Eureka Math

Engage NY Eureka Math 3rd Grade Module 7 Lesson 15 Answer Key

Eureka Math Grade 3 Module 7 Lesson 15 Pattern Sheet Answer Key

Multiply.

multiply by 9 (1–5)
Answer:

Explanation:
9 × 1 = 9
9 × 2 = 18
9 × 3 = 27
9 × 4 = 36
9 × 5 = 45.

Eureka Math Grade 3 Module 7 Lesson 15 Problem Set Answer Key

Question 1.
Mrs. Kozlow put a border around a 5-foot by 6-foot rectangular bulletin board. How many feet of border did Mrs. Kozlow use?
Answer:
Mrs. Kozlow used 22ft of border.

Explanation:

Length of the side of the rectangular bulletin board = 6ft
Width of the side of the rectangular bulletin board = 5ft
Perimeter of the rectangular bulletin board = 2 (Length + Width)
= 2 ( 6ft + 5ft )
= 2 × 11ft
= 22ft.

Question 2.
Jason built a model of the Pentagon for a social studies project. He made each outside wall 33 centimeters long. What is the perimeter of Jason’s model pentagon?
Answer:
Perimeter of the Jason’s model Pentagon = 165 centimeters.

Explanation:

Length of the Side of the Jason’s model Pentagon = 33 centimeters
Perimeter of the Jason’s model Pentagon = 5 × Side
= 5 × 33 centimeters
= 165 centimeters.

Question 3.
The Holmes family plants a rectangular 8-yard by 9-yard vegetable garden. How many yards of fencing do they need to put a fence around the garden?
Answer:
Perimeter of the rectangular vegetable garden = 34-yard.

Explanation:

Length of the rectangular vegetable garden=  9-yard
Width of the rectangular vegetable garden = 8-yard
Perimeter of the rectangular vegetable garden = 2 ( Length + Width )
= 2 (9-yard + 8-yard)
= 2 × 17-yard
= 34-yard.

Question 4.
Marion paints a 5-pointed star on her bedroom wall. Each side of the star is 18 inches long. What is the perimeter of the star?

Answer:
Perimeter of the Star = 180 inches.

Explanation:

Length of the side of the Star = 18 inches
Number of sides of Star = 10
Perimeter of the Star = 10 × Side
= 10 ×18 inches
=180 inches.

Question 5.
The soccer team jogs around the outside of the soccer field twice to warm up. The rectangular field measures 60 yards by 100 yards. What is the total number of yards the team jogs?
Answer:
The total number of yards the team jogs = 640 yards.

Explanation:

Length of the rectangular soccer field = 100 yards
Width of the rectangular soccer field = 60 yards
Perimeter of the rectangular soccer field = 2 ( Length + Width )
= 2 ( 100 yards + 60 yards )
= 2 × 160 yards
= 320 yards.
Number of rounds the soccer team jogs around the outside of the soccer field to warm up = twice
The total number of yards the team jogs = 320 yards × 2
= 640 yards.

Question 6.
Troop 516 makes 3 triangular flags to carry at a parade. They sew ribbon around the outside edges of the flags. The flags’ side lengths each measure 24 inches. How many inches of ribbon does the troop use?
Answer:
The troop used  216inches  of ribbon.

Explanation:

Number of triangular flags Troop 516 makes = 3 or Three
Length of the triangular flags Troop 516 makes = 24inches
Perimeter of the triangular flags Troop 516 makes =  3 × Side
= 3 × 24inches
= 72inches.
Perimeter of the 3 triangular flags Troop 516 makes = 3 × 72inches
= 216inches.

Eureka Math Grade 3 Module 7 Lesson 15 Exit Ticket Answer Key

Marlene ropes off a square section of her yard where she plants grass. One side length of the square measures 9 yards. What is the total length of rope Marlene uses?
Answer:
The total length of rope Marlene uses = 36 yards.

Explanation:

Length of the side of the square = 9 yards.
Perimeter of the square = 4 × Side
= 4 × 9 yards
= 36 yards.

Eureka Math Grade 3 Module 7 Lesson 15 Homework Answer Key

Question 1.
Miguel glues a ribbon border around the edges of a 5-inch by 8-inch picture to create a frame. What is the total length of ribbon Miguel uses?
Answer:
The total length of ribbon Miguel uses = 26-inch.

Explanation:

Length of the rectangular picture = 8-inch
Width of the rectangular picture = 5-inch
Perimeter of the rectangular picture = 2 ( Length + Width )
= 2 ( 8-inch + 5-inch )
= 2 × 13-inch
= 26-inch.

Question 2.
A building at Elmira College has a room shaped like a regular octagon. The length of each side of the room is 5 feet. What is the perimeter of this room?
Answer:
The perimeter of this room = 40 feet.

Explanation:

length of each side of the regular octagon room = 5 feet
Perimeter of the regular octagon room = 8 × Side
= 8 × 5 feet
= 40 feet.

Question 3.
Manny fences in a rectangular area for his dog to play in the backyard. The area measures 35 yards by 45 yards. What is the total length of fence that Manny uses?
Answer:
The total length of fence that Manny uses = 160 yards.

Explanation:

Length of the rectangular fence  = 45 yards
Width of the rectangular fence  = 35 yards
Perimeter of rectangular fence  = 2 ( Length + Width )
= 2 (45 yards + 35 yards )
= 2 × 80 yards
= 160 yards.

Question 4.
Tyler uses 6 craft sticks to make a hexagon. Each craft stick is 6 inches long. What is the perimeter of Tyler’s hexagon?
Answer:
Perimeter of the Tyler’s hexagon = 36 inches.

Explanation:

Length of the side of the Tyler’s hexagon = 6 inches
Perimeter of the Tyler’s hexagon = 6 × Side
= 6 × 6 inches
= 36 inches.

Question 5.
Francis made a rectangular path from her driveway to the porch. The width of the path is 2 feet. The length is 28 feet longer than the width. What is the perimeter of the path?
Answer:
Perimeter of the rectangular path = 60 feet.

Explanation:

Length of the rectangular path = 28 feet
Width of the rectangular path = 2 feet
Perimeter of the rectangular path = 2 ( Length  + Width )
= 2 ( 28 feet + 2 feet )
= 2 × 30 feet
= 60 feet.

Question 6.
The gym teacher uses tape to mark a 4-square court on the gym floor as shown. The outer square has side lengths of 16 feet. What is the total length of tape the teacher uses to mark Square A?

Answer:
Perimeter of the Square court = 64 ft.
The total length of tape the teacher uses to mark Square A = 16 ft.

Explanation:

Length of the Square court = 16 ft
Perimeter of the Square court =  4 × Side
= 4 × 16 ft
= 64 ft.
The total length of tape the teacher uses to mark Square A = Perimeter of the Square court ÷ 4
= 64 ft ÷ 4
= 16 ft.

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

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 3 Sprint Answer Key

A
Add Equal Groups

Question 1.
2 + 2 =
2 + 2 = 4,

Explanation:
Given 2 + 2 we add 2 with 2,
we get 4 as 2 + 2 = 4.

Question 2.
2 twos =
2 twos = 4,

Explanation:
Given 2 twos means 2 by 2 times as
2 X 2 we get 4 so 2 twos = 4.

Question 3.
5 + 5 =
5 + 5 = 10,

Explanation:
Given 5 + 5 we add 5 with 5,
we get 10 as 5 + 5 = 10.

Question 4.
2 fives =
2 fives = 10,

Explanation:
Given 2 fives means 2 by 5 as
2 X 5 we get 10 so 2 fives = 10.

Question 5.
2 + 2 + 2 =
2 + 2 + 2 = 6,

Explanation:
Given 2 + 2 + 2 we add 2 thrice
we get 6 as 2 + 2 + 2 = 6.

Question 6.
3 twos =
3 twos = 6,

Explanation:
Given 3 twos means 3 by 2 as
3 X 2 we get 6 so 3 twos = 6.

Question 7.
2 + 2 + 2 + 2 =
2 + 2 + 2 + 2 = 8,

Explanation:
Given 2 + 2 + 2 + 2 we add 2 four times
we get 8 as 2 + 2 + 2 + 2 = 8.

Question 8.
4 twos =
4 twos = 8,

Explanation:
Given 4 twos means 4 by 2 as
4 X 2 we get 8 so 4 twos = 8.

Question 9.
5 + 5 + 5 =
5 + 5 + 5 = 15,

Explanation:
Given 5 + 5 + 5 we add 5 thrice
we get 15 as 5 + 5 + 5 = 15.

Question 10.
3 fives =
3 fives = 15,

Explanation:
Given 3 fives means 3 by 5 as
3 X 5 we get 15 so 3 fives = 15.

Question 11.
5 + 5 + 5 + 5 =
5 + 5 + 5 + 5 = 20,

Explanation:
Given 5 + 5 + 5 + 5 we add 5 four times
we get 20 as 5 + 5 + 5 + 5 = 20.

Question 12.
4 fives =
4 fives = 20,

Explanation:
Given 4 fives means 4 by 5 as
4 X 5 we get 20 so 4 fives = 20.

Question 13.
2 fours=
2 fours = 8,

Explanation:
Given 2 fours means 2 by 4 as
2 X 4 we get 8 so 2 fours = 8.

Question 14.
4 + 4 =
4 + 4 = 8,

Explanation:
Given 4 + 4 we add 4 with 4,
we get 8 as 4 + 4 = 8.

Question 15.
2 threes =
2 threes = 6,

Explanation:
Given 2 threes means 2 by 3 as
2 X 3 we get 6 so 2 threes = 6.

Question 16.
3 + 3 =
3 + 3 = 6,

Explanation:
Given 3 + 3 we add 3 with 3,
we get 6 as 3 + 3 = 6.

Question 17.
2 sixes =
2 sixes = 12,

Explanation:
Given 2 sixes means 2 by 6 as
2 X 6 we get 12 so 2 sixes = 12.

Question 18.
6 + 6 =
6 + 6 = 12,

Explanation:
Given 6 + 6 we add 6 with 6,
we get 12 as 6 + 6 = 12.

Question 19.
5 twos =
5 twos = 10,

Explanation:
Given 5 twos means 5 by 2 as
5 X 2 we get 10 so 5 twos = 10.

Question 20.
2 + 2 + 2 + 2 + 2 =
2 + 2 + 2 + 2 + 2 = 10,

Explanation:
Given 2 + 2 + 2 + 2 + 2  we add 2 five times
we get 10 as 2 + 2 + 2 + 2 + 2 = 10.

Question 21.
5 fives =
5 fives = 25,

Explanation:
Given 5 fives means 5 by 5 as
5 X 5 we get 25 so 5 fives = 25.

Question 22.
5 + 5 + 5 + 5 + 5 =
5 + 5 + 5 + 5 + 5 = 25,

Explanation:
Given 5 + 5 + 5 + 5 + 5  we add 5 five times
we get 25 as 5 + 5 + 5 + 5 + 5 = 25.

Question 23.
7 + 7 =
7 + 7 = 14,

Explanation:
Given 7 + 7 we add 7 with 7,
we get 14 as 7 + 7 = 14.

Question 24.
2 sevens =
2 sevens = 14,

Explanation:
Given 2 sevens means 2 by 7 as
2 X 7 we get 14 so 2 sevens = 14.

Question 25.
9 + 9 =
9 + 9 = 18,

Explanation:
Given 9 + 9 we add 9 with 9,
we get 18 as 9 + 9 = 18.

Question 26.
2 nines =
2 nines = 18,

Explanation:
Given 2 nines means 2 by 9 as
2 X 9 we get 18 so 2 nines = 18.

Question 27.
8 + 8 =
8 + 8 = 16,

Explanation:
Given 8 + 8 we add 8 with 8,
we get 16 as 8 + 8 = 16.

Question 28.
2 eights =
2 eights = 16,

Explanation:
Given 2 eights means 2 by 8 as
2 X 8 we get 16 so 2 eights = 16.

Question 29.
3 + 3 + 3 =
3 + 3 + 3 = 9,

Explanation:
Given 3 + 3 + 3 we add 3 thrice
we get 9 as 3 + 3 + 3 = 9.

Question 30.
3 threes =
3 threes = 9,

Explanation:
Given 3 threes means 3 by 3 as
3 X 3 we get 9 so 3 threes = 9.

Question 31.
4 + 4 + 4 =
4 + 4 + 4 = 12,

Explanation:
Given 4 + 4 + 4 we add 4 thrice
we get 12 as 4 + 4 + 4 = 12.

Question 32.
3 fours =
3 fours = 12,

Explanation:
Given 3 fours means 3 by 4 as
3 X 4 we get 12 so 3 fours = 12.

Question 33.
3 + 3 + 3 + 3 =
3 + 3 + 3 + 3 = 12,

Explanation:
Given 3 + 3 + 3 + 3 we add 3 four times
we get 12 as 3 + 3 + 3 + 3= 12.

Question 34.
4 threes =
4 threes = 12,

Explanation:
Given 4 threes means 4 by 3 as
4 X 3 we get 12 so 4 threes = 12.

Question 35.
4 fives =
4 fives = 20,

Explanation:
Given 4 fives means 4 by 5 as
4 X 5 we get 20 so 4 fives = 20.

Question 36.
5 + 5 + 5 + 5 =
5 + 5 + 5 + 5 = 20,

Explanation:
Given 5 + 5 + 5 + 5 we add 5 four times
we get 20 as 5 + 5 + 5 + 5 = 20.

Question 37.
3 sixes =
3 sixes = 18,

Explanation:
Given 3 sixes means 3 by 6 as
3 X 6 we get 18 so 3 sixes = 18.

Question 38.
6 + 6 + 6 =
6 + 6 + 6 = 18,

Explanation:
Given 6 + 6 + 6 we add 6 thrice
we get 18 as 6 + 6 + 6 = 18.

Question 39.
3 eights =
3 eights = 24,

Explanation:
Given 3 eights means 3 by 8 as
3 X 8 we get 24 so 3 eights = 24.

Question 40.
8 + 8 + 8 =
8 + 8 + 8 = 24,

Explanation:
Given 8 + 8 + 8 we add 8 thrice
we get 24 as 8 + 8 + 8 = 24.

Question 41.
3 sevens =
3 sevens = 21,

Explanation:
Given 3 sevens means 3 by 7 as
3 X 7 we get 21 so 3 sevens = 21.

Question 42.
7 + 7 + 7 =
7 + 7 + 7 = 21,

Explanation:
Given 7 + 7 + 7 we add 7 thrice
we get 21 as 7 + 7 + 7 = 21.

Question 43.
3 nines =
3 nines = 27,

Explanation:
Given 3 nines means 3 by 9 as
3 X 9 we get 27 so 3 nines = 27.

Question 44.
9 + 9 + 9 =
9 + 9 + 9 = 27,

Explanation:
Given 9 + 9 + 9 we add 9 thrice
we get 27 as 9 + 9 + 9 = 27.

B
Add Equal Groups

Question 1.
5 + 5 =
5 + 5 = 10,

Explanation:
Given 5 + 5 we add 5 with 5,
we get 10 as 5 + 5 = 10.

Question 2.
2 fives =
2 fives = 10,

Explanation:
Given 2 fives means 2 by 5 as
2 X 5 we get 10 so 2 fives = 10.

Question 3.
2 + 2 =
2 + 2 = 4,

Explanation:
Given 2 + 2 we add 2 with 2,
we get 4 as 2 + 2 = 4.

Question 4.
2 twos =
2 twos = 4,

Explanation:
Given 2 twos means 2 by 2 times as
2 X 2 we get 4 so 2 twos = 4.

Question 5.
5 + 5 + 5 =
5 + 5 + 5 = 15,

Explanation:
Given 5 + 5 + 5 we add 5 thrice
we get 15 as 5 + 5 + 5 = 15.

Question 6.
3 fives =
3 fives = 15,

Explanation:
Given 3 fives means 3 by 5 as
3 X 5 we get 15 so 3 fives = 15.

Question 7.
5 + 5 + 5 + 5 =
5 + 5 + 5 + 5 = 20,

Explanation:
Given 5 + 5 + 5 + 5 we add 5 four times
we get 20 as 5 + 5 + 5 + 5 = 20.

Question 8.
4 fives =
4 fives = 20,

Explanation:
Given 4 fives means 4 by 5 as
4 X 5 we get 20 so 4 fives = 20.

Question 9.
2 + 2 + 2 =
2 + 2 + 2 = 6,

Explanation:
Given 2 + 2 + 2 we add 2 thrice
we get 6 as 2 + 2 + 2 = 6.

Question 10.
3 twos =
3 twos = 6,

Explanation:
Given 3 twos means 3 by 2 as
3 X 2 we get 6 so 3 twos = 6.

Question 11.
2 + 2 + 2 + 2 =
2 + 2 + 2 + 2 = 8,

Explanation:
Given 2 + 2 + 2 + 2 we add 2 four times
we get 8 as 2 + 2 + 2 + 2 = 8.

Question 12.
4 twos =
4 twos = 8,

Explanation:
Given 4 twos means 4 by 2 as
4 X 2 we get 8 so 4 twos = 8.

Question 13.
2 threes =
2 threes = 6,

Explanation:
Given 2 threes means 2 by 3 as
2 X 3 we get 6 so 2 threes = 6.

Question 14.
3 + 3 =
3 + 3 = 6,

Explanation:
Given 3 + 3 we add 3 with 3,
we get 6 as 3 + 3 = 6.

Question 15.
2 sixes =
2 sixes = 12,

Explanation:
Given 2 sixes means 2 by 6 as
2 X 6 we get 12 so 2 sixes = 12.

Question 16.
6 + 6 =
6 + 6 = 12,

Explanation:
Given 6 + 6 we add 6 with 6,
we get 12 as 6 + 6 = 12.

Question 17.
2 fours =
2 fours = 8,

Explanation:
Given 2 fours means 2 by 4 as
2 X 4 we get 8 so 2 fours = 8.

Question 18.
4 + 4 =
4 + 4 = 8,

Explanation:
Given 4 + 4 we add 4 with 4,
we get 8 as 4 + 4 = 8.

Question 19.
5 fives =
5 fives = 25,

Explanation:
Given 5 fives means 5 by 5 as
5 X 5 we get 25 so 5 fives = 25.

Question 20.
5 + 5 + 5 + 5 + 5 =
5 + 5 + 5 + 5 + 5 = 25,

Explanation:
Given 5 + 5 + 5 + 5 + 5 we add 5 five times
we get 25 as 5 + 5 + 5 + 5 + 5 = 25.

Question 21.
5 twos =
5 twos = 10,

Explanation:
Given 5 twos means 5 by 2 as
5 X 2 we get 10 so 5 twos = 10.

Question 22.
2 + 2 + 2 + 2 + 2 =
2 + 2 + 2 + 2 + 2 = 10,

Explanation:
Given 2 + 2 + 2 + 2 + 2 we add 2 five times
we get 10 as 2 + 2 + 2 + 2 + 2 = 10.

Question 23.
8 + 8 =
8 + 8 = 16,

Explanation:
Given 8 + 8 we add 8 with 8,
we get 16 as 8 + 8 = 16.

Question 24.
2 eights =
2 eights = 16,

Explanation:
Given 2 eights means 2 by 8 as
2 X 8 we get 16 so 2 eights = 16.

Question 25.
7 + 7 =
7 + 7 = 14,

Explanation:
Given 7 + 7 we add 7 with 7,
we get 14 as 7 + 7 = 14.

Question 26.
2 sevens =
2 sevens = 14,

Explanation:
Given 2 sevens means 2 by 7 as
2 X 7 we get 14 so 2 sevens = 14.

Question 27.
9 + 9 =
9 + 9 = 18,

Explanation:
Given 9 + 9 we add 9 with 9,
we get 18 as 9 + 9 = 18.

Question 28.
2 nines =
2 nines = 18,

Explanation:
Given 2 nines means 2 by 9 as
2 X 9 we get 18 so 2 nines = 18.

Question 29.
3 + 3 + 3 + 3 =
3 + 3 + 3 + 3 = 12,

Explanation:
Given 3 + 3 + 3 + 3 we add 3 four times
we get 12 as 3 + 3 + 3 + 3= 12.

Question 30.
4 threes =
4 threes = 12,

Explanation:
Given 4 threes means 4 by 3 as
4 X 3 we get 12 so 4 threes = 12.

Question 31.
4 + 4 + 4 =
4 + 4 + 4 = 12,

Explanation:
Given 4 + 4 + 4 we add 4 thrice
we get 12 as 4 + 4 + 4 = 12.

Question 32.
3 fours =
3 fours = 12,

Explanation:
Given 3 fours means 3 by 4 as
3 X 4 we get 12 so 3 fours = 12.

Question 33.
3 + 3 + 3 =
3 + 3 + 3 = 9,

Explanation:
Given 3 + 3 + 3 we add 3 thrice
we get 9 as 3 + 3 + 3 = 9.

Question 34.
3 threes =
3 threes = 9,

Explanation:
Given 3 threes means 3 by 3 as
3 X 3 we get 9 so 3 threes = 9.

Question 35.
4 fives =
4 fives = 20,

Explanation:
Given 4 fives means 4 by 5 as
4 X 5 we get 20 so 4 fives = 20.

Question 36.
5 + 5 + 5 + 5 =
5 + 5 + 5 + 5 = 20,

Explanation:
Given 5 + 5 + 5 + 5 we add 5 four times
we get 20 as 5 + 5 + 5 + 5 = 20.

Question 37.
3 sevens =
3 sevens = 21,

Explanation:
Given 3 sevens means 3 by 7 as
3 X 7 we get 21 so 3 sevens = 21.

Question 38.
7 + 7 + 7 =
7 + 7 + 7 = 21,

Explanation:
Given 7 + 7 + 7 we add 7 thrice
we get 21 as 7 + 7 + 7 = 21.

Question 39.
3 nines =
3 nines = 27,

Explanation:
Given 3 nines means 3 by 9 as
3 X 9 we get 27 so 3 nines = 27.

Question 40.
9 + 9 + 9 =
9 + 9 + 9 = 27,

Explanation:
Given 9 + 9 + 9 we add 9 thrice
we get 27 as 9 + 9 + 9 = 27.

Question 41.
3 sixes =
3 sixes = 18,

Explanation:
Given 3 sixes means 3 by 6 as
3 X 6 we get 18 so 3 sixes = 18.

Question 42.
6 + 6 + 6 =
6 + 6 + 6 = 18,

Explanation:
Given 6 + 6 + 6 we add 6 thrice
we get 18 as 6 + 6 + 6 = 18.

Question 43.
3 eights =
3 eights = 24,

Explanation:
Given 3 eights means 3 by 8 as
3 X 8 we get 24 so 3 eights = 24.

Question 44.
8 + 8 + 8 =
8 + 8 + 8 = 24,

Explanation:
Given 8 + 8 + 8 we add 8 thrice
we get 24 as 8 + 8 + 8 = 24.

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

Solve Problems 1–4 using the pictures provided for each problem.

Question 1.
There are 5 flowers in each bunch.
How many flowers are in 4 bunches?

a. Number of groups: _____4______
Size of each group: _____5________
b. 4 × 5 = ___20_______
c. There are ___20______ flowers altogether.

a. Number of groups are 4,
b. 4 X 5 = 20,
c. There are 20 flowers altogether.

Explanation:
Given there are 5 flowers in each bunch.
a. Number of groups are 4 and
size of each group:  5 as shown in the picture,

b. Multiplication sentence is 4 X 5 = 20,

c. As there are 4 groups of 5 flowers each we have
4 X 5 = 20 flowers altogether.

Question 2.
There are ___3____ candies in each box.
How many candies are in 6 boxes?

a. Number of groups: _____6______
Size of each group: ______3_______
b. 6 × ____3______ = ____18_______
c. There are ___18______ candies altogether.

a. Number of groups: 6 and
size of each group: 3
b. 6 X 3 = 18
c. There are 18 candies altogether.

Explanation:
Given there are 3 candies in each box,
a. Number of groups are 6 and
size of each group is 3 as shown in the picture,

b. Multiplication sentence is 6 X 3 = 18,

c. As there are 6 groups of 3 candies each we have
6 X 3 = 18 candies altogether.

Question 3.
There are 4 oranges in each row. How many
oranges are there in ___3___ rows?
a. Number of rows: ______3_____ Size of each row: ____4______

b. _____3______ × 4 = _____12______
c. There are ____12_____ oranges altogether.

a. Number of rows: 3 and size of each row: 4,
b. 3 X 4 = 12,
c. There are 12 oranges altogether.

Explanation:
Given there are 4 oranges in each row,
a. Number of rows are 3 and
size of each row is 4 as shown in the picture,

b. Multiplication sentence is 3 X 4 = 12,

c. As there are 3 rows of 4 oranges each we have
3 X 4 = 12 oranges altogether.

Question 4.
There are ____2____ loaves of bread in each row.
How many loaves of bread are there in 5 rows?
a. Number of rows: _____5______ Size of each row: _____2_____
b. ____5______ × _____2_____ = ____10_______

c. There are ___10__ loaves of bread altogether.

a. Number of rows: 5 and size of each row: 2,
b. 5 X 2 = 10,
c. There are 10 loaves of bread altogether.

Explanation:
Given there are 2 loaves of bread in each row
a. Number of rows are 5 and
size of each row is 2 as shown in the picture,

b. Multiplication sentence is 5 X 2 = 10,

c. As there are 5 rows of 2 loaves of bread each
we have 5 X 2 = 10 loaves of bread altogether.

Question 5.
a. Write a multiplication equation for the array shown below.

b. Draw a number bond for the array where each
part represents the amount in one row.

a. Multiplication equation : 4 X 3 = 12,

Explanation:
As shown in the picture there are 4 rows of 3 X
in each row, So the multiplication equation is 4 X 3 = 12.

b.

Explanation:
Drawn a number bond for the array where each
part represents the amount in one row.

Question 6.
Draw an array using factors 2 and 3.
Then, show a number bond where each part
represents the amount in one row.

Explanation:
Drawn an array using factors 2 and 3 as
2 X 3 = 6, 2 rows of 3 each and showed
a number bond where each part represents
the amount in one row as shown in the picture above.

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

Draw an array that shows 5 rows of 3 squares.
Then, show a number bond where each part
represents the amount in one row.

Explanation:
Drawn an array that shows 5 rows of 3 squares as
5 X 3 = 15 and showed a number bond where
each part represents the amount in one row as
shown in the picture above.

Solve Problems 1–4 using the pictures provided for each problem.

Question 1.
There are 5 pineapples in each group.
How many pineapples are there in 5 groups?

a. Number of groups: _______5______ Size of each group: ____5________
b. 5 × 5 = _____25_______
c. There are ____25______ pineapples altogether.

a. Number of groups : 5 and size of each group: 5,
b. 5 X 5 = 25,
c. There are 25 pineapples altogether.

Explanation:
Given there are 5 pineapples in each group
a. Number of groups are 5 and
size of each group is 5 as shown in the picture,

b. Multiplication sentence is 5 X 5 = 25,

c. As there are 5 rows of 5 pineapples each
we have 5 X 5 = 25 pineapples altogether.

Question 2.
There are ____4___ apples in each basket.
How many apples are there in 6 baskets?

a. Number of groups: _______6______ Size of each group: _____4_______
b. 6 × ____4______ = _____24_______
c. There are ____24______ apples altogether.

a. Number of groups : 6 and size of each group: 4,
b. 6 X 4 = 24,
c. There are 24 apples altogether.

Explanation:
Given there are 6 baskets with 4 apples
in each basket  as shown in the picture
a. Number of groups are 6 and
size of each group is 4,

b. Multiplication sentence is 6 X 4 = 24,

c. As there are 4 apples each in 6 baskets
we have 6 X 4 = 24 apples altogether.

Question 3.
There are 4 bananas in each row. How many
bananas are there in _4_ rows?
a. Number of rows: _____4______ Size of each row: ____4______
b. _____4____ × 4 = ___16_______
c. There are _____16_____ bananas altogether.

a. Number of rows : 4 and size of each row: 4,
b. 4 X 4 = 16,
c. There are 16 bananas altogether.

Explanation:
Given there are 4 rows of 4 bananas
in each row as shown in the picture
a. Number of rows are 4 and
size of each row is 4,

b. Multiplication sentence is 4 X 4 = 16,

c. As there are 4 bananas in 4 rows each
we have 4 X 4 = 16 bananas altogether.

Question 4.
There are _____3______ peppers in each row.
How many peppers are there in 6 rows?
a. Number of rows: ______6_____ Size of each row: ____3______
b. ____6_____ × ____3______ = ____18______
c. There are ____18______ peppers altogether.

a. Number of rows : 6 and size of each row: 3,
b. 6 X 3 = 18,
c. There are 18 peppers altogether.

Explanation:
Given there are 6 rows of 3 peppers
in each row as shown in the picture
a. Number of rows are 6 and
size of each row is 3,

b. Multiplication sentence is 6 X 3 = 18,

c. As there are 3 peppers in 6 rows each
we have 6 X 3 = 18 peppers altogether.

Question 5.
Draw an array using factors 4 and 2.
Then, show a number bond where each part
represents the amount in one row.

Explanation:
Drawn an array that shows factors 4 and 2 as
4 X 2 = 8, 4 rows of 2 each and showed a
number bond where each part represents the
amount in one row as shown in the picture above.

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Engage NY Eureka Math 3rd Grade Module 7 Lesson 6 Answer Key

Eureka Math Grade 3 Module 7 Answer Key

Eureka Math Grade 3 Module 7 Lesson 6 Problem Set Answer Key

Use a ruler and a right angle tool to help you draw the figures with the attributes given below.
Question 1.
Draw a triangle with 1 right angle.
Answer:

Explanation:
ABC is an right angled triangle.

Question 2.
Draw a quadrilateral with 4 right angles and sides that are all 2 inches long.
Answer:

Explanation:
ABCD is a quadrilateral with 4 right angles and sides that are all 2 cm long.

Question 3.
Draw a quadrilateral with at least 1 set of parallel sides. Trace the parallel sides green.
Answer:

Explanation:
ABCD is a quadrilateral of having AD and BC Parallel sides.

Question 4.
Draw a pentagon with at least 2 equal sides. Label the 2 equal side lengths of your shape.
Answer:

Explanation:
Pentagon is an polygon having five sides.
ABCDE is having same length of sides of ED and BC of 2cm.

Question 5.
Draw a hexagon with at least 2 equal sides. Label the 2 equal side lengths of your shape.
Answer:

Explanation:
Hexagon is an polygon having six sides.
AB and ED are having the sides of same length of 3cm each.

Question 6.
Sam says that he drew a polygon with 2 sides and 2 angles. Can Sam be correct? Use pictures to help you explain your answer.
Answer:
Sam cannot be correct because we cannot draw a polygon with 2 sides and 2 angles. We need at least three lines to draw a polygon with three angles.

Explanation:

Eureka Math Grade 3 Module 7 Lesson 6 Exit Ticket Answer Key

Use a ruler and a right angle tool to help you draw a shape that matches the attributes of Jeanette’s shape. Label your drawing to explain your thinking.

Jeanette says her shape has 4 right angles and 2 sets of parallel sides. It is not a regular quadrilateral.
Answer:

Explanation:
ABCD is a Quadrilateral, where AB and DC are two parallel sides and  A and D are two right angles.

Eureka Math Grade 3 Module 7 Lesson 6 Homework Answer Key

Use a ruler and a right angle tool to help you draw the figures with the given attributes below.

Question 1.
Draw a triangle that has no right angles.
Answer:

Explanation:
Triangle ABC is an acute triangle.

Question 2.
Draw a quadrilateral that has at least 2 right angles.
Answer:

Explanation:
ABCD is an quadrilateral having at least 2 right angles.

Question 3.
Draw a quadrilateral with 2 equal sides. Label the 2 equal side lengths of your shape.
Answer:

Explanation:
ABCD is an quadrilateral of four sides, where AD and BC having same length of 2cms each.

Question 4.
Draw a hexagon with at least 2 equal sides. Label the 2 equal side lengths of your shape.
Answer:

Explanation:
Hexagon polygon has six sides in it.
ABCDEF is a Hexagon having six sides where AB and DE are having same length of 2cms each.

Question 5.
Draw a pentagon with at least 2 equal sides. Label the 2 equal side lengths of your shape.
Answer:

Explanation:
Pentagon is a figure having 5 sides in it.
ABCDE is an pentagon having five sides, where AB and BC sides of same length of 2cms each.

Question 6.
Cristina describes her shape. She says it has 3 equal sides that are each 4 centimeters in length. It has no right angles. Do your best to draw Cristina’s shape, and label the side lengths.
Answer:

Explanation:
Cristina figure described as below:
ABC is an equilateral Triangle of equal lengths sides of 4cm each which does not have any right angles in it.

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

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

Multiply.

multiply by 5 (6–10)
Answer:

Explanation:
5 × 1 = 5
5 × 2 = 10
5 × 3 = 15
5 × 4 = 20
5 × 5 = 25
5 × 6 = 30
5 × 7 = 35
5 × 8 = 40
5 × 9 = 45
5 × 10 = 50.

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

Question 1.
Use tetrominoes to create at least two different rectangles. Then, color the grid below to show how you created your rectangles. You may use the same tetromino more than once.

Answer:

Explanation:
The rectangles in the given cubes chart are two different shapes with same number of cubes.
They are in two different colors.

Question 2.
Use tetrominoes to create at least two squares, each with an area of 36 square units. Then, color the grid below to show how you created your squares. You may use the same tetromino more than once.

Answer:

Explanation:
Two different colored Squares of same length of cubes = 6 units.

a. Write an equation to show the area of a square above as the sum of the areas of the tetrominoes you used to make the square.
Answer:
Side of the Square = 6 units.
Area of Square = Side × Side

Explanation:

b. Write an equation to show the area of a square above as the product of its side lengths.
Answer:
Area of Square = 36 square units.

Explanation:
Side of the Square = 6 units.
Area of Square = Side × Side
= 6 × 6
=  36 square units.

Question 3.
a. Use tetrominoes to create at least two different rectangles, each with an area of 12 square units. Then, color the grid below to show how you created the rectangles. You may use the same tetromino more than once.

Answer:

Explanation:
Two different rectangles are drawn with the same area of 12 square units.

b. Explain how you know the area of each rectangle is 12 square units.
Answer:
Area of the rectangle = 12 square units.

Explanation:
Area of the rectangle = Length × width
—> First Yellow color rectangle:
Length = 4 units
Width = 3 units
Area of the rectangle = Length × width
= 4 × 3
= 12 square units.
—>Second  Orange color rectangle:
Length = 6 units
Width = 2 units
Area of the rectangle = Length × width
= 6 × 2
= 12 square units.

Question 4.
Marco created a rectangle with tetrominoes and traced its outline in the space below. Use tetrominoes to re-create it. Estimate to draw lines inside the rectangle below to show how you re-created Marco’s rectangle.

Answer:

Explanation:
Marco created a rectangle with tetrominoes and traced its outline is re-created and shown above.

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

Use your tetrominoes to make a rectangle that has an area of 20 square units. Then, color the grid to show how you made your rectangle. You may use the same tetromino more than once.

Answer:
Area of the rectangle = 20 square units.

Explanation:
Length of the rectangle drawn = 5 units
Width of the rectangle drawn = 4 units
Area of the rectangle = Length × Width
= 5 × 4
= 20 square units.

Eureka Math Grade 3 Module 7 Lesson 7 Homework Answer Key

Question 1.
Color tetrominoes on the grid to create three different rectangles. You may use the same tetromino more than once.

Answer:

Explanation:
There are three different shaped rectangle drawn above.

Question 2.
Color tetrominoes on the grid below to:
a. Create a square with an area of 16 square units.
Answer:
Area of Square = 16 Square units.

Explanation:
Side of the Square = 4 units
Area of Square = side × side
= 4 × 4
= 16 Square units.

b. Create at least two different rectangles, each with an area of 24 square units.
You may use the same tetromino more than once.

Answer:

Explanation:
Two different rectangles are drawn with the same area of 24 square units.

Question 3.
Explain how you know the rectangles you created in Problem 2(b) have the correct area.
Answer:
Area of the rectangles = 24 square units.

Explanation:
—> First rectangle:
Length of the First rectangle = 8 units
Width of the First rectangle = 3 units
Area of the First rectangle = Length × Width
= 8 × 3
= 24 square units.
—> Second rectangle:
Length of the Second rectangle = 8 units
Width of the Second rectangle = 3 units
Area of the Second rectangle = Length × Width
= 6 × 4
= 24 square units.

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

Eureka Math Grade 3 Module 1 Answer Key

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

Question 1.
Draw an array that shows 5 rows of 3.

Explanation:
Drawn an array that shows 5 rows of 3 as
5 X 3 as shown above.

Question 2.
Draw an array that shows 3 rows of 5.

Explanation:
Drawn an array that shows 3 rows of 5 as
3 X 5 as shown above.

Question 3.
Write multiplication expressions for the arrays in
Problems 1 and 2. Let the first factor in each expression
represent the number of rows. Use the commutative property
to make sure the equation below is true.

Explanation:
Wrote multiplication expressions for the arrays in
Problems 1 and 2 as 5 X 3, 3 X 5 and let the first factor
in each expression represent the number of rows.
Used the commutative property to make sure the equation
below is true as 5 X 3 = 15, 3 X 5 = 15,
So,  5 X 3 = 3 X 5 is true equals to 15.

Question 4.
Write a multiplication sentence for each expression.
You might skip-count to find the totals. The first one is done for you.
a. 2 threes: 2 × 3 = 6
b. 3 twos: _________________
c. 3 fours: ________________
d. 4 threes: ________________
e. 3 sevens: ________________
f. 7 threes: ________________
g. 3 nines: _________________
h. 9 threes: ________________
i. 10 threes: _______________

b. 3 twos : 3 x 2 = 6,
Explanation:
Given 3 twos, 3 multiplied by 2 gives 6,
So, 3 X 2 = 6.

c. 3 fours: 3 X 4 = 12,
Explanation:
Given 3 fours, 3 multiplied by 4 gives 12,
So, 3 X 4 = 12.

d. 4 threes: 4 X 3 = 12,
Explanation:
Given 4 threes, 4 multiplied by 3 gives 12,
So, 4 X 3 = 12.

e. 3 sevens:  3 x 7 = 21,
Explanation:
Given 3 sevens, 3 multiplied by 7 gives 21,
So, 3 X 7 = 21.

f. 7 threes:  7 X 3 = 21,
Explanation:
Given 7 threes, 7 multiplied by 3 gives 21,
So, 7 X 3 = 21.

g. 3 nines: 3 X 9 = 27,
Explanation:
Given 3 nines, 3 multiplied by 9 gives 27,
So, 3 X 9 = 27.

h. 9 threes: 9 X 3 = 27,
Explanation:
Given 9 threes, 9 multiplied by 3 gives 27,
So, 9 X 3 = 27.

i. 10 threes: 10 x 3 =30,
Explanation:
Given 10 threes, 10 multiplied by 3 gives 30,
So, 10 X 3 = 30.

Question 5.
Find the unknowns that make the equations true.
Then, draw a line to match related facts.
a. 3 + 3 + 3 + 3 + 3 = _________
b. 3 × 9 = _________
c. 7 threes + 1 three = _________
d. 3 × 8 = _________
e. _________ = 5 × 3
f. 27 = 9 × _________

a. 3 + 3 + 3 + 3 + 3 = 15,

Explanation:
Given 3 + 3 + 3 + 3 + 3  adding 3, 5 times gives 15,
So 3 + 3 + 3 + 3 + 3 = 15.

b. 3 × 9 = 27,

Explanation:
Given 3 X 9, 3 multiplied by 9 gives 27,
So 3 X 9 = 27.

c. 7 threes + 1 three = 8 threes = 24,

Explanation:
Given 7 threes + 1 three = 7 X 3 + 1 X 3 = 21 + 3 = 24,
So, 7 threes + 1 three = 8 threes = 24.

d. 3 X 8 = 24,

Explanation:
Given 3 X 8, 3 multiplied by 8 gives 24,
So 3 X 8 = 24.

e. 15 = 5 × 3,

Explanation:
Given 5 X 3, 5 multiplied by 3 gives 15,
So 15 = 5 X 3.

f. 27 = 9 × 3

Explanation:
Given 27 = 9 X ___, Lets take missing number as x,
27 = 9 X x, So x = 27 ÷ 9 = 3, So 27 = 9 X 3.

Question 6.
Isaac picks 3 tangerines from his tree every day for 7 days.
a. Use circles to draw an array that represents the
tangerines Isaac picks.

Explanation:
Given Isaac picks 3 tangerines from his tree
every day for 7 days.
a. Used circles to draw an array that represents the
tangerines Isaac picks as 3 X 7 as shown above.

b. How many tangerines does Isaac pick in 7 days?
Write and solve a multiplication sentence to find the total.

Isaac picks 21 tangerines in 7 days,
Multiplication sentence to find the total is 3 x 7 = 21.

Explanation:
Wrote and solved a multiplication sentence to
find the total as 3 x 7 = 21,
therefore Isaac picks 21 tangerines in 7 days.

c. Isaac decides to pick 3 tangerines every day for 3 more days.
Draw x’s to show the new tangerines on the array in Part (a).

Explanation:
Given Isaac decides to pick 3 tangerines every day for 3 more days.
Drawn  x’s to show the new tangerines on the array in Part (a) as
shown above in the picture.

d. Write and solve a multiplication sentence to
find the total number of tangerines Isaac picks.

Multiplication sentence is 3 x 10 =
(3 X 7) + (3 X 3) = 21 + 9 = 30,
The total number of tangerines Isaac picks are 30,

Explanation:
Wrote and solved a multiplication sentence as
3 x 10 = (3 X 7) + (3 X 3) = 21 + 9 = 30,
Therefore the total number of tangerines Isaac picks are 30.

Question 7.
Sarah buys bottles of soap. Each bottle costs $2.
a. How much money does Sarah spend if she buys 3 bottles of soap?
_____$2_____ × ____3______ = $___6_____
b. How much money does Sarah spend if she buys 6 bottles of soap?
_____$2_____ × _____6_____ = $__12______

a. Sarah spends $6 if she buys 3 bottles of soap,

Explanation:
Given Sarah buys bottles of soap. Each bottle costs $2,
So money spent by Sarah if she buys 3 bottles of soap are
$2 X 3 = $6.

b. Sarah spends $12 if she buys 6 bottles of soap,

Explanation:
Given Sarah buys bottles of soap. Each bottle costs $2,
So money spent by Sarah if she buys 6 bottles of soap are
$2 X 6 = $12.

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

Mary Beth organizes stickers on a page in her sticker book.
She arranges them in 3 rows and 4 columns.
a. Draw an array to show Mary Beth’s stickers.
b. Use your array to write a multiplication sentence to
find Mary Beth’s total number of stickers.
c. Label your array to show how you skip-count to
solve your multiplication sentence.
d. Use what you know about the commutative property
to write a different multiplication sentence for your array.

a.
Explanation:
Given Mary Beth organizes stickers on a page in
her sticker book. She arranges them in 3 rows and
4 columns.
Drawn an array as 3 X 4 to show Mary Beth’s stickers.

b. Mary Beth’s total number of stickers are 12,

Explanation:
Used my array to write a multiplication sentence as 3 X 4 = 12,
in finding Mary Beth’s total number of stickers.

c.

Explanation:
Labeled my array to show skip-count to 12 and
solved my multiplication sentence as 3 x 4 = 12.

d. Commutative property to write a different
multiplication sentence for my array is 3 X 4 = 4 x 3,

Explanation:
Used what I know about the commutative property
to write a different multiplication sentence for my array as
3 X 4 = 4 x 3.

Eureka Math Grade 3 Module 1 Lesson 8 Homework Answer Key

Question 1.
Draw an array that shows 6 rows of 3.

Explanation:
Drawn an array that shows 6 rows of 3 as
6 X 3 as shown above.

Question 2.
Draw an array that shows 3 rows of 6.

Explanation:
Drawn an array that shows 3 rows of 6 as
3 X 6 as shown above.

Question 3.
Write multiplication expressions for the arrays in
Problems 1 and 2. Let the first factor in each expression
represent the number of rows. Use the commutative property
to make sure the equation below is true.


Explanation:
Wrote multiplication expressions for the arrays in
Problems 1 and 2 as 6 X 3, 3 X 6 and let the first factor
in each expression represent the number of rows.
Used the commutative property to make sure the equation
below is true as 6 X 3 = 18, 3 X 6 = 18,
So,  6 X 3 = 3 X 6 is true equals to 18.

Question 4.
Write a multiplication sentence for each expression.
You might skip-count to find the totals. The first one is done for you.
a. 5 threes: 5 × 3 = 15
b. 3 fives: __________________
c. 6 threes: ________________
d. 3 sixes: ___________________
e. 7 threes: __________________
f. 3 sevens: __________________
g. 8 threes: ________________
h. 3 nines: _________________
i. 10 threes: _______________

b. 3 fives: 3 X 5 = 15,

Explanation:
Given 3 fives, 3 multiplied by 5 gives 15,
So, 3 X 5 = 15.

c. 6 threes: 6 X 3 = 18,

Explanation:
Given 6 threes, 6 multiplied by 3 gives 18,
So, 6 X 3 = 18.

d. 3 sixes: = 3 X 6 = 18,

Explanation:
Given 3 sixes, 3 multiplied by 6 gives 18,
So, 3 X 6 = 18.

e. 7 threes: 7 X 3 = 21,

Explanation:
Given 7 threes, 7 multiplied by 3 gives 21,
So, 7 X 3 = 21.

f. 3 sevens: 3 X 7 = 21,

Explanation:
Given 3 sevens, 3 multiplied by 7 gives 21,
So, 3 X 7 = 21.

g. 8 threes: 8 X 3 = 24,

Explanation:
Given 8 threes, 8 multiplied by 3 gives 24,
So, 8 X 3 = 24.

h. 3 nines: 3 X 9 = 27,

Explanation:
Given 3 nines, 3 multiplied by 9 gives 27,
So, 3 X 9 = 27.

i. 10 threes: 10 X 3 = 30,

Explanation:
Given 10 threes, 10 multiplied by 3 gives 30,
So, 10 X 3 = 30.

Question 5.
Find the unknowns that make the equations true. Then, draw a line to match related facts.
a. 3 + 3 + 3 + 3 + 3 + 3 = _________
b. 3 × 5 = _________
c. 8 threes + 1 three = _________
d. 3 × 9 = _________
e. _________ = 6 × 3
f. 15 = 5 × _________

a. 3 + 3 + 3 + 3 + 3 + 3 = 18,

Explanation:
Given 3 + 3 + 3 + 3 + 3 + 3 adding 3, 6 times gives 18,
So 3 + 3 + 3 + 3 + 3 + 3 = 18.

b. 3 × 5 = 15,

Explanation:
Given 3 X 5, 3 multiplied by 5 gives 15,
So 3 X 5 = 15.

c. 8 threes + 1 three =9 threes = 27,

Explanation:
Given 8 threes + 1 three = 8 X 3 + 1 X 3 = 24 + 3 = 27,
So, 8 threes + 1 three = 9 threes = 27.

d. 3 X 9 = 27,

Explanation:
Given 3 X 9, 3 multiplied by 9 gives 27,
So 3 X 9 = 27.

e. 18 = 6 × 3,

Explanation:
Given 6 X 3, 6 multiplied by 3 gives 18,
So 18 = 6 X 3.

f. 15 = 5 × 3,

Explanation:
Given 15 = 5 X ___, Lets take missing number as x,
15 = 5 X x, So x = 15 ÷ 5 = 3, So 15 = 5 X 3.

Question 6.
Fernando puts 3 pictures on each page of his photo album.
He puts pictures on 8 pages.
a. Use circles to draw an array that represents the total number
of pictures in Fernando’s photo album.
b. Use your array to write and solve a multiplication sentence
to find Fernando’s total number of pictures.
c. Fernando adds 2 more pages to his book. He puts 3 pictures
on each new page. Draw x’s to show the new pictures on
the array in Part (a).
d. Write and solve a multiplication sentence to find the new
total number of pictures in Fernando’s album.
a.

Explanation:
Given Fernando puts 3 pictures on each page of his photo album.
He puts pictures on 8 pages.
a. Used circles to draw an array that represents the total number
of pictures in Fernando’s photo album as 3 X 8.

b. Multiplication sentence for Fernando’s total number
of pictures is 3 X 8 = 24,

Explanation:
Used my array to write and solve a multiplication sentence
to find Fernando’s total number of pictures as 3 X 8 = 24.

c.

Explanation:
Fernando adds 2 more pages to his book. He puts 3 pictures
on each new page. Drawn x’s to show the new pictures on
the array in Part (a) as shown above.

d. Multiplication sentence is 3 x 10 =
(3 X 8) + (3 X 2) = 24 + 6 = 30,
The new total number of pictures in Fernando’s album are 30,

Explanation:
Wrote and solved a multiplication sentence as
3 x 10 = (3 X 8) + (3 X 2) = 24 + 6 = 30,
Therefore the new total number of pictures in Fernando’s album are 30.

Question 7.
Ivania recycles. She gets 3 cents for every can she recycles.
a. How much money does Ivania make if she recycles 4 cans?
____3______ × ___4_______ = ___12_____ cents
b. How much money does Ivania make if she recycles 7 cans?
____3______ × ____7______ = ____21____ cents

a. Ivania makes 12 cents if she recycles 4 cans,

Explanation:
Given Ivania recycles and she gets 3 cents for every can she recycles,
So money Ivania makes if she recycles 4 cans is
3 cents X 4 = 12 cents.

b. Ivania makes 21 cents if she recycles 7 cans,

Explanation:
Given Ivania recycles and she gets 3 cents for every can she recycles,
So money Ivania makes if she recycles 7 cans is
3 cents X 7 = 21 cents.

Engage NY Eureka Math 3rd Grade Module 7 Lesson 14 Answer Key

Eureka Math Grade 3 Module 7 Lesson 14 Pattern Sheet Answer Key

Multiply.

multiply by 8 (6–10)
Answer:

Explanation:
8 × 5 = 40
8 × 6 = 48
8 × 7 = 56
8 × 8 = 64
8 × 9 = 72
8 × 10 = 80.

Eureka Math Grade 3 Module 7 Lesson 14 Problem Set Answer Key

Question 1.
Label the unknown side lengths of the regular shapes below. Then, find the perimeter of each shape.
a.

Perimeter = _______ in
Answer:
Length of the side of the ABCDEFG Octogen = 64in.

Explanation:

Length of the side of the ABCDEFG Octogen = 8in
Perimeter of the ABCDEFG Octogen = 8 × side
= 8 × 8in
= 64in.

b.

Perimeter = _______ ft
Answer:
Perimeter of the ABC Triangle = 21ft.

Explanation:

Length of the side of the ABC Triangle = 7ft
Perimeter of the ABC Triangle = 3 × Side
= 3 × 7ft
= 21ft.

c.

Perimeter = _______ m
Answer:
Perimeter of the ABCD Square = 36m.

Explanation:

Length of the Side of the ABCD Square = 9m
Perimeter of the ABCD Square = 4 × Side
= 4 × 9m
= 36m.

d.

Perimeter = _______ in
Answer:
Perimeter of the ABCDE Pentagon = 36in.

Explanation:

Length of the side of the ABCDE Pentagon = 6in
Perimeter of the ABCDE Pentagon = 6 × Side
= 6 × 6in
= 36in.

Question 2.
Label the unknown side lengths of the rectangle below. Then, find the perimeter of the rectangle.

Perimeter = _______ cm
Answer:
Perimeter of the side of the ABCD Rectangle = 18cm.

Explanation:

Length of the side of the ABCD Rectangle = 7cm
Width of the side of the ABCD Rectangle = 2cm
Perimeter of the side of the ABCD Rectangle = 2 (Length + Width)
= 2 ( 7cm + 2cm )
= 2 × 9cm
= 18cm.

Question 3.
David draws a regular octagon and labels a side length as shown below. Find the perimeter of David’s octagon.

Answer:
Perimeter of the ABCDEFGH Octogen = 36cm.

Explanation:

Length of the side of the ABCDEFGH Octogen = 6cm
Perimeter of the ABCDEFGH Octogen = 6 × Side
= 6 × 6cm
= 36cm.

Question 4.
Paige paints an 8-inch by 9-inch picture for her mom’s birthday. What is the total length of wood that Paige needs to make a frame for the picture?
Answer:
Perimeter of the Paige’s paints = 34inch.

Explanation:
Length of the Paige’s paints = 9inch
Width of the Paige’s paints = 8inch
Perimeter of the Paige’s paints = 2 (Length + Width )
= 2 (9inch + 8inch)
= 2 × 17inch
= 34inch.

Question 5.
Mr. Spooner draws a regular hexagon on the board. One of the sides measures 4 centimeters. Giles and Xander find the perimeter. Their work is shown below. Whose work is correct? Explain your answer.
Giles’s Work
Perimeter = 4 cm + 4 cm + 4 cm + 4 cm + 4 cm + 4 cm
Perimeter = 24 cm

Xander’s Work
Perimeter = 6 × 4 cm
Perimeter = 24 cm
Answer:
Perimeter of the Hexagon = 24 centimeters.
Both Giles and Xander’s work is correct because they got the calculation value correct even though the methodology was different of each.

Explanation:
Length of the side of the Hexagon = 4 centimeters
Perimeter of the Hexagon = 6 × Side
= 6 × 4 centimeters
= 24 centimeters.
Both are correct.

Eureka Math Grade 3 Module 7 Lesson 14 Exit Ticket Answer Key

Travis traces a regular pentagon on his paper. Each side measures 7 centimeters. He also traces a regular hexagon on his paper. Each side of the hexagon measures 5 centimeters. Which shape has a greater perimeter? Show your work.
Answer:
Perimeter of the Pentagon is greater than the Perimeter of the regular hexagon because the measurement value of the Pentagon is more than the measurement value of the regular hexagon.

Explanation:

Length of the side of the Pentagon = 7 centimeters
Perimeter of the Pentagon = 5 × Side
= 5 × 7 centimeters
= 35 centimeters.
Length of the side of the regular hexagon = 5 centimeters
Perimeter of the regular hexagon = 6 × Side
= 6 × 5 centimeters
= 30 centimeters.

Eureka Math Grade 3 Module 7 Lesson 14 Homework Answer Key

Question 1.
Label the unknown side lengths of the regular shapes below. Then, find the perimeter of each shape.
a.

Perimeter = _______ in
Answer:
Perimeter of the ABC Triangle = 16in.

Explanation:

Length of the Side of the ABC Triangle = 4in
Perimeter of the ABC Triangle = Side + Side + Side
= AB + BC + CA
= 4in + 4in + 4in
= 12in + 4in
=16in.

b.

Perimeter = _______ cm

Answer:
Perimeter of the square = 32cm.

Explanation:

Length of the Side of the square = 8cm
Perimeter of the square = 4 × Side
= 4 × 8cm
= 32cm.

c.

Perimeter = _______ m
Answer:
Perimeter of the Octagon = 72m.

Explanation:

Length of the side of the Octagon = 9m
Perimeter of the Octagon = 8 × Side
= 8 × 9m
= 72m.

d.

Perimeter = _______ in
Answer:
Perimeter of the Hexagon = 36in.

Explanation:

Length of the side of the Hexagon = 6in
Perimeter of the Hexagon = 6 × Side
= 6 × 6in
= 36in.

Question 2.
Label the unknown side lengths of the rectangle below. Then, find the perimeter of the rectangle.

Perimeter = _______ cm
Answer:
Perimeter of the rectangle =  2 ( Length + Width )
= 2 ( 9cm + 4cm )
= 2 × 13cm
= 26cm.

Explanation:
Length of the rectangle = 9cm
Width of the rectangle = 4cm
Perimeter of the rectangle =  2 ( Length + Width )
= 2 ( 9cm + 4cm )
= 2 × 13cm
= 26cm.

Question 3.
Roxanne draws a regular pentagon and labels a side length as shown below. Find the perimeter of Roxanne’s pentagon.

Answer:
Perimeter of the Roxanne’s pentagon = 5 × Side
= 5 × 7cm
= 35cm.

Explanation:
Length of the side of the Roxanne’s pentagon = 7cm
Perimeter of the Roxanne’s pentagon = 5 × Side
= 5 × 7cm
= 35cm.

Question 4.
Each side of a square field measures 24 meters. What is the perimeter of the field?
Answer:
Perimeter of the Square field = 4 × Side
= 4 × 24meters
= 96meters.

Explanation:
Length of the side of the Square field = 24 meters
Perimeter of the Square field = 4 × Side
= 4 × 24meters
= 96meters.

Question 5.
What is the perimeter of a rectangular sheet of paper that measures 8 inches by 11 inches?
Answer:
Perimeter of the rectangular sheet of paper = 2 ( Length  + Width )
= 2 (11inches + 8inches)
= 38inches.

Explanation:
Length of the rectangular sheet of paper = 11inches
Width of the rectangular sheet of paper = 8inches
Perimeter of the rectangular sheet of paper = 2 ( Length  + Width )
= 2 (11inches + 8inches)
= 2 × 19inches
= 38inches.

Engage NY Eureka Math 3rd Grade Module 1 Lesson 7 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 7 Problem Set Answer Key

Question 1.
a. Draw an array that shows 6 rows of 2.
b. Write a multiplication sentence where the
first factor represents the number of rows.
________ × ________ = ________
a.

Explanation:
Drawn an array that shows 6 rows of 2 as
6 X 2 as shown above in the picture.

b. Multiplication sentence where the first factor represents
the number of rows as 6 X 2 = 12.

Explanation:
Wrote a multiplication sentence where the first factor represents
the number of rows as 6 X 2 = 12.

Question 2.
a. Draw an array that shows 2 rows of 6.
b. Write a multiplication sentence where the first factor
represents the number of rows.
____2____ × ____6____ = ___12_____
a.

Explanation:
Drawn an array that shows 2 rows of 6 as
2 X 6 as shown above in the picture.

b. Multiplication sentence where the first factor represents
the number of rows as 2 X 6 = 12.

Explanation:
Wrote a multiplication sentence where the first factor represents
the number of rows as 2 X 6 = 12.

Question 3.
a. Turn your paper to look at the arrays in Problems 1
and 2 in different ways. What is the same and what is
different about them?
b. Why are the factors in your multiplication sentences
in a different order?

a. Same is the result because both have the same value as 12,
Different is number of rows and number of columns are
not the same,

Explanation:
Turned my paper to look at the arrays in Problems 1
and 2 in different ways.
Is the same because both got same results as 12
and different about them is in problems 1 and 2
the number of rows and number of columns are not the same
as one is 6 rows X 2 columns  and in other it is  2 rows X 6 columns.

b. The factors in my multiplication sentences are
in a different order because first we are taking
6 rows X 2 columns and in other it is  2 rows X 6 columns.

Explanation:
The factors in my multiplication sentences are
in a different order because first we are taking
6 rows X 2 columns  and in other it is  2 rows X 6 columns,
So, rows and columns differ in the both.

Question 4.
Write a multiplication sentence for each expression.
You might skip-count to find the totals.

Explanation:
Wrote a multiplication sentence for each expression as
shown above, might have skipped-count to find the totals.

Question 5.
Write and solve multiplication sentences where the
second factor represents the size of the row.

Explanation:
Wrote and solved multiplication sentences where the
second factor represents the size of the row as
2 X 4 = 8 and 4 X 2= 8.

Question 6.
Ms. Nenadal writes 2 × 7 = 7 × 2 on the board.
Do you agree or disagree? Draw arrays to help
explain your thinking.

Yes, I agree Ms. Nenadal writings as 2 X 7 = 7 X 2,

Explanation:
Given Ms. Nenadal writes 2 X 7 = 7 X 2 on the board,
Yes, I agree because 2 X 7 = 14 and 7 x 2 = 14,
Drawn arrays to help and explained my thinking
as shown above.

Question 7.
Find the missing factor to make each equation true.

Explanation:
Found the missing factor to make each equations true as below
5 X 2 = 2 X 5,
As each equation is true, so the missing factor in 2 X ___ is 5
as LHS is equal to RHS, So 5 X 2 = 2 X 5.

2 X 8 = 8 X 2,
As each equation is true, so the missing factor in ___ X 8 is 2
as LHS is equal to RHS, So 2 X 8 = 8 X 2.

2 X 10 = 10 X 2,
As each equation is true, so the missing factor in ___ X 2  is 10
as LHS is equal to RHS, So 2 X 10 = 10 X 2.

2 X 9 = 9 X 2,
As each equation is true, so the missing factor in 2 X ___ is 9
as LHS is equal to RHS, So 2 X 9 = 9 X 2.

Question 8.
Jada gets 2 new packs of erasers. Each pack has 6 erasers in it.
a. Draw an array to show how many erasers Jada has altogether.
b. Write and solve a multiplication sentence to describe the array.
c. Use the commutative property to write and solve a different
multiplication sentence for the array.

a.

Jada has 12 erasers altogether,

Explanation:
Given Jada gets 2 new packs of erasers. Each pack has 6 erasers in it.
So, drawn an array to show 2 X 6 = 12 number of erasers
Jada has altogether.

b. Multiplication sentence is 2 X 6 = 12, the array has
12 erasers.

Explanation:
Wrote and solved a multiplication sentence as
2 X 6 = 12 erasers, describing the array as 2 packs in rows
and 6 erasers in columns, So 2 packs X 6 erasers = 12 erasers.

c.  Commutative property for multiplication sentence
for the array is 2 X 6 = 6 X 2 = 12.

Explanation:
Used the commutative property to write and solve a different
multiplication sentence for the array as 2 X 6 = 6 x 2 = 12 erasers.

Eureka Math Grade 3 Module 1 Lesson 7 Exit Ticket Answer Key

Do you agree or disagree with the statement in the box?
Draw arrays and use skip-counting to explain your thinking.

Yes, I agree with the statement in the box as 2 x 5 = 5 X 2,

Explanation:
Yes, I agree or disagree with the statement in the box
as 2 X 5 = 10 = 5 X 10,
Drawn arrays and used skip-counting as 2 X 5 = 10 = 5 X 10.

Eureka Math Grade 3 Module 1 Lesson 7 Homework Answer Key

Question 1.
a. Draw an array that shows 7 rows of 2.
b. Write a multiplication sentence where the first factor
represents the number of rows.
____7____ × ___2____ = ____14____

Explanation:
Drawn an array that shows 7 rows of 2 as
7 X 2 as shown above in the picture.

b. Multiplication sentence where the first factor represents
the number of rows as 7 X 2 = 14.

Explanation:
Wrote a multiplication sentence where the first factor represents
the number of rows as 7 X 2 = 14.

Question 2.
a. Draw an array that shows 2 rows of 7.
b. Write a multiplication sentence where the first factor
represents the number of rows.
____2____ × ___7_____ = ___14____

Explanation:
Drawn an array that shows 2 rows of 7 as
2 X 7 as shown above in the picture.

b. Multiplication sentence where the first factor represents
the number of rows as 2 X 7 = 14.

Explanation:
Wrote a multiplication sentence where the first factor represents
the number of rows as 2 X 7 = 14.

Question 3.
a. Turn your paper to look at the arrays in Problems 1 and 2
in different ways. What is the same and what is different about them?
b. Why are the factors in your multiplication sentences in a different order?

a. Same is the result because both have the same value as 14,
Different is number of rows and number of columns are
not the same,

Explanation:
Turned my paper to look at the arrays in Problems 1
and 2 in different ways.
Is the same because both got same results as 14
and different about them is in problems 1 and 2
the number of rows and number of columns are not the same
as one is 7 rows X 2 columns  and in other it is  2 rows X 7 columns.

b. The factors in my multiplication sentences are
in a different order because first we are taking
7 rows X 2 columns and in other it is  2 rows X 7 columns.

Explanation:
The factors in my multiplication sentences are
in a different order because first we are taking
7 rows X 2 columns  and in other it is  2 rows X 7 columns,
So, rows and columns differ in the both.

Question 4.
Write a multiplication sentence to match the number of groups.
Skip-count to find the totals. The first one is done for you.


Explanation:
Wrote a multiplication sentence to match the number of groups
as shown above, might have skipped-count to find the totals.

Question 5.
Write and solve multiplication sentences where
the second factor represents the size of the row.


Explanation:
Wrote and solved multiplication sentences where the
second factor represents the size of the row as
2 X 6 = 12 and 6 X 2= 12.

Question 6.
Angel writes 2 × 8 = 8 × 2 in his notebook.
Do you agree or disagree? Draw arrays to help
explain your thinking.

Explanation:
Given Angel writes 2 X 8 = 8 X 2 in his notebook,
Yes, I agree because 2 X 8 = 16 and 8 x 2 = 16,
Drawn arrays to help and explained my thinking
as shown above.

Question 7.
Find the missing factor to make each equation true.


Explanation:
Found the missing factor to make each equations true as below
2 X 6 = 6 X 2,
As each equation is true, so the missing factor in 6 X ___ is 2
as LHS is equal to RHS, So 2 X 6 = 6 X 2.

7 X 2 = 2 X 7,
As each equation is true, so the missing factor in ___ X 2 is 7
as LHS is equal to RHS, So 7 X 2 = 2 X 7.

9 X 2 = 2 X 9,
As each equation is true, so the missing factor in ___ X 9 is 2
as LHS is equal to RHS, So 9 X 2 = 2 X 9.

2 X 10 = 10 X 2,
As each equation is true, so the missing factor in 2 X ___ is 10
as LHS is equal to RHS, So 2 X 10 = 10 X 2.

Question 8.
Tamia buys 2 bags of candy.
Each bag has 7 pieces of candy in it.
a. Draw an array to show how many pieces of
candy Tamia has altogether.
b. Write and solve a multiplication sentence to
describe the array.
c. Use the commutative property to write and solve a
different multiplication sentence for the array.

a.

Tamia has 14 candy’s altogether,

Explanation:
Given Tamia buys 2 bags of candy. Each bag has
7 pieces of candy in it.
So, drawn an array to show 2 X 7 = 14 number of candy’s
Tamia has altogether.

b. Multiplication sentence is 2 X 7 = 14, the array has
14 candy’s.

Explanation:
Wrote and solved a multiplication sentence as
2 X 7 = 14 candy’s, describing the array as 2 bags in rows
and 7 candy’s in columns, So 2 bags X 7 candy’s = 14 candy’s.

c.  Commutative property for multiplication sentence
for the array is 2 X 7 = 7 X 2 = 14.

Explanation:
Used the commutative property to write and solve a different
multiplication sentence for the array as 2 X 7 = 7 x 2 = 14 candy’s.

Engage NY Eureka Math 3rd Grade Module 7 Lesson 13 Answer Key

Eureka Math Grade 3 Module 7 Lesson 13 Pattern Sheet Answer Key

Multiply.

multiply by 8 (1–5)
Answer:

Explanation:
8 × 1 = 8
8 × 2 = 16
8 × 3 = 24
8 × 4 = 32
8 × 5 = 40.

Eureka Math Grade 3 Module 7 Lesson 13 Problem Set Answer Key

Question 1.
Find the perimeter of the following shapes.
a.

P = 3 in + 8 in + 3 in + 8 in
= _________ in
Answer:
Perimeter of the ABCD Rectangle = Side + Side + Side + Side
= AB + BC +CD + DA
= 8in + 3in + 8in + 3in
= 22in.

Explanation:

Length of the AB side of the Rectangle = 8in
Length of the BC side of the Rectangle = 3in
Length of the CA side of the Rectangle = 8in
Length of the DA side of the Rectangle = 3in
Perimeter of the ABCD Rectangle = Side + Side + Side + Side
= AB + BC +CD + DA
= 8in + 3in + 8in + 3in
= 11in + 8in + 3in
= 19in + 3in
= 22in.

b.

P = ____ cm + ____ cm + ____ cm + ____ cm
= _________ cm
Answer:
Perimeter of the ABCD Square = Side + Side + Side + Side
= AB + BC +CD + DA
= 4cm + 4cm + 4cm + 4cm
= 16cm.

Explanation:

Length of the AB side of the ABCD Square = 4cm
Length of the BC side of the ABCD Square = 4cm
Length of the CD side of the ABCD Square = 4cm
Length of the DA side of the ABCD Square = 4cm
Perimeter of the ABCD Square = Side + Side + Side + Side
= AB + BC +CD + DA
= 4cm + 4cm + 4cm + 4cm
= 8cm + 4cm + 4cm
= 12cm + 4cm
= 16cm.

c.

P = ____ cm + ____ cm + ____ cm
= _________ cm
Answer:
Perimeter of the ABC Triangle = Side + Side + Side
= AB + BC + CA
= 9cm + 11cm + 6cm
= 26cm.

Explanation:

Length of the AB side of the ABC Triangle =  9cm
Length of the BC side of the ABC Triangle = 11cm
Length of the CA side of the ABC Triangle = 6cm
Perimeter of the ABC Triangle = Side + Side + Side
= AB + BC + CA
= 9cm + 11cm + 6cm
= 20cm + 6cm
= 26cm.

d.

P = ____ m + ____ m + ____ m + ____ m
= _________ m
Answer:
Perimeter of the ABCD Trapezium = Side + Side + Side + Side
= AB + BC + CD + DA
= 15m + 9m + 5m + 7m
= 36m.

Explanation:

Length of the AB side of the ABCD Trapezium = 15cm
Length of the BC side of the ABCD Trapezium = 9cm
Length of the CD side of the ABCD Trapezium = 5cm
Length of the DA side of the ABCD Trapezium = 7cm
Perimeter of the ABCD Trapezium = Side + Side + Side + Side
= AB + BC + CD + DA
= 15m + 9m + 5m + 7m
= 24m + 5m + 7m
= 29m + 7m
= 36m.

e.

P = ____ in + ____ in + ____ in + ____ in + ____ in
= _________ in
Answer:
Perimeter of the ABCDE Pentagon = Side + Side + Side + Side + Side
= AB + BC + CD + DE + EA
= 9in + 2in + 2in + 9in + 3in
= 25in.

Explanation:

Length of the AB side of the ABCDE Pentagon = 9in
Length of the BC side of the ABCDE Pentagon = 2in
Length of the CD side of the ABCDE Pentagon = 2in
Length of the DE side of the ABCDE Pentagon = 9in
Length of the EA side of the ABCDE Pentagon = 3in
Perimeter of the ABCDE Pentagon = Side + Side + Side + Side + Side
= AB + BC + CD + DE + EA
= 9in + 2in + 2in + 9in + 3in
= 11in + 2in + 9in + 3in
= 13in + 9in + 3in
= 22in + 3in
= 25in.

Question 2.
Alan’s rectangular swimming pool is 10 meters long and 16 meters wide. What is the perimeter?

Answer:
Perimeter of the ABCD Rectangle = Side + Side + Side + Side
= AB + BC + CD + DA
= 16m + 10m + 16m + 10m
= 52m.

Explanation:

Length of the AB side of the ABCD Rectangle = 16m
Length of the BC side of the ABCD Rectangle = 10m
Length of the CA side of the ABCD Rectangle = 16m
Length of the DA side of the ABCD Rectangle = 10m
Perimeter of the ABCD Rectangle = Side + Side + Side + Side
= AB + BC + CD + DA
= 16m + 10m + 16m + 10m
= 26m + 16m + 10m
= 42m + 10m
= 52m.

Question 3.
Lila measures each side of the shape below.

a. What is the perimeter of the shape?
b. Lila says the shape is a pentagon. Is she correct? Explain why or why not.
Answer:
a. The perimeter of the ABCDE shape = Side + Side + Side + Side + Side
= AB + BC + CD + DE + EA
= 9in + 6in + 3in + 2in + 4in
= 24in.

b. Lila is correct because Pentagon is a figure which has five sides in it and her figure is a five sided shape.

Explanation:

a. The perimeter of the ABCDE shape = ???
Length of the AB side of the ABCDE shape = 9in
Length of the BC side of the ABCDE shape = 6in
Length of the CD side of the ABCDE shape = 3in
Length of the DE side of the ABCDE shape = 2in
Length of the EA side of the ABCDE shape = 4in
The perimeter of the ABCDE shape = Side + Side + Side + Side + Side
= AB + BC + CD + DE + EA
= 9in + 6in + 3in + 2in + 4in
= 15in + 3in + 2in + 4in
= 18in + 2in + 4in
= 20in + 4in
= 24in.

b. Pentagon is a figure which has  five side in it. Lila is correct.

Eureka Math Grade 3 Module 7 Lesson 13 Exit Ticket Answer Key

Which shape below has the greater perimeter? Explain your answer.

Answer:
Perimeter of shape A = 14in
Perimeter of shape B = 15in
Perimeter of shape B is greater than the Perimeter of shape A because the measurement value of Perimeter of shape B is more than the the measurement value of Perimeter of shape A.

Explanation:

Perimeter of shape A = Side + Side + Side + Side + Side + Side + Side
= AB + BC + CD + DE + EF + FG + GA
= 2in + 2in + 2in + 2in + 2in + 2in + 2in
= 4in + 2in + 2in + 2in + 2in + 2in
= 6in + 2in + 2in + 2in + 2in
= 8in + 2in + 2in + 2in
= 10in + 2in + 2in
= 12in + 2in
= 14in.
Perimeter of shape B = Side + Side + Side + Side + Side
= HI + IJ + JK + KL +LH
= 4in + 2in + 2in + 4in + 3in
= 6in + 2in + 4in + 3in
= 8in +  4in + 3in
= 12in + 3in
= 15in.

Eureka Math Grade 3 Module 7 Lesson 13 Homework Answer Key

Question 1.
Find the perimeters of the shapes below. Include the units in your equations. Match the letter inside each shape to its perimeter to solve the riddle. The first one has been done for you.

What kind of meals do math teachers eat?

Answer:
Perimeter of q Triangle shape = 21in.
Perimeter of r Pentagon shape = 36ft.
Perimeter of s Parallelogram shape = 24cm.
Perimeter of a Trapezium shape = 28yd.
Perimeter of m rhombus shape = 16in.
Perimeter of e Rectangular shape = 26cm.
Perimeter of u quadrilateral shape = 20m.
Perimeter of l Pentagon shape = 15m.

Square meals kind of meals  math teachers eats.

Explanation:

Perimeter of q Triangle shape = Side + Side + Side
= 7in + 7in + 7in
= 14in + 7in
= 21in.

Perimeter of r Pentagon shape = Side + Side + Side + Side + Side
= 6ft + 9ft + 6ft + 6ft + 9ft
= 15ft + 6ft + 6ft + 9ft
= 21ft + 6ft + 9ft
= 27ft + 9ft
= 36ft.

Perimeter of s Parallelogram shape = Side + Side + Side + Side
= 7cm + 5cm + 7cm + 5cm
= 12cm + 7cm + 5cm
= 19cm + 5cm
= 24cm.

Perimeter of a Trapezium shape = Side + Side + Side + Side
= 9yd + 7yd + 5yd + 7yd
= 16yd + 5yd + 7yd
= 21yd +  7yd
= 28yd.

Perimeter of m rhombus shape = Side + Side + Side + Side
= 4in + 4in + 4in + 4in
= 8in  + 4in + 4in
= 12in + 4in
= 16in.

Perimeter of e Rectangular shape = Side + Side + Side + Side
= 8cm + 5cm + 8cm + 5cm
= 13cm + 8cm + 5cm
= 21cm + 5cm
= 26cm.

Perimeter of u quadrilateral shape = Side + Side + Side + Side
= 6m + 4m + 7m + 3m
= 10m + 7m + 3m
= 17m + 3m
= 20m.

Perimeter of l Pentagon shape= Side + Side + Side + Side + Side
= 4m + 2m + 2m + 4m + 3m
= 6m + 2m + 4m + 3m
= 8m + 4m + 3m
= 12m + 3m
= 15m.

Question 2.
Alicia’s rectangular garden is 33 feet long and 47 feet wide. What is the perimeter of Alicia’s garden?

Answer:
Perimeter of Alicia’s rectangular garden = 160ft.

Explanation:

Length of the ABCD Alicia’s rectangular garden = 33ft
Width of the ABCD Alicia’s rectangular garden = 47ft
Perimeter of ABCD Alicia’s rectangular garden = 2 (Length + Width)
= 2 ( 33ft + 47ft )
= 2 × 80ft
= 160ft.

Question 3.
Jaque’s measured the side lengths of the shape below.

a. Find the perimeter of Jaques’s shape.
b. Jaques says his shape is an octagon. Is he right? Why or why not?
Answer:
a. Perimeter of Jaques’s shape =  Side + Side + Side + Side + Side + Side + Side + Side
= AB + BC + CD + DE + EF + FG + GH +HA
= 7 in + 3 in + 4 in+ 5 in + 4 in + 3 in + 2 in +5 in
= 33in.

Explanation:
a.

Perimeter of Jaques’s shape =  Side + Side + Side + Side + Side + Side + Side + Side
= AB + BC + CD + DE + EF + FG + GH +HA
= 7 in + 3 in + 4 in+ 5 in + 4 in + 3 in + 2 in +5 in
= 10 in + 4 in + 5 in + 4 in + 3 in + 2 in +5 in
= 14 in + 5 in + 4 in + 3 in + 2 in +5 in
= 19 in + 4 in + 3 in + 2 in +5 in
= 23 in + 3 in + 2 in +5 in
= 26 in + 2 in+5 in
= 28 in + 5 in
= 33 in.

b.  Yes, he is correct. Jaques says his shape is an octagon because in geometry , an octagon is an eight-sided  polygon or 8-gon. His shape has eight sides.

Explanation:

His shape has eight sides. In geometry , an octagon is an eight-sided  polygon or 8-gon.

Engage NY Eureka Math 3rd Grade Module 1 Lesson 6 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 6 Problem Set Answer Key

Question 1.
Rick puts 15 tennis balls into cans. Each can holds 3 balls.
Circle groups of 3 to show the balls in each can.

Rick needs ____5___ cans.
___5___ × 3 = 15
15 ÷ 3 = __5____

Rick needs 5 cans,

Explanation:
Given Rick puts 15 tennis balls into cans and each can
holds 3 balls. Circled groups of 3 to show the balls in each can.
Rick needs 15 ÷ 3 = 5, 5 X 3 = 15,
Therefore Rick needs 5 cans.

Question 2.
Rick uses 15 tennis balls to make 5 equal groups.
Draw to show how many tennis balls are in each group.
There are ____3___ tennis balls in each group.
5 × __3____ = 15
15 ÷ 5 = ___3___

There are 3 tennis balls in each group.

Explanation:
Given Rick uses 15 tennis balls to make 5 equal groups.
Drawn circles to show 3 number of tennis balls are in
each group as 15 ÷ 5 = 3, 5 X 3 = 15,
Therefore, there are 3 tennis balls in each group.

Question 3.
Use an array to model Problem 1.
a. ___5___ × 3 = 15
15 ÷ 3 = ___5___
The number in the blanks represents
_____5 groups___________________.

Explanation:
Used an array to model Problem 1 as
5 X 3 = 15 or 15 ÷ 3 = 5,
The number in the blanks represents groups as
5 groups of balls.

b. 5 × __3____ = 15
15 ÷ 5 = __3____
The number in the blanks represents
____________3 balls in each group________.

The number in the blanks represents 3 balls in each group,

Explanation:
Rick uses 15 tennis balls to make 5 equal groups,
15 ÷ 5 = 3 or 5 X 3= 15, The number in the blanks
represents 3 balls in each group.

Question 4.
Deena makes 21 jars of tomato sauce. She puts 7 jars in
each box to sell at the market. How many boxes does Deena need?
21 ÷ 7 = __3____
___3___ × 7 = 21
What is the meaning of the unknown factor and quotient?
_________3______________

Deena needs 3 boxes,

Explanation:
Given Deena makes 21 jars of tomato sauce and she
puts 7 jars in each box to sell at the market.
So number of boxes Deena needs are 21 ÷ 7 = 3, (3 X 7 = 21)
Therefore, Deena needs 3 boxes.

Question 5.
The teacher gives the equation 4 × __3__ = 12. Charlie
finds the answer by writing and solving 12 ÷ 4 = __3__.
Explain why Charlie’s method works.

Charlie method is 12 ÷ 4 = 3,

Explanation:
Given The teacher gives the equation 4 × ____ = 12.
Charlie finds the answer by writing and solving as
12 ÷ 4 = 3,
because multiplication and division are closely related,
given that division is the inverse operation of multiplication.
When we divide, we look to separate into equal groups,
while multiplication involves joining equal groups.
If we divide this product by one of the factors,
we get the other factor as a result. So Charlie uses
division method to solve missing factor in the given
equation as 3. So the equation is 4 X 3 = 12.

Question 6.
The blanks in Problem 5 represent the size of the groups.
Draw an array to represent the equations.

Explanation:
Drawn an array that represents the equations
as  4 X 3 = 12.

Eureka Math Grade 3 Module 1 Lesson 6 Exit Ticket Answer Key

Cesar arranges 12 notecards into rows of 6 for his presentation.
Draw an array to represent the problem.
12 ÷ 6 = ___2_____
____2____ × 6 = 12
What do the unknown factor and quotient represent?
___2 notecards in each row____

Explanation:
Given Cesar arranges 12 notecards into rows of
6 for his presentation.
Drawn an array to represent the problem as
12 ÷ 6 = 2, 2 X 6 = 12, means 6 rows  and 2 columns
as shown above, The unknown factor and quotient 2
represents 2 notecards in each row.

Eureka Math Grade 3 Module 1 Lesson 6 Homework Answer Key

Question 1.
Mr. Hannigan puts 12 pencils into boxes. Each box holds 4 pencils.
Circle groups of 4 to show the pencils in each box.

Mr. Hannigan needs ___3____ boxes.
___3___ × 4 = 12
12 ÷ 4 = __3____

Mr. Hannigan needs 3 boxes,

Explanation:
Given Mr. Hannigan puts 12 pencils into boxes.
Each box holds 4 pencils.
Circled groups of 4 to show the pencils in each box
as shown above, So Mr. Hannigan needs 12 ÷ 4 = 3 boxes,
( 3 x 4 = 12).

Question 2.
Mr. Hannigan places 12 pencils into 3 equal groups.
Draw to show how many pencils are in each group.
There are ___4____ pencils in each group.
3 × ___4___ = 12
12 ÷ 3 = ___4___

There are 4 pencils in each group,

Explanation:
Given Mr. Hannigan places 12 pencils into 3 equal groups,
Drawn to show 4 number of pencils are in each group
as 12 ÷ 3 = 4, 3 X 4 = 12, Therefore 4 pencils are there in each group.

Question 3.
Use an array to model Problem 1.
a. ___3___ × 4 = 12
12 ÷ 4 = ___3___
The number in the blanks represents
___________3 groups______________.
a.

The number in the blanks represents 3 groups,

Explanation:
Used an array to model Problem 1 as 3 x 4 = 12,
or 12 ÷ 4 = 3, therefore the number in the blanks
represents 3 groups.

b. 3 × ___4___ = 12
12 ÷ 3 = __4____
The number in the blanks represents
___________4 pencils in each group_____________.

The number in the blanks represents 4 pencils in each group,

Explanation:
Given Mr. Hannigan places 12 pencils into 3 equal groups.
12 ÷ 3 = 4 or 3 X 4= 12, The number in the blanks
represents 4 pencils in each group.

Question 4.
Judy washes 24 dishes. She then dries and stacks the dishes
equally into 4 piles. How many dishes are in each pile?
24 ÷ 4 = ___6____
4 × ___6_____ = 24
What is the meaning of the unknown factor and quotient?
__________6 dishes are in each pile____________________

There are 6 dishes in each pile,

Explanation:
Given Judy washes 24 dishes and she then dries and
stacks the dishes equally into 4 piles.
So number of dishes in each pile are 24 ÷ 4 = 6, or
4 X 6 = 24, The meaning of the unknown factor and
quotient is 6 dishes are there in each pile.

Question 5.
Nate solves the equation _____ × 5 = 15 by writing and
solving 15 ÷ 5 = ____. Explain why Nate’s method works.

Nate solves the equation as 3 X 5 = 15,

Explanation:
Given Nate solves the equation _____ × 5 = 15 by writing and
solving as 15 ÷ 5 = 3, Nate’s method is correct because in
the given equation ___ X 5 = 15 , we bring 5 to other side it becomes as
15 ÷ 5 = 3 now we check as 3 X 5 it becomes 15 only,
So Nate’s method work.

Question 6.
The blanks in Problem 5 represent the number of groups.
Draw an array to represent the equations.

Explanation:
Drawn an array to represent the equations as
3 X 5 = 15 or 15 ÷ 5 = 3 as shown in the picture above.

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

Eureka Math Grade 3 Module 1 Answer Key

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

Question 1.
Fill in the blanks to make true statements.

a. 3 groups of five = ___15______
3 fives = ____15_____
3 × 5 = ___15______

3 groups of five = 15,
3 fives = 15,
3 X 5 = 15,

Explanation:
Given expressions as
3 groups of five means 3 X 5 = 15,
3 fives = 3 x 5 = 15 and
3 X 5 = 15 as 3 is multiplied by 5 we get 15.

b. 3 + 3 + 3 + 3 + 3 = __15___
5 groups of three = ___15______
5 × 3 = ____15_____

3 + 3 + 3 + 3 + 3 = 15,
5 groups of three = 15,
5 × 3 = 15,

Explanation:
Given expressions as
3 + 3 + 3 + 3 + 3 = 15 as 3 is added 5 times,
5 groups of three = 15 means 5  X 3 = 15,
5 X 3 = 15 as 5 is multiplied by 3 we get 15.

c. 6 + 6 + 6 + 6 = ___24________
____4___ groups of six = ___24_______
4 × __6____ = ____24______

6 + 6 + 6 + 6 = 24,
4 groups of six = 24,
4 X 6 = 24,

Explanation:
Given expressions as
6 + 6 + 6 + 6 = 24 as 6 is added 4 times,
4 groups of six = 24 means 4  X 6 = 24,
4 X 6 = 24 as 4 is multiplied by 6 we get 24.

d. 4 +__4__ + __4__ + __4__ + _4___ + __4__ = ___24______
6 groups of ____4____ = ____24_______
6 × __4____ = ____24______

4 + 4 + 4 + 4 + 4 + 4 = 24,
6 groups of 4 = 24,
6 x 4 = 24,

Explanation:
Given expressions as
4 + ___ + ___ +___ +___ +___ =  means by seeing the picture
4 is added 6 times as 4 + 4 + 4 + 4 + 4 + 4 we get 24,
6 groups of ______ =  means again by seeing the picture
6 X 4 we get 24, So 6 groups of 4 = 24,
6 X 4 = 24 as 6 is multiplied by 4 we get 24.

Question 2.
The picture below shows 2 groups of apples. Does the picture
show 2 × 3? Explain why or why not.

No, the picture does not show 2 X 3,

Explanation:
Given the picture below shows 2 groups of apples but
in the 2 groups the number of apples are not the same
as in group -1 we have 3 apples, group 2 we have 2 apples
in total there are 5 apples and 2 X 3 = 6 whose value is not
same as shown in the picture ,
So, the picture does not show 2 X 3.

Question 3.
Draw a picture to show 2 × 3 = 6.

Explanation:
Drawn a picture above to show 2 X 3 = 6 .

Question 4.
Caroline, Brian, and Marta share a box of chocolates.
They each get the same amount. Circle the chocolates below
to show 3 groups of 4. Then, write a repeated addition sentence
and a multiplication sentence to represent the picture.

Addition sentence is 4 + 4 + 4 = 12,
Multiplication sentence is 3 X 4 = 12,

Explanation:
Given Caroline, Brian, and Marta share a box of chocolates and
they each get the same amount as 12 ÷ 3 = 4 each.
Circled the chocolates above to show 3 groups of 4 and
wrote a repeated addition sentence as  4 + 4 + 4 = 12 and
a multiplication sentence to represent the picture as 3 X 4 = 12.

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

Question 1.
The picture below shows 4 groups of 2 slices of watermelon.
Fill in the blanks to make true repeated addition and
multiplication sentences that represent the picture.

True repeated addition is 2 + 2 + 2 + 2 = 8,
Multiplication sentences is 4 X 2 = 8.

Explanation:
Given the picture above shows 4 groups of
2 slices of watermelon.
Filled in the blanks to make true repeated addition
as 2 + 2 + 2 + 2 = 8 and multiplication sentences that
represent the picture as 4 X 2 = 8.

Question 2.
Draw a picture to show 3 + 3 + 3 = 9. Then,
write a multiplication sentence to represent the picture.

Addition sentence is  3 + 3 + 3 = 9,
Multiplication sentences is 3 X 3 = 9.

Explanation:
Drawn a picture above as  3 groups of 3 birds  to show
3 + 3 + 3 = 9 and writing a multiplication sentence to
represent the picture as 3 X 3 = 9 birds in total.

Eureka Math Grade 3 Module 1 Lesson 1 Homework Answer Key

Question 1.
Fill in the blanks to make true statements.

a. 4 groups of five = ____20_____
4 fives = ___20______
4 × 5 = ____20_____

4 groups of five = 20,
4 fives = 20,
4 X 5 = 20,

Explanation:
Given expressions as
4 groups of five means 4 X 5 = 20,
4 fives = 4 x 5 = 20 and
4 X 5 = 20 as 4 is multiplied by 5 we get 20.

b. 5 groups of four = _________
5 fours = _________
5 × 4 = _________

5 groups of four = 20,
5 fours = 20,
5 X 4 = 20,

Explanation:
Given expressions as
5 groups of four means 5 X 4 = 20,
5 fours = 5 x 4 = 20 and
5 X 4 = 20 as 5 is multiplied by 4 we get 20.

c. 6 + 6 + 6 = ___18________
____3___ groups of six = ____18______
3 × __6____ = ____18______

6 + 6 + 6  = 18,
3 groups of 6 = 18,
3 x 6 = 18,

Explanation:
Given expressions as 6 + 6 + 6 = 18 as
6 is added 3 times we get 18,
_____ groups of six =  means again by seeing the picture
3 groups of 6 =  3 X 6 we get 18, So 3 groups of 6 = 18,
3 X 6 = 18 as 3 is multiplied by 6 we get 18.

d. 3 + __3__ + __3__ + __3__ + __3__ + __3__ = ___18____
6 groups of ____3____ = _____18______
6 × __3__ = ____18______

3 + 3 + 3 + 3 + 3 + 3 = 18,
6 groups of 3 = 18,
6 x 3 = 18,

Explanation:
Given expressions as
3 + ___ + ___ +___ +___ +___ =  means by seeing the picture
3 is added 6 times as 3 + 3 + 3 + 3 + 3 + 3 we get 18,
6 groups of ______ =  means again by seeing the picture
6 X 3 we get 18, So 6 groups of 3 = 18,
6 X 3 = 18 as 6 is multiplied by 3 we get 18.

Question 2.
The picture below shows 3 groups of hot dogs.
Does the picture show 3 × 3? Explain why or why not.

Yes, the picture shows 3 X 3,

Explanation:
Given the picture below shows 3 groups of of hot dogs,
In the 3 groups the number of hot dogs are the same as 3,
So in total we have 3 X 3 = 9 hotdogs in the picture
which matches with the given equation as 3 X 3 whose
value is also same 9, So, the picture shows 3 X 3.

Question 3.
Draw a picture to show 4 × 2 = 8,

Explanation:
Drawn picture of flowers as shown above to show 4 X 2 = 8.

Question 4.
Circle the pencils below to show 3 groups of 6.
Write a repeated addition and a multiplication sentence
to represent the picture.


Repeated addition sentence is 6 + 6 + 6 = 18,
Multiplication sentence is 3 X 6 = 18,

Explanation:
Circled the pencils above to show 3 groups of 6.
Wrote a repeated addition as 6 + 6 + 6 = 18 and
a multiplication sentence as 3 X 6 = 18 to represent the picture.

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.

multiply by 7 (6–10)
Answer:

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.

Perimeter = _____cm +_____cm +_____cm +_____cm
= _______ cm
Answer:
Perimeter of the ABCD given shape = Side + Side + Side + Side
= 2cm + 2cm + 2cm + 2cm
= 8cm.

Explanation:

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.

Perimeter = _____________________
= _______ cm
Answer:
Perimeter of the given ABCDEF Hexagon shape = Side + Side + Side + Side + Side + Side
= 3cm + 3cm + 3cm + 3cm + 3cm + 3cm
= 18cm.

Explanation:

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.

Perimeter = _____________________
= _______ cm
Answer:
Perimeter of the given ABCD Parallelogram shape = Side + Side + Side + Side
= 4cm + 4cm + 4cm + 4cm
= 16cm.

Explanation:

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.

Perimeter = _____________________
= _______ cm
Answer:
Perimeter of the ABC Triangle = Side + Side + Side
= 5cm + 5cm + 5cm
= 15cm.

Explanation:

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.

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:

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.

Answer:
Perimeter of ABCD Carson’s shape = Side + Side + Side + Side
= AB + BC + CD+ DA
= 2cm + 2cm + 2cm + 2cm
=8cm.

Explanation:

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?

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:

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.

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:

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.

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:

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.

Perimeter = _____cm +_____cm +_____cm
= _______ cm
Answer:
Perimeter of ABC triangle = Side + Side + Side
= AB + BC + CA
= 3cm + 5cm + 4cm
= 12cm.

Explanation:

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.

Perimeter = _____________________
= _______ cm
Answer:
Perimeter of ABCD rectangle = Side + Side + Side + Side
= AB + BC + CD + DA
= 6cm + 4cm + 6cm + 4cm
= 20cm.

Explanation:

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.

Perimeter = _____________________
= _______ cm
Answer:
Perimeter of the ABCD Quadrilateral = Side + Side + Side + Side
= AB + BC + CD + DA
= 3cm + 4cm + 5cm + 4cm
= 16cm.

Explanation:

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.

Perimeter = _____________________
= _______ cm
Answer:
Perimeter of the ABCD Parallelogram = Side + Side + Side + Side
= AB + BC + CD + DA
= 5cm + 5cm + 5cm + 5cm
= 20cm.

Explanation:

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.

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:

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.

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:

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.

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:

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?

Answer:
NO, I disagree with Jamal. Jamal answer is incorrect because actual Perimeter of his ABCD rectangle is 30cm not 32cm.

Explanation:

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.

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.

Divide 6 tomatoes into groups of 3.
There are ____2_____ groups of 3 tomatoes.
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.

Divide 8 lollipops into groups of 2.
There are ____4___ groups.
8 ÷ 2 = ___4____

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.

Divide 10 stars into groups of 5.
10 ÷ 5 = ___2____

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.

Divide the shells to show 12 ÷ 3 = ____4____,
where the unknown represents the number of groups.
How many groups are there? ___4_____

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.

a. Write a division sentence where the answer
represents the number of Rachel’s bags.
b. Draw a number bond to represent the problem.

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.

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.

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.

12 ÷ 6 = ___2____

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.

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.

Divide 4 triangles into groups of 2.
There are _____2____ groups of 2 triangles.
4 ÷ 2 = 2

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.

Divide 9 eggs into groups of 3.
There are ___3____ groups.
9 ÷ 3 = __3_____

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.

Divide 12 buckets of paint into groups of 3.
12 ÷ 3 = __4_____

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.

Group the squares to show 15 ÷ 5 = __3___,
where the unknown represents the number of groups.
How many groups are there? ____3____

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.

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.

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.

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.

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.

Engage NY Eureka Math 3rd Grade Module 7 Lesson 11 Answer Key

Eureka Math Grade 3 Module 7 Lesson 11 Problem Set Answer Key

Question 1.
Follow the directions below using the shape you created yesterday.
a. Tessellate your shape on a blank piece of paper.
b. Color your tessellation to create a pattern.
c. Outline the perimeter of your tessellation with a highlighter.
d. Use a string to measure the perimeter of your tessellation.
Answer:
a.

b.

c.

d. A string is used to measure the perimeter of the figure.

Explanation:
The figure asked to draw has been drawn considering all the given points to follow while drawing it.

Question 2.
Compare the perimeter of your tessellation to a partner’s. Whose tessellation has a greater perimeter? How do you know?
Answer:
My tessellation has the greater perimeter than my friend’s tessellation perimeter.

Explanation:
My tessellation has the greater perimeter than my friend’s tessellation perimeter because I have compared both the tessellation perimeter with the string. My tessellation string was down that compared to my friend’s tessellation.

Question 3.
How could you increase the perimeter of your tessellation?
Answer:
I could increase the perimeter of my tessellation by tessellating it by adding some more rows to my shape.

Explanation:
I could increase the perimeter of my tessellation by tessellating  more shapes. If I wish to tessellating my figure   I would tesselate by adding some more rows to my shape.

Question 4.
How would overlapping your shape when you tessellated change the perimeter of your tessellation?
Answer:
If I overlap my figure, it wouldn’t  fit together I think. If I overlap my figure, I think the perimeter will decrease of the figure.

Explanation:
Overlapping any figure means covering the figure with same sized figure or different sized figure. If I overlap my figure, I think the perimeter will decrease of my figure that of the overlapped figure.

Eureka Math Grade 3 Module 7 Lesson 11 Exit Ticket Answer Key

Estimate to draw at least four copies of the given regular hexagon to make a new shape, without gaps or overlaps. Outline the perimeter of your new shape with a highlighter. Shade in the area with a colored pencil.

Answer:

Explanation:
The area of the new shape is shaded with Yellow color.
The perimeter of the new shape is highlighted with Red color.

Eureka Math Grade 3 Module 7 Lesson 11 Homework Answer Key

Question 1.
Samson tessellates regular hexagons to make the shape below.

a. Outline the perimeter of Samson’s new shape with a highlighter.
Answer:

Explanation:
Perimeter is the distance around the outside of a shape. Perimeter of the given figure is highlighted in red highlighter.

b. Explain how Samson could use a string to measure the perimeter of his new shape.
Answer:
Samson could use a string to measure the perimeter of his new shape by placing it all-around the outer layer of the new shape.

Explanation:

Perimeter is the distance around the outside of a shape.

c. How many sides does his new shape have?
Answer:
The new shape has 18 or eighteen sides in it.

Explanation:

The sides of the new shape are counted one by one.

d. Shade in the area of his new shape with a colored pencil.
Answer:

Explanation:
The new shape area has be shaded with light blue colored pencil in it.

Question 2.
Estimate to draw at least four copies of the given triangle to make a new shape, without gaps or overlaps. Outline the perimeter of your new shape with a highlighter. Shade in the area with a colored pencil.

Answer:

Explanation:
The Area of the new shape is shaded with light grey colored pencil in it.
The Perimeter of the new shape is highlighted with red color.

Question 3.
The marks on the strings below show the perimeters of Shyla’s and Frank’s shapes. Whose shape has a greater perimeter? How do you know?

Answer:
The perimeter of Frank’s shape is having greater perimeter than Shyla’s shape because the  string measurement of Frank’s shape is lower than the Shyla’s shape string measurement.

Explanation:

The string measurement of Frank’s shape has further crossed  the string measurement of Shyla’s shape.

Question 4.
India and Theo use the same shape to create the tessellations shown below.

a. Estimate to draw the shape India and Theo used to make their tessellations.
Answer:

Explanation:
The shape India and Theo used to make their tessellations is been draw in the figure highlighted in the yellow color.

b. Theo says both tessellations have the same perimeter. Do you think Theo is right? Why or why not?
Answer:
Yes, Theo says correct because the shape used to draw the tessellation of both India and Theo’s are the same.  The number of shapes used by both are the same in their respective own  tessellation yet the position differs one is in vertical presentation and horizontal presentation .

Explanation:

The number of shapes used by both are the same in their respective own  tessellation yet the position differs one is in vertical presentation and horizontal presentation .

Engage NY Eureka Math 3rd Grade Module 1 Lesson 14 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 14 Sprint Answer Key

A
Multiply or Divide by 3

Question 1.
2 × 3 =
2 X 3 = 6,

Explanation:
Given 2 X 3 we multiply 2 with 3,
we get 6 as 2 X 3 = 6.

Question 2.
3 × 3 =
3 X 3 = 9,

Explanation:
Given 3 X 3 we multiply 3 with 3,
we get 9 as 3 X 3 = 9.

Question 3.
4 × 3 =
4 X 3 = 12,

Explanation:
Given 4 X 3 we multiply 4 with 3,
we get 12 as 4 X 3 = 12.

Question 4.
5 × 3 =
5 X 3 = 15,

Explanation:
Given 5 X 3 we multiply 5 with 3,
we get 15 as 5 X 3 = 15.

Question 5.
1 × 3 =
1 X 3 = 6,

Explanation:
Given 1 X 3 we multiply 1 with 3,
we get 3 as 1 X 3 = 3.

Question 6.
6 ÷ 3 =
6 ÷ 3 = 2,

Explanation:
Given 6 ÷ 3 we divide 6 by 3,
we get 2 as 6 ÷ 3 = 2.

Question 7.
9 ÷ 3 =
9 ÷ 3 = 3,

Explanation:
Given 9 ÷ 3 we divide 9 by 3,
we get 3 as 9 ÷ 3 = 3.

Question 8.
15 ÷ 3 =
15 ÷ 3 = 5,

Explanation:
Given 15 ÷ 3 we divide 15 by 3,
we get 5 as 15 ÷ 3 = 5.

Question 9.
3 ÷ 1 =
3 ÷ 1 = 3,

Explanation:
Given 3 ÷ 1 we divide 3 by 1,
we get 3 as 3 ÷ 1 = 3.

Question 10.
12 ÷ 3 =
12 ÷ 3 = 4,

Explanation:
Given 12 ÷ 3 we divide 12 by 3,
we get 4 as 12 ÷ 3 = 4.

Question 11.
6 × 3 =
6 X 3 = 18,

Explanation:
Given 6 X 3 we multiply 6 with 3,
we get 18 as 6 X 3 = 18.

Question 12.
7 × 3 =
7 X 3 = 21,

Explanation:
Given 7 X 3 we multiply 7 with 3,
we get 21 as 7 X 3 = 21.

Question 13.
8 × 3 =
8 X 3 = 24,

Explanation:
Given 8 X 3 we multiply 8 with 3,
we get 24 as 8 X 3 = 24.

Question 14.
9 × 3 =
9 X 3 = 27,

Explanation:
Given 9 X 3 we multiply 9 with 3,
we get 27 as 9 X 3 = 27.

Question 15.
10 × 3 =
10 X 3 = 30,

Explanation:
Given 10 X 3 we multiply 10 with 3,
we get 30 as 10 X 3 = 30.

Question 16.
24 ÷ 3 =
24 ÷ 3 = 8,

Explanation:
Given 24 ÷ 3 we divide 24 by 3,
we get 8 as 24 ÷ 3 = 8.

Question 17.
21 ÷ 3 =
21 ÷ 3 = 7,

Explanation:
Given 21 ÷ 3 we divide 21 by 3,
we get 7 as 27 ÷ 3 = 7.

Question 18.
27 ÷ 3 =
27 ÷ 3 = 9,

Explanation:
Given 27 ÷ 3 we divide 27 by 3,
we get 9 as 27 ÷ 3 = 9.

Question 19.
18 ÷ 3 =
18 ÷ 3 = 6,

Explanation:
Given 18 ÷ 3 we divide 18 by 3,
we get 6 as 18 ÷ 3 = 6.

Question 20.
30 ÷ 3 =
30 ÷ 3 = 10,

Explanation:
Given 30 ÷ 3 we divide 30 by 3,
we get 10 as 30 ÷ 3 = 10.

Question 21.
__ × 3 = 15
5 X 3 = 15,

Explanation:
Given __ X 3 = 15, Let us take missing number
as x, So x X  3 = 15, means x = 15 ÷ 3 = 5,
therefore 5 X 3 = 15.

Question 22.
__ × 3 = 12
4 X 3 = 12,
Explanation:
Given __ X 3 = 12, Let us take missing number
as x, So x X  3 = 12, means x = 12 ÷ 3 = 4,
therefore 4 X 3 = 12.

Question 23.
__ × 3 = 30
10 X 3 = 30,

Explanation:
Given __ X 3 = 30, Let us take missing number
as x, So x X  3 = 30, means x = 30 ÷ 3 = 10,
therefore 10 X 3 = 30.

Question 24.
__ × 3 = 6
2 X 3 = 6,

Explanation:
Given __ X 3 = 6, Let us take missing number
as x, So x X  3 = 6, means x = 6 ÷ 3 = 2,
therefore 2 X 3 = 6.

Question 25.
__ × 3 = 9
3 X 3 = 9,

Explanation:
Given __ X 3 = 9, Let us take missing number
as x, So x X  3 = 9, means x = 9 ÷ 3 = 3,
therefore 3 X 3 = 9.

Question 26.
30 ÷ 3 =
30 ÷ 3 = 10,

Explanation:
Given 30 ÷ 3 we divide 30 by 3,
we get 10 as 30 ÷ 3 = 10.

Question 27.
15 ÷ 3 =
15 ÷ 3 = 5,

Explanation:
Given 15 ÷ 3 we divide 15 by 3,
we get 5 as 15 ÷ 3 = 5.

Question 28.
3 ÷ 1 =
3 ÷ 1 = 3,

Explanation:
Given 3 ÷ 1 we divide 3 by 1,
we get 3 as 3 ÷ 1 = 3.

Question 29.
6 ÷ 3 =
6 ÷ 3 = 2,

Explanation:
Given 6 ÷ 3 we divide 6 by 3,
we get 2 as 6 ÷ 3 = 2.

Question 30.
9 ÷ 3 =
9 ÷ 3 = 3,

Explanation:
Given 9 ÷ 3 we divide 9 by 3,
we get 3 as 9 ÷ 3 = 3.

Question 31.
__ × 3 = 18
6 X 3 = 18,

Explanation:
Given __ X 3 = 18, Let us take missing number
as x, So x X  3 = 18, means x = 18 ÷ 3 = 6,
therefore 6 X 3 = 18.

Question 32.
__ × 3 = 21
7 X 3 = 21,

Explanation:
Given __ X 3 = 21, Let us take missing number
as x, So x X  3 = 21, means x = 21 ÷ 3 = 7,
therefore 7 X 3 = 21.

Question 33.
__ × 3 = 27
9 X 3 = 27,

Explanation:
Given __ X 3 = 27, Let us take missing number
as x, So x X  3 = 27, means x = 27 ÷ 3 = 9,
therefore 9 X 3 = 27.

Question 34.
__ × 3 = 24
8 X 3 = 24,

Explanation:
Given __ X 3 = 24, Let us take missing number
as x, So x X  3 = 24, means x = 24 ÷ 3 = 8,
therefore 8 X 3 = 24.

Question 35.
21 ÷ 3 =
21 ÷ 3 = 7,

Explanation:
Given 21 ÷ 3 we divide 21 by 3,
we get 7 as 21 ÷ 3 = 7.

Question 36.
27 ÷ 3 =
27 ÷ 3 = 9,

Explanation:
Given 27 ÷ 3 we divide 27 by 3,
we get 9 as 27 ÷ 3 = 9.

Question 37.
18 ÷ 3 =
18 ÷ 3 = 6,

Explanation:
Given 18 ÷ 3 we divide 18 by 3,
we get 6 as 18 ÷ 3 = 6.

Question 38.
24 ÷ 3 =
24 ÷ 3 = 8,

Explanation:
Given 24 ÷ 3 we divide 24 by 3,
we get 8 as 24 ÷ 3 = 8.

Question 39.
11 × 3 =
11 X 3 = 33,

Explanation:
Given 11 X 3 we multiply 11 with 3,
we get 33 as 11 X 3 = 33.

Question 40.
33 ÷ 3 =
33 ÷ 3 = 11,

Explanation:
Given 33 ÷ 3 we divide 33 by 3,
we get 11 as 33 ÷ 3 = 11.

Question 41.
12 × 3 =
12 X 3 = 36,

Explanation:
Given 12 X 3 we multiply 12 with 3,
we get 36 as 12 X 3 = 36.

Question 42.
36 ÷ 3 =
36 ÷ 3 = 12,

Explanation:
Given 36 ÷ 3 we divide 36 by 3,
we get 12 as 36 ÷ 3 = 12.

Question 43.
13 × 3 =
13 X 3 = 39,

Explanation:
Given 13 X 3 we multiply 13 with 3,
we get 39 as 13 X 3 = 39.

Question 44.
39 ÷ 3 =
39 ÷ 3 = 13,

Explanation:
Given 39 ÷ 3 we divide 39 by 3,
we get 13 as 39 ÷ 3 = 13.

B
Multiply or Divide by 3

Question 1.
1 × 3 =
1 X 3 = 3,

Explanation:
Given 1 X 3 we multiply 1 with 3,
we get 3 as 1 X 3 = 3.

Question 2.
2 × 3 =
2 X 3 = 6,

Explanation:
Given 2 X 3 we multiply 2 with 3,
we get 6 as 2 X 3 = 6.

Question 3.
3 × 3 =
3 X 3 = 9,

Explanation:
Given 3 X 3 we multiply 3 with 3,
we get 9 as 3 X 3 = 9.

Question 4.
4 × 3 =
4 X 3 = 12,

Explanation:
Given 4 X 3 we multiply 4 with 3,
we get 12 as 4 X 3 = 12.

Question 5.
5 × 3 =
5 X 3 = 15,

Explanation:
Given 5 X 3 we multiply 5 with 3,
we get 15 as 5 X 3 = 15.

Question 6.
9 ÷ 3 =
9 ÷ 3 = 3,

Explanation:
Given 9 ÷ 3 we divide 9 by 3,
we get 3 as 9 ÷ 3 = 3.

Question 7.
6 ÷ 3 =
6 ÷ 3 = 2,

Explanation:
Given 6 ÷ 3 we divide 6 by 3,
we get 2 as 6 ÷ 3 = 2.

Question 8.
12 ÷ 3 =
12 ÷ 3 = 4,

Explanation:
Given 12 ÷ 3 we divide 12 by 3,
we get 4 as 12 ÷ 3 = 4.

Question 9.
3 ÷ 1 =
3 ÷ 1 = 3,

Explanation:
Given 3 ÷ 1 we divide 3 by 1,
we get 3 as 3 ÷ 1 = 3.

Question 10.
15 ÷ 3 =
15 ÷ 3 = 5,

Explanation:
Given 15 ÷ 3 we divide 15 by 3,
we get 5 as 15 ÷ 3 = 5.

Question 11.
10 × 3 =
10 X 3 = 30,

Explanation:
Given 10 X 3 we multiply 10 with 3,
we get 30 as 10 X 3 = 30.

Question 12.
6 × 3 =
6 X 3 = 18,

Explanation:
Given 6 X 3 we multiply 6 with 3,
we get 18 as 6 X 3 = 18.

Question 13.
7 × 3 =
7 X 3 = 21,

Explanation:
Given 7 X 3 we multiply 7 with 3,
we get 21 as 7 X 3 = 21.

Question 14.
8 × 3 =
8 X 3 = 24,

Explanation:
Given 8 X 3 we multiply 8 with 3,
we get 24 as 8 X 3 = 24.

Question 15.
9 × 3 =
9 X 3 = 27,

Explanation:
Given 9 X 3 we multiply 9 with 3,
we get 27 as 9 X 3 = 27.

Question 16.
21 ÷ 3 =
21 ÷ 3 = 7,

Explanation:
Given 21 ÷ 3 we divide 21 by 3,
we get 7 as 21 ÷ 3 = 7.

Question 17.
18 ÷ 3 =
18 ÷ 3 = 6,

Explanation:
Given 18 ÷ 3 we divide 18 by 3,
we get 6 as 18 ÷ 3 = 6.

Question 18.
24 ÷ 3 =
24 ÷ 3 = 8,

Explanation:
Given 24 ÷ 3 we divide 24 by 3,
we get 8 as 24 ÷ 3 = 8.

Question 19.
30 ÷ 3 =
30 ÷ 3 = 10,

Explanation:
Given 30 ÷ 3 we divide 30 by 3,
we get 10 as 30 ÷ 3 = 10.

Question 20.
27 ÷ 3 =
27 ÷ 3 = 9,

Explanation:
Given 27 ÷ 3 we divide 27 by 3,
we get 9 as 27 ÷ 3 = 9.

Question 21.
__ × 3 = 12
4 X 3 = 12,

Explanation:
Given __ X 3 = 12, Let us take missing number
as x, So x X  3 = 12, means x = 12 ÷ 3 = 4,
therefore 4 X 3 = 12.

Question 22.
__ × 3 = 15
5 X 3 = 15,

Explanation:
Given __ X 3 = 15, Let us take missing number
as x, So x X  3 = 15, means x = 15 ÷ 3 = 5,
therefore 5 X 3 = 15.

Question 23.
__ × 3 = 6
2 X 3 = 6,

Explanation:
Given __ X 3 = 6, Let us take missing number
as x, So x X  3 = 6, means x = 6 ÷ 3 = 2,
therefore 2 X 3 = 6.

Question 24.
__ × 3 = 30
10 X 3 = 30,

Explanation:
Given __ X 3 = 30, Let us take missing number
as x, So x X  3 = 30, means x = 30 ÷ 3 = 10,
therefore 10 X 3 = 30.

Question 25.
__ × 3 = 9
3 X 3 = 9,

Explanation:
Given __ X 3 = 9, Let us take missing number
as x, So x X  3 = 9, means x = 9 ÷ 3 = 3,
therefore 3 X 3 = 9.

Question 26.
6 ÷ 3 =
6 ÷ 3 = 2,

Explanation:
Given 6 ÷ 3 we divide 6 by 3,
we get 2 as 6 ÷ 3 = 2.

Question 27.
3 ÷ 1 =
3 ÷ 1 = 3,

Explanation:
Given 3 ÷ 1 we divide 3 by 1,
we get 3 as 3 ÷ 1 = 3.

Question 28.
30 ÷ 3 =
30 ÷ 3 = 10,

Explanation:
Given 30 ÷ 3 we divide 30 by 3,
we get 10 as 30 ÷ 3 = 10.

Question 29.
15 ÷ 3 =
15 ÷ 3 = 5,

Explanation:
Given 15 ÷ 3 we divide 15 by 3,
we get 5 as 15 ÷ 3 = 5.

Question 30.
9 ÷ 3 =
9 ÷ 3 = 3,

Explanation:
Given 9 ÷ 3 we divide 9 by 3,
we get 3 as 9 ÷ 3 = 3.

Question 31.
__ × 3 = 18
6 X 3 = 18,

Explanation:
Given __ X 3 = 18, Let us take missing number
as x, So x X  3 = 18, means x = 18 ÷ 3 = 6,
therefore 6 X 3 = 18.

Question 32.
__ × 3 = 24
8 X 3 = 24,

Explanation:
Given __ X 3 = 24, Let us take missing number
as x, So x X  3 = 24, means x = 24 ÷ 3 = 8,
therefore 8 X 3 = 24.

Question 33.
__ × 3 = 27
9 X 3 = 27,

Explanation:
Given __ X 3 = 27, Let us take missing number
as x, So x X  3 = 27, means x = 27 ÷ 3 = 9,
therefore 9 X 3 = 27.

Question 34.
__ × 3 = 21
7 X 3 = 21,

Explanation:
Given __ X 3 = 21, Let us take missing number
as x, So x X  3 = 21, means x = 21 ÷ 3 = 7,
therefore 7 X 3 = 21.

Question 35.
24 ÷ 3 =
24 ÷ 3 = 8,

Explanation:
Given 24 ÷ 3 we divide 24 by 3,
we get 8 as 24 ÷ 3 = 8.

Question 36.
27 ÷ 3 =
27 ÷ 3 = 9,

Explanation:
Given 27 ÷ 3 we divide 27 by 3,
we get 9 as 27 ÷ 3 = 9.

Question 37.
18 ÷ 3 =
18 ÷ 3 = 6,

Explanation:
Given 18 ÷ 3 we divide 18 by 3,
we get 6 as 18 ÷ 3 = 6.

Question 38.
21 ÷ 3 =
21 ÷ 3 = 7,

Explanation:
Given 21 ÷ 3 we divide 21 by 3,
we get 7 as 21 ÷ 3 = 7.

Question 39.
11 × 3 =
11 X 3 = 33,

Explanation:
Given 11 X 3 we multiply 11 with 3,
we get 33 as 11 X 3 = 33.

Question 40.
33 ÷ 3 =
33 ÷ 3 = 11,

Explanation:
Given 33 ÷ 3 we divide 33 by 3,
we get 11 as 33 ÷ 3 = 11.

Question 41.
12 × 3 =
12 X 3 = 36,

Explanation:
Given 12 X 3 we multiply 12 with 3,
we get 36 as 12 X 3 = 36.

Question 42.
36 ÷ 3 =
36 ÷ 3 = 12,

Explanation:
Given 36 ÷ 3 we divide 36 by 3,
we get 12 as 36 ÷ 3 = 12.

Question 43.
13 × 3 =
13 X 3 = 39,

Explanation:
Given 13 X 3 we multiply 13 with 3,
we get 39 as 13 X 3 = 39.

Question 44.
39 ÷ 3 =
39 ÷ 3 = 13,

Explanation:
Given 39 ÷ 3 we divide 39 by 3,
we get 13 as 39 ÷ 3 = 13.

Eureka Math Grade 3 Module 1 Lesson 14 Problem Set Answer Key

Question 1.
Skip-count by fours. Match each answer to the appropriate expression.

Skipped-counts by fours.
Matched each answer to the appropriate expression as
4 = 1 X 4,
8 = 2 X 4,
12 = 3 X 4,
16 = 4 X 4,
20 = 5 X 4,
24 = 6 X 4,
28 = 7 X 4,
32 = 8 X 4,
36 = 9 X 4,
40 = 10 X 4.

Question 2.
Mr. Schmidt replaces each of the 4 wheels on 7 cars.
How many wheels does he replace? Draw and label a tape diagram to solve.
Mr. Schmidt replaces _____28______ wheels.

Mr. Schmidt replaces 28 wheels on 7 cars,

Explanation:
Given Mr. Schmidt replaces each of the 4 wheels on 7 cars.
Drawn and labeled a tape diagram to solve how many wheels
he replaced as shown above, So number of wheels replaced are
7 X 4 = 28, therefore, Mr. Schmidt replaces _____28______ wheels.

Question 3.
Trina makes 4 bracelets. Each bracelet has 6 beads.
Draw and label a tape diagram to show the
total number of beads Trina uses.

Trina uses 24 beads to make 4 bracelets,

Explanation:
Given Trina makes 4 bracelets. Each bracelet has 6 beads.
Draw and label a tape diagram to show the
total number of beads Trina uses as 4 X 6 =24 beads.

Question 4.
Find the total number of sides on 5 rectangles.

Total number of sides on 5 rectangles are 20 side,

Explanation:
We know a rectangle has 4 sides, So number of sides on
5 rectangles are 5 X 4 = 20 sides.

Eureka Math Grade 3 Module 1 Lesson 14 Exit Ticket Answer Key

Arthur has 4 boxes of chocolates. Each box has 6 chocolates inside. How many chocolates does Arthur have altogether? Draw and label a tape diagram to solve.

Eureka Math Grade 3 Module 1 Lesson 14 Homework Answer Key

Question 1.
Skip-count by fours. Match each answer to the appropriate expression.

Skipped-counts by fours.
Matched each answer to the appropriate expression as
4 = 1 X 4,
8 = 2 X 4,
12 = 3 X 4,
16 = 4 X 4,
20 = 5 X 4,
24 = 6 X 4,
28 = 7 X 4,
32 = 8 X 4,
36 = 9 X 4,
40 = 10 X 4.

Question 2.
Lisa places 5 rows of 4 juice boxes in the refrigerator.
Draw an array and skip-count to find the total number of juice boxes.
There are ____20_______ juice boxes in total.

The total number of juice boxes are 20,

Explanation:
Given Lisa places 5 rows of 4 juice boxes in the refrigerator.
Drawn an array and skipped-count for finding the total
number of juice boxes are as 5 X 4 = 20 juice boxes in total.

Question 3.
Six folders are placed on each table. How many folders are
there on 4 tables? Draw and label a tape diagram to solve.

There are 24 folders on 4 tables,

Explanation:
Given Six folders are placed on each table. So number of
folders are there on 4 tables are 4 X 6 = 24,
Drawn and labeled a tape diagram to solve as shown above,
Therefore, there are 24 folders on 4 tables.

Question 4.
Find the total number of corners on 8 squares.

Total number of corners on 8 squares is 32,

Explanation:
We know a square has 4 corners,

So 8 squares will have 8 X 4 = 32 corners,
therefore, total number of corners on 8 squares is 32.