Based on My Classmate’s Work: 
Answer:
Eureka Math Grade 3 Module 7 Lesson 30 Exit Ticket Answer Key
Jayden solves the problem as shown below.
The recreation center soccer field measures 35 yards by 65 yards. Chris dribbles the soccer ball around the field 4 times. What is the total number of yards Chris dribbles the ball?
Question 1.
What strategies did Jayden use to solve this problem?
Answer:
First took the required measurements of the recreation center soccer field and calculated its perimeter using formula. Finally , calculated the total number of yards Chris dribbles the ball
Explanation:
First, Jayden took the information about measurements of the rectangle.
Later Jayden calculated its required perimeter of the recreation center soccer field using the formula.
Last Jayden calculated the total number of yards Chris dribbles the ball.
Question 2.
What did Jayden do well?
Answer:
Jayden has did the calculation well and solved the problem in a systematic way.
Explanation:
Jayden solved the problem every well in a systematic and understandable way.
Eureka Math Grade 3 Module 7 Lesson 30 Homework Answer Key
Use this form to critique Student A’s problemsolving work on the next page.
Student:  Student A  Problem Number:  
Strategies Student A Used:  
Things Student A Did Well:  
Suggestions for Improvement:  
Strategies I Would Like to Try Based on Student A’s Work: 
Answer:
Question 1.
Katherine puts 2 squares together to make the rectangle below. The side lengths of the squares measure 8 inches.
a. What is the perimeter of Katherine’s rectangle?
b. What is the area of Katherine’s rectangle?
c. Katherine draws 2 of the rectangles in Problem 1 side by side. Her new, larger rectangle is shown below. What is the area of the new, larger rectangle?
Answer:
a. Perimeter of the ACDF Katherine’s rectangle = 32 in.
b. Area of the ACDF Katherine’s rectangle = 64 square in.
c. Area of the OQRT new, Katherine’s larger rectangle = 64 square in.
Explanation:
a.
Length of the AC side of ACDF Katherine’s rectangle = 8 in
Length of the CD side of ACDF Katherine’s rectangle = 8 in
Length of the DF side of ACDF Katherine’s rectangle = 8 in
Length of the AF side of ACDF Katherine’s rectangle = 8 in
Perimeter of ACDF Katherine’s rectangle = Length of the AC side of ACDF Katherine’s rectangle + Length of the CD side of ACDF Katherine’s rectangle + Length of the DF side of ACDF Katherine’s rectangle + Length of the AF side of ACDF Katherine’s rectangle
= 8 in + 8 in + 8 in + 8 in
= 16 in + 8 in + 8 in
= 24 in + 8 in
= 32 in.
b. Length of the AC side of the ACDF Katherine’s rectangle = 8 in
Width of the side FG of the ACDF Katherine’s rectangle = 8 in
Area of the ACDF Katherine’s rectangle = Length × Width
= 8 in × 8 in
= 64 square in.
c.
Length of the OQ side of the OQRT new, Katherine’s larger rectangle = 8 in
Width of the QR side of the OQRT new, Katherine’s larger rectangle = 8 in
Area of the OQRT new, Katherine’s larger rectangle = Length × Width
= 8 in × 8 in
= 64 square in.