Eureka Math

Engage NY Eureka Math 3rd Grade Module 1 Lesson 16 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 16 Pattern Sheet Answer Key

Multiply.

multiply by 4 (6–10)


Explanation:
Multiplied by 4 (6–10) as shown above.

Eureka Math Grade 3 Module 1 Lesson 16 Problem Set Answer Key

Question 1.
Label the array. Then, fill in the blanks below to make true number sentences.

6 x 4 = 24

Explanation:
Labeled the array and filled the blanks to make true
number sentences as (6 X 4) = (5 X 4) + (1 X 4) =
20 + 4 = 24.  6 X 4 we write as ((5 + 1 ) X 4).
So 6 X 4 = 24.

7 X 4 = 28,

Explanation:
Labeled the array and filled the blanks to make true
number sentences as (7 X 4) = (5 X 4) + (2 X 4) =
20 + 8 = 28. 7 X 4 we write as ((5 + 2) X 4).
So, 7 X 4 = 28.

8 X 4 = 32,

Explanation:
Labeled the array and filled the blanks to make true
number sentences as (8 X 4) = (5 X 4) + (3 X 4) =
20 + 12 = 32. 8 X 4 we write as ((5 + 3) X 4).
So, 8 X 4 = 32.

9 X 4 = 36,

Explanation:
Labeled the array and filled the blanks to make true
number sentences as (9 X 4) = (5 X 4) + (4 X 4) =
20 + 16 = 36. 9 X 4 we write as ((5 + 4) X 4).
So, 9 X 4 = 36.

Question 2.
Match the equal expressions.

Explanation:
Matched the equal expressions as
(5 X 4) + (3 X 4) = (5 + 3) X 4 = 8 X 4 = 32,
(5 X 4) + (1 X 4) = (5 + 1) X 4 = 6 X 4 = 24,
(5 X 4) + (4 X 4) = (5 + 4) X 4 = 9 X 4 = 36,
(5 X 4) + (2 X 4) = (5 + 2) X 4 = 7 X 4 = 28.

Question 3.
Nolan draws the array below to find the answer to the
multiplication expression 10 × 4. He says, “10 × 4 is just
double 5 × 4.” Explain Nolan’s strategy.

Nolan’s strategy is 10 X 4 = 2 X (5 X 4) = 2 X 20 = 40,
or 10 X 4 = 40,

Explanation:
Given Nolan draws the array  to find the answer to the
multiplication expression 10 × 4. Nolan says, “10 × 4 is just
double 5 × 4.” Nolan’s strategy is 10 X 4 = 2 X (5 X 4) =
2 X 20 = 40 is same as 10 X 4 = 40.
So, “10 × 4 is just double 5 × 4.”

Eureka Math Grade 3 Module 1 Lesson 16 Exit Ticket Answer Key

Destiny says, “I can use 5 × 4 to find the answer to 7 × 4.”
Use the array below to explain Destiny’s strategy using words and numbers.

Explanation:
Given Destiny says, “I can use 5 × 4 to find the answer to 7 × 4.”
Used the array to explain Destiny’s strategy using words and numbers
as to find 7 X 4 we write 7 as (5 + 2) and multiply with 4 as
(5 + 2) X 4 = (5 X 4) + (2 X 4) = 20 + 8 = 28, So we used 5 X 4 to
find the answer for 7 X 4.

Eureka Math Grade 3 Module 1 Lesson 16 Homework Answer Key

Question 1.
Label the array. Then, fill in the blanks below to make true number sentences.


Explanation:
Labeled the array and filled in the blanks to make true
number sentences as 6 X 4 = (5 X 4) + (1 X 4) = 20 + 4 = 24,
or 6 X 4 = 24.

Explanation:
Labeled the array and filled in the blanks to make true
number sentences as 8 X 4 = (5 X 4) + (3 X 4) = 20 + 12 = 32,
or 8 X 4 = 32.

Question 2.
Match the multiplication expressions with their answers.

Explanation:
Matched the multiplication expressions with their answers as
4 X 6 = 24,
4 X 7 = 28,
4 X 8 = 32,
4 X 9 = 36.

Question 3.
The array below shows one strategy for solving 9 × 4.
Explain the strategy using your own words.

Explanation:
Given the array shown one strategy for solving 9 × 4.
The strategy used is 9 X 4 = (5 X 4) + (4 X 4) = 20 + 16 = 36,
9 X 4 = 36.

Engage NY Eureka Math 3rd Grade Module 1 Lesson 19 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 19 Problem Set Answer Key

Question 1.
Label the array. Then, fill in the blanks to make true number sentences.
a. 36 ÷ 3 = 12__

a. 36 ÷ 3 = 12,

Explanation:
36 ÷ 3 can be written as (30 ÷ 3) + (6 ÷ 3) = 10 + 2 = 12 or
36 ÷ 3 = 12. Filled the blanks to make true number sentences.

b. 25 ÷ 5 = _5_

b. 25 ÷ 5 = 5,

Explanation:
25 ÷ 5 can be written as (20 ÷ 5) + (5 ÷ 5) = 4  + 1 = 5 or
25 ÷ 5 = 5. Filled the blanks to make true number sentences.

c. 28 ÷ 4 = __7___

28 ÷ 4 =7,

Explanation:
28 ÷ 4 can be written as (20 ÷ 4) + (8 ÷ 4) = 5  + 2 = 7 or
28 ÷ 4 = 7. Filled the blanks to make true number sentences.

d. 32 ÷ 4 = _8_

32 ÷ 4 = 8,

Explanation:
32 ÷ 4 can be written as (20 ÷ 4) + (12 ÷ 4) = 5  + 3 = 8 or
32 ÷ 4 = 8. Filled the blanks to make true number sentences.

Question 2.
Match the equal expressions.

Matched the equal expressions as

Explanation:
Matching the equal expressions as
24 ÷ 2 = (20 ÷ 2) + (4 ÷  2),
24 is written as (20 + 4) ÷ 2.
36 ÷ 3 = (30 ÷ 3) + (6 ÷ 3),
36 is written as (30 + 6) ÷ 3.
39 ÷  3 = (30 ÷  3) + (9 ÷  3),
39 is written as (30 + 9) ÷ 3.
26 ÷ 2 = ( 20 ÷ 2) + (6 ÷ 2),
26 is written as (20 + 6) ÷ 2.

Question 3.
Nell draws the array below to find the answer to 24 ÷ 2. Explain Nell’s strategy.

Nell’s strategy is 24 ÷ 2 = (12 ÷ 2) + (12 ÷ 2),
6 + 6 = 12, So, 24 ÷ 2 = 12,

Explanation:
To find answer for 24 ÷ 2 Nell draws an array
his strategy is he divided 24 as 12 + 12 then has
divided with 2, So equation for Nell’s drawing is
24 ÷ 2 = (12 ÷ 2) + (12 ÷ 2), 6 + 6 = 12.
Therefore 24 ÷ 2 = 12.

Eureka Math Grade 3 Module 1 Lesson 19 Exit Ticket Answer Key

Complete the equations below to solve 22 ÷ 2 = __11___.

22 ÷ 2 = 11,

Explanation:

We solve 22 ÷ 2 as below first we write 22 as 20 + 2
and divide by 2, (20 + 2) ÷ 2 = (20 ÷ 2) +(2 ÷ 2) =
10 + 1 = 11, So 22 ÷ 2 = 11.

Eureka Math Grade 3 Module 1 Lesson 19 Homework Answer Key

Question 1.
Label the array. Then, fill in the blanks to make true number sentences.
a. 18 ÷ 3 = _6_


18 ÷ 3 = 6,

Explanation:
Labelled the array, filled the blanks to make
true number sentences as 18 ÷ 3 = ( 9 ÷ 3) + (9 ÷ 3) =
3 + 3 = 6 or 18 ÷ 3 = 6.
We wrote 18 ÷ 3 as (9 + 9) ÷ 3.

b. 21 ÷ 3 = _7_


21 ÷ 3 = 7,

Explanation:
Labelled the array, filled the blanks to make
true number sentences as 21÷ 3 = (15 ÷ 3) + (6 ÷ 3) =
5 + 2 = 7 or 21 ÷ 3 = 7.
We wrote 21 ÷ 3 as (15 + 6) ÷ 3.

c. 24 ÷ 4 = _6_

24 ÷ 4 = 6,

Explanation:
Labelled the array, filled the blanks to make
true number sentences as 24 ÷ 4 = (20÷ 4) + (4 ÷ 4) =
5 + 1 = 6 or 24÷ 4 = 6.
We wrote 24 ÷ 4 as (20 + 4) ÷ 4.

d. 36 ÷ 4 = _9_


36 ÷ 4 = 9,

Explanation:
Labelled the array, filled the blanks to make
true number sentences as 36 ÷ 4 = (20÷ 4) + (16 ÷ 4) =
5 + 4 = 9 or 36÷ 4 = 9.
We wrote 36 ÷ 4 as (20 + 16) ÷ 4.

Question 2.
Match equal expressions.

Matched the equal expressions as

Explanation:
Matching the equal expressions as
28 ÷ 2 = (20 ÷ 2) + (8 ÷  2),
28 is written as (20 + 8) ÷ 2.
33 ÷ 3 = (30 ÷ 3) + (3 ÷ 3),
33 is written as (30 + 3) ÷ 3.
36 ÷  3 = (30 ÷  3) + (6 ÷  3),
36 is written as (30 + 6) ÷ 3.
26 ÷ 2 = ( 20 ÷ 2) + (6 ÷ 2),
26 is written as (20 + 6) ÷ 2.

Question 3.
Alex draws the array below to find the answer to 35 ÷ 5. Explain Alex’s strategy.

Alex’s strategy is 35 ÷ 5 = (20 ÷ 5) + (15 ÷ 5) =
4 + 3 = 7, So, 35 ÷ 5 = 7,

Explanation:
To find answer for 35 ÷ 5 Alex draws an array
his strategy is he divided 35 as 20 + 15 then has
divided with 5, So equation for Alex’s drawing is
35 ÷ 5 = (20 ÷ 5) + (15 ÷ 5), 4 + 3 = 7,
Therefore 35 ÷ 5 = 7.

Engage NY Eureka Math 3rd Grade Module 1 Lesson 18 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 18 Sprint Answer Key

A
Add or Subtract Using 5

Question 1.
0 + 5 =
0 + 5 = 5,

Explanation:
Given 0 + 5 = adding 5 to 0 we get 5,
So, 0 + 5 = 5.

Question 2.
5 + 5 =
5 + 5 = 10,

Explanation:
Given 5 + 5 = adding 5 to 5 we get 10,
So, 5 + 5 = 10.

Question 3.
10 + 5 =
10 + 5 = 15,

Explanation:
Given 10 + 5 = adding 5 to 10 we get 15,
So, 10 + 5 = 15.

Question 4.
15 + 5 =
15 + 5 = 20,

Explanation:
Given 15 + 5 = adding 5 to 15 we get 20,
So, 15 + 5 = 20.

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

Explanation:
Given 20 + 5 = adding 5 to 20 we get 25,
So, 20 + 5 = 5.

Question 6.
25 + 5 =
25 + 5 = 30,

Explanation:
Given 25 + 5 = adding 5 to 25 we get 30,
So, 25 + 5 = 30.

Question 7.
30 + 5 =
30 + 5 = 35,

Explanation:
Given 30 + 5 = adding 5 to 30 we get 35,
So, 30 + 5 = 35.

Question 8.
35 + 5 =
35 + 5 = 40,

Explanation:
Given 35 + 5 = adding 5 to 35 we get 40,
So, 35 + 5 = 40.

Question 9.
40 + 5 =
40 + 5 = 45,

Explanation:
Given 40 + 5 = adding 5 to 40 we get 45,
So, 40 + 5 = 45.

Question 10.
45 + 5 =
45 + 5 = 50,

Explanation:
Given 45 + 5 = adding 5 to 45 we get 50,
So, 45 + 5 = 50.

Question 11.
50 – 5 =
50 – 5 = 45,

Explanation:
Given 50 – 5 = subtracting 5 from 50 we get 45,
So, 50 – 5 = 45.

Question 12.
45 – 5 =
45 – 5 = 40,

Explanation:
Given 45 – 5 = subtracting 5 from 45 we get 40,
So, 45 – 5 = 40.

Question 13.
40 – 5 =
40 – 5 = 35,

Explanation:
Given 40 – 5 = subtracting 5 from 40 we get 35,
So, 40 – 5 = 35.

Question 14.
35 – 5 =
35 – 5 = 30,

Explanation:
Given 35 – 5 = subtracting 5 from 35 we get 30,
So, 35 – 5 = 30.

Question 15.
30 – 5 =
30 – 5 = 25,

Explanation:
Given 30 – 5 = subtracting 5 from 30 we get 25,
So, 30 – 5 = 25.

que25 16.
25 – 5 =
25 – 5 = 20,

Explanation:
Given 25 – 5 = subtracting 5 from 25 we get 20,
So, 25 – 5 = 20.

Question 17.
20 – 5 =
20 – 5 = 15,

Explanation:
Given 20 – 5 = subtracting 5 from 20 we get 15,
So, 20 – 5 = 15.

Question 18.
15 – 5 =
15 – 5 = 10,

Explanation:
Given 15 – 5 = subtracting 5 from 15 we get 10,
So, 15 – 5 = 10.

Question 19.
10 – 5 =
10 – 5 = 5,

Explanation:
Given 10 – 5 = subtracting 5 from 10 we get 5,
So, 10 – 5 = 5.

Question 20.
5 – 5 =
5 – 5 = 0,

Explanation:
Given 5 – 5 = subtracting 5 from 5 we get 0,
So, 5 – 5 = 0.

Question 21.
5 + 0 =
5 + 0 = 5,

Explanation:
Given 5 + 0 = adding 5 to 0 we get 5,
So, 5 + 0 = 5.

Question 22.
5 + 5 =
5 + 5 = 10,

Explanation:
Given 5 + 5 = adding 5 to 5 we get 10,
So, 5 + 5 = 10.

Question 23.
10 + 5 =
10 + 5 = 15,

Explanation:
Given 10 + 5 = adding 5 to 10 we get 15,
So, 10 + 5 = 15.

Question 24.
15 + 5 =
15 + 5 = 20,

Explanation:
Given 15 + 5 = adding 5 to 15 we get 20,
So, 15 + 5 = 20.

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

Explanation:
Given 20 + 5 = adding 5 to 20 we get 25,
So, 20 + 5 = 25.

Question 26.
25 + 5 =
25 + 5 = 30,

Explanation:
Given 25 + 5 = adding 5 to 25 we get 30,
So, 25 + 5 = 30.

Question 27.
30 + 5 =
30 + 5 = 35,

Explanation:
Given 30 + 5 = adding 5 to 30 we get 35,
So, 30 + 5 = 35.

Question 28.
35 + 5 =
35 + 5 = 40,

Explanation:
Given 35 + 5 = adding 5 to 35 we get 40,
So, 35 + 5 = 40.

Question 29.
40 + 5 =
40 + 5 = 45,

Explanation:
Given 40 + 5 = adding 5 to 40 we get 45,
So, 40 + 5 = 45.

Question 30.
45 + 5 =
45 + 5 = 50,

Explanation:
Given 45 + 5 = adding 5 to 45 we get 50,
So, 45 + 5 = 50.

Question 31.
0 + 50 =
0 + 50 = 50,

Explanation:
Given 0 + 50 = adding 0 to 50 we get 50,
So, 0 + 50 = 50.

Question 32.
50 + 50 =
50 + 50 = 100,

Explanation:
Given 50 + 50 = adding 50 to 50 we get 100,
So, 50 + 50 = 100.

Question 33.
50 + 5 =
50 + 5 = 55,

Explanation:
Given 50 + 5 = adding 5 to 50 we get 55,
So, 50 + 5 = 55.

Question 34.
55 + 5 =
55 + 5 = 60,

Explanation:
Given 55 + 5 = adding 5 to 55 we get 60,
So, 55 + 5 = 55.

Question 35.
60 – 5 =
0 + 50 = 50,

Explanation:
Given 0 + 50 = adding 0 to 50 we get 50,
So, 0 + 50 = 50.

Question 36.
55 – 5 =
55 – 5 = 50,

Explanation:
Given 55 – 5 = subtracting 5 from 55 we get 50,
So, 55 – 5 = 50.

Question 37.
60 + 5 =
60 + 5 = 65,

Explanation:
Given 60 + 5 = adding 5 to 60 we get 65,
So, 60 + 5 = 15.

Question 38.
65 + 5 =
65 + 5 = 70,

Explanation:
Given 65 + 5 = adding 5 to 65 we get 70,
So, 65 + 5 = 70.

Question 39.
70 – 5 =
70 – 5 = 65,

Explanation:
Given 70 – 5 = subtracting 5 from 70 we get 65,
So, 70 – 5 = 65.

Question 40.
65 – 5 =
65 – 5 = 60,

Explanation:
Given 65 – 5 = subtracting 5 from 65 we get 60,
So, 65 – 5 = 60.

Question 41.
100 + 50 =
100 + 50 = 150,

Explanation:
Given 100 + 50 = adding 50 to 100 we get 150,
So, 100 + 50 = 150.

Question 42.
150 + 50 =
150 + 50 = 200,

Explanation:
Given 150 + 50 = adding 50 to 150 we get 200,
So, 150 + 50 = 200.

Question 43.
200 – 50 =
200 – 50 = 150,

Explanation:
Given 200 – 50 = subtracting 50 from 200 we get 150,
So, 200 – 50 = 150.

Question 44.
150 – 50 =
150 – 50 = 100,

Explanation:
Given 150 – 50 = subtracting 50 from 150 we get 100,
So, 150 – 50 = 100.

B
Add or Subtract Using 5

Question 1.
5 + 0 =
5 + 0 = 5,

Explanation:
Given 5 + 0 = adding 5 to 0 we get 5,
So, 5 + 0 = 5.

Question 2.
5 + 5 =
5 + 5 = 10,

Explanation:
Given 5 + 5 = adding 5 to 5 we get 10,
So, 5 + 5 = 10.

Question 3.
5 + 10 =
5 + 10 = 15,

Explanation:
Given 5 + 10 = adding 5 to 10 we get 15,
So, 5 + 10 = 15.

Question 4.
5 + 15 =
5 + 15 = 20,

Explanation:
Given 5 + 15 = adding 5 to 15 we get 20,
So, 5 + 15 = 20.

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

Explanation:
Given 5 + 20 = adding 5 to 20 we get 25,
So, 5 + 20 = 25.

Question 6.
5 + 25 =
5 + 25 = 30,

Explanation:
Given 5 + 25 = adding 5 to 25 we get 30,
So, 5 + 25 = 30.

Question 7.
5 + 30 =
5 + 30 = 35,

Explanation:
Given 5 + 30 = adding 5 to 30 we get 35,
So, 5 + 30 = 35.

Question 8.
5 + 35 =
5 + 35 = 40,

Explanation:
Given 5 + 35 = adding 5 to 35 we get 40,
So, 5 + 35 = 40.

Question 9.
5 + 40 =
5 + 40 = 45,

Explanation:
Given 5 + 40 = adding 5 to 40 we get 45,
So, 5 + 40 = 45.

Question 10.
5 + 45 =
5 + 45 = 50,

Explanation:
Given 5 + 45 = adding 5 to 45 we get 50,
So, 5 + 45 = 50.

Question 11.
50 – 5 =
50 – 5 = 45,

Explanation:
Given 50 – 5 = subtracting 5 from 50 we get 45,
So, 50 – 5 = 45.

Question 12.
45 – 5 =
45 – 5 = 40,

Explanation:
Given 45 – 5 = subtracting 5 from 45 we get 40,
So, 45 – 5 = 40.

Question 13.
40 – 5 =
40 – 5 = 35,

Explanation:
Given 40 – 5 = subtracting 5 from 40 we get 35,
So, 40 – 5 = 35.

Question 14.
35 – 5 =
35 – 5 = 30,

Explanation:
Given 35 – 5 = subtracting 5 from 35 we get 30,
So, 35 – 5 = 30.

Question 15.
30 – 5 =
30 – 5 = 25,

Explanation:
Given 30 – 5 = subtracting 5 from 30 we get 25,
So, 30 – 5 = 25.

Question 16.
25 – 5 =
25 – 5 = 20,

Explanation:
Given 25 – 5 = subtracting 5 from 25 we get 20,
So, 25 – 5 = 20.

Question 17.
20 – 5 =
20 – 5 = 15,

Explanation:
Given 20 – 5 = subtracting 5 from 20 we get 15,
So, 20 – 5 = 15.

Question 18.
15 – 5 =
15 – 5 = 10,

Explanation:
Given 15 – 5 = subtracting 5 from 15 we get 10,
So, 15 – 5 = 10.

Question 19.
10 – 5 =
10 – 5 = 5,

Explanation:
Given 10 – 5 = subtracting 5 from 10 we get 5,
So, 10 – 5 = 5.

Question 20.
5 – 5 =
5 – 5 = 0,

Explanation:
Given 5 – 5 = subtracting 5 from 5 we get 0,
So, 5 – 5 = 0.

Question 21.
0 + 5 =
0 + 5 = 5,

Explanation:
Given 0 + 5 = adding 5 to 0 we get 5,
So, 0 + 5 = 5.

Question 22.
5 + 5 =
5 + 5 = 10,

Explanation:
Given 5 + 5 = adding 5 to 5 we get 10,
So, 5 + 5 = 10.

Question 23.
10 + 5 =
10 + 5 = 15,

Explanation:
Given 10 + 5 = adding 5 to 10 we get 15,
So, 10 + 5 = 15.

Question 24.
15 + 5 =
15 + 5 = 20,

Explanation:
Given 15 + 5 = adding 5 to 15 we get 20,
So, 15 + 5 = 20.

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

Explanation:
Given 20 + 5 = adding 5 to 20 we get 25,
So, 20 + 5 = 25.

Question 26.
25 + 5 =
25 + 5 = 30,

Explanation:
Given 25 + 5 = adding 5 to 25 we get 30,
So, 25 + 5 = 30.

Question 27.
30 + 5 =
30 + 5 = 35,

Explanation:
Given 30 + 5 = adding 5 to 30 we get 35,
So, 30 + 5 = 35.

Question 28.
35 + 5 =
35 + 5 = 40,

Explanation:
Given 35 + 5 = adding 5 to 35 we get 40,
So, 35 + 5 = 40.

Question 29.
40 + 5 =
40 + 5 = 45,

Explanation:
Given 40 + 5 = adding 5 to 40 we get 45,
So, 40 + 5 = 45.

Question 30.
45 + 5 =
45 + 5 = 50,

Explanation:
Given 45 + 5 = adding 5 to 45 we get 50,
So, 45 + 5 = 50.

Question 31.
50 + 0 =
50 + 0 = 50,

Explanation:
Given 50 + 5 = adding 0 to 50 we get 50,
So, 50 + 0 = 50.

Question 32.
50 + 50 =
50 + 50 = 100,

Explanation:
Given 50 + 50 = adding 50 to 50 we get 100,
So, 50 + 50 = 100.

Question 33.
5 + 50 =
5 + 50 = 55,

Explanation:
Given 5 + 50 = adding 5 to 50 we get 55,
So, 5 + 50 = 55.

Question 34.
5 + 55 =
5 + 55 = 60,

Explanation:
Given 5 + 55 = adding 5 to 55 we get 60,
So, 5 + 55 = 60.

Question 35.
60 – 5 =
60 – 5 = 55,

Explanation:
Given 60 – 5 = subtracting 5 from 60 we get 55,
So, 60 – 5 = 55.

Question 36.
55 – 5 =
55 – 5 = 50,

Explanation:
Given 55 – 5 = subtracting 5 from 55 we get 50,
So, 55 – 5 = 50.

Question 37.
5 + 60 =
5 + 60 = 65,

Explanation:
Given 5 + 60 = adding 5 to 60 we get 65,
So, 5 + 60 = 65.

Question 38.
5 + 65 =
5 + 65 = 70,

Explanation:
Given 5 + 65 = adding 5 to 65 we get 70,
So, 5 + 65 = 70.

Question 39.
70 – 5 =
70 – 5 = 65,

Explanation:
Given 70 – 5 = subtracting 5 from 70 we get 65,
So, 65 – 5 = 60.

Question 40.
65 – 5 =
65 – 5 = 60,

Explanation:
Given 65 – 5 = subtracting 5 from 65 we get 60,
So, 65 – 5 = 60.

Question 41.
50 + 100 =
50 + 100 = 150,

Explanation:
Given 50 + 100 = adding 50 to 100 we get 150,
So, 50 + 100 = 150.

Question 42.
50 + 150 =
50 + 150 = 200,

Explanation:
Given 50 + 150 = adding 50 to 150 we get 200,
So, 50 + 150 = 200.

Question 43.
200 – 50 =
200 – 50 = 150,

Explanation:
Given 200 – 50 = subtracting 50 from 200 we get 150,
So, 200 – 50 = 150.

Question 44.
150 – 50 =
150 – 50 = 100,

Explanation:
Given 150 – 50 = subtracting 50 from 150 we get 100,
So, 150 – 50 = 100.

Eureka Math Grade 3 Module 1 Lesson 18 Problem Set Answer Key

Question 1.
8 × 10 = __80____

5 tens + ______3 tens_________ = 8 tens
(5 × 10) + (___3___ × 10) = 8 × 10
50 + ___30______ = ___80______
8 × 10 = ____80________

8 × 10 = 80

5 tens + 3 tens = 8 tens
(5 × 10) + (3 × 10) = 8 × 10
50 + 30 = 80
8 × 10 = 80

Explanation:
Given 8 X 10 =
We can write 8 X 10 as 5 tens + 3 tens,
(5 X 10) + (3 X 10) = 8 X 10,
50 + 30 = 80 or 8 X 10 = 80.

Question 2.
7 × 4 = ___28___

5 fours + ______2 fours____ = 7 fours
(5 × 4) + (___2___ × 4) = 7 × 4
20 + ____8_____ = ____28_____
7 × 4 = _____28_______

7 x 4 = 28,

5 fours + 2 fours = 7 fours
(5 × 4) + (2 × 4) = 7 × 4
20 + 8 = 28
7 × 4 = 28,

Explanation:
Given 7 X 4 =
We can write 7 x 4 as 5 fours + 2 fours,
(5 X 4) + (2 X 4) = 7 X 4,
20 + 8 = 28 or 7 X 4 = 28.

Question 3.
9 × 10 = ___90___

5 tens + _____4 tens__________ = 9 tens
(5 × 10) + (___4___ × 10) = 9 × 10
___50______ + ___40______ = ___ 90______
9 × 10 = _____90_______

9 X 10 = 90,

5 tens + 4 tens = 9 tens
(5 × 10) + (4 × 10) = 9 × 10
50 + 40 =  90
9 × 10 = 90

Explanation:
Given 9 X 10 =
We can write 9 X 10 as 5 tens + 4 tens,
(5 X 10) + (4 X 10) = 9 X 10,
50 + 40 = 90 or 9 X 10 = 90.

Question 4.
10 × 10 = __100____

____5 tens_________ + ___5 tens_________ = 10 tens
(___5___ × 10) + (__5____ × 10) = 10 × 10
___50_____ + ___50______ = ___100______
10 × 10 = ____100________

10 X 10 = 100

5 tens + 5 tens = 10 tens
(5 × 10) + (5 × 10) = 10 × 10
50 + 50 = 100
10 × 10 = 100

Explanation:
Given 10 X 10 =
We can write 10 X 10  as 5 tens + 5 tens,
(5 X 10) + (5 X 10) = 10 X 10,
50 + 50 = 100 or 10 X 10 = 100.

Question 5.
There are 7 teams in the soccer tournament.
Ten children play on each team. How many children are
playing in the tournament? Use the break apart and
distribute strategy, and draw a number bond to solve.
There are ____70______ children playing in the tournament.

There are 70 children playing in the tournament,
7 X 10 =

5 tens + 2 tens = 7 tens
(5 × 10) + (2 × 10) = 7 × 10
50 + 20 = 70
7 × 10 = 70

Explanation:
Given there are 7 teams in the soccer tournament.
Ten children play on each team. So number of children
playing in the tournament are 7 X 10 = 70,
Used the break apart and distribute strategy,
drawn a number bond to solve 7 X 10 =
wrote 7 as 5 tens + 2 tens = 7 tens,
(5 × 10) + (2 × 10) = 7 × 10,
50 + 20 = 70,
7 × 10 = 70,
Therefore, there are 70 children playing in the tournament.

Question 6.
What is the total number of sides on 8 triangles?

Total number of 24 sides on 8 triangles,

Explanation:
Given to find the total number of sides on 8 triangles,
as we know a triangle has 3 sides so 8 triangles means
3 sides X  8  = 24 sides,
therefore, total number of 24 sides on 8 triangles.

Question 7.
There are 12 rows of bottled drinks in the vending machine.
Each row has 10 bottles. How many bottles are in the vending machine?

There are 120 bottles in total in the vending machine,

Explanation:
Given there are 12 rows of bottled drinks in the vending machine
and each row has 10 bottles, So total number of bottles in
the vending machine are 12 X 10 = 120 bottles.

Eureka Math Grade 3 Module 1 Lesson 18 Exit Ticket Answer Key

Dylan used the break apart and distribute strategy to
solve a multiplication problem. Look at his work below,
write the multiplication problem Dylan solved,
and complete the number bond.

Explanation:
Given Dylan used the break apart and distribute strategy to
solve a multiplication problem. Looking at his work,
wrote the multiplication problem Dylan solved,
and completed the number bond as shown above,
as 6 X 4 = (5 X 4) + (1 X 4) = 20 + 4 =24, or 6 X 4 = 24.

Eureka Math Grade 3 Module 1 Lesson 18 Homework Answer Key

Question 1.
Match.

Matched as shown below :

Explanation:
Matched expressions as
7 tens = 5 tens + 2 tens = (5 X 10) + (2 X 10) = 70,
8 fours = 5 fours + 3 fours = (5 X 4) + (3 X 4) = 32,
9 tens =  6 tens + 3 tens = (6 X 10) + (3 X 10) = 90,
7 threes = 5 threes + 2 threes = (5 X 3) + (2 X 3) = 21.

Question 2.
9 × 4 = ___36___

(_5_ × 4) + (_4_ × 4) = 9 × 4
_20__ + _16_ = _36_
9 × 4 = _36_.

9 X 4 = 36,

(5 X 4 + 4 X 4) = 9 X 4,
20 + 16 = 36,
9 X 4 = 36,

Explanation:
Given 9 X 4 =
We can write 9 X 4 as 5 fours + 4 fours,
(5 X 4) + (4 X 4) = 9 X 4,
20 + 16 = 36 or 9 X 4 = 36.

Question 3.
Lydia makes 10 pancakes. She tops each pancake with
4 blueberries. How many blueberries does Lydia use in all?
Use the break apart and distribute strategy, and
draw a number bond to solve.
Lydia uses __40__ blueberries in all.

Lydia uses 40 blueberries in all,

Explanation:
Given Lydia makes 10 pancakes. She tops each pancake with
4 blueberries. So number of  blueberries Lydia use in all are
4 X 10 = 40, Used the break apart and distribute strategy as
(2 X 10) + (2 X 10) = 20 + 20 = 40 and drawn a
number bond to solve 4 x 10 as 40.

Question 4.
Steven solves 7 × 3 using the break apart and distribute strategy.
Show an example of what Steven’s work might look like below.

7 X 3 = 21,

Example: Steven has 7 friends and each friend has
3 balloons, how many total balloons Steven will collect
from his friends?

Explanation:
Given Steven has 7 friends and each friend has
3 balloons, how many total balloons Steven will collect
from his friends?
Steven solves 7 × 3 using the
break apart and distribute strategy shown above as
7 X 3 = (4 X 3) + (3 X 3) = 12 + 9 = 21,
or 7 X 3 = 21.
So Steven collects 21 balloons in all.

Question 5.
There are 7 days in 1 week. How many days are there in 10 weeks?

There are 70 days in 10 weeks,

Explanation:
Given there are 7 days in 1 week. So number of days
in 10 weeks are 7 X 10 = 70 days.
therefore there are 70 days in 10 weeks.

Engage NY Eureka Math 3rd Grade Module 7 Lesson 21 Answer Key

Eureka Math Grade 3 Module 7 Lesson 21 Sprint Answer Key

A
Multiply or Divide by 3

Answer:

Question 1.
2 × 3 =
Answer:
2 × 3 = 6.

Question 2.
3 × 3 =
Answer:
3 × 3 = 9.

Question 3.
4 × 3 =
Answer:
4 × 3 = 12.

Question 4.
5 × 3 =
Answer:
5 × 3 = 15.

Question 5.
1 × 3 =
Answer:
1 × 3 = 3.

Question 6.
6 ÷ 3 =
Answer:
6 ÷ 3 = 2.

Question 7.
9 ÷ 3 =
Answer:
9 ÷ 3 = 3.

Question 8.
15 ÷ 3 =
Answer:
15 ÷ 3 = 5.

Question 9.
3 ÷ 3 =
Answer:
3 ÷ 3 = 1.

Question 10.
12 ÷ 3 =
Answer:
12 ÷ 3 = 4.

Question 11.
6 × 3 =
Answer:
6 × 3 = 18.

Question 12.
7 × 3 =
Answer:
7 × 3 = 21.

Question 13.
8 × 3 =
Answer:
8 × 3 = 24.

Question 14.
9 × 3 =
Answer:
9 × 3 = 27.

Question 15.
10 × 3 =
Answer:
10 × 3 = 30.

Question 16.
24 ÷ 3 =
Answer:
24 ÷ 3 = 8.

Question 17.
21 ÷ 3 =
Answer:
21 ÷ 3 = 7.

Question 18.
27 ÷ 3 =
Answer:
27 ÷ 3 = 9.

Question 19.
18 ÷ 3 =
Answer:
18 ÷ 3 = 6.

Question 20.
30 ÷ 3 =
Answer:
30 ÷ 3 = 10.

Question 21.
___ × 3 = 15
Answer:
_5__ × 3 = 15.

Question 22.
___ × 3 = 3
Answer:
__1_ × 3 = 3.

Question 23.
___ × 3 = 30
Answer:
_10__ × 3 = 30.

Question 24.
___ × 3 = 6
Answer:
_2__ × 3 = 6.

Question 25.
___ × 3 = 9
Answer:
_3__ × 3 = 9.

Question 26.
30 ÷ 3 =
Answer:
30 ÷ 3 = 10.

Question 27.
15 ÷ 3 =
Answer:
15 ÷ 3 = 5.

Question 28.
3 ÷ 3 =
Answer:
3 ÷ 3 = 1.

Question 29.
6 ÷ 3 =
Answer:
6 ÷ 3 = 2.

Question 30.
9 ÷ 3 =
Answer:
9 ÷ 3 = 3.

Question 31.
___ × 3 = 18
Answer:
_6__ × 3 = 18.

Question 32.
___ × 3 = 21
Answer:
_7__ × 3 = 21.

Question 33.
___ × 3 = 27
Answer:
_9__ × 3 = 27.

Question 34.
___ × 3 = 24
Answer:
_8__ × 3 = 24.

Question 35.
21 ÷ 3 =
Answer:
21 ÷ 3 = 7.

Question 36.
27 ÷ 3 =
Answer:
27 ÷ 3 = 9.

Question 37.
18 ÷ 3 =
Answer:
18 ÷ 3 = 6.

Question 38.
24 ÷ 3 =
Answer:
24 ÷ 3 = 8.

Question 39.
11 × 3 =
Answer:
11 × 3 = 33.

Question 40.
33 ÷ 3 =
Answer:
33 ÷ 3 = 11.

Question 41.
12 × 3 =
Answer:
12 × 3 = 36.

Question 42.
36 ÷ 3 =
Answer:
36 ÷ 3 = 12.

Question 43.
13 × 3 =
Answer:
13 × 3 = 39.

Question 44.
39 ÷ 3 =
Answer:
39 ÷ 3 = 13.

B
Multiply or Divide by 3

Answer:

Question 1.
1 × 3 =
Answer:
1 × 3 = 3.

Question 2.
2 × 3 =
Answer:
2 × 3 = 6.

Question 3.
3 × 3 =
Answer:
3 × 3 = 9.

Question 4.
4 × 3 =
Answer:
4 × 3 = 12.

Question 5.
5 × 3 =
Answer:
5 × 3 = 15.

Question 6.
9 ÷ 3 =
Answer:
9 ÷ 3 = 3.

Question 7.
6 ÷ 3 =
Answer:
6 ÷ 3 = 2.

Question 8.
12 ÷ 3 =
Answer:
12 ÷ 3 = 4.

Question 9.
3 ÷ 3 =
Answer:
3 ÷ 3 = 1.

Question 10.
15 ÷ 3 =
Answer:
15 ÷ 3 = 5.

Question 11.
10 × 3 =
Answer:
10 × 3 = 30.

Question 12.
6 × 3 =
Answer:
6 × 3 = 18.

Question 13.
7 × 3 =
Answer:
7 × 3 = 21.

Question 14.
8 × 3 =
Answer:
8 × 3 = 24.

Question 15.
9 × 3 =
Answer:
9 × 3 = 27.

Question 16.
21 ÷ 3 =
Answer:
21 ÷ 3 = 7.

Question 17.
18 ÷ 3 =
Answer:
18 ÷ 3 = 6.

Question 18.
24 ÷ 3 =
Answer:
24 ÷ 3 = 8.

Question 19.
30 ÷ 3 =
Answer:
30 ÷ 3 = 10.

Question 20.
27 ÷ 3 =
Answer:
27 ÷ 3 = 9.

Question 21.
___ × 3 = 3
Answer:
__1_ × 3 = 3.

Question 22.
___ × 3 = 15
Answer:
_5__ × 3 = 15.

Question 23.
___ × 3 = 6
Answer:
_2__ × 3 = 6.

Question 24.
___ × 3 = 30
Answer:
_10__ × 3 = 30.

Question 25.
___ × 3 = 9
Answer:
_3__ × 3 = 9.

Question 26.
6 ÷ 3 =
Answer:
6 ÷ 3 = 2.

Question 27.
3 ÷ 3 =
Answer:
3 ÷ 3 = 1.

Question 28.
30 ÷ 3 =
Answer:
30 ÷ 3 = 10.

Question 29.
15 ÷ 3 =
Answer:
15 ÷ 3 = 5.

Question 30.
9 ÷ 3 =
Answer:
9 ÷ 3 = 3.

Question 31.
___ × 3 = 18
Answer:
_6__ × 3 = 18.

Question 32.
___ × 3 = 24
Answer:
_8__ × 3 = 24.

Question 33.
___ × 3 = 27
Answer:
_9__ × 3 = 27.

Question 34.
___ × 3 = 21
Answer:
_7__ × 3 = 21.

Question 35.
24 ÷ 3 =
Answer:
24 ÷ 3 = 8.

Question 36.
27 ÷ 3 =
Answer:
27 ÷ 3 = 9.

Question 37.
18 ÷ 3 =
Answer:
18 ÷ 3 = 6.

Question 38.
21 ÷ 3 =
Answer:
21 ÷ 3 = 7.

Question 39.
11 × 3 =
Answer:
11 × 3 = 33.

Question 40.
33 ÷ 3 =
Answer:
33 ÷ 3 = 11.

Question 41.
12 × 3 =
Answer:
12 × 3 = 36.

Question 42.
36 ÷ 3 =
Answer:
36 ÷ 3 = 12.

Question 43.
13 × 3 =
Answer:
13 × 3 = 39.

Question 44.
39 ÷ 3 =
Answer:
39 ÷ 3 = 13.

Eureka Math Grade 3 Module 7 Lesson 21 Problem Set Answer Key

Question 1.
On your centimeter grid paper, shade and label as many rectangles as you can with a perimeter of 16 centimeters.
a. Sketch the rectangles below, and label the side lengths.
b. Find the area of each rectangle you drew above.
Answer:
a.
b. Area of the ABCD Rectangle = 7 square cm.
Area of the EFGH Rectangle = 12 square cm.
Area of the IJKL Rectangle = 15 square cm.

Explanation:
a.
b.
Figure 1 : ABCD Rectangle
Length of the ABCD Rectangle = 7cm
Width of the ABCD Rectangle = 1cm
Area of the ABCD Rectangle = Length  × Width
= 7cm × 1cm
= 7 square cm.

Figure 2 : EFGH Rectangle
Length of the EFGH Rectangle = 6cm
Width of the EFGH Rectangle = 2cm
Area of the EFGH Rectangle = Length  × Width
= 6cm × 2cm
= 12 square cm.

Figure 3 : IJKL Rectangle
Length of the IJKL Rectangle = 5cm
Width of the IJKL Rectangle = 3cm
Area of the IJKL Rectangle = Length  × Width
= 5cm × 3cm
= 15 square cm.

Question 2.
On your centimeter grid paper, shade and label as many rectangles as you can with a perimeter of 18 centimeters.
a. Sketch the rectangles below, and label the side lengths.
b. Find the area of each rectangle you drew above.
Answer:
a.
b. Area of the ABCD Rectangle = 8 square cm.
Area of the EFGH Rectangle = 14 square cm.
Area of the IJKL Rectangle = 18 square cm.
Area of the MNOP Rectangle = 20 square cm.

Explanation:
a.

b.

Figure 1 : ABCD Rectangle
Length of the ABCD Rectangle = 8cm
Width of the ABCD Rectangle = 1cm
Area of the ABCD Rectangle = Length  × Width
= 8cm × 1cm
= 8 square cm.

Figure 2 : EFGH Rectangle
Length of the EFGH Rectangle = 7cm
Width of the EFGH Rectangle = 2cm
Area of the EFGH Rectangle = Length  × Width
= 7cm × 2cm
= 14 square cm.

Figure 3 : IJKL Rectangle
Length of the IJKL Rectangle = 6cm
Width of the IJKL Rectangle = 3cm
Area of the IJKL Rectangle = Length  × Width
= 6cm × 3cm
= 18 square cm.

Figure 4 : MNOP Rectangle
Length of the MNOP Rectangle = 5cm
Width of the MNOP Rectangle = 4cm
Area of the MNOP Rectangle = Length  × Width
= 5cm × 4cm
= 20 square cm.

Question 3.
Use centimeter grid paper to shade in as many rectangles as you can with the given perimeters.
a. Use the charts below to show how many rectangles you shaded for each given perimeter. You might not use all the spaces in the charts.

b. Did you make a square with either of the given perimeters? How do you know?
Answer:
Yes, I  have made a square with lengths of the given perimeters of 20cm. I know it  because a rectangle with the side length all  equal to 5cm and a rectangle with all equal sides is to be known as Square.

Explanation:
Yes, because a rectangle with side length all  equal to 5cm and a rectangle with all equal sides is to be known as Square.

Question 4.
Macy and Gavin both draw rectangles with perimeters of 16 centimeters. Use words and pictures to explain how it is possible for Macy’s and Gavin’s rectangles to have the same perimeters but different areas.
Answer:

Area of the ABCD Macy’s rectangle = 7 Square cm.
Perimeter of the ABCD Macy’s rectangle = 16cms.
Area of the EFGH Gavin’s rectangle = 12 Square cm.
Perimeter of the EFGH Gavin’s rectangle = 16cms.
Here, Macy and Gavin has drawn two different rectangles with different lengths  perimeters same 16cms yet different areas. Well, its possible to draw as many as rectangles with perimeters of 16 centimeters but different areas.

Explanation:

Figure1: Macy’s rectangle:
Length of the ABCD Macy’s rectangle = 7cm
Width of the ABCD Macy’s rectangle = 1cm
Area of the ABCD Macy’s rectangle = Length × Width
= 7cm ×  1cm
= 7 Square cm.
Perimeter of the ABCD Macy’s rectangle =  2 ( Length + Width )
= 2 ( 7cm + 1cm)
= 2 ×  8cm
= 16 cms.

Figure2: Gavin’s rectangle:
Length of the EFGH Gavin’s rectangle = 7cm
Width of the EFGH Gavin’s rectangle = 1cm
Area of the EFGH Gavin’s rectangle = Length × Width
= 6cm ×  2cm
= 12 Square cm.
Perimeter of the EFGH Gavin’s rectangle =  2 ( Length + Width )
= 2 ( 6cm + 2cm)
= 2 ×  8cm
= 16 cms .

Eureka Math Grade 3 Module 7 Lesson 21 Exit Ticket Answer Key

On the grid below, shade and label at least two different rectangles with a perimeter of 20 centimeters.

Answer:

Explanation:
I have drawn three rectangles having lengths as ABCD Rectangle ( 8cm,2cm) , EFGH Rectangle  ( 6cm,4cm)  and IJKL Rectangle ( 7cm,3cm).

Eureka Math Grade 3 Module 7 Lesson 21 Homework Answer Key

Question 1.
Margo finds as many rectangles as she can with a perimeter of 14 centimeters.
a. Shade Margo’s rectangles on the grid below. Label the length and width of each rectangle.

b. Find the areas of the rectangles in part (a) above.
c. The perimeters of the rectangles are the same. What do you notice about the areas?
Answer:
a.

b. Area of the ABCD Rectangle = 12 square cm.
Area of the EFGH Rectangle = 10 square cm.
Area of the IJKL Rectangle = 6 square cm.

c. Perimeters of the three rectangles are the same having 14 cms, where as their areas are completely different from each other.

Explanation:
a.

b. Figure 1 : ABCD Rectangle
Length of the ABCD Rectangle = 4cm
Width of the ABCD Rectangle = 3cm
Area of the ABCD Rectangle = Length  × Width
= 4cm × 3cm
= 12 square cm.

Figure 2 : EFGH Rectangle
Length of the EFGH Rectangle = 5cm
Width of the EFGH Rectangle = 2cm
Area of the EFGH Rectangle = Length  × Width
= 5cm × 2cm
= 10 square cm.

Figure 3 : IJKL Rectangle
Length of the IJKL Rectangle = 6cm
Width of the IJKL Rectangle = 1cm
Area of the IJKL Rectangle = Length  × Width
= 6cm × 1cm
= 6 square cm.

c. Perimeter of the ABCD  rectangle =  2 ( Length + Width )
= 2 ( 4cm + 3cm)
= 2 ×  7cm
= 14 cms.

Perimeter of the EFGH rectangle =  2 ( Length + Width )
= 2 ( 5cm + 2cm)
= 2 ×  7cm
= 14 cms.

Perimeter of the IJKL rectangle =  2 ( Length + Width )
= 2 ( 6cm + 1cm)
= 2 ×  7cm
= 14 cms.
Perimeters of the three rectangles are the same having 14 cms, where as their areas are completely different from each other.
Perimeter requires only the lengths of the four sides of rectangle to be added where as for area we require length and width of the rectangle.

Question 2.
Tanner uses unit squares to build rectangles that have a perimeter of 18 units. He creates the chart below to record his findings.
a. Complete Tanner’s chart. You might not use all the spaces in the chart.

b. Explain how you found the widths and lengths in the chart above.
Answer:
The chart in itself has the labelling heading of the columns as length and width and their areas.
If not always length value is going to be more than the value of width in the measurements of rectangle.

Explanation:
The chart in itself has the labelling heading of the columns as length and width and their areas. If not always length value is going to be more than the value of width in the measurements of rectangle. This is how I found the widths and lengths in the chart above.

Question 3.
Jason and Dina both draw rectangles with perimeters of 12 centimeters, but their rectangles have different areas. Explain with words, pictures, and numbers how this is possible.
Answer:

Area of the Jason’s ABCD Rectangle = 8 square cm.
Perimeter of the Jason’s ABCD Rectangle = 12 cm.
Area of the Dina’s EFGH Rectangle = 5 square cm.
Perimeter of the Dina’s EFGH Rectangle = 12 cm.
Perimeter requires only the lengths of the four sides of rectangle to be added where as for area we require length and width of the rectangle. Jason’s ABCD rectangle and Dina’s EFGH rectangle are having same perimeter yet different areas.

Explanation:
Jason’s ABCD Rectangle:
Length of the Jason’s ABCD Rectangle = 4cm
Width of the Jason’s ABCD Rectangle = 2cm
Area of the Jason’s ABCD Rectangle = Length × Width
= 4cm ×  2cm
= 8 square cm.
Perimeter of the Jason’s ABCD Rectangle = 2 ( Length + Width )
= 2 ( 4cm + 2cm )
= 2 × 6cm
= 12 cm.

Dina’s EFGH Rectangle:
Length of the Dina’s EFGH Rectangle = 5cm
Width of the Dina’s EFGH Rectangle = 1cm
Area of the Dina’s EFGH Rectangle = Length × Width
= 5cm ×  1cm
= 5 square cm.
Perimeter of the Dina’s EFGH Rectangle = 2 ( Length + Width )
= 2 ( 5cm + 1cm )
= 2 × 6cm
= 12 cm.
Jason’s ABCD rectangle and Dina’s EFGH rectangle are having same perimeter yet different areas.

Eureka Math Grade 3 Module 7 Lesson 21 Data Sheet Answer Key

Use the data you gathered from Problem Sets 20 and 21 to complete the charts to show how many rectangles you can create with a given perimeter. You might not use all the spaces in the charts.

Engage NY Eureka Math 3rd Grade Module 1 Lesson 17 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 17 Sprint Answer Key

A
Multiply or Divide by 4

Explanation:
Multiplied or divided the given number by 4,
as shown above.

B
Multiply or Divide by 4

Eureka Math Grade 3 Module 1 Lesson 17 Problem Set Answer Key

Question 1.
Use the array to complete the related equations.

Explanation:
Used the array to complete the related equations as shown above and as
1 X 4 = 4,   4 ÷ 4 = 1,
2 X 4 = 8,   8 ÷ 4 = 2,
3 X 4 = 12, 12 ÷ 4 = 3,
4 X 4 = 16, 16 ÷ 4 = 4,
5 X 4 = 20, 20 ÷ 4 = 5,
6 X 4 = 24, 24 ÷ 4 = 6,
7 X 4 = 28, 28 ÷ 4 = 7,
8 X 4 = 32, 32 ÷ 4 = 8,
9 X 4 = 36, 36 ÷ 4 = 9,
10 X 4 = 40, 40 ÷ 4 = 10.

Question 2.
The baker packs 36 bran muffins in boxes of 4.
Draw and label a tape diagram to find the number of boxes he packs.

The baker packs 9 boxes of muffins,

Explanation:
The baker packs 36 bran muffins in boxes of 4.
Drawn and labeled a tape diagram,
The number of boxes baker packs is 36 ÷ 4 = 9 boxes.
Therefore, the baker packs 9 boxes of muffins.

Question 3.
The waitress arranges 32 glasses into 4 equal rows.
How many glasses are in each row?

There are 8 glasses in each row,

Explanation:
Given the waitress arranges 32 glasses into 4 equal rows.
Number of glasses in each row are 32 ÷ 4 = 8 (8 X 4 = 32),
Therefore, there are 8 glasses in each row.

Question 4.
Janet paid $28 for 4 notebooks. Each notebook costs the same amount.
What is the cost of 2 notebooks?

The cost of 2 notebooks is $14,

Explanation:
Given Janet paid $28 for 4 notebooks and each
notebook costs the same amount, So cost of each
notebook is $28 ÷ 4 = $7 (4 X 7 = 28),
Now cost for 2 notebooks is 2 X $7 = $14.

Eureka Math Grade 3 Module 1 Lesson 17 Exit Ticket Answer Key

Question 1.
Mr. Thomas organizes 16 binders into stacks of 4.
How many stacks does he make? Draw and label a
number bond to solve.

Mr. Thomas makes 4 stacks of 4,

Explanation:
Given Mr. Thomas organizes 16 binders into stacks of 4.
So number of stacks does he make are 16 ÷ 4 = 4 ( 4 X 4 = 16),
Drawn and labeled a number bond to solve number of stacks
Mr. Thomas makes as shown above in the picture.

Question 2.
The chef uses 28 avocados to make 4 batches of guacamole.
How many avocados are in 2 batches of guacamole?
Draw and label a tape diagram to solve.

In 2 batches of guacamole the chef uses 14 avocados,

Explanation:
Given the chef uses 28 avocados to make 4 batches of guacamole.
So number of avocados in one batch of guacamole are
28 ÷ 4 =7 (7 X 4 = 28),
Now number of avocados are in 2 batches of guacamole are
2 X 7 = 14 avocados.
Drawn and labeled a tape diagram as shown above.

Eureka Math Grade 3 Module 1 Lesson 17 Homework Answer Key

Question 1.
Use the array to complete the related equations.

Explanation:
Used the array to complete the related equations as shown above and as
1 X 4 = 4,   4 ÷ 4 = 1,
2 X 4 = 8,   8 ÷ 4 = 2,
3 X 4 = 12, 12 ÷ 4 = 3,
4 X 4 = 16, 16 ÷ 4 = 4,
5 X 4 = 20, 20 ÷ 4 = 5,
6 X 4 = 24, 24 ÷ 4 = 6,
7 X 4 = 28, 28 ÷ 4 = 7,
8 X 4 = 32, 32 ÷ 4 = 8,
9 X 4 = 36, 36 ÷ 4 = 9,
10 X 4 = 40, 40 ÷ 4 = 10.

Question 2.
The teacher puts 32 students into groups of 4.
How many groups does she make? Draw and label a
tape diagram to solve.

Teacher makes 8 groups,

Explanation :
Given the teacher puts 32 students into groups of 4.
Teacher makes 32 ÷ 4 = 8 (4 X 8 = 32) groups,
Drawn and labeled tape diagram to solve number of
groups teacher has made.

Question 3.
The store clerk arranges 24 toothbrushes into 4 equal rows.
How many toothbrushes are in each row?

In each row there are 6 toothbrushes,

Explanation:
Given the store clerk arranges 24 toothbrushes into 4 equal rows.
So number of toothbrushes in each row are 24 ÷ 4 = 6 (6 X 4 = 24).

Question 4.
An art teacher has 40 paintbrushes. She divides them equally
among her 4 students. She finds 8 more brushes and divides
these equally among the students, as well. How many brushes
does each student receive?

Each student will receive 12 paintbrushes,

Explanation:
Given An art teacher has 40 paintbrushes. She divides them equally
among her 4 students. She finds 8 more brushes and divides
these equally among the students, as well.
So number of brushes she has 40 + 8 = 48 brushes,
Now number of brushes each student will receive are
48 ÷ 4 = 12 ( 12 X 4 = 48).
Therefore, each student will receive 12 paintbrushes.

Engage NY Eureka Math 3rd Grade Module 7 Lesson 20 Answer Key

Eureka Math Grade 3 Module 7 Lesson 20 Sprint Answer Key

A
Multiply or Divide by 2

Answer:

Question 1.
2 × 2 =
Answer:
2 × 2 = 4.

Question 2.
3 × 2 =
Answer:
3 × 2 = 6.

Question 3.
4 × 2 =
Answer:
4 × 2 = 8.

Question 4.
5 × 2 =
Answer:
5 × 2 = 10.

Question 5.
1 × 2 =
Answer:
1 × 2 = 2.

Question 6.
4 ÷ 2 =
Answer:
4 ÷ 2 = 2.

Question 7.
6 ÷ 2 =
Answer:
6 ÷ 2 = 3.

Question 8.
10 ÷ 2 =
Answer:
10 ÷ 2 = 5.

Question 9.
2 ÷ 1 =
Answer:
2 ÷ 1 = 2.

Question 10.
8 ÷ 2 =
Answer:
8 ÷ 2 = 4.

Question 11.
6 × 2 =
Answer:
6 × 2 = 12.

Question 12.
7 × 2 =
Answer:
7 × 2 = 14.

Question 13.
8 × 2 =
Answer:
8 × 2 = 16.

Question 14.
9 × 2 =
Answer:
9 × 2 = 18.

Question 15.
10 × 2 =
Answer:
10 × 2 = 20.

Question 16.
16 ÷ 2 =
Answer:
16 ÷ 2 = 8.

Question 17.
14 ÷ 2 =
Answer:
14 ÷ 2 = 7.

Question 18.
18 ÷ 2 =
Answer:
18 ÷ 2 = 9.

Question 19.
12 ÷ 2 =
Answer:
12 ÷ 2 = 6.

Question 20.
20 ÷ 2 =
Answer:
20 ÷ 2 = 10.

Question 21.
___ × 2 = 10
Answer:
_5__ × 2 = 10.

Question 22.
___ × 2 = 12
Answer:
_6__ × 2 = 12.

Question 23.
___ × 2 = 20
Answer:
_10__ × 2 = 20.

Question 24.
___ × 2 = 4
Answer:
_2__ × 2 = 4.

Question 25.
___ × 2 = 6
Answer:
__3_ × 2 = 6.

Question 26.
20 ÷ 2 =
Answer:
20 ÷ 2 = 10.

Question 27.
10 ÷ 2 =
Answer:
10 ÷ 2 = 5.

Question 28.
2 ÷ 1 =
Answer:
2 ÷ 1 = 2.

Question 29.
4 ÷ 2 =
Answer:
4 ÷ 2 = 2.

Question 30.
6 ÷ 2 =
Answer:
6 ÷ 2 = 3.

Question 31.
___ × 2 = 12
Answer:
_6__ × 2 = 12.

Question 32.
___ × 2 = 14
Answer:
_7__ × 2 = 14.

Question 33.
___ × 2 = 18
Answer:
_9__ × 2 = 18.

Question 34.
___ × 2 = 16
Answer:
_8__ × 2 = 16.

Question 35.
14 ÷ 2 =
Answer:
14 ÷ 2 = 7.

Question 36.
18 ÷ 2 =
Answer:
18 ÷ 2 = 9.

Question 37.
12 ÷ 2 =
Answer:
12 ÷ 2 = 6.

Question 38.
16 ÷ 2 =
Answer:
16 ÷ 2 = 8.

Question 39.
11 × 2 =
Answer:
11 × 2 = 22.

Question 40.
22 ÷ 2 =
Answer:
22 ÷ 2 = 11.

Question 41.
12 × 2 =
Answer:
12 × 2 = 24.

Question 42.
24 ÷ 2 =
Answer:
24 ÷ 2 = 12.

Question 43.
14 × 2 =
Answer:
14 × 2 = 28.

Question 44.
28 ÷ 2 =
Answer:
28 ÷ 2 = 14.

B
Multiply or Divide by 2

Answer:

Question 1.
1 × 2 =
Answer:
1 × 2 = 2.

Question 2.
2 × 2 =
Answer:
2 × 2 = 4.

Question 3.
3 × 2 =
Answer:
3 × 2 = 6.

Question 4.
4 × 2 =
Answer:
4 × 2 = 8.

Question 5.
5 × 2 =
Answer:
5 × 2 = 10.

Question 6.
6 ÷ 2 =
Answer:
6 ÷ 2 = 3.

Question 7.
4 ÷ 2 =
Answer:
4 ÷ 2 = 2.

Question 8.
8 ÷ 2 =
Answer:
8 ÷ 2 = 4.

Question 9.
2 ÷ 1 =
Answer:
2 ÷ 1 = 2.

Question 10.
10 ÷ 2 =
Answer:
10 ÷ 2 = 5.

Question 11.
10 × 2 =
Answer:
10 × 2 = 20.

Question 12.
6 × 2 =
Answer:
6 × 2 = 12.

Question 13.
7 × 2 =
Answer:
7 × 2 = 14.

Question 14.
8 × 2 =
Answer:
8 × 2 = 16.

Question 15.
9 × 2 =
Answer:
9 × 2 = 18.

Question 16.
14 ÷ 2 =
Answer:
14 ÷ 2 = 7.

Question 17.
12 ÷ 2 =
Answer:
12 ÷ 2 = 6.

Question 18.
16 ÷ 2 =
Answer:
16 ÷ 2 = 8.

Question 19.
20 ÷ 2 =
Answer:
20 ÷ 2 = 10.

Question 20.
18 ÷ 2 =
Answer:
18 ÷ 2 = 9.

Question 21.
___ × 2 = 12
Answer:
_6__ × 2 = 12.

Question 22.
___ × 2 = 10
Answer:
_5__ × 2 = 10.

Question 23.
___ × 2 = 4
Answer:
_2__ × 2 = 4.

Question 24.
___ × 2 = 20
Answer:
_10__ × 2 = 20.

Question 25.
___ × 2 = 6
Answer:
_3__ × 2 = 6.

Question 26.
4 ÷ 2 =
Answer:
4 ÷ 2 = 2.

Question 27.
2 ÷ 1 =
Answer:
2 ÷ 1 = 2.

Question 28.
20 ÷ 2 =
Answer:
20 ÷ 2 = 10.

Question 29.
10 ÷ 2 =
Answer:
10 ÷ 2 = 5.

Question 30.
6 ÷ 2 =
Answer:
6 ÷ 2 = 3.

Question 31.
___ × 2 = 12
Answer:
_6__ × 2 = 12.

Question 32.
___ × 2 = 16
Answer:
_8__ × 2 = 16.

Question 33.
___ × 2 = 18
Answer:
_9__ × 2 = 18.

Question 34.
___ × 2 = 14
Answer:
_7__ × 2 = 14.

Question 35.
16 ÷ 2 =
Answer:
16 ÷ 2 = 8.

Question 36.
18 ÷ 2 =
Answer:
18 ÷ 2 = 9.

Question 37.
12 ÷ 2 =
Answer:
12 ÷ 2 = 6.

Question 38.
14 ÷ 2 =
Answer:
14 ÷ 2 = 7.

Question 39.
11 × 2 =
Answer:
11 × 2 = 22.

Question 40.
22 ÷ 2 =
Answer:
22 ÷ 2 = 11.

Question 41.
12 × 2 =
Answer:
12 × 2 = 24.

Question 42.
24 ÷ 2 =
Answer:
24 ÷ 2 = 12.

Question 43.
13 × 2 =
Answer:
13 × 2 = 26.

Question 44.
26 ÷ 2 =
Answer:
26 ÷ 2 = 13.

Eureka Math Grade 3 Module 7 Lesson 20 Problem Set Answer Key

Question 1.
Use your square unit tiles to build as many rectangles as you can with a perimeter of 12 units.
a. Estimate to draw your rectangles below. Label the side lengths of each rectangle.
b. Explain your strategy for finding rectangles with a perimeter of 12 units.
c. Find the areas of all the rectangles in part (a) above.
d. The perimeters of all the rectangles are the same. What do you notice about their areas?
Answer:
a.

b. Well, I have choose the multiples of 12 to draw the rectangles. Perimeter of the rectangles = 12units.

c. Area of the ABCD rectangle = 12 square units.
Area of the EFGH rectangle = 12 square units.
Area of the IJKL rectangle = 12 square units.

d. Perimeter of ABCD Rectangle = 16units.
Perimeter of EFGH Rectangle = 14units.
Perimeter of IJKL Rectangle = 26units.
Well, areas of all rectangles are same yet their perimeters are completely different to one and the another rectangle.

Explanation:
a.

b. Perimeter of the rectangles = 12units.
Well, I have choose the multiples of 12 to draw the rectangles.

c.
Figure 1: ABCD Rectangle:
Length of the ABCD rectangle = 6units
Width of the ABCD rectangle = 2units
Area of the ABCD rectangle = Length × Width
= 6units × 2units
= 12 square units.

Figure 2: EFGH Rectangle:
Length of the EFGH rectangle = 4units
Width of the EFGH rectangle = 3units
Area of the EFGH rectangle = Length × Width
= 4units × 3units
= 12 square units.

Figure 3: IJKL Rectangle:
Length of the IJKL rectangle = 12units
Width of the IJKL rectangle = 1units
Area of the IJKL rectangle = Length × Width
= 12units × 1units
= 12 square units.

d.
Figure 1: ABCD Rectangle:
Length of the ABCD rectangle = 6units
Width of the ABCD rectangle = 2units
Perimeter of ABCD Rectangle = 2 ( Length + Width  )
= 2 ( 6units + 2units )
= 2 × 8units
= 16 units.

Figure 2: EFGH Rectangle:
Length of the EFGH rectangle = 4units
Width of the EFGH rectangle = 3units
Perimeter of EFGH Rectangle = 2 ( Length + Width  )
= 2 ( 4units + 3units )
= 2 × 7units
= 14 units.

Figure 3: IJKL Rectangle:
Length of the IJKL rectangle = 12units
Width of the IJKL rectangle = 1units
Perimeter of IJKL Rectangle = 2 ( Length + Width  )
= 2 ( 12units + 1units )
= 2 × 13units
= 26 units.
Well, areas of all rectangles are same yet their perimeters are completely different to one and the another rectangle.

Question 2.
Use your square unit tiles to build as many rectangles as you can with a perimeter of 14 units.
a. Estimate to draw your rectangles below. Label the side lengths of each rectangle.
b. Find the areas of all the rectangles in part (a) above.
c. Given a rectangle’s perimeter, what other information do you need to know about the rectangle to find its area?
Answer:
a.
b. Area of the ABCD rectangle = 30 square units.
Area of the EFGH rectangle = 18 square units.

c.  Perimeter of any figure is basically gives the length of the figure. Area of rectangle is the region covered by the rectangle in a two-dimensional plane. Area requires length and width for its calculation.

Explanation:
a.

b. Figure 1 : ABCD Rectangle:
Length of the ABCD rectangle = 14units
Width of the ABCD rectangle = 1unit
Area of the ABCD rectangle =  Length × Width
= 14units × 1units
= 14 square units.

Figure 2 :EFGH Rectangle:
Length of the EFGH rectangle = 7units
Width of the EFGH rectangle = 2unit
Area of the EFGH rectangle = Length × Width
=  7units × 2units
= 14 square units.

c. Perimeter of any figure is basically gives the length of the figure. Area of rectangle is the region covered by the rectangle in a two-dimensional plane. Area requires length and width for its calculation.

Eureka Math Grade 3 Module 7 Lesson 20 Exit Ticket Answer Key

Use your square unit tiles to build as many rectangles as you can with a perimeter of 8 units.
a. Estimate to draw your rectangles below. Label the side lengths of each rectangle.
b. Find the areas of the rectangles in part (a) above.
Answer:
a.

b. Area of the ABCD rectangle = 8 square units.
Area of the EFGH rectangle = 8 square units.

Explanation:
a.

b. Figure 1  : ABCD Rectangle:
Length of the ABCD rectangle = 4units
Width of the ABCD rectangle = 2unit
Area of the ABCD rectangle = Length × Width
=  4units × 2units
= 8 square units.

Figure 2 : EFGH Rectangle:
Length of the EFGH rectangle = 8units
Width of the EFGH rectangle = 1unit
Area of the EFGH rectangle = Length × Width
=  8units × 1units
= 8 square units.

Eureka Math Grade 3 Module 7 Lesson 20 Homework Answer Key

Question 1.
Cut out the unit squares at the bottom of the page. Then, use them to make as many rectangles as you can with a perimeter of 10 units.
a. Estimate to draw your rectangles below. Label the side lengths of each rectangle.
b. Find the areas of the rectangles in part (a) above.

Answer:
a.

b. Area of the ABCD rectangle = 10 square units.
Area of the EFGH rectangle = 10 square units.

Explanation:
a.

b. Figure 1  : ABCD Rectangle:
Length of the ABCD rectangle = 10units
Width of the ABCD rectangle = 1unit
Area of the ABCD rectangle =  Length × Width
= 10units × 1units
= 10 square units.

Figure 2: EFGH Rectangle:
Length of the EFGH rectangle = 5units
Width of the EFGH rectangle = 2units
Area of the EFGH rectangle = Length × Width
= 5units × 2units
= 10 square units.

Question 2.
Gino uses unit square tiles to make rectangles with a perimeter of 14 units. He draws his rectangles as shown below. Using square unit tiles, can Gino make another rectangle that has a perimeter of 14 units? Explain your answer.

Answer:

Perimeter of ABCD Rectangle = 14units.
Perimeter of EFGH Rectangle = 14units.

Explanation:

Figure 1: ABCD Rectangle
Length of the ABCD Rectangle = 6units
Width of the ABCD Rectangle = 1units
Perimeter of ABCD Rectangle = 2 (Length + Width)
= 2 ( 6units + 1unit )
= 2 × 7units
= 14units.

Figure 2: EFGH Rectangle
Length of the EFGH Rectangle = 4units
Width of the EFGH Rectangle = 3units
Perimeter of EFGH Rectangle = 2 (Length + Width)
= 2 ( 4units + 3unit )
= 2 × 7units
= 14units.

Question 3.
Katie draws a square that has a perimeter of 20 centimeters.
a. Estimate to draw Katie’s square below. Label the length and width of the square.
b. Find the area of Katie’s square.
c. Estimate to draw a different rectangle that has the same perimeter as Katie’s square.
d. Which shape has a greater area, Katie’s square or your rectangle?
Answer:
a.
b. Area of the ABCD Katie’s Square = 25 square units.

c.

Perimeter of the EFGH Rectangle = 20units.

d. Area of the ABCD Katie’s Square = 25 square units.
Area of the EFGH Rectangle = 24 square units.
Area of the ABCD Katie’s Square is greater than the Area of the EFGH Rectangle.

Explanation:
a.
b.  Length of the side of the ABCD Katie’s Square = 5units
Area of the ABCD Katie’s Square = Side × Side
= 5units × 5units
= 25 square units.

c.
Length of the EFGH Rectangle = 6units
Width of the EFGH Rectangle = 4units
Perimeter of the EFGH Rectangle = 2 ( Length + Width )
= 2 ( 6units + 4units )
= 2 × 10units
= 20units.

d. Length of the side of the ABCD Katie’s Square = 5units
Area of the ABCD Katie’s Square = Side × Side
= 5units × 5units
= 25 square units.

Length of the EFGH Rectangle = 6units
Width of the EFGH Rectangle = 4units
Area of the EFGH Rectangle = Length × Width
= 6units × 4units
= 24 square units.

Eureka Math Grade 3 Module 7 Lesson 20 Data Sheet Answer Key

Use the data you gathered from Problem Sets 20 and 21 to complete the charts to show how many rectangles you can create with a given perimeter. You might not use all the spaces in the charts.

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

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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 math pre calculus answer key | Eureka Math Pre Calculus Answer Key

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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.