Eureka Math Grade 3 Module 1 Lesson 8 Answer Key

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.
Eureka Math Grade 3 Module 1 Lesson 8 Answer Key-1
Explanation:
Drawn an array that shows 5 rows of 3 as 5 × 3 as shown above.

Question 2.
Draw an array that shows 3 rows of 5.
Eureka Math Grade 3 Module 1 Lesson 8 Answer Key-2
Explanation:
Drawn an array that shows 3 rows of 5 as 3 × 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.
Eureka Math Grade 3 Module 1 Lesson 8 Problem Set Answer Key 1

Eureka Math Grade 3 Module 1 Lesson 8 Answer Key-3
Explanation:
Wrote multiplication expressions for the arrays in Problems 1 and 2 as 5 × 3, 3 × 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 × 3 = 15, 3 × 5 = 15,
So,  5 × 3 = 3 × 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 × 2 = 6.

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

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

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

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

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

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

i. 10 threes: 10 x 3 =30,
Explanation:
Given 10 threes, 10 multiplied by 3 gives 30,
So, 10 × 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 × 9, 3 multiplied by 9 gives 27,
So 3 × 9 = 27.

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

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

d. 3 × 8 = 24,

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

e. 15 = 5 × 3,

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

f. 27 = 9 × 3

Explanation:
Given 27 = 9 × ___, Lets take missing number as x,
27 = 9 × x, So x = 27 ÷ 9 = 3, So 27 = 9 × 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.
Eureka Math Grade 3 Module 1 Lesson 8 Answer Key-4
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 × 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).
Eureka Math Grade 3 Module 1 Lesson 8 Answer Key-5
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 × 7) + (3 × 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 × 7) + (3 × 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 × 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 × 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. Eureka Math Grade 3 Module 1 Lesson 8 Answer Key-6
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 × 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 × 4 = 12, in finding Mary Beth’s total number of stickers.

c.
Eureka Math Grade 3 Module 1 Lesson 8 Answer Key-7
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 × 4 = 4 x 3,

Explanation:
Used what I know about the commutative property to write a different multiplication sentence for my array as
3 × 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.
Eureka Math Grade 3 Module 1 Lesson 8 Answer Key-8
Explanation:
Drawn an array that shows 6 rows of 3 as 6 × 3 as shown above.

Question 2.
Draw an array that shows 3 rows of 6.
Eureka Math Grade 3 Module 1 Lesson 8 Answer Key-9
Explanation:
Drawn an array that shows 3 rows of 6 as 3 × 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.
Eureka Math 3rd Grade Module 1 Lesson 8 Homework Answer Key 2
Eureka Math Grade 3 Module 1 Lesson 8 Answer Key-10
Explanation:
Wrote multiplication expressions for the arrays in Problems 1 and 2 as 6 × 3, 3 × 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 × 3 = 18, 3 × 6 = 18,
So,  6 × 3 = 3 × 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 × 5 = 15,

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

c. 6 threes: 6 × 3 = 18,

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

d. 3 sixes: = 3 × 6 = 18,

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

e. 7 threes: 7 × 3 = 21,

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

f. 3 sevens: 3 × 7 = 21,

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

g. 8 threes: 8 × 3 = 24,

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

h. 3 nines: 3 × 9 = 27,

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

i. 10 threes: 10 × 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 × 5, 3 multiplied by 5 gives 15,
So 3 × 5 = 15.

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

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

d. 3 × 9 = 27,

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

e. 18 = 6 × 3,

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

f. 15 = 5 × 3,

Explanation:
Given 15 = 5 × ___, Lets take missing number as x,
15 = 5 × x, So x = 15 ÷ 5 = 3, So 15 = 5 × 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 the total number of pictures in Fernando’s album.
a.
Eureka Math Grade 3 Module 1 Lesson 8 Answer Key-11

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.
Eureka Math Grade 3 Module 1 Lesson 8 Answer Key-12
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× 8) + (3 × 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 × 8) + (3 × 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 × 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 × 7 = 21 cents.

Eureka Math Grade 3 Module 7 Lesson 14 Answer Key

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.
Engage NY Math 3rd Grade Module 7 Lesson 14 Pattern Sheet Answer Key p 1
multiply by 8 (6–10)
Answer:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-14-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 14 Pattern Sheet Answer Key

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.
Engage NY Math Grade 3 Module 7 Lesson 14 Problem Set Answer Key pr 1
Perimeter = _______ in
Answer:
Length of the side of the ABCDEFG Octogen = 64in.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-14-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 14 Problem Set Answer Key-1a
Length of the side of the ABCDEFG Octogen = 8in
Perimeter of the ABCDEFG Octogen = 8 × side
= 8 × 8in
= 64in.

b.
Engage NY Math Grade 3 Module 7 Lesson 14 Problem Set Answer Key pr 2
Perimeter = _______ ft
Answer:
Perimeter of the ABC Triangle = 21ft.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-14-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 14 Problem Set Answer Key-1b
Length of the side of the ABC Triangle = 7ft
Perimeter of the ABC Triangle = 3 × Side
= 3 × 7ft
= 21ft.

c.
Engage NY Math Grade 3 Module 7 Lesson 14 Problem Set Answer Key pr 3
Perimeter = _______ m
Answer:
Perimeter of the ABCD Square = 36m.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-14-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 14 Problem Set Answer Key-1c
Length of the Side of the ABCD Square = 9m
Perimeter of the ABCD Square = 4 × Side
= 4 × 9m
= 36m.

d.
Engage NY Math Grade 3 Module 7 Lesson 14 Problem Set Answer Key pr 4
Perimeter = _______ in
Answer:
Perimeter of the ABCDE Pentagon = 36in.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-14-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 14 Problem Set Answer Key-1d
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.
Engage NY Math Grade 3 Module 7 Lesson 14 Problem Set Answer Key pr 5
Perimeter = _______ cm
Answer:
Perimeter of the side of the ABCD Rectangle = 18cm.

Explanation:

Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-14-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 14 Problem Set Answer Key-2
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.
Engage NY Math Grade 3 Module 7 Lesson 14 Problem Set Answer Key pr 6
Answer:
Perimeter of the ABCDEFGH Octogen = 36cm.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-14-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 14 Problem Set Answer Key-3
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:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-14-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 14 Exit Ticket Answer Key
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.
Eureka Math Grade 3 Module 7 Lesson 14 Homework Answer Key h 1
Perimeter = _______ in
Answer:
Perimeter of the ABC Triangle = 16in.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-14-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 14 Homework Answer Key-1a
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.
Eureka Math Grade 3 Module 7 Lesson 14 Homework Answer Key h 2
Perimeter = _______ cm

Answer:
Perimeter of the square = 32cm.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-14-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 14 Homework Answer Key-1b
Length of the Side of the square = 8cm
Perimeter of the square = 4 × Side
= 4 × 8cm
= 32cm.

c.
Eureka Math Grade 3 Module 7 Lesson 14 Homework Answer Key h 3
Perimeter = _______ m
Answer:
Perimeter of the Octagon = 72m.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-14-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 14 Homework Answer Key-1c
Length of the side of the Octagon = 9m
Perimeter of the Octagon = 8 × Side
= 8 × 9m
= 72m.

d.
Eureka Math Grade 3 Module 7 Lesson 14 Homework Answer Key h 4
Perimeter = _______ in
Answer:
Perimeter of the Hexagon = 36in.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-14-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 14 Homework Answer Key-1d
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.
Eureka Math Grade 3 Module 7 Lesson 14 Homework Answer Key h 5
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.
Eureka Math Grade 3 Module 7 Lesson 14 Homework Answer Key h 6
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.