## Engage NY Eureka Math 3rd Grade Module 7 Lesson 23 Answer Key

### Eureka Math Grade 3 Module 7 Lesson 23 Sprint Answer Key

A
Multiply or Divide by 5

Question 1.
2 × 5 =
2 × 5 = 10.

Question 2.
3 × 5 =
3 × 5 = 15.

Question 3.
4 × 5 =
4 × 5 = 20.

Question 4.
5 × 5 =
5 × 5 = 25.

Question 5.
1 × 5 =
1 × 5 = 5.

Question 6.
10 ÷ 5 =
10 ÷ 5 = 2.

Question 7.
15 ÷ 5 =
15 ÷ 5 = 3.

Question 8.
25 ÷ 5 =
25 ÷ 5 = 5.

Question 9.
5 ÷ 5 =
5 ÷ 5 = 1.

Question 10.
20 ÷ 5 =
20 ÷ 5 = 4.

Question 11.
6 × 5 =
6 × 5 = 30.

Question 12.
7 × 5 =
7 × 5 = 35.

Question 13.
8 × 5 =
8 × 5 = 40.

Question 14.
9 × 5 =
9 × 5 = 45.

Question 15.
10 × 5 =
10 × 5 = 50.

Question 16.
40 ÷ 5 =
40 ÷ 5 = 8.

Question 17.
35 ÷ 5 =
35 ÷ 5 = 7.

Question 18.
45 ÷ 5 =
45 ÷ 5 = 9.

Question 19.
30 ÷ 5 =
30 ÷ 5 = 6.

Question 20.
50 ÷ 5 =
50 ÷ 5 = 10.

Question 21.
___ × 5 = 25
5 × 5 = 25.

Question 22.
___ × 5 = 5
1 × 5 = 5.

Question 23.
___ × 5 = 50
10 × 5 = 50.

Question 24.
___ × 5 = 10
2 × 5 = 10.

Question 25.
___ × 5 = 15
3 × 5 = 15.

Question 26.
50 ÷ 5 =
50 ÷ 5 = 10.

Question 27.
25 ÷ 5 =
25 ÷ 5 = 5.

Question 28.
5 ÷ 5 =
5 ÷ 5 = 1.

Question 29.
10 ÷ 5 =
10 ÷ 5 = 2.

Question 30.
15 ÷ 5 =
15 ÷ 5 = 3.

Question 31.
___ × 5 = 30
6 × 5 = 30.

Question 32.
___ × 5 = 35
7 × 5 = 35.

Question 33.
___ × 5 = 45
9 × 5 = 45.

Question 34.
___ × 5 = 40
8 × 5 = 40.

Question 35.
35 ÷ 5 =
35 ÷ 5 = 7.

Question 36.
45 ÷ 5 =
45 ÷ 5 = 9.

Question 37.
30 ÷ 5 =
30 ÷ 5 = 6.

Question 38.
40 ÷ 5 =
40 ÷ 5 = 8.

Question 39.
11 × 5 =
11 × 5 = 55.

Question 40.
55 ÷ 5 =
55 ÷ 5 = 11.

Question 41.
15 ÷ 5 =
15 ÷ 5 = 3.

Question 42.
60 ÷ 5 =
60 ÷ 5 = 12.

Question 43.
12 × 5 =
12 × 5 = 60.

Question 44.
70 ÷ 5 =
70 ÷ 5 = 14.

B
Multiply or Divide by 5

Question 1.
1 × 5 =
1 × 5 = 5.

Question 2.
2 × 5 =
2 × 5 = 10.

Question 3.
3 × 5 =
3 × 5 = 15.

Question 4.
4 × 5 =
4 × 5 = 20.

Question 5.
5 × 5 =
5 × 5 = 25.

Question 6.
15 ÷ 5 =
15 ÷ 5 = 3.

Question 7.
10 ÷ 5 =
10 ÷ 5 = 2.

Question 8.
20 ÷ 5 =
20 ÷ 5 = 4.

Question 9.
5 ÷ 5 =
5 ÷ 5 = 1.

Question 10.
25 ÷ 5 =
25 ÷ 5 = 5.

Question 11.
10 × 5 =
10 × 5 = 50.

Question 12.
6 × 5 =
6 × 5 = 30.

Question 13.
7 × 5 =
7 × 5 = 35.

Question 14.
8 × 5 =
8 × 5 = 40.

Question 15.
9 × 5 =
9 × 5 = 45.

Question 16.
35 ÷ 5 =
35 ÷ 5 = 7.

Question 17.
30 ÷ 5 =
30 ÷ 5 = 6.

Question 18.
40 ÷ 5 =
40 ÷ 5 = 8.

Question 19.
50 ÷ 5 =
50 ÷ 5 = 10.

Question 20.
45 ÷ 5 =
45 ÷ 5 = 9.

Question 21.
___ × 5 = 5
1 × 5 = 5.

Question 22.
___ × 5 = 25
5 × 5 = 25.

Question 23.
___ × 5 = 10
2 × 5 = 10.

Question 24.
___ × 5 = 50
10 × 5 = 50.

Question 25.
___ × 5 = 15
3 × 5 = 15.

Question 26.
10 ÷ 5 =
10 ÷ 5 = 50.

Question 27.
5 ÷ 5 =

Question 28.
50 ÷ 5 =

Question 29.
25 ÷ 5 =

Question 30.
15 ÷ 5 =

Question 31.
___ × 5 = 15

Question 32.
___ × 5 = 20

Question 33.
___ × 5 = 45

Question 34.
___ × 5 = 35

Question 35.
40 ÷ 5 =

Question 36.
45 ÷ 5 =

Question 37.
30 ÷ 5 =

Question 38.
35 ÷ 5 =

Question 39.
11 × 5 =

Question 40.
55 ÷ 5 =

Question 41.
12 × 5 =

Question 42.
60 ÷ 5 =

Question 43.
13 × 5 =

Question 44.
65 ÷ 5 =

### Eureka Math Grade 3 Module 7 Lesson 23 Problem Set Answer Key

Question 1.
Gale makes a miniature stop sign, a regular octagon, with a perimeter of 48 centimeters for the town he built with blocks. What is the length of each side of the stop sign?
The length of each side of the stop sign = 6 centimeters.

Explanation:
Perimeter of Gale a miniature stop sign, a regular octagon = 48 centimeters
Number of sides of Gale a miniature stop sign, a regular octagon = 8
The length of each side of the stop sign = Perimeter of Gale a miniature stop sign, a regular octagon  ÷ Number of sides of Gale a miniature stop sign, a regular octagon
= 48 centimeters ÷  8
= 6 centimeters.

Question 2.
Travis bends wire to make rectangles. Each rectangle measures 34 inches by 12 inches. What is the total length of the wire needed for two rectangles?
The total length of the wire needed for two rectangles = 184 inches.

Explanation:
Length of the Travis’s rectangle = 34 inches
Width of the Travis’s rectangle = 12 inches
Perimeter of the Travis’s rectangle = 2 ( Length + Width)
= 2 ( 34 inches + 12 inches )
= 2 × 46 inches
= 92 inches.
The total length of the wire needed for two rectangles = 2 ×  Perimeter of the Travis’s rectangle
= 2 ×  92 inches
= 184 inches.

Question 3.
The perimeter of a rectangular bathroom is 32 feet. The width of the room is 8 feet. What is the length of the room?
The length of the room = 8 feet.

Explanation:
Perimeter of a rectangular bathroom = 32 feet.
Width of a rectangular bathroom = 8 feet.
Length of a rectangular bathroom = ??
Perimeter of a rectangular bathroom = 2 ( Length + Width)
=> 32 feet = (2 × Length) + (2 × 8 feet)
=> 32 feet = (2 × Length) + 16 feet
=> 32 feet – 16 feet = (2 × Length)
=> 16 feet = (2 × Length)
=> 16 feet ÷ 2 = Length
=> 8 feet = Length

Question 4.
Raj uses 6-inch square tiles to make a rectangle, as shown below. What is the perimeter of the rectangle in inches?

Perimeter of the rectangle in inches =96-inches.

Explanation:

Length of the side square tiles Raj uses to make a rectangle = 6-inch
Number of side of square tiles in the rectangle = 16
Perimeter of the rectangle in inches = Length of the side square tiles Raj uses to make a rectangle × Number of side of square tiles in the rectangle
= 6-inch × 16
= 96-inches.

Question 5.
Mischa makes a 4-foot by 6-foot rectangular banner. She puts ribbon around the outside edges. The ribbon costs $2 per foot. What is the total cost of the ribbon? Answer: Total cost of the ribbon =$20.

Explanation:
Length of the rectangular banner = 6-foot
Width of the rectangular banner = 4-foot
Perimeter of the rectangular banner = 2 ( Length + Width)
= 2 ( 6-foot + 4-foot )
= 2 × 10-foot
= 20- foot.
Cost of the ribbon per foot = $2 Total cost of the ribbon = Perimeter of the rectangular banner × Cost of the ribbon per foot = 20- foot ×$2
= \$40.

Question 6.
Colton buys a roll of wire fencing that is 120 yards long. He uses it to fence in his 18-yard by 24-yard rectangular garden. Will Colton have enough wire fencing left over to fence in a 6-yard by 8-yard rectangular play space for his pet rabbit?
Yes, Colton has wire fencing left over to fence for rectangular play space for his pet rabbit.

Explanation:
Total length of wire Colton purchased = 120 yards.
Length of the rectangular garden = 24-yard
Width of the rectangular garden = 18-yard
Perimeter of the rectangular garden = 2 ( Length + Width)
= 2 ( 24-yard + 18-yard )
= 2 × 42-yard
= 84-yard.
Wire used = Total length of wire Colton purchased – Perimeter of the rectangular garden
= 120 yards – 84 yards
= 36 yards.
Length of the rectangular play space for his pet rabbit = 8-yard
Width of the rectangular play space for his pet rabbit = 6-yard
Perimeter of the rectangular play space for his pet rabbit = 2 ( Length + Width)
= 2 ( 8-yard + 6-yard )
= 2 × 14-yard
= 28-yard.
Wire left = Wire used – Perimeter of the rectangular play space for his pet rabbit
= 36 yards – 28 yards
= 8 yards.

### Eureka Math Grade 3 Module 7 Lesson 23 Exit Ticket Answer Key

Adriana traces a regular triangle to create the shape below. The perimeter of her shape is 72 centimeters. What are the side lengths of the triangle?

The side lengths of the triangle = 9 centimeters.

Explanation:

Perimeter of Adriana’s shape = 72 centimeters
Number of sides of the triangle shape = 8
The side lengths of the triangle = Perimeter of Adriana’s shape  ÷ Number of sides of the triangle shape
= 72 centimeters ÷ 8
= 9 centimeters.

### Eureka Math Grade 3 Module 7 Lesson 23 Homework Answer Key

Question 1.
Rosie draws a square with a perimeter of 36 inches. What are the side lengths of the square?
The side lengths of the square = 9 inches.

Explanation:
Perimeter of the Square = 36 inches
Number of sides in the square = 4
The side lengths of the square = Perimeter of the Square  ÷ Number of sides in the square
= 36 inches ÷ 4
= 9 inches.

Question 2.
Judith uses craft sticks to make two 24-inch by 12-inch rectangles. What is the total perimeter of the 2 rectangles?
Total perimeter of the 2 rectangles = 144 inches.

Explanation:
Length of the rectangle = 24-inch
Width of the rectangle = 12-inch
Perimeter of the rectangle = 2 ( Length + Width)
= 2 ( 24-inch + 12-inch )
= 2 × 36-inch
= 72 inches.
Total perimeter of the 2 rectangles = Perimeter of the rectangle × 2
= 72 inches × 2
= 144 inches.

Question 3.
An architect draws a square and a rectangle, as shown below, to represent a house that has a garage. What is the total perimeter of the house with its attached garage?

Perimeter of the attached garage = 250 ft.

Explanation:

Length of the side AB in the attached garage = 55 ft
Length of the side BC in the attached garage = 10 ft
Length of the side CD in the attached garage = 30 ft
Length of the side DE in the attached garage = 30 ft
Length of the side EF in the attached garage = 30 ft
Length of the side FG in the attached garage = 55 ft
Length of the side GA in the attached garage = 40 ft
Perimeter of the attached garage = Length of the side AB in the attached garage  +  Length of the side BC in the attached garage + Length of the side CD in the attached garage + Length of the side DE in the attached garage + Length of the side EF in the attached garage + Length of the side FG in the attached garage + Length of the side GA in the attached garage
= 55 ft + 10 ft + 30 ft + 30 ft + 30 ft + 55 ft + 40 ft
= 65 ft + 30 ft + 30 ft + 30 ft + 55 ft + 40 ft
= 95 ft + 30 ft + 30 ft + 55 ft + 40 ft
= 125 ft + 30 ft + 55 ft + 40 ft
= 155 ft + 55 ft + 40 ft
= 210 ft + 40ft
= 250 ft.

Question 4.
Manny draws 3 regular pentagons to create the shape shown below. The perimeter of 1 of the pentagons is 45 inches. What is the perimeter of Manny’s new shape?

Perimeter of Manny’s new shape = 99 inches.

Explanation:

Perimeter of the regular pentagons Manny draws = 45 inches
Number of side in the regular pentagons Manny draws = 5
Length of the side of the regular pentagons Manny draws = Perimeter of the regular pentagons Manny draws ÷ Number of side in the regular pentagons Manny draws
= 45 inches ÷ 5
= 9 inches.
Number of side Manny’s new shape  has = 11 ( Side AB + Side BC +Side CD + Side DE + Side EF +Side FG + Side GH + Side HI + Side IJ +Side JK + Side KA )
Perimeter of Manny’s new shape = Number of side Manny’s new shape has × Length of the side of the regular pentagons Manny draws
= 11 × 9 inches
= 99 inches.

Question 5.
Johnny uses 2-inch square tiles to make a square, as shown below. What is the perimeter of Johnny’s square?

Perimeter of Johnny’s square = 24 inches.

Explanation:

Number of side in the Johnny’s square = 12
Length of  side  in the Johnny’s square = 2 inches
Perimeter of Johnny’s square = Number of side in the Johnny’s square  × Length of side in the Johnny’s square
= 12 × 2 inches
= 24 inches.

Question 6.
Lisa tapes three 7-inch by 9-inch pieces of construction paper together to make a happy birthday sign for her mom. She uses a piece of ribbon that is 144 inches long to make a border around the outside edges of the sign. How much ribbon is leftover?

Ribbon  leftover = 76 inches.

Explanation:

Length of the Lisa’s piece of construction paper = 9-inch
Width of the Lisa’s piece of construction paper = 7-inch
Perimeter of the pieces of construction paper used to make a happy birthday sign for her mom = Side AB + Side BC + Side CD + Side DE + Side EF + Side FG + Side GH +Side HA
= 9-inch + 9-inch + 9-inch + 7-inch + 9-inch + 9-inch + 9-inch + 7-inch
= 18-inch + 9-inch + 7-inch + 9-inch + 9-inch + 9-inch + 7-inch
= 27-inch + 7-inch + 9-inch + 9-inch + 9-inch + 7-inch
= 34-inch + 9-inch + 9-inch + 9-inch + 7-inch
= 43-inch + 9-inch + 9-inch + 7-inch
= 52-inch + 9-inch + 7-inch
= 61-inch + 7-inch
= 68-inch.
Piece of ribbon Lisa’s  has = 144 inches
Ribbon  leftover = Piece of ribbon Lisa’s  has  – Perimeter of the pieces of construction paper used to make a happy birthday sign for her mom
= 144 inches – 68-inch.
= 76 inches.