Eureka Math Grade 3 Module 7 Lesson 12 Answer Key

Engage NY Eureka Math 3rd Grade Module 7 Lesson 12 Answer Key

Eureka Math Grade 3 Module 7 Lesson 12 Pattern Sheet Answer Key

Multiply.
Engage NY Math 3rd Grade Module 7 Lesson 12 Pattern Sheet Answer Key p 1
multiply by 7 (6–10)
Answer:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka Math 3 Module 7 Lesson 12 Pattern Sheet Answer Key

Explanation:
7 × 6 = 42
7 × 7 = 49
7 × 8 = 56
7 × 9 = 63
7 × 10 = 70.

Eureka Math Grade 3 Module 7 Lesson 12 Problem Set Answer Key

Question 1.
Measure and label the side lengths of the shapes below in centimeters. Then, find the perimeter of each shape.
a.
Engage NY Math Grade 3 Module 7 Lesson 12 Problem Set Answer Key pr 1
Perimeter = _____cm +_____cm +_____cm +_____cm
= _______ cm
Answer:
Perimeter of the ABCD given shape = Side + Side + Side + Side
= 2cm + 2cm + 2cm + 2cm
= 8cm.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka Math 3 Module 7 Lesson 12 Pattern Sheet Answer Key-1a
Length of the AB side of the ABCD shape = 2cm
Length of the BC side of the ABCD shape = 2cm
Length of the CD side of the ABCD shape = 2cm
Length of the DA side of the ABCD shape = 2cm
Perimeter of the given ABCD shape = Side + Side + Side + Side
= AB + BC + CD+ DA
= 2cm + 2cm + 2cm + 2cm
= 4cm + 2cm + 2cm
= 6cm + 2cm
= 8cm.

 

b.
Engage NY Math Grade 3 Module 7 Lesson 12 Problem Set Answer Key pr 2
Perimeter = _____________________
= _______ cm
Answer:
Perimeter of the given ABCDEF Hexagon shape = Side + Side + Side + Side + Side + Side
= 3cm + 3cm + 3cm + 3cm + 3cm + 3cm
= 18cm.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka Math 3 Module 7 Lesson 12 Pattern Sheet Answer Key-1b
Length of the AB side of the given ABCDEF Hexagon shape = 3cm.
Length of the BC side of the given ABCDEF Hexagon shape = 3cm.
Length of the CD side of the given ABCDEF Hexagon shape = 3cm.
Length of the DE side of the given ABCDEF Hexagon shape = 3cm.
Length of the EF side of the given ABCDEF Hexagon shape = 3cm.
Length of the FA side of the given ABCDEF Hexagon shape = 3cm.
Perimeter of the given ABCDEF Hexagon shape = Side + Side + Side + Side + Side + Side
= AB + BC + CD + DE + EF + FA
= 3cm + 3cm + 3cm + 3cm + 3cm + 3cm
= 6cm +3cm + 3cm + 3cm + 3cm
= 9cm +3cm + 3cm + 3cm
= 12cm +3cm + 3cm
= 15cm + 3cm
= 18cm.

c.
Engage NY Math Grade 3 Module 7 Lesson 12 Problem Set Answer Key pr 3
Perimeter = _____________________
= _______ cm
Answer:
Perimeter of the given ABCD Parallelogram shape = Side + Side + Side + Side
= 4cm + 4cm + 4cm + 4cm
= 16cm.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka Math 3 Module 7 Lesson 12 Pattern Sheet Answer Key-1c
Length of the AB side of the given ABCD Parallelogram shape = 4cm
Length of the BC side of the given ABCD Parallelogram shape = 4cm
Length of the CD  side of the given ABCD Parallelogram shape = 4cm
Length of the DA side of the given ABCD Parallelogram shape = 4cm
Perimeter of the given ABCD Parallelogram shape = Side + Side + Side + Side
=AB + BC + CD +DA
= 4cm + 4cm + 4cm + 4cm
= 8cm + 4cm + 4cm
= 12cm + 4cm
=16cm.

d.
Engage NY Math Grade 3 Module 7 Lesson 12 Problem Set Answer Key pr 4
Perimeter = _____________________
= _______ cm
Answer:
Perimeter of the ABC Triangle = Side + Side + Side
= 5cm + 5cm + 5cm
= 15cm.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 12 Problem Set Answer Key-1d
Length of the AB side of the ABC Triangle = 5cm
Length of the BC side of the ABC Triangle = 5cm
Length of the CA side of the ABC Triangle = 5cm
Perimeter of the ABC Triangle = Side + Side + Side
= AB + BC + CA
= 5cm + 5cm + 5cm
= 10cm + 5cm
= 15cm.

e.
Engage NY Math Grade 3 Module 7 Lesson 12 Problem Set Answer Key pr 5
Perimeter = _____________________
= _______ cm
Answer:
Perimeter of the given ABCDEF figure = Side + Side + Side + Side + Side + Side
= AB + BC + CD + DE+ EF + FA
= 5.5cm+ 1cm + 3cm + 2cm + 3cm + 3cm
= 17.5cm

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 12 Problem Set Answer Key-1e
Length of AB side if the given ABCDEF figure = 5.5cm
Length of BC side if the given ABCDEF figure = 1cm
Length of CD side if the given ABCDEF figure = 3cm
Length of DE side if the given ABCDEF figure = 2cm
Length of EF side if the given ABCDEF figure = 3cm
Length of FA side if the given ABCDEF figure = 3cm
Perimeter of the given ABCDEF figure = Side + Side + Side + Side + Side + Side
= AB + BC + CD + DE+ EF + FA
= 5.5cm+ 1cm + 3cm + 2cm + 3cm + 3cm
= 6.5cm + 3cm + 2cm + 3cm + 3cm
= 9.5cm + 2cm + 3cm + 3cm
= 11.5cm + 3cm + 3cm
= 14.5cm + 3cm
= 17.5cm

Question 2.
Carson draws two triangles to create the new shape shown below. Use a ruler to find the side lengths of Carson’s shape in centimeters. Then, find the perimeter.
Engage NY Math Grade 3 Module 7 Lesson 12 Problem Set Answer Key pr 6
Answer:
Perimeter of ABCD Carson’s shape = Side + Side + Side + Side
= AB + BC + CD+ DA
= 2cm + 2cm + 2cm + 2cm
=8cm.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka Math 3 Module 7 Lesson 12 Pattern Sheet Answer Key-2

Length of AB side of ABCD Carson’s shape = 2cm
Length of BC side of ABCD Carson’s shape = 2cm
Length of CD side of ABCD Carson’s shape = 2cm
Length of DA side of ABCD Carson’s shape = 2cm
Perimeter of ABCD Carson’s shape = Side + Side + Side + Side
= AB + BC + CD+ DA
= 2cm + 2cm + 2cm + 2cm
=4cm + 2cm + 2cm
= 6cm + 2cm
=8cm.

Question 3.
Hugh and Daisy draw the shapes shown below. Measure and label the side lengths in centimeters. Whose shape has a greater perimeter? How do you know?
Engage NY Math Grade 3 Module 7 Lesson 12 Problem Set Answer Key pr 7
Answer:
Perimeter of ABCDE Hugh’s shape = Side + Side + Side + Side + Side
= AB + BC + CD + DE + EA
= 3cm + 3cm + 3cm + 3cm + 3cm
= 15cm.
Perimeter of FGHI Daisy’s shape =  Side + Side + Side + Side
= FG+ GH+ HI+ IF
=  4cm + 5cm + 4cm + 3cm
= 16cm.
Perimeter of FGHI Daisy’s shape is greater than the Perimeter of ABCDE Hugh’s shape because the  measurement value is greater than the other one’s shape.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 12 Homework Answer Key-3..
Length of the AB side of ABCDE Hugh’s shape = 3cm
Length of the BC side of ABCDE Hugh’s shape = 3cm
Length of the CD side of ABCDE Hugh’s shape = 3cm
Length of the DE side of ABCDE Hugh’s shape = 3cm
Length of the EA side of ABCDE Hugh’s shape = 3cm
Perimeter of ABCDE Hugh’s shape = Side + Side + Side + Side + Side
= AB + BC + CD + DE + EA
= 3cm + 3cm + 3cm + 3cm + 3cm
= 6cm + 3cm + 3cm + 3cm
= 9cm + 3cm + 3cm
= 12cm + 3cm
= 15cm.

Length of the FG side of FGHI Daisy’s shape = 4cm
Length of the GH side of FGHI Daisy’s shape = 5cm
Length of the HI side of FGHI Daisy’s shape = 4cm
Length of the IF side of FGHI Daisy’s shape = 3cm
Perimeter of FGHI Daisy’s shape =  Side + Side + Side + Side
= FG+ GH+ HI+ IF
=  4cm + 5cm + 4cm + 3cm
= 9cm + 4cm +3cm
= 13cm + 3cm
= 16cm.

Question 4.
Andrea measures one side length of the square below and says she can find the perimeter with that measurement. Explain Andrea’s thinking. Then, find the perimeter in centimeters.
Engage NY Math Grade 3 Module 7 Lesson 12 Problem Set Answer Key pr 8
Answer:
Andrea’s figure is a Square. So, her figure’s  all sides are going to be equal. Yes, she is correct in her thinking.
Perimeter of ABCD Andrea’s Square = Side + Side + Side + Side
= 4cm + 4cm + 4cm+ 4cm
= 16cm.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 12 Problem Set Answer Key-4
Andrea’s figure is a Square. So, her figure’s  all sides are going to be equal. Yes, she is correct in her thinking, ” measuring one side length of the square below, she can find the perimeter with that measurement.”
Length of AB side of Andrea’s Square = 4cm
Perimeter of ABCD Andrea’s Square = Side + Side + Side + Side
= 4cm + 4cm + 4cm+ 4cm
= 8cm + 4cm + 4cm
=12cm + 4cm
= 16cm.

Eureka Math Grade 3 Module 7 Lesson 12 Exit Ticket Answer Key

Measure and label the side lengths of the shape below in centimeters. Then, find the perimeter.
Eureka Math 3rd Grade Module 7 Lesson 12 Exit Ticket Answer Key t 1
Perimeter = __________________________________________
= _______ cm
Answer:
Perimeter of the given ABCDEFGHIJK Shape =  Side + Side + Side + Side + Side + Side + Side + Side + Side + Side + Side + Side
= AB + BC+ CD + DE + EF + FG + GH + HI + IJ + JK +KL + LA
= 4cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
= 32cm.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 12 Exit Ticket Answer Key
Length of the AB side of the given ABCDEFGHIJK Shape = 4cm
Length of the BC side of the given ABCDEFGHIJK Shape = 2cm
Length of the CD side of the given ABCDEFGHIJK Shape = 2cm
Length of the DE side of the given ABCDEFGHIJK Shape = 4cm
Length of the EF side of the given ABCDEFGHIJK Shape = 2cm
Length of the FG side of the given ABCDEFGHIJK Shape = 2cm
Length of the GH side of the given ABCDEFGHIJK Shape = 4cm
Length of the HI side of the given ABCDEFGHIJK Shape = 2cm
Length of the IJ side of the given ABCDEFGHIJK Shape =  2cm
Length of the JK side of the given ABCDEFGHIJK Shape = 4cm
Length of the KL side of the given ABCDEFGHIJK Shape = 2cm
Length of the LA side of the given ABCDEFGHIJK Shape = 2cm
Perimeter of the given ABCDEFGHIJK Shape =  Side + Side + Side + Side + Side + Side + Side + Side + Side + Side + Side + Side
= AB + BC+ CD + DE + EF + FG + GH + HI + IJ + JK +KL + LA
= 4cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
= 6cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
= 8cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
= 12cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
= 14cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
= 16cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
= 20 cm + 2cm + 2cm + 4cm + 2cm + 2cm
= 22cm + 2cm + 4cm + 2cm + 2cm
= 24cm + 4cm + 2cm + 2cm
= 28cm + 2cm + 2cm
= 30cm + 2cm
= 32cm.

Eureka Math Grade 3 Module 7 Lesson 12 Homework Answer Key

Question 1.
Measure and label the side lengths of the shapes below in centimeters. Then, find the perimeter of each shape.
a.
Eureka Math Grade 3 Module 7 Lesson 12 Homework Answer Key h 1
Perimeter = _____cm +_____cm +_____cm
= _______ cm
Answer:
Perimeter of ABC triangle = Side + Side + Side
= AB + BC + CA
= 3cm + 5cm + 4cm
= 12cm.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 12 Homework Answer Key-1a
Length of the AB side of ABC triangle = 3cm
Length of the BC side of ABC triangle = 5cm
Length of the CA side of ABC triangle = 4cm
Perimeter of ABC triangle = Side + Side + Side
= AB + BC + CA
= 3cm + 5cm + 4cm
= 8cm + 4cm
= 12cm.

b.
Eureka Math Grade 3 Module 7 Lesson 12 Homework Answer Key h 2
Perimeter = _____________________
= _______ cm
Answer:
Perimeter of ABCD rectangle = Side + Side + Side + Side
= AB + BC + CD + DA
= 6cm + 4cm + 6cm + 4cm
= 20cm.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 12 Homework Answer Key-1b
Length of the AB side of ABCD rectangle = 6cm
Length of the BC side of ABCD rectangle = 4cm
Length of the CD side of ABCD rectangle = 6cm
Length of the DA side of ABCD rectangle = 4cm
Perimeter of ABCD rectangle = Side + Side + Side + Side
= AB + BC + CD + DA
= 6cm + 4cm + 6cm + 4cm
= 10cm + 6cm + 4cm
= 16cm + 4cm
= 20cm.

 

c.
Eureka Math Grade 3 Module 7 Lesson 12 Homework Answer Key h 3
Perimeter = _____________________
= _______ cm
Answer:
Perimeter of the ABCD Quadrilateral = Side + Side + Side + Side
= AB + BC + CD + DA
= 3cm + 4cm + 5cm + 4cm
= 16cm.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 12 Homework Answer Key-1c
Length of the AB side in the ABCD Quadrilateral = 3cm
Length of the BC side in the ABCD Quadrilateral = 4cm
Length of the CD side in the ABCD Quadrilateral = 5cm
Length of the DA side in the ABCD Quadrilateral = 4cm
Perimeter of the ABCD Quadrilateral = Side + Side + Side + Side
= AB + BC + CD + DA
= 3cm + 4cm + 5cm + 4cm
= 7cm + 5cm + 4cm
= 12cm + 4cm
= 16cm.

d.
Eureka Math Grade 3 Module 7 Lesson 12 Homework Answer Key h 4
Perimeter = _____________________
= _______ cm
Answer:
Perimeter of the ABCD Parallelogram = Side + Side + Side + Side
= AB + BC + CD + DA
= 5cm + 5cm + 5cm + 5cm
= 20cm.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 12 Homework Answer Key-1d
Length of the AB side of the ABCD parallelogram = 5cm
Length of the BC side of the ABCD parallelogram = 5cm
Length of the CD side of the ABCD parallelogram = 5cm
Length of the DA side of the ABCD parallelogram = 5cm
Perimeter of the ABCD Parallelogram = Side + Side + Side + Side
= AB + BC + CD + DA
= 5cm + 5cm + 5cm + 5cm
= 10cm + 5cm + 5cm
= 15cm + 5cm
= 20cm.

e.
Eureka Math Grade 3 Module 7 Lesson 12 Homework Answer Key h 5
Perimeter = _____________________
= _______ cm
Answer:
Perimeter of the given ABCDEFGH figure = Side + Side + Side + Side + Side + Side + Side + Side
= AB + BC + CD + DE + EF + FG + GH + HA
= 2cm + 2cm + 3.5cm + 2cm + 2cm + 2.5cm + 7.5cm + 2.5cm
= 24cm.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 12 Homework Answer Key-1e
Length of AB side of the given ABCDEFGH figure = 2cm
Length of BC side of the given ABCDEFGH figure = 2cm
Length of CD side of the given ABCDEFGH figure = 3.5cm
Length of DE side of the given ABCDEFGH figure = 2cm
Length of EF side of the given ABCDEFGH figure = 2cm
Length of FG side of the given ABCDEFGH figure = 2.5cm
Length of GH side of the given ABCDEFGH figure = 7.5cm
Length of HA side of the given ABCDEFGH figure = 2.5cm
Perimeter of the given ABCDEFGH figure = Side + Side + Side + Side + Side + Side + Side + Side
= AB + BC + CD + DE + EF + FG + GH + HA
= 2cm + 2cm + 3.5cm + 2cm + 2cm + 2.5cm + 7.5cm + 2.5cm
= 4cm + 3.5cm + 2cm + 2cm + 2.5cm + 7.5cm + 2.5cm
= 7.5cm + 2cm + 2cm + 2.5cm + 7.5cm + 2.5cm
= 9.5cm + 2cm + 2.5cm + 7.5cm + 2.5cm
= 11.5cm + 2.5cm + 7.5cm + 2.5cm
= 14cm + 7.5cm + 2.5cm
= 21.5cm + 2.5cm
= 24cm.

Question 2.
Melinda draws two trapezoids to create the hexagon shown below. Use a ruler to find the side lengths of Melinda’s hexagon in centimeters. Then, find the perimeter.
Eureka Math Grade 3 Module 7 Lesson 12 Homework Answer Key h 6
Answer:
Perimeter of ABCDEF Melinda’s hexagon = Side + Side + Side + Side + Side + Side
= AB + BC + CD + DE + EF +FA
= 3cm + 3cm + 3cm + 3cm + 3cm + 3cm
= 18cm.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 12 Homework Answer Key-2
Length of the AB side of ABCDEF Melinda’s hexagon = 3cm
Length of the BC side of ABCDEF Melinda’s hexagon = 3cm
Length of the CD side of ABCDEF Melinda’s hexagon = 3cm
Length of the DE side of ABCDEF Melinda’s hexagon = 3cm
Length of the EF side of ABCDEF Melinda’s hexagon = 3cm
Length of the FA side of ABCDEF Melinda’s hexagon = 3cm
Perimeter of the ABCDEF Melinda’s hexagon = Side + Side + Side + Side + Side + Side
= AB + BC + CD + DE + EF +FA
= 3cm + 3cm + 3cm + 3cm + 3cm + 3cm
= 6cm + 3cm + 3cm + 3cm + 3cm
= 9cm + 3cm + 3cm + 3cm
= 12cm + 3cm + 3cm
= 15cm + 3cm
= 18cm.

Question 3.
Victoria and Eric draw the shapes shown below. Eric says his shape has a greater perimeter because it has more sides than Victoria’s shape. Is Eric right? Explain your answer.
Eureka Math Grade 3 Module 7 Lesson 12 Homework Answer Key h 7
Answer:
Yes, Eric is correct because DEFG Square has more sides than the ABC Victoria’s triangle shape.
Perimeter of DEFG Eric’s Square shape is greater than the Perimeter of ABC Victoria’s triangle shape.
Perimeter of ABC Victoria’s triangle shape = 12cm

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 12 Homework Answer Key-3
Length of AB Victoria’s triangle shape = 3cm
Length of BC Victoria’s triangle shape = 5cm
Length of CA Victoria’s triangle shape =4cm
Perimeter of ABC Victoria’s triangle shape = Side + Side + Side
= 3cm +  4cm + 5cm
= 7cm + 5cm
= 12cm.

Length of DEFG Eric’s Square shape = 4cm
Perimeter of DEFG Eric’s Square shape = Side × Side
= 4cm × 4cm
= 16cm.
Perimeter of Eric’s Square shape is greater than the Perimeter of Victoria’s triangle shape.

Question 4.
Jamal uses his ruler and a right angle tool to draw the rectangle shown below. He says the perimeter of his rectangle is 32 centimeters. Do you agree with Jamal? Why or why not?
Eureka Math Grade 3 Module 7 Lesson 12 Homework Answer Key h 8
Answer:
NO, I disagree with Jamal. Jamal answer is incorrect because actual Perimeter of his ABCD rectangle is 30cm not 32cm.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 12 Homework Answer Key-4
Length of the ABCD Rectangle Jamal drawn = 7.5cm
Width of the ABCD Rectangle Jamal drawn = 4cm
Perimeter of ABCD Rectangle Jamal drawn = Length × Width
= 7.5 × 4
= 30 cm.
Jamal answer is incorrect because actual Perimeter of his ABCD rectangle is 30cm not 32cm.

Eureka Math Grade 3 Module 1 Lesson 5 Answer Key

Engage NY Eureka Math 3rd Grade Module 1 Lesson 5 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 5 Problem Set Answer Key

Question 1.
Eureka Math Grade 3 Module 1 Lesson 5 Problem Set Answer Key 1
Divide 6 tomatoes into groups of 3.
There are ____2_____ groups of 3 tomatoes.

Answer:
6 ÷ 3 = 2
There are 2 groups of 3 tomatoes,

Explanation:
Dividing 6 tomatoes into groups of 3 we get
6 ÷ 3 = 2 groups
So there are 2 groups of 3 tomatoes.

Question 2.
Eureka Math Grade 3 Module 1 Lesson 5 Problem Set Answer Key 2
Divide 8 lollipops into groups of 2.
There are ____4___ groups.
8 ÷ 2 = ___4____
Eureka Math Grade 3 Module 1 Lesson 5 Answer Key-1

Answer:

There are 4 groups,

Explanation:
Dividing 8 lollipops into groups of 2 as
8 ÷ 2 = 4 we get 4 groups,
So, there are 4 groups of 2.

Question 3.
Eureka Math Grade 3 Module 1 Lesson 5 Problem Set Answer Key 3
Divide 10 stars into groups of 5.
10 ÷ 5 = ___2____
Eureka Math Grade 3 Module 1 Lesson 5 Answer Key-2

Answer:

There are 2 groups,

Explanation:
Dividing 10 stars into groups of 5 as
10 ÷ 5 = 2 we get 2 groups,
So, there are 2 groups of 5.

Question 4.
Eureka Math Grade 3 Module 1 Lesson 5 Problem Set Answer Key 4
Divide the shells to show 12 ÷ 3 = ____4____,
where the unknown represents the number of groups.
How many groups are there? ___4_____
Eureka Math Grade 3 Module 1 Lesson 5 Answer Key-3

Answer:

There are 4 groups,

Explanation:
Dividing 12 shells into groups of 3 as
12 ÷ 3 = 4 we get 4 groups,
So, there are 4 groups of 3.

Question 5.
Rachel has 9 crackers. She puts 3 crackers in each bag.
Circle the crackers to show Rachel’s bags.
Eureka Math Grade 3 Module 1 Lesson 5 Problem Set Answer Key 5
a. Write a division sentence where the answer represents the number of Rachel’s bags.
b. Draw a number bond to represent the problem.
Eureka Math Grade 3 Module 1 Lesson 5 Answer Key-4

Answer:

Division sentence 9 ÷ 3 = 3,
Number of Rachel’s bags are 3,

Explanation:
Given Rachel has 9 crackers.
She puts 3 crackers in each bag.
Circled the crackers to show Rachel’s bags,
a. Division sentence 9 ÷ 3 = 3,
therefore, number of Rachel’s bags are 3.
b.
Eureka Math Grade 3 Module 1 Lesson 5 Answer Key-5
Explanation:
Drawn a number bond to represent the problem
as shown above

Question 6.
Jameisha has 16 wheels to make toy cars.
She uses 4 wheels for each car.
a. Use a count-by to find the number of cars Jameisha can build. Make a drawing to match your counting.
b. Write a division sentence to represent the problem.

a.
Eureka Math Grade 3 Module 1 Lesson 5 Answer Key-6
Answer:
Number of cars Jameisha builds are 4,

Explanation:
Given Jameisha has 16 wheels to make toy cars.
She uses 4 wheels for each car.
Used a count-by to find the number of cars
Jameisha can build as 16 ÷ 4 = 4 cars,
Made a drawing to match my counting as shown above.

b. Division sentence to represent the problem is 16 ÷ 4 = 4 cars,

Explanation:
Given Jameisha has 16 wheels to make toy cars.
She uses 4 wheels for each car, So the division sentence to represent the problem is 16 ÷ 4 = 4 cars.

Eureka Math Grade 3 Module 1 Lesson 5 Exit Ticket Answer Key

Question 1.
Divide 12 triangles into groups of 6.
Engage NY Math 3rd Grade Module 1 Lesson 5 Exit Ticket Answer Key 6
12 ÷ 6 = ___2____
Eureka Math Grade 3 Module 1 Lesson 5 Answer Key-7
Answer:
There are 2 groups of 6,

Explanation:
Dividing 12 triangles into groups of 6 as 12 ÷ 6 = 2 we get 2 groups,
So, there are 2 groups of 6.

Question 2.
Spencer buys 20 strawberries to make smoothies.
Each smoothie needs 5 strawberries.
Use a count-by to find the number of smoothies
Spencer can make.
Make a drawing to match your counting.

Eureka Math Grade 3 Module 1 Lesson 5 Answer Key-8
Answer:
Spencer makes 4 smoothies,

Explanation:
Given Spencer buys 20 strawberries to make smoothies.
Each smoothie needs 5 strawberries.
Used a count-by the number of smoothies
Spencer can make are 20 ÷ 5 = 4,
Made a drawing to match my counting as 5 X 4 = 20 strawberries as shown above in the picture.

Eureka Math Grade 3 Module 1 Lesson 5 Homework Answer Key

Question 1.
Eureka Math 3rd Grade Module 1 Lesson 5 Homework Answer Key 7
Divide 4 triangles into groups of 2.
There are _____2____ groups of 2 triangles.
4 ÷ 2 = 2

Answer:
There are 2 groups of 2 triangles,

Explanation:
Dividing 4 triangles into groups of 2 as
4 ÷ 2 = 2 we get 2 groups,
So, there are 2 groups of 2 triangles.

Question 2.
Eureka Math 3rd Grade Module 1 Lesson 5 Homework Answer Key 8
Divide 9 eggs into groups of 3.
There are ___3____ groups.
9 ÷ 3 = __3_____
Eureka Math Grade 3 Module 1 Lesson 5 Answer Key-9

Answer:
There are 3 groups of 3 eggs,

Explanation:
Dividing 9 eggs into groups of 3 as
9 ÷ 3 = 3 we get 3 groups,
So, there are 3 groups of 3 eggs.

Question 3.
Eureka Math 3rd Grade Module 1 Lesson 5 Homework Answer Key 9
Divide 12 buckets of paint into groups of 3.
12 ÷ 3 = __4_____
Eureka Math Grade 3 Module 1 Lesson 5 Answer Key-10
Answer:
There are 4 groups of 3 paint buckets,

Explanation:
Divided 12 buckets of paint into groups of 3 as
12 ÷ 3 = 4 we get 4 groups of paint buckets,
So, there are 4 groups of 3 paint buckets.

Question 4.
Eureka Math 3rd Grade Module 1 Lesson 5 Homework Answer Key 10
Group the squares to show 15 ÷ 5 = __3___,
where the unknown represents the number of groups.
How many groups are there? ____3____
Eureka Math Grade 3 Module 1 Lesson 5 Answer Key-11
Answer:
There are 3 groups of 5 squares,

Explanation:
Grouped the squares to show 15 ÷ 5 = 3,
where the unknown represents the number of groups as 3 groups of 5 squares.

Question 5.
Daniel has 12 apples. He puts 6 apples in each bag.
Circle the apples to find the number of bags Daniel makes.
Eureka Math 3rd Grade Module 1 Lesson 5 Homework Answer Key 11
a. Write a division sentence where the answer represents the number of Daniel’s bags.
b. Draw a number bond to represent the problem.
a.
Eureka Math Grade 3 Module 1 Lesson 5 Answer Key-12
Daniel’s has 2 bags of 6 apples each,
Division sentence : 12 ÷ 6 = 2 bags,

Explanation:
Daniel has 12 apples and he puts 6 apples in each bag.
Circled the apples to find the number of bags Daniel makes as 12 ÷ 6 = 2 bags,
a. Writing a division sentence where the answer represents the number of Daniel’s bags as 12 ÷ 6 = 2 bags.

b.
Eureka Math Grade 3 Module 1 Lesson 5 Answer Key-13
Explanation:
Drawn a number bond to represent the problem as shown above in the picture.

Question 6.
Jacob draws cats. He draws 4 legs on each cat for a total of 24 legs.
a. Use a count-by to find the number of cats Jacob draws. Make a drawing to match your counting.
b. Write a division sentence to represent the problem.

a.
Eureka Math Grade 3 Module 1 Lesson 5 Answer Key-14
Answer:
The number of cats Jacob draws are 6,

Explanation:
Given Jacob draws cats. He draws 4 legs on each cat for a total of 24 legs.
a. Used a count-by to find the number of cats Jacob draws as 24 ÷ 4 = 6 cats,
Made a drawing to match my counting as shown above.

b. Division sentence is 24 ÷ 4 = 6 cats,

Explanation:
Wrote a division sentence to represent the problem as
24 ÷ 4 = 6 cats.