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.

13 Times Table Chart

13 times multiplication tables in table format and image format is given here. So, it makes it easy for you to remember the values. Download 13 table charts for free and prepare well. The 13th table is helpful to perform the multiplication of numbers easily. You can save your time in competitive eams by learning these multiplication tables.

How to Read 13 Table?

Check the reading of the 13 multiplication table here.

One time thirteen is 13.

Two times thirteen is 26.

Three times thirteen is 39.

Four times thirteen is 52.

Five times thirteen is 65.

Six times thirteen is 78.

Seven times thirteen is 91.

Eight times thirteen is 104.

Nine times thirteen is 117.

Ten times thirteen is 130.

Importance of Multiplication Tables

Multiplication tables play an essential role in mathematics. It is the foundation of elementary maths. By learning the table chart, you will get self-confidence while doing multiplications. You can keep the information at your fingertips that help you to solve the questions quickly. Multiplication tables will enhance your memory power and improve the calculations speed.

Tables from 2 to 20 help in performing the simple arithmetic operations. So that you can save time and do calculations easily. Without learning the 13 times table, you can also calculate the multiplicative of 13 by performing the arithmetic multiplication operations.

Multiplication Table of 13 up to 20

Studying 13 Multiplication Table is an essential skill to solve the division and multiplication questions. Check out the below table to know how to write a 13 times table chart.

Tips and Tricks to Learn 13 Times Table

Here we are giving the easy tips that are helpful to remember the 13th table. Follow the below tricks and learn the multiplication tables quickly.

• To remember the 13 times table, first, we need to memorize the 3 times table. So, the multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, . . .
• For getting the multiples of 13, add natural numbers to the ten’s digit of the 3 multiples. Therefore, 13 times table is obtained as (1 + 0)3 = 13, (2 + 0)6 = 26 , (3 + 0)9 = 39, (4 + 1)2 = 52, (5 + 1)5 = 65, (6 + 1)8 = 78, (7 + 2)1 = 91, (8 + 2)4 = 104, (9 + 2)7 = 117, (10 + 3)0 = 130, . . .
• 13 does not have any rules that make the multiplication of 13 table easier to memorize, then there is a structure for every 10 multiples of 13. They are 13, 26, 39, 52, 65, 78, 91, 104, 117, 130. In all these multiples, the last digit i.e units place digit is repeating. So, one can remember this logic to memorize the table.

Get More Math Tables:

Solved Examples on 13 Times Table Multiplication

Example 1:

Using the table of 13, calculate 13 times 13 plus 13?

Solution:

From the given data

We can express the given data in the form of Mathematical Expression

= (13 x 13) + 13

= 169 + 13

= 182

Therefore, 13 times 13 plus 13 is 182.

Example 2:

If David’s father has to pay the amount “12 less than 13 times 15” in dollars. Using the table of 13, find how much he needs to pay?

Solution:

From the given data,

The mathematical expression of 12 less than 13 times 15 = (13 x 15) – 12

= 195 – 12

= 183

therefore, David’s father is required to pay $183.

Example 3: 

Families in a colony are going on a picnic. If 13 people ride in each car and there are 5 cars, then how many people are going on a picnic?

Solution:

Given that,

The number of people going on picnic = 13

Number of cars = 5

Then, multiply the number of people on each car, total number of cars on the picnic to get the total number of persons going for the picnic.

The number of persons going on the picnic = 13 x 5

= 65

Therefore, 65 people going on a picnic.

Example 4:

Using the 13 times table, check whether 13 times 7 minus 1 plus 10 is 100?

Solution:

Firstly, let us express the given statement in the form of mathematical expression

13 times 7 minus 1 plus 10 = (13 x 7) – 1 + 10

= (91) + 9

= 100

Hence, 13 times 7 minus 1 plus 10 is 100.

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.

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.

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.

Multiplication Tables for 0 to 25

Below is the list of Math Tabes for 0 to 25 to make your math calculations easier and faster. These Time Tables from 0 to 25 help your child to skillfully tackle complex problems too with ease. Remembering the Multiplication Tables not just brings out self-confidence in you but also keeps information prepared at your fingertips so that you can use it whenever required. On Mastering the 0 to 25 Multiplication Charts children can enhance their speed and accuracy.

Importance of Math Tables or Multiplication Tables

Math Tables are the basic blocks of arithmetic calculations. Learning these Multiplication Tables can stimulate the memory power in kids and develop observation skills. Things learned at an early age can be retained for a long time. Refer to the below modules to know why one should learn Maths Tables and they are as such

• Multiplication Tables are quite important as they support in student’s mathematical learning.
• Students can get a strong grasp of facts associated with multiplication.
• Makes it easier to solve problems in Mathematics.
• The one who tends to be good at these Times Tables will be self-confident while learning new math concepts.

Tips to Memorize Multiplication Times Tables

Most of you might struggle to memorize the math tables. However, we have curated few effective tips that help you to learn the Math Times Tables easily. They are in the following fashion

• Practice Skip Counting i.e. if you start with a number 2 keep adding 2 every time you count. That is you would say 2, 4, 6, 8, 10, 12, 14…..
• Recite the multiplication tables in order at least once a day until you remember them. Also, learn reverse recitation of tables.
• If you have any difficulty in remembering you can always opt for writing and memorizing. Make it a habit in your daily routine and try to memorize once.
• Apply multiplication tables to real life and try to understand them using real-life scenarios.
• Identify the Multiplication Table Patterns so that you can remember them much faster.

Learn 2 to 9 Times Multiplication Tricks

FAQs on Maths Times Tables

1. What is the easiest way to memorize the multiplication tables?

One of the easiest ways to memorize the Math Tables is through addition. We know the number of times a number is multiplied with another number it is added to itself the same number of times repeatedly.

2. How to memorize Multiplication Times Table?

In order to memorize the Multiplication Times Table start reciting it verbally on a regular basis or write it down on paper. In fact, you can solve questions involving multiplication charts to retain them for a long time.

 3. How Can I Learn Tables Fast?

Start with the easiest ones and then work on them until you feel confident. Learn the Tips & Tricks to memorize the Math Tables easily. Recite them as many times as possible or write to remember them fast.

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.

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.

3 Times Multiplication Chart

Learning Table of 3 is quite essential to do your math calculations in a much faster way. Utilize the Table of 3 to understand and gain a deeper insight into the multiplication operation. For your idea, we have given the 3 Times Multiplication Table in image format. You can download and save it without paying us a single penny. Use this Table of 3 Chart for quick reference and try to recite it every day and memorize it easily.

How to Read Three Times Multiplication Table?

One time three is 3

Two times three is 6

Three times three is 9

Four times three is 12

Five times three is 15

Six times three is 18

Seven times three is 21

Eight times three is 24

Nine times three is 27

Ten times three is 30

Eleven times three is 33

Twelve times three is 36

How to write a Multiplication Table of 3? | Three Times Table upto 20

Learn how to write Table of 3 by checking out the below table. We have outlined the 3 Times Table up to 20 natural numbers so that you can get deeper knowledge on the multiplication which would be beneficial in the long run.

Why should you Learn the Multiplication Table of 3?

Go through the below lines to know the importance of the Three Table. They are along the lines

• Learning the Table of 3 is a necessary skill to solve problems on long division and multiplication.
• You can perform your metal math calculations right in your head and improve your problem-solving skills.
• By Learning the Multiplication Table of 3, you can understand the pattern of multiples.
• Ensures faster calculations and saves a lot of time.

Tips to Memorize 3 Times Table

Follow the below-listed tips & tricks to keep in mind to learn and memorize the Multiplication Table of 3. They are in the below fashion

• • • • Apparently, there are any rules that make the table of 3 easy to remember. However, there is a pattern for every 10 multiples of 3 i.e. 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.
      • The last digit of these multiples will always repeat. You can remember these digits to memorize the multiples of 3 easily.
      • Another way to learn 3 up to 10 is through skip counting.

    

By referring to the above image you can understand that the units digit are nothing but counting numbers written three in one column in descending order from right to left. 0 is left outside the grid. Later, place the digit 0 before the units digit in row 1 and digit 1 in the second row and digit 2 in the third row. Also, place 3 before the digit 0 outside the grid. That’s it you have got the first ten multiples of 3.

Get More Math Tables:

Solved Examples Involving 3 Times Table

1. Using 3 times table, evaluate 3 times 6 minus 3?

Solution:

From the given statement we can write the mathematical expression as follows

= 3*6-3

= 18-3

=15

Thus, 3 times 6 minus 3 is 15

2. Using the table of 3, Find the value of 3 plus 3 times 7 minus 5?

Solution:

First, let us write 3 plus 3 times 7 minus 5 mathematically

= 3+3*7-5

= 3+21-5

= 24-5

= 19

Hence, 3 plus 3 times 7 minus 5 is 19

3. Find the value of 4 plus 3 times 6?

Solution:

Let us write the given statement mathematically

= 4+3*6

= 4+18

= 22

Hence, 4 plus 3 times 6 is 22

4. Using the table of 3, find the value of 3 times 5 minus 3 times 2?

Solution:

Let us write the given statement mathematically

= 3*5-3*2

= 15-6

=9

Hence, 3 times 5 minus 3 times 2 is 9.

Multiplication Table of 2

Here is the Multiplication Chart of Two in image format. Use it as a quick reference to make your mental math calculations much easier. Learning Table of 2 Promotes a better understanding of the multiplication operation. Knowing the basic math tables you can solve the maths problems in no time. You can save a lot of time in your competitive exams by learning the 2 Times Multiplication Table.

How to Read Table of Two?

One time two is 2

Two times two is 4

Three times two is 6

Four times two is 8

Five times two is 10

Six times two is 12

Seven times two is 14

Eight times two is 16

Nine times two is 18

Ten times two is 20

Eleven times two is 22

Twelve times two is 24

Multiplication Table of Two upto 20 | How to Write 2 Times Table?

Below is the Table of 2 up to 20 Natural Numbers. You can avail the Multiplication Table of Two here and use it whenever you need to perform quick calculations. Refer to the 2 Times Table over here to memorize it quickly and recite it on a regular basis so that you can retain it for a long time.

Get More Math Tables:

How to Learn Multiplication Table of Two?

Table of 2 is quite basic and one of the easiest tables to learn and memorize. Let us learn how to obtain and memorize the multiplication table of 2 in the below sections. Basic things to observe in the Multiplication Table of Two are as such

• 2 Times Table follows the pattern 2, 4, 6, 8, 0 in the units place.
• It will always have even numbers
• It is simple as we can learn using skip counting i.e. 2, 4, 6, 8, 10, 12……

Tips to Memorize Two Times Table

Follow the tips & tricks to memorize the multiplication table of 2 by referring to the below modules. They are along the lines

• Two Times Table always follows the pattern of the even numbers i.e. 2, 4, 6, 8, 10, ….
• Another simple way to memorize the Table of Two is through addition.
• You can memorize the Two Table taking the help of the 1 Times Table. All you have to do is simply add the natural numbers to multiples of 1 to get the Two Table.
  • 1×1+1 = 2
  • 1×2+2 =4
  • 1×3+3 =6
  • 1×4+4 =8
  • 1×5+5=10
  • 1×6+6 =12
  • 1×7+7=14
  • 1×8+8=16
  • 1×9+9=18
  • 1×10+10=20 and so on.

Solved Examples Involving Multiplication Table of 2

1. If Reethu eats 2 chocolates per day, Using the table of 2 find how many chocolates will she have eaten at the end of the 5th day?

Solution:

Reetu eats a number of chocolates per day = 2

Number of Chocolates Reethu eats at the end of the 5th day = 2*5

= 10

Therefore, Reethu eats 10 chocolates at the end of the 5th day.

2. Using 2 times table, find 2 times 3 times 4 times 2?

Solution:

Expressing the given statement in the form of mathematical expression we have

= 2*3*4*2

= 48

Therefore, 2 times 3 times 4 times 2 is 48.

3. James wants to buy 6 Pencils. If the cost of one Pencil is $2, using the table of 2, find the cost of 6 Pencils?

Solution:

The Cost of one Pencil is $2

No. of Pencils James wants to buy = 6

Cost of 6 Pencils = $2*6

= $12

Therefore, the cost of 6 Pencils is $12.

Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem

Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem will help you to complete your homework in time without any mistakes. The main aim of the ccssmathanswers.com site is to provide quick and simple methods to all the students of 8th grade. The solutions to all the questions in Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem are prepared by the math experts. Tap the links and practice the problems provided in the HMH Go Math 8th Grade Solution Key Chapter 12 The Pythagorean Theorem.

Chapter 12- Lesson 1: 

• Guided Practice – The Pythagorean Theorem – Page No. 378
• 12.1 Independent Practice – The Pythagorean Theorem – Page No. 379
• FOCUS ON HIGHER ORDER THINKING – The Pythagorean Theorem – Page No. 380

Chapter 12- Lesson 2: 

• Guided Practice – Converse of the Pythagorean Theorem – Page No. 384
• 12.2 Independent Practice – Converse of the Pythagorean Theorem – Page No. 385
• Converse of the Pythagorean Theorem – Page No. 386

Chapter 12- Lesson 3: 

• Guided Practice – Distance Between Two Points – Page No. 390
• 12.3 Independent Practice – Distance Between Two Points – Page No. 391
• Distance Between Two Points – Page No. 392
• Ready to Go On? – Model Quiz – Page No. 393
• Selected Response – Mixed Review – Page No. 394

Guided Practice – The Pythagorean Theorem – Page No. 378

Question 1.
Find the length of the missing side of the triangle

a² + b² = c² → 24² + ? = c² → ? = c²
The length of the hypotenuse is _____ feet.
_____ feet

Answer: The length of the hypotenuse is 26 feet.

Explanation: According to Pythagorean Theorem, we shall consider values of a = 24ft, b = 10ft.
Therefore c = √(a² +b²)
c = √(24² + 10²)
= √(576 + 100)
= √676 = 26ft

Question 2.
Mr. Woo wants to ship a fishing rod that is 42 inches long to his son. He has a box with the dimensions shown.

a. Find the square of the length of the diagonal across the bottom of the box.
________ inches

Answer: 1700 inches.

Explanation: Here we consider the length of the diagonal across the bottom of the box as d.
Therefore, according to Pythagorean Theorem
W² + l² = d²
40² + 10² = d²
1600 + 100 = d²
1700 = d²

Question 2.
b. Find the length from a bottom corner to the opposite top corner to the nearest tenth. Will the fishing rod fit?
________ inches

Answer: 42.42 inches.

Explanation: We denote by r, the length from the bottom corner to the opposite top corner. We use our Pythagorean formula to find r.
h² + s² = r²
10² + 1700 = r²
100 + 1700 = r²
1800 = r²,    r = √1800 => 42.42 inches

ESSENTIAL QUESTION CHECK-IN

Question 3.
State the Pythagorean Theorem and tell how you can use it to solve problems.

Answer:
Pythagorean Theorem: In a right triangle, the sum of squares of the legs a and b is equal to the square of the hypotenuse c.
a² + b² = c²
We can use it to find the length of a side of a right triangle when the lengths of the other two sides are known.

12.1 Independent Practice – The Pythagorean Theorem – Page No. 379

Find the length of the missing side of each triangle. Round your answers to the nearest tenth.

Question 4.

________ cm

Answer: 8.9 cm.

Explanation: According to Pythagorean theorem we consider values of a = 4cm, b = 8cm.
c² = a² + b²
= 4² + 8²
= 16 + 64
c²= 80, c= √80 => 8.944
After rounding to nearest tenth value c= 8.9cm

Question 5.

________ in.

Answer: 11.5 in.

Explanation: According to Pythagorean theorem we consider values of b = 8in, c= 14in
c² = a² + b²
14² = a² + 8²
196 = a² + 64
a² = 196 – 64
a  = √132 => 11.4891
a = 11.5 in

Question 6.
The diagonal of a rectangular big-screen TV screen measures 152 cm. The length measures 132 cm. What is the height of the screen?
________ cm

Answer: 75.4 cm

Explanation: Let’s consider the diagonal of the TV screen as C = 152cm, length as A = 132 cm, and height of the screen as B.

As C² = A² + B²
   152² = 132² + B²
23,104 = 17,424 + B²
B² = 23,104 – 17,424
B = √5680 => 75.365
So the height of the screen B = 75.4cm

Question 7.
Dylan has a square piece of metal that measures 10 inches on each side. He cuts the metal along the diagonal, forming two right triangles. What is the length of the hypotenuse of each right triangle to the nearest tenth of an inch?
________ in.

Answer: 14.1in.

Explanation:

Using the Pythagorean Theorem, we have:
a² + b² = c²
10² + 10² = c²
100 + 100 = c²
200 = c²
We are told to round the length of the hypotenuse of each right triangle to the nearest tenth of an inch, therefore: c = 14.1in

Question 8.
Represent Real-World Problems A painter has a 24-foot ladder that he is using to paint a house. For safety reasons, the ladder must be placed at least 8 feet from the base of the side of the house. To the nearest tenth of a foot, how high can the ladder safely reach?
________ ft

Answer: 22.6 ft.

Explanation: Consider the below diagram. Length of the ladder C = 24ft, placed at a distance from the base B = 8ft, let the safest height be A.

By using Pythagorean Theorem:
C² = A² + B²
24² = A² + 8²
576 = A² + 64
A² = 576 – 64 => 512
A = √512 => 22.627
After rounding to nearest tenth, value of A = 22.6ft

Question 9.
What is the longest flagpole (in whole feet) that could be shipped in a box that measures 2 ft by 2 ft by 12 ft?

________ ft

Answer: The longest flagpole (in whole feet) that could be shipped in this box is 12 feet.

Explanation: From the above diagram we have to find the value of r, which gives us the length longest flagpole that could be shipped in the box. Where width w = 2ft, height h = 2ft and length l = 12ft.

First find s, the length of the diagonal across the bottom of the box.
w² + l² = s²
2² + 12² = s²
4 + 144 = s²
148 = s²
We use our expression for s to find r, since triangle with sides s, r, and h also form a right-angle triangle.
h² + s² = r²
2² + 148 = r²
4 + 148 = r²
152 = r²
r = 12.33ft.

Question 10.
Sports American football fields measure 100 yards long between the end zones, and are 53 \(\frac{1}{3}\) yards wide. Is the length of the diagonal across this field more or less than 120 yards? Explain.
____________

Answer: The diagonal across this field is less than 120 yards.

Explanation: From the above details we will get a diagram as shown below.

We are given l = 100 and w = 53  =  . If we denote with d the diagonal of the field, using the Pythagorean Theorem, we have:
l² + w² = d²
100² + (160/3)² = d²
10000 + (25600/9) = d²
9*10000 + 9*(25600/9) = 9* d²
90000 + 25600 = 9 d²
(115600/9) = d²
(340/9) = d²
d = 113.3
Hence the diagonal across this field is less than 120 yards.

Question 11.
Justify Reasoning A tree struck by lightning broke at a point 12 ft above the ground as shown. What was the height of the tree to the nearest tenth of a foot? Explain your reasoning.

________ ft

Answer: The total height of the tree was 52.8ft

Explanation:

By using the Pythagorean Theorem
a² + b²  = c²
12² + 39² = c²
144 + 1521 = c²
1665 = c²
We are told to round the length of the hypotenuse to the nearest tenth of a foot, therefore: c = 40.8ft.
Therefore, the total height of the tree was:
height = a+c
height = 12 +40.8
height = 52.8 feet

FOCUS ON HIGHER ORDER THINKING – The Pythagorean Theorem – Page No. 380

Question 12.
Multistep Main Street and Washington Avenue meet at a right angle. A large park begins at this corner. Joe’s school lies at the opposite corner of the park. Usually Joe walks 1.2 miles along Main Street and then 0.9 miles up Washington Avenue to get to school. Today he walked in a straight path across the park and returned home along the same path. What is the difference in distance between the two round trips? Explain.
________ mi

Answer: Joe walks 1.2 miles less if he follows the straight path across the park.

Explanation: Using the Pythagorean Theorem, we find the distance from his home to school following the straight path across the park:
a² + b²  = c²
1.2² + 0.9² = c²
1.44 + 0.81 = c²
2.25 = c²
1.5 = c
Therefore, the distance of Joe’s round trip following the path across the park is 3 miles (d_home-school + d_school-home = 1.5 + 1.5). Usually, when he walks along Main Street and Washington Avenue, the distance of his round trip is 4.2 miles (d_home-school + d_school-home = (1.2 + 0.9) + (0.9+1.2)). As we can see, Joe walks 1.2 miles less if he follows the straight path across the park.

Question 13.
Analyze Relationships An isosceles right triangle is a right triangle with congruent legs. If the length of each leg is represented by x, what algebraic expression can be used to represent the length of the hypotenuse? Explain your reasoning.

Answer: c = x√ 2

Explanation: From the Pythagorean Theorem, we know that if a and b are legs and c is the hypotenuse, then a² + b²  = c². In our case, the length of each leg is represented by x, therefore we have:
a² + b²  = c²
x² + x² = c²
2x² = c²
c = x√ 2

Question 14.
Persevere in Problem Solving A square hamburger is centered on a circular bun. Both the bun and the burger have an area of 16 square inches.

a. How far, to the nearest hundredth of an inch, does each corner of the burger stick out from the bun? Explain.
________ in

Answer: Each corner of the burger sticks out 0.57 inches from the bun.

Explanation: Frist, we need to find the radius r of the circular bun. We know that its area A is 16 square inches, therefore:

A = πr²
16 = 3.14*r²
r² = (16/3.14)
r = 2.26

Then, we need to find the side s of the square hamburger. We know that its area A is 16 square inches, therefore:
A = s²
16 = s²
s = 4
Using the Pythagorean Theorem, we have to find diagonal d of the square hamburger:
s² + s² = d²
4² + 4² = d²
16 + 16 = d²
32 = d²
d = 5.66
To find how far does each corner of the burger stick out from the bun, we denote this length by a and we get:
a = (d/2) – r => (5.66/2) – 2.26
a = 0.57.
Therefore, Each corner of the burger sticks out 0.57 inches from the bun.

Question 14.
b. How far does each bun stick out from the center of each side of the burger?
________ in

Answer: Each bun sticks out 0.26 inches from the center of each side of the burger.

Explanation:

We found that r = 2.26 and s = 4. To find how far does each bun stick out from the center of each side of the burger, we denote this length by b and we get:
b = r – (s/2) = 2.26 – (4/2)
b = 0.26 inches.

Question 14.
c. Are the distances in part a and part b equal? If not, which sticks out more, the burger or the bun? Explain.

Answer: The distances a and b are not equal. From the calculations, we found that the burger sticks out more than the bun.

Guided Practice – Converse of the Pythagorean Theorem – Page No. 384

Question 1.
Lashandra used grid paper to construct the triangle shown.

a. What are the lengths of the sides of Lashandra’s triangle?
_______ units, _______ units, _______ units,

Answer: The length of Lashandra’s triangle is 8 units, 6 units, 10 units.

Question 1.
b. Use the converse of the Pythagorean Theorem to determine whether the triangle is a right triangle.

The triangle that Lashandra constructed is / is not a right triangle.
_______ a right triangle

Answer: Lashandra’s triangle is right angled triangle as it satisfied Pythagorean theorem

Explanation:
Verifying with Pythagorean formula a² + b²  = c²
8^{2 +} 6² = 10²
64 + 36 =100
100 = 100.

Question 2.
A triangle has side lengths 9 cm, 12 cm, and 16 cm. Tell whether the triangle is a right triangle.
Let a = _____, b = _____, and c = ______.

By the converse of the Pythagorean Theorem, the triangle is / is not a right triangle.
_______ a right triangle

Answer: The given triangle is not a right-angled triangle

Explanation: Verifying with Pythagorean formula a² + b²  = c²
9² + 12² = 16²
81 + 144 = 256
225 ≠ 256.
Hence given dimensions are not from the right angled triangle.

Question 3.
The marketing team at a new electronics company is designing a logo that contains a circle and a triangle. On one design, the triangle’s side lengths are 2.5 in., 6 in., and 6.5 in. Is the triangle a right triangle? Explain.
_______

Answer: It is a right-angled triangle.

Explanation: Let a = 2.5, b = 6 and c= 6.5
Verifying with Pythagorean formula a² + b²  = c²
2.5² + 6² = 6.5²
6.25 + 36 = 42.25
42.25 = 42.25.
Hence it is a right-angled triangle.

ESSENTIAL QUESTION CHECK-IN

Question 4.
How can you use the converse of the Pythagorean Theorem to tell if a triangle is a right triangle?

Answer: Knowing the side lengths, we substitute them in the formula a² + b²  = c², where c contains the biggest value. If the equation holds true, then the given triangle is a right triangle. Otherwise, it is not a right triangle.

12.2 Independent Practice – Converse of the Pythagorean Theorem – Page No. 385

Tell whether each triangle with the given side lengths is a right triangle.

Question 5.
11 cm, 60 cm, 61 cm
______________

Answer: Since 11² + 60² = 61², the triangle is a right-angled triangle.

Explanation: Let a = 11, b = 60 and c= 61
Using the converse of the Pythagorean Theorem a² + b²  = c²
11² + 60² = 61²
121 + 3600 = 3721
3721 = 3721.
Since 11² + 60² = 61², the triangle is a right-angled triangle.

Question 6.
5 ft, 12 ft, 15 ft
______________

Answer: Since 5² + 12² ≠ 15², the triangle is not a right-angled triangle.

Explanation: Let a = 5, b = 12 and c= 15
Using the converse of the Pythagorean Theorem a² + b²  = c²
 5² + 12² = 15²
25 + 144 = 225
169 ≠ 225.
Since 5² + 12² ≠ 15², the triangle is not a right-angled triangle.

Question 7.
9 in., 15 in., 17 in.
______________

Answer: Since 9² + 15² ≠ 17², the triangle is not a right-angled triangle.

Explanation: Let a = 9, b = 15 and c= 17
Using the converse of the Pythagorean Theorem a² + b²  = c²
9² + 15² = 17²
81 + 225 = 225
306 ≠ 225.
Since 9² + 15² ≠ 17², the triangle is not a right-angled triangle.

Question 8.
15 m, 36 m, 39 m
______________

Answer: Since 15² + 36² = 39², the triangle is a right-angled triangle.

Explanation: Let a = 15, b = 36 and c= 39
Using the converse of the Pythagorean Theorem a² + b²  = c²
15² + 36² = 39²
225 + 1296 = 1521
1521 = 1521.
Since 15² + 36² = 39², the triangle is a right-angled triangle.

Question 9.
20 mm, 30 mm, 40 mm
______________

Answer: Since 20² + 30² ≠ 40², the triangle is not a right-angled triangle.

Explanation: Let a = 20, b = 30 and c= 40
Using the converse of the Pythagorean Theorem a² + b²  = c²
20² + 30² = 40²
400 + 900 = 1600
1300 ≠ 1600.
Since 20² + 30² ≠ 40², the triangle is not a right-angled triangle.

Question 10.
20 cm, 48 cm, 52 cm
______________

Answer: Since 20² + 48² = 52², the triangle is a right-angled triangle.

Explanation: Let a = 20, b = 48 and c= 52
Using the converse of the Pythagorean Theorem a² + b²  = c²
20² + 48² = 52²
400 + 2304 = 2704
2704 = 2704.
Since 20² + 48² = 52², the triangle is a right-angled triangle.

Question 11.
18.5 ft, 6 ft, 17.5 ft
______________

Answer: Since 6² + 17.5² = 18.5², the triangle is a right-angled triangle.

Explanation: Let a = 6, b = 17.5 and c= 18.5
Using the converse of the Pythagorean Theorem a² + b²  = c²
6² + 17.5² = 18.5²
36 + 306.25 = 342.25
342.5 = 342.25.
Since 6² + 17.5² = 18.5², the triangle is a right-angled triangle.

Question 12.
2 mi, 1.5 mi, 2.5 mi
______________

Answer: Since 2² + 1.5² = 2.5², the triangle is a right-angled triangle.

Explanation: Let a = 2, b = 1.5 and c= 2.5
Using the converse of the Pythagorean Theorem a² + b²  = c²
 2² + 1.5² = 2.5²
4 + 2.25 = 6.25
6.25 = 6.25.
Since  2² + 1.5² = 2.5², the triangle is a right-angled triangle.

Question 13.
35 in., 45 in., 55 in.
______________

Answer: Since 35² + 45² ≠ 55², the triangle is not a right-angled triangle.

Explanation: Let a = 35, b = 45 and c= 55
Using the converse of the Pythagorean Theorem a² + b²  = c²
35² + 45² = 55²
1225 + 2025 = 3025
3250 ≠ 3025.
Since 35² + 45² ≠ 55², the triangle is not a right-angled triangle.

Question 14.
25 cm, 14 cm, 23 cm
______________

Answer: Since  14² + 23² ≠ 25², the triangle is not a right-angled triangle.

Explanation: Let a = 14, b = 23 and c= 25 (longest side)
Using the converse of the Pythagorean Theorem a² + b²  = c²
14² + 23² = 25²
196 + 529 = 625
725 ≠ 625.
Since  14² + 23² ≠25², the triangle is not a right-angled triangle.

Question 15.
The emblem on a college banner consists of the face of a tiger inside a triangle. The lengths of the sides of the triangle are 13 cm, 14 cm, and 15 cm. Is the triangle a right triangle? Explain.
________

Answer: Since  13² + 14² ≠ 15², the triangle is not a right-angled triangle.

Explanation: Let a = 13, b = 14 and c= 15
Using the converse of the Pythagorean Theorem a² + b²  = c²
13² + 14² = 15²
169 + 196 = 225
365 ≠ 225.
Since  13² + 14² ≠ 15², the triangle is not a right-angled triangle.

Question 16.
Kerry has a large triangular piece of fabric that she wants to attach to the ceiling in her bedroom. The sides of the piece of fabric measure 4.8 ft, 6.4 ft, and 8 ft. Is the fabric in the shape of a right triangle? Explain.
________

Answer: The triangular piece of fabric that Kerry has is in the shape of a right angle since it follows the Pythagorean theorem.

Explanation: Let a = 4.8, b = 6.4 and c= 8
Using the converse of the Pythagorean Theorem a² + b²  = c²
4.8² + 6.4² = 8²
23.04 + 40.96 = 64
64 = 64.
Since 4.8² + 6.4² = 8², the triangle is a right-angled triangle.

Question 17.
A mosaic consists of triangular tiles. The smallest tiles have side lengths 6 cm, 10 cm, and 12 cm. Are these tiles in the shape of right triangles? Explain.
________

Answer: Since 6² + 10² ≠ 12², by the converse of the Pythagorean Theorem, we say that the tiles are not in the shape of right-angled triangle.

Explanation: Let a = 6, b = 10 and c= 12
Using the converse of the Pythagorean Theorem a² + b²  = c²
 6² + 10² = 12²
36 + 100 = 144
136 ≠ 144.
Since 6² + 10² ≠ 12², by the converse of the Pythagorean Theorem, we say that the tiles are not in the shape of right-angled triangle.

Question 18.
History In ancient Egypt, surveyors made right angles by stretching a rope with evenly spaced knots as shown. Explain why the rope forms a right angle.

Answer: The rope has formed a right-angled triangle because the length of its sides follows Pythagorean Theorem.

Explanation: The knots are evenly placed at equal distances
The lengths in terms of knots are a=4 knots, b = 3knots, c = 5 knots
Therefore a² + b²  = c²
4² + 3²  = 5²
16+9 = 25
25 = 25.
Hence rope has formed a right-angled triangle because the length of its sides follows Pythagorean Theorem.

Converse of the Pythagorean Theorem – Page No. 386

Question 19.
Justify Reasoning Yoshi has two identical triangular boards as shown. Can he use these two boards to form a rectangle? Explain.

Answer: Since it was proved that both can form a right-angled triangle, we can form a rectangle by joining them.

Explanation: Given both triangles are identical, if both are right-angled triangles then we can surely join to form a rectangle.
Let’s consider a = 0.75, b= 1 and c=1.25.
By using converse Pythagorean Theorem a² + b²  = c²
0.75² + 1² = 1.25²
0.5625 + 1 = 1.5625
1.5625 = 1.5625.
Since it was proved that both can form right angled triangle, we can form a rectangle by joining them.

Question 20.
Critique Reasoning Shoshanna says that a triangle with side lengths 17 m, 8 m, and 15 m is not a right triangle because 17² + 8² = 353, 15² = 225, and 353 ≠ 225. Is she correct? Explain
_______

Answer: She is not right, A triangle with sides 15, 8, and 17 is a right-angled triangle.

Explanation: Lets consider a =15, b= 8 and c = 17 (which is long side)
We will verify by using converse Pythagorean Theorem a² + b²  = c²
15² + 8² = 17²
225 + 64 = 289
289 = 289.
Since the given dimensions satisfied Pythagorean Theorem, we can say it is a right-angled triangle. In the given above statement what Shoshanna did was c² + b² = a², which is not the correct definition of the Pythagorean Theorem.

FOCUS ON HIGHER ORDER THINKING

Question 21.
Make a Conjecture Diondre says that he can take any right triangle and make a new right triangle just by doubling the side lengths. Is Diondre’s conjecture true? Test his conjecture using three different right triangles.
_______

Answer: Yes, Diondre’s conjecture is true. By doubling the sides of a right triangle would create a new right triangle.

Explanation: Given a right triangle, the Pythagorean Theorem holds. Therefore, a² + b²  = c²
If we double the side lengths of that triangle, we get:
(2a)² + (2b)²  = (2c)²
4a² + 4b² = 4c²
4(a² + b²) = 4c²
a² + b²  = c²                    
As we can see doubling the sides of a right triangle would create a new right triangle.We can test that by using three different right triangles.

The triangle with sides a = 6, b = 8 and c = 10 is a right triangle. We double its sides and check if the new triangle is a right triangle. After doubling value of a = 12, b = 16 and c = 20.
12² + 16² = 20²
144 + 256 = 400
400 = 400
Hence proved!
Since 12² + 16² = 20², the new triangle is a right triangle by the converse of the Pythagorean Theorem.

The triangle with sides a = 3, b = 4 and c = 5 is a right triangle. We double its sides and check if the new triangle is a right triangle. After doubling value of a = 6, b = 8 and c = 10.
6² + 8² = 10²
36 + 64 = 100
100 = 100
Hence proved!
Since 6² + 8² = 10², the new triangle is a right triangle by the converse of the Pythagorean Theorem.

The triangle with sides a = 12, b = 16 and c = 20 is a right triangle. We double its sides and check if the new triangle is a right triangle. After doubling value of a = 24, b = 32 and c = 40.
24² + 32² = 40²
576 + 1024 = 1600
1600 = 1600
Hence proved!
Since 24² + 32² = 40², the new triangle is a right triangle by the converse of the Pythagorean Theorem.

Question 22.
Draw Conclusions A diagonal of a parallelogram measures 37 inches. The sides measure 35 inches and 1 foot. Is the parallelogram a rectangle? Explain your reasoning.
_______

Answer: Since 12² + 35² = 37², the triangle is right triangle. Therefore, the given parallelogram is a rectangle.

Explanation: A rectangle is a parallelogram where the interior angles are right angles. To prove if the given parallelogram is a rectangle, we need to prove that the triangle formed by the diagonal of the parallelogram and two sides of it, is a right triangle. Converting all the values into inches, we have a = 12, b = 35 and c = 37. Using the converse of the Pythagorean Theorem, we have:
a² + b²  = c²
12² + 35² = 37²
144 + 1225 = 1369
1369 = 1369.
Since 12² + 35² = 37², the triangle is right triangle. Therefore, the given parallelogram is a rectangle.

Question 23.
Represent Real-World Problems A soccer coach is marking the lines for a soccer field on a large recreation field. The dimensions of the field are to be 90 yards by 48 yards. Describe a procedure she could use to confirm that the sides of the field meet at right angles.

Answer: To confirm that the sides of the field meet at right angles, she could measure the diagonal of the field and use the converse of the Pythagorean Theorem. If a² + b²  = c² (where a = 90, b = 48 and c is the length of the diagonal), then the triangle is right triangle. This method can be used for every corner to decide if they form right angles or not.

Guided Practice – Distance Between Two Points – Page No. 390

Question 1.
Approximate the length of the hypotenuse of the right triangle to the nearest tenth using a calculator.

_______ units

Answer: The length of the hypotenuse of the right triangle to the nearest tenth is 5.8 units.

Explanation: From the above figure let’s take
Length of the vertical leg = 3 units
Length of the horizontal leg = 5 units
let length of the hypotenuse = c
By using Pythagorean Theorem a² + b²  = c²
c² = 3² + 5²
c² = 9 +25
c = √34 => 5.830.
Therefore Length of the hypotenuse of the right triangle to the nearest tenth is 5.8 units.

Question 2.
Find the distance between the points (3, 7) and (15, 12) on the coordinate plane.
_______ units

Answer: Distance between points on the coordinate plane is 13

Explanation: So (x₁, y₁) = (3,7) and  (x₂, y₂) = (15, 12)
distance formula d = √( x₂ – x₁)² + √( y₂ – y₁)²
d = √(15 -3)² + √(12 – 7)²
d = √12² + 5²
d = √144 + 25
d = √169 => 13
Therefore distance between points on the coordinate plane is 13.

Question 3.
A plane leaves an airport and flies due north. Two minutes later, a second plane leaves the same airport flying due east. The flight plan shows the coordinates of the two planes 10 minutes later. The distances in the graph are measured in miles. Use the Pythagorean Theorem to find the distance shown between the two planes.

_______ miles

Answer: The distance between the two planes is 103.6 miles.

Explanation:
Length of the vertical d_v = √(80 -1)² + √(1-1)²
= √79² => 79.
Length of the horizontal d_{h =} √(68 -1)² + √(1-1)²
= √67² => 67.
Distance between the two planes D = √(79² + 67²)
= √(6241+4489) => √10730
= 103.5857 => 103.6 miles.

ESSENTIAL QUESTION CHECK-IN

Question 4.
Describe two ways to find the distance between two points on a coordinate plane.

Answer:

Explanation: We can draw a right triangle whose hypotenuse is the segment connecting the two points and then use the Pythagorean Theorem to find the length of that segment. We can also the Distance formula to find the length of that segment.

For example, plot three points; (1,2), (20,2) and (20,12)

Using the Pythagorean Theorem:

The length of the horizontal leg is the absolute value of the difference between the x-coordinates of the points (1,2) and (20,2).
|1 – 20| = 19
The length of the horizontal leg is 19.

The length of the vertical leg is the absolute value of the difference between the y-coordinates of the points (20,2) and (20,12).
|2 – 12| = 10
The length of the vertical leg is 10.

Let a = 19, b = 10 and let c represent the hypotenuse. Find c.
a² + b²  = c²
19² + 10²  = c²
361 + 100 = c²
461 = c²
distance is 21.5 = c

Using the Distance formula:
d= √( x₂ – x₁)² + √( y₂ – y₁)²
The length of the horizontal leg is between (1,2) and (20,2).
d= √( x₂ – x₁)² + √( y₂ – y₁)²
  =  √(20 -1)² + √(2-2)²
= √(19)² + √(0)²
= √361 => 19
The length of the vertical leg is between (20,2) and (20,12).
d= √( x₂ – x₁)² + √( y₂ – y₁)²
  =  √(20 -20)² + √(12-2)²
= √(0)² +√(10)²
= √100 => 10
The length of the diagonal leg is between (1,2) and (20,12).
d= √( x₂ – x₁)² + √( y₂ – y₁)²
  =  √(20 -1)² + √(12-2)²
= √(19)² + √(10)²
= √(361+100) => √461 = 21.5

12.3 Independent Practice – Distance Between Two Points – Page No. 391

Question 5.
A metal worker traced a triangular piece of sheet metal on a coordinate plane, as shown. The units represent inches. What is the length of the longest side of the metal triangle? Approximate the length to the nearest tenth of an inch using a calculator. Check that your answer is reasonable.

_______ in.

Answer: The length of the longest side of the metal triangle to the nearest tenth is 7.8 units.

Explanation: From the above figure let’s take
Length of the vertical leg = 6 units
Length of the horizontal leg = 5 units
let length of the hypotenuse = c
By using Pythagorean Theorem a² + b²  = c²
c² = 6² + 5²
c² = 36 +25
c = √61 => 7.8
Therefore Length of the longest side of the metal triangle to the nearest tenth is 7.8 units.

Question 6.
When a coordinate grid is superimposed on a map of Harrisburg, the high school is located at (17, 21) and the town park is located at (28, 13). If each unit represents 1 mile, how many miles apart are the high school and the town park? Round your answer to the nearest tenth.
_______ miles

Answer: The high school and the town park are 13.6 miles apart.

Explanation: The coordinates of the high school are said to be (17,21), where as the coordinates of the park  are (28,13). In a coordinate plane, the distance d between the points (17,21) and (28,13) is:

d= √( x₂ – x₁)² + √( y₂ – y₁)²
  =  √(28 -17)² + √(13-21)²
= √(11)² + √(-8)²
= √(121+64) => √185 = 13.6014

Rounding the answer to the nearest tenth:
d = 13.6.
Taking into consideration that each unit represents 1 mile, the high school and town park are 13.6 miles apart.

Question 7.
The coordinates of the vertices of a rectangle are given by R(- 3, – 4), E(- 3, 4), C (4, 4), and T (4, – 4). Plot these points on the coordinate plane at the right and connect them to draw the rectangle. Then connect points E and T to form diagonal \(\overline { ET } \).
a. Use the Pythagorean Theorem to find the exact length of \(\overline { ET } \).

Answer: The diagonal ET is about 10.63 units long.

Explanation:
Taking into consideration the triangle TRE, the length of the vertical leg (ER) is 8 units. The length of the horizontal leg (RT) is 7 units. Let a = 8 and b =7. Let c represent the length of the hypotenuse, the diagonal ET. We use the Pythagorean Theorem to find c.
a² + b²  = c²
c² = 8² + 7²
c² = 64 +49
c = √113 => 10,63.
The diagonal ET is about 10.63 units long.

Question 7.
b. How can you use the Distance Formula to find the length of \(\overline { ET } \) ? Show that the Distance Formula gives the same answer.

Answer: The diagonal ET is about 10.63 units long. As we can see the answer is the same as the one we found using the Pythagorean Theorem.

Explanation: Using the distance formula, in a coordinate plane, the distance d between the points E(-3,4) and T(4, -4) is:
d= √( x₂ – x₁)² + √( y₂ – y₁)²
  =  √(4 – (-3))² + √(- 4 – 4)²
= √(7)² + √(-8)²
= √(49+64) => √113 = 10.63.
The diagonal ET is about 10.63 units long. As we can see the answer is the same as the one we found using the Pythagorean Theorem.

Question 8.
Multistep The locations of three ships are represented on a coordinate grid by the following points: P(- 2, 5), Q(- 7, – 5), and R(2, – 3). Which ships are farthest apart?

Answer: Ships P and Q are farthest apart

Explanation: Distance Formula: In a coordinate plane, the distance d between two points (x₁,y₁) and (x₂,y₂) is:

d= √( x₂ – x₁)² + √( y₂ – y₁)²
The distance d₁ between the two points P(-2,5) and Q(-7,-5) is:
d_{1 =} √( x_Q – x_P)² + √( y_Q – y_P)²
= √(-7 – (-2))² + √(- 5 – 5)²
= √(-5)² + √(-10)²
= √(25+100) => √125 = 11.18

The distance d₂ between the two points Q(-7,-5) and R(2,-3) is:
d_{3 =} √( x_R – x_Q)² + √( y_R – y_Q)²
  = √(2 – (-7))² + √(- 3 – 5)²
= √(9)² + √(2)²
= √(81+4) => √85 = 9.22

The distance d₃ between the two points P(-2,5) and R(2,-3) is:
d_{3 =} √( x_R – x_P)² + √( y_R – y_P)²
= √(2 – (-2))² + √(- 3 – 5)²
= √(4)² + √(-8)²
= √(16+64) => √80 = 8.94.
As we can see, the greatest distance is d₁ 11.8, which means that ships P and Q are farthest apart.

Distance Between Two Points – Page No. 392

Question 9.
Make a Conjecture Find as many points as you can that are 5 units from the origin. Make a conjecture about the shape formed if all the points 5 units from the origin were connected.

Answer: (0,5), (3,4), (4,3),(5,0),(4,-3),(3,-4),(0,-5),(-3,-4),(-4,-3),(-5,0),(-4,3),(-3,4).

Explanation: Some of the points that are 5 units away from the origin are: (0,5), (3,4), (4,3),(5,0),(4,-3),(3,-4),(0,-5),(-3,-4),(-4,-3),(-5,0),(-4,3),(-3,4) etc, If all the points 5 units away from the origin are connected, a circle would be formed.

Question 10.
Justify Reasoning The graph shows the location of a motion detector that has a maximum range of 34 feet. A peacock at point P displays its tail feathers. Will the motion detector sense this motion? Explain.

Answer: Considering each unit represents 1 foot, the motion detector, and peacock are 33.5 feet apart. Since the motion detector has a maximum range of 34 feet, it means that it will sense the motion of the peacock’s feathers.

Explanation: The coordinates of the motion detector are said to be (0,25), whereas the coordinates of the peacock are (30,10). In a coordinate plane, the distance d between the points (0,25) and (30,10) is:
d ₌ √( x₂ – x₁)² + √( y₂ – y₁)²
= √(30 – 0)² + √(10 – 25)²
= √(30)² + √(-15)²
= √(900+225) => √1125.
Rounding answer to the nearest tenth:
d = 33.5 feet.
Considering each unit represents 1 foot, the motion detector and peacock are 33.5 feet apart. Since the motion detector has a maximum range of 34 feet, it means that it will sense the motion of the peacock’s feathers.

FOCUS ON HIGHER ORDER THINKING

Question 11.
Persevere in Problem Solving One leg of an isosceles right triangle has endpoints (1, 1) and (6, 1). The other leg passes through the point (6, 2). Draw the triangle on the coordinate plane. Then show how you can use the Distance Formula to find the length of the hypotenuse. Round your answer to the nearest tenth.

Answer: 7.1 units.

Explanation:

One leg of an isosceles right triangle has endpoints (1,1) and (6,1), which means that the leg is 5 units long. Since the triangle is isosceles, the other leg should be 5 units long too, therefore the endpoints of the second leg that passes through the point (6,2) are (6,1) and (6,6).
In the coordinate plane, the length of the hypotenuse is the distance d between the points (1,1) and (6,6).
d ₌ √( x₂ – x₁)² + √( y₂ – y₁)²
= √(6 – 1)² + √(6 – 1)²
= √(5)² + √(5)²
= √(25+25) => √50.
Rounding answer to nearest tenth:
d = 7.1.
The hypotenuse is around 7.1 units long.

Question 12.
Represent Real-World Problems The figure shows a representation of a football field. The units represent yards. A sports analyst marks the locations of the football from where it was thrown (point A) and where it was caught (point B). Explain how you can use the Pythagorean Theorem to find the distance the ball was thrown. Then find the distance.

_______ yards

Answer: The distance between point A and B is 37 yards

Explanation:

To find the distance between points A and B, we draw segment AB and label its length d. Then we draw vertical segment AC and Horizontal segment CB. We label the lengths of these segments a and b. triangle ACB is a right triangle with hypotenuse AB.
Since AC is vertical segment, its length, a, is the difference between its y-coordinates. Therefore, a = 26 – 14 = 12 units.
Since CB is horizontal segment, its length b is the difference between its x-coordinates. Therefore, b = 75 – 40 = 35units.
We use the Pythagorean Theorem to find d, the length of segment AB.
d² = a² + b²
d² = 12² + 35²
d² = 144 + 1225
d² = 1369 => d = √1369 => 37
The distance between point A and B is 37 yards

Ready to Go On? – Model Quiz – Page No. 393

12.1 The Pythagorean Theorem

Find the length of the missing side.

Question 1.

________ meters

Answer: Length of missing side is 28m

Explanation: Lets consider value of a = 21 and c = 35.
Using Pythagorean Theorem a² + b²  = c²
21² + b² = 35²                                            
441 + b² = 1225
b²= 784 => b = √784 = 28.
Therefore length of missing side is 28m.

Question 2.

________ ft

Answer: Length of missing side is 34ft

Explanation: Let’s consider value of a = 16 and b = 30.
Using Pythagorean Theorem a² + b²  = c²
16² + 30² = c²                                              
256 + 900 = c²
c²= 1156 => c = √1156 = 34.
Therefore length of missing side is 34ft.

12.2 Converse of the Pythagorean Theorem

Tell whether each triangle with the given side lengths is a right triangle.

Question 3.
11, 60, 61
____________

Answer: Since 11² + 60² = 61², by the converse of the Pythagorean Theorem, we say that the given sides are in the shape of right-angled triangle.

Explanation: Let a = 11, b = 60 and c= 61
Using the converse of the Pythagorean Theorem a² + b²  = c²
11² + 60² = 61²
121 + 3600 = 3721
3721 = 3721
Since 11² + 60² = 61², by the converse of the Pythagorean Theorem, we say that the given sides are in the shape of right-angled triangle.                      
Question 4.
9, 37, 40
____________

Answer: Since  9² + 37² ≠ 40², by the converse of the Pythagorean Theorem, we say that the given sides are not in the shape of right-angled triangle.

Explanation: Let a = 9, b = 37 and c= 40
Using the converse of the Pythagorean Theorem a² + b²  = c²
9² + 37² = 40²
81 + 1369 = 1600
1450 ≠ 3721.
Since  9² + 37² ≠ 40², by the converse of the Pythagorean Theorem, we say that the given sides are not in the shape of right-angled triangle.

Question 5.
15, 35, 38
____________

Answer: Since 15² + 35² ≠ 38², by the converse of the Pythagorean Theorem, we say that the given sides are not in the shape of right-angled triangle.

Explanation: Let a = 15, b = 35 and c= 38
Using the converse of the Pythagorean Theorem a² + b²  = c²
15² + 35² = 38²
225 + 1225 = 1444
1450 ≠ 1444
Since 15² + 35² ≠ 38², by the converse of the Pythagorean Theorem, we say that the given sides are not in the shape of right-angled triangle.                                                                        

Question 6.
28, 45, 53
____________

Answer: Since 28² + 45² = 53², by the converse of the Pythagorean Theorem, we say that the given sides are in the shape of right-angled triangle.

Explanation: Let a = 28, b = 45 and c= 53
Using the converse of the Pythagorean Theorem a² + b²  = c²
28² + 45² = 53²
784 + 2025 = 2809
2809 = 2809
Since 28² + 45² = 53², by the converse of the Pythagorean Theorem, we say that the given sides are in the shape of right-angled triangle.                                
Question 7.
Keelie has a triangular-shaped card. The lengths of its sides are 4.5 cm, 6 cm, and 7.5 cm. Is the card a right triangle?
____________

Answer: Since 4.5² + 6² = 7.5², by the converse of the Pythagorean Theorem, we say that the given sides are in the shape of right-angled triangle.

Explanation: Let a = 4.5, b = 6 and c= 7.5
Using the converse of the Pythagorean Theorem a² + b²  = c²
4.5² + 6² = 7.5²
20.25 + 36 = 56.25
56.25= 56.25
Since 4.5² + 6² = 7.5², by the converse of the Pythagorean Theorem, we say that the given sides are in the shape of right-angled triangle.                                                                            

12.3 Distance Between Two Points

Find the distance between the given points. Round to the nearest tenth.

Question 8.
A and B
________ units

Answer: Distance between A and B is 6.7 units

Explanation: A= (-2,3) and B= (4,6)

Distance between A and B is d ₌ √( x₂ – x₁)² + √( y₂ – y₁)²
= √(4 – (-2)² + √(6 – 3)²
= √(6)² + √(3)²
= √(36+9) => √45 = 6.7 units

Question 9.
B and C
________ units

Answer: Distance between B and C is 7.07 units

Explanation: B= (4,6) and C= (3,1)

Distance between B and C is d ₌ √( x₂ – x₁)² + √( y₂ – y₁)²
= √(4 – 3)² + √(6 – (-1))²
= √(1)² + √(7)²
= √(1+49) => √50 = 7.07 units

Question 10.
A and C
________ units

Answer: Distance between A and C is 6.403 units

Explanation: A= (-2,3) and C= (3, -1)

Distance between A and C is d ₌ √( x₂ – x₁)² + √( y₂ – y₁)²
= √(3 – (-2)² + √(-1 – 3)²
= √(5)² + √(-4)²
= √(25+16) => √41 = 6.403 units

ESSENTIAL QUESTION

Question 11.
How can you use the Pythagorean Theorem to solve real-world problems?

Answer: We can use the Pythagorean Theorem to find the length of a side of a right triangle when we know the lengths of the other two sides. This application is usually used in architecture or other physical construction projects. For example, it can be used to find the length of a ladder, if we know the height of the wall and distance on the ground from the wall of the ladder.

Selected Response – Mixed Review – Page No. 394

Question 1.
What is the missing length of the side?

A. 9 ft
B. 30 ft
C. 39 ft
D. 120 ft

Answer: C

Explanation:
Given a= 80 ft
b= ?
c= 89 ft
As a²+b²=c ²
80²+b²= 89²
6,400+b²= 7,921
b²= 7,921-6,400
b= √1,521
b= 39 ft.

Question 2.
Which relation does not represent a function?
Options:
A. (0, 8), (3, 8), (1, 6)
B. (4, 2), (6, 1), (8, 9)
C. (1, 20), (2, 23), (9, 26)
D. (0, 3), (2, 3), (2, 0)

Answer: D

Explanation: The value of X is the same for 2 points and 2 values of Y [(2, 3), (2, 0)]. The value of X is repeated for a function to exist, no two points can have the same X coordinates.

Question 3.
Two sides of a right triangle have lengths of 72 cm and 97 cm. The third side is not the hypotenuse. How long is the third side?
Options:
A. 25 cm
B. 45 cm
C. 65 cm
D. 121 cm

Answer: C

Explanation:
Given a= 72 cm
b= ?
c= 97 cm
As a²+b²=c ²
72²+b²= 97²
5,184+b²= 9,409
b²= 9,409-5,184
b= √4,225
b= 65 cm.

Question 4.
To the nearest tenth, what is the distance between point F and point G?

Options:
A. 4.5 units
B. 5.0 units
C. 7.3 units
D. 20 units

Answer: A.

Explanation:
Given F= (-1,6) =(x1,y1).
G= (3,4) = (x2,y2).
The difference between F&G points is
d= √(x2-x1)² + (y2-y1)²
=  √(3 – (-1))² + (4 – 6)²
 = √(4)² + (-2)²
= √16+4
= √20
= 4.471
= 4.5 units.

Question 5.
A flagpole is 53 feet tall. A rope is tied to the top of the flagpole and secured to the ground 28 feet from the base of the flagpole. What is the length of the rope?
Options:
A. 25 feet
B. 45 feet
C. 53 feet
D. 60 feet

Answer: D

Explanation:

By Pythagorean theorem
a²+b²=c ²
53²+28²= C²
2,809+784= C²
C² = 9,409-5,184
C² = 3,593
C= √3,593
C= 59.94 feet
=60 feet.

Question 6.
Which set of lengths are not the side lengths of a right triangle?
Options:
A. 36, 77, 85
B. 20, 99, 101
C. 27, 120, 123
D. 24, 33, 42

Answer: D.

Explanation:
Check if side lengths in option A form a right triangle.
Let a= 36, b= 77, c= 85
By Pythagorean theorem
a²+b²=c ²
36²+77²= 85²
1,296+ 5,929= 7,225
7,225= 7,225
As 36²+77²= 85² the triangle is a right triangle.

Check if side lengths in option B form a right triangle.
Let a= 20, b= 99, c= 101
By Pythagorean theorem
a²+b²=c ²
20²+99²= 101²
400+ 9,801= 10,201
10,201= 10,201
As 20²+99²= 101² the triangle is a right triangle.

Check if side lengths in option B form a right triangle.
Let a= 27, b= 120, c= 123
By Pythagorean theorem
a²+b²=c ²
27²+120²= 123²
729+ 14,400= 15,129
15,129= 15,129
As 27²+120²= 123² the triangle is a right triangle.

Check if side lengths in option B form a right triangle.
Let a= 27, b= 120, c= 123
By Pythagorean theorem
a²+b²=c ²
24²+33²= 42²
576+ 1,089= 1,764.
1,665= 1,764
As 24²+33² is not equal to 42² the triangle is a right triangle.

Question 7.
A triangle has one right angle. What could the measures of the other two angles be?
Options:
A. 25° and 65°
B. 30° and 15°
C 55° and 125°
D 90° and 100°

Answer: A

Explanation:
The sum of all the angles of a triangle is 180
<A+<B+<C= 180°
<A+<B+ 90°= 180°
<A+<B= 180°-90°
<A+<B= 90, here we will verify with the given options.
25°+65°= 90°
So, the measure of the other two angles are 25° and 65°

Mini-Task

Question 8.
A fallen tree is shown on the coordinate grid below. Each unit represents 1 meter.

a. What is the distance from A to B?
_______ meters

Answer: 13.34  m.

Explanation:
A= (-5,3)
B= (8,0)
Distance between A & B is
D= √{8-(-5)² + (0-3)²
= √(13)² + (-3)²
= √169+9
= √178
= 13.34  m.

Question 8.
b. What was the height of the tree before it fell?
_______ meters

Answer: 16.3 m.

Explanation:
Length of the broken part= 13.3 m
Length of vertical part= 3 m
Total Length = 13.3 m + 3 m
= 16.3 m.

Final Words

In addition to the exercise problems, we have provided the solutions for the review questions. So all the students are requested to test your knowledge and solve the problems provided at the end of this chapter. Refer HMH Go Math Grade 8 Answer Keu and try to score the highest marks in the exams. Hope you liked the explanations provided in this chapter. Stay tuned to get the solutions according to the list of the chapters of all the grades.

Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume

Download HMH Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume and learn offline. With the help of these Go Math 6th Grade Solution Key Chapter 11 Surface Area and Volume, you can score good marks in the exams. The topics include 3-D figures and Nets, Explore Surface Area Using Nets, Surface Area of Prisms, and so on. This will also help to build a strong foundation of all these concepts for secondary level classes.

Lesson 1: Three-Dimensional Figures and Nets

• Share and Show – Page No. 599
• Problem Solving + Applications – Page No. 600
• Three-Dimensional Figures and Nets – Page No. 601
• Lesson Check – Page No. 602

Lesson 2: Investigate • Explore Surface Area Using Nets

• Share and Show – Page No. 605
• What’s the Error? – Page No. 606
• Explore Surface Area Using Nets – Page No. 607
• Lesson Check – Page No. 608

Lesson 3: Algebra • Surface Area of Prisms

• Share and Show – Page No. 611
• Unlock the Problem – Page No. 612
• Surface Area of Prisms – Page No. 613
• Lesson Check – Page No. 614

Lesson 4: Algebra • Surface Area of Pyramids

• Share and Show – Page No. 617
• Problem Solving + Applications – Page No. 618
• Surface Area of Pyramids – Page No. 619
• Lesson Check – Page No. 620

 Mid-Chapter Checkpoint

• Mid-Chapter Checkpoint – Vocabulary – Page No. 621
• Page No. 622

Lesson 5: Investigate • Fractions and Volume

• Share and Show – Page No. 625
• Problem Solving + Applications – Page No. 626
• Fractions and Volume – Page No. 627
• Lesson Check – Page No. 628

Lesson 6: Algebra • Volume of Rectangular Prisms

• Share and Show – Page No. 631
• Aquariums – Page No. 632
• Volume of Rectangular Prisms – Page No. 633
• Lesson Check – Page No. 634

Lesson 7: Problem Solving • Geometric Measurements

• Share and Show – Page No. 637
• On Your Own – Page No. 638
• Problem Solving Geometric Measurements – Page No. 639
• Lesson Check – Page No. 640

Chapter 11 Review/Test

• Chapter 11 Review/Test – Page No. 641
• Page No. 642
• Page No. 643
• Page No. 644
• Page No. 645
• Page No. 646

Share and Show – Page No. 599

Identify and draw a net for the solid figure.

Question 1.

Answer: The base Square or Rectangle, and lateral faces are Triangle and the figure is a Square pyramid or Rectangular pyramid.

Explanation:

Question 2.

Answer: Cube or Rectangular prism.

Explanation: The base is a square or rectangle and lateral faces are squares are rectangle. The figure is a Cube or Rectangular prism.

Identify and sketch the solid figure that could be formed by the net.

Question 3.

Answer: Triangular pyramid.

Explanation: The net has four triangles, so it is a triangular pyramid.

Question 4.

Answer: Cube

Explanation: The net has six squares.

On Your Own

Identify and draw a net for the solid figure.

Question 5.

Answer: Triangular prism.

Explanation: The base is a rectangle and the lateral faces are triangle and rectangles, so it is a triangular prism.

Question 6.

Answer:  Rectangular Prism.

Explanation: The base is a rectangle and the lateral faces are squares and rectangles. And it is a Rectangular prism.

Problem Solving + Applications – Page No. 600

Solve.

Question 7.
The lateral faces and bases of crystals of the mineral galena are congruent squares. Identify the shape of a galena crystal.

Answer: Cube

Explanation: The shape of the galena is Cube.

Question 8.
Rhianon draws the net below and labels each square. Can Rhianon fold her net into a cube that has letters A through G on its faces? Explain.

Answer: No, she cannot fold her net into a cube. Rhianon’s net has seven squares but there are only six squares in a net of a cube.

Question 9.
Describe A diamond crystal is shown. Describe the figure in terms of the solid figures you have seen in this lesson.

Answer: We can see that Diamond crystal consists of two square pyramids with congruent bases and the pyramids are reversed and placed base to base.

Question 10.
Sasha makes a triangular prism from paper.
The bases are _____.
The lateral faces are _____.

Answer:
The bases are Triangle
The lateral faces are Rectangle

Three-Dimensional Figures and Nets – Page No. 601

Identify and draw a net for the solid figure.

Question 1.

Answer: Rectangular Prism

Explanation:

Question 2.

Answer: Cube, Rectangular prism

Explanation:

Question 3.

Answer: Square Pyramid

Explanation:

Question 4.

Answer: Triangular Prism

Explanation:

Problem Solving

Question 5.
Hobie’s Candies are sold in triangular-pyramidshaped boxes. How many triangles are needed to make one box?

Answer: 4

Explanation: As triangled pyramids have four faces.

Question 6.
Nina used plastic rectangles to make 6 rectangular prisms. How many rectangles did she use?

Answer: 36

Explanation:

Question 7.
Describe how you could draw more than one net to represent the same three-dimensional figure. Give examples.

Answer:

Explanation:

Lesson Check – Page No. 602

Question 1.
How many vertices does a square pyramid have?

Answer: 5

Explanation:

Question 2.
Each box of Fred’s Fudge is constructed from 2 triangles and 3 rectangles. What is the shape of each box?

Answer: Triangular Prism

Explanation:

Spiral Review

Question 3.
Bryan jogged the same distance each day for 7 days. He ran a total of 22.4 miles. The equation 7d = 22.4 can be used to find the distance d in miles he jogged each day. How far did Bryan jog each day?

Answer: 3.2 miles

Explanation: As given equation 7d= 22.4,
d= 22.4÷7
= 3.2 miles.

Question 4.
A hot-air balloon is at an altitude of 240 feet. The balloon descends 30 feet per minute. What equation gives the altitude y, in feet, of the hot-air balloon after x minutes?

Answer: Y= 240- 30X.

Explanation: Given altitude Y, and the ballon was descended 30 feet per minute. So the equation is Y= 240- 30X.

Question 5.
A regular heptagon has sides measuring 26 mm and is divided into 7 congruent triangles. Each triangle has a height of 27 mm. What is the area of the heptagon?

Answer: 351 mm²

Explanation: Area of heptagon= 1/2 b×h
= 1/2 (26)×(27)
= 13×27
= 351 mm²

Question 6.
Alexis draws quadrilateral STUV with vertices S(1, 3), T(2, 2), U(2, –3), and V(1, –2). What name best classifies the quadrilateral?

Answer: Parallelogram

Explanation:

Share and Show – Page No. 605

Use the net to find the surface area of the prism.

Question 1.

Answer:

Explanation: First we must find the area of each face
A= 4×3= 12
B= 4×3= 12
C= 5×4= 20
D= 5×4= 20
E= 5×3= 15
F= 5×3= 15
So, the surface area is 12+12+20+20+15+15= 94 cm²

Find the surface area of the rectangular prism.

Question 2.

Answer: 222 cm²

Explanation: Area of a rectangular prism is 2(wl+hl+hw) = 2(7×9+ 3×9+ 3×7)
= 2(63+27+21)
= 2(111)
= 222 cm²

Question 3.

Answer:

Explanation: Area of a rectangular prism is 2(wl+hl+hw) = 2(10×10+ 10×10+ 10×10)
= 2(100+100+100)
= 2(300)
= 600 cm²

Question 4.

Answer: 350 cm²

Explanation: Area of a rectangular prism is 2(wl+hl+hw) = 2(15×5+ 5×5+ 15×5)
= 2(75+25+75)
= 2(175)
= 350 cm²

Problem Solving + Applications

Question 5.
A cereal box is shaped like a rectangular prism. The box is 20 cm long by 5 cm wide by 30 cm high. What is the surface area of the cereal box?

Answer: 1700 cm²

Explanation: The length of the box is 20 cm, the wide is 5 cm and the height is 30 cm. So surface area of the cereal box is 2(wl+hl+hw)= 2(20×5+30×20+30×5)
= 2(100+600+150)
= 2(850)
= 1700 cm²

Question 6.
Darren is painting a wooden block as part of his art project. The block is a rectangular prism that is 12 cm long by 9 cm wide by 5 cm high. Describe the rectangles that make up the net for the prism.

Answer:

Question 7.
In Exercise 6, what is the surface area, in square meters, that Darren has to paint?

Answer: 416 cm²

Explanation: Surface area = 2(wl+hl+hw)
= 2(9×12+5×12+ 5×9)
= 2(108+60+45)
= 2(213)
= 416 cm²

What’s the Error? – Page No. 606

Question 8.
Emilio is designing the packaging for a new MP3 player. The box for the MP3 player is 5 cm by 3 cm by 2 cm. Emilio needs to find the surface area of the box.
Look at how Emilio solved the problem. Find his error.
STEP 1: Draw a net.

STEP 2: Find the areas of all the faces and add them.
Face A: 3 × 2 = 6 cm².
Face B: 3 × 5 = 15 cm².
Face C: 3 × 2 = 6 cm².
Face D: 3 × 5 = 15 cm².
Face E: 3 × 5 = 15 cm².
Face F: 3 × 5 = 15 cm².
The surface area is 6 + 15 + 6 + 15 + 15 + 15 = 72 cm².
Correct the error. Find the surface area of the prism.

Answer: Emilio drew the net incorrectly Face D and F should have been 2 cm by 5 cm, not 3 cm by 5 cm

Explanation:
Face A: 3×2= 6 cm²
Face B: 3×5= 15 cm²
Face C: 3×2= 6 cm²
Face D: 2×5= 10 cm²
Face E: 3×5= 15 cm²
Face F: 2×5= 10 cm²
So, the surface area of the prism area is 6+15+6+10+15+10= 62 cm^2.

Question 9.
For numbers 9a–9d, select True or False for each statement.

9a. The area of face A is 10 cm².
9b. The area of face B is 10 cm².
9c. The area of face C is 40 cm².
9d. The surface area of the prism is 66 cm².

9a. The area of face A is 10 cm².

Answer: True

Explanation: The area of face A is 2×5= 10 cm².

9b. The area of face B is 10 cm².

Answer: False

Explanation: The area of face B is 2×8= 16  cm².

9c. The area of face C is 40 cm².

Answer: The area of face C is 8×5= 40 cm².

9d. The surface area of the prism is 66 cm².

Answer: 160 cm².

Explanation: The surface area of the prism is
= 2×10+2×10+2×40
= 20+20+80
= 160 cm².

Explore Surface Area Using Nets – Page No. 607

Use the net to find the surface area of the rectangular prism.

Question 1.

_______ square units

Answer: 52 square units.

Explanation:
The area of face A is 6 squares.
The area of face B is 8 squares.
The area of face C is 6 squares.
The area of face D is 12 squares.
The area of face E is 8 squares.
The area of face F is 12 squares.
The surface area is 6+8+6+12+8+12= 52 square units.

Question 2.

_______ square units

Answer: 112 square units.

Explanation:
The area of face A is 16 squares.
The area of face B is 8 squares.
The area of face C is 32 squares.
The area of face D is 16 squares.
The area of face E is 32 squares.
The area of face F is 8 squares.
The surface area is 112 square units.

Question 3.

Answer: 102 mm²

Explanation: Area= 2(wl+hl+hw)
= 2(3×7+3×7+3×3)
= 2(21+21+9)
= 2(51)
= 102 mm²

Question 4.

_______ in.²

Answer: 58 in.²

Explanation: Area= 2(wl+hl+hw)
= 2(5×1+ 4×1+ 4×5)
= 2(5+4+20)
= 2(29)
= 58 in.²

Question 5.

_______ ft²

Answer: 77 ft²

Explanation: Area= 2(wl+hl+hw)
= 2(6.5×2+3×2+3×6.5)
= 2(13+6+19.5)
= 2(38.5)
= 77 ft²

Problem Solving

Question 6.
Jeremiah is covering a cereal box with fabric for a school project. If the box is 6 inches long by 2 inches wide by 14 inches high, how much surface area does Jeremiah have to cover?
_______ in.²

Answer: 248 in.²

Explanation: Surface area of a cereal box is 2(wl+hl+hw)
= 2(2×6+14×6+14×2)
= 2(12+84+28)
= 2(124)
= 248 in.²

Question 7.
Tia is making a case for her calculator. It is a rectangular prism that will be 3.5 inches long by 1 inch wide by 10 inches high. How much material (surface area) will she need to make the case?
_______ in.²

Answer: 97 in.²

Explanation: Surface Area= 2(wl+hl+hw)
= 2(1×3.5+ 10×3.5+ 10×1)
= 2(3.5+35+10)
= 2(48.5)
= 97 in.²

Question 8.
Explain in your own words how to find the surface area of a rectangular prism.

Answer: To find the surface area we must know the width, length, and height of the prism and then we can apply the formula which is
Surface area= 2(width ×length)+ 2(length×height)+ 2(height×width)
= 2(width ×length+ length×height+ 2(height×width)

Lesson Check – Page No. 608

Question 1.
Gabriela drew a net of a rectangular prism on centimeter grid paper. If the prism is 7 cm long by 10 cm wide by 8 cm high, how many grid squares does the net cover?
_______ cm²

Answer: 412 cm^2.

Explanation: Surface area is 2(wl+hl+hw)
= 2(10×7+8×7+8×10)
= 2(70+56+80)
= 2(206)
= 412 cm^2.

Question 2.
Ben bought a cell phone that came in a box shaped like a rectangular prism. The box is 5 inches long by 3 inches wide by 2 inches high. What is the surface area of the box?
_______ in.²

Answer: 62 in.²

Explanation: Surface area is 2(wl+hl+hw)
= 2(3×5+2×5+2×3)
= 2(15+10+6)
= 2(31)
= 62 in.²

Spiral Review

Question 3.
Katrin wrote the inequality x + 56 < 533. What is the solution of the inequality?

Answer: X<477.

Explanation: X+56<533
= X<533-56
= X<477.

Question 4.
The table shows the number of mixed CDs y that Jason makes in x hours.

Which equation describes the pattern in the table?

Answer: y= 5x

Explanation:
y/x = 10/2= 15/4= 3
y= 5x
The pattern is y is x multipled by 5.

Question 5.
A square measuring 9 inches by 9 inches is cut from a corner of a square measuring 15 inches by 15 inches. What is the area of the L-shaped figure that is formed?
_______ in.²

Answer: 144 in.²

Explanation: The area of a square A= a², so we will find the area of each square.
Area= 9²
= 9×9
= 81 in.²
And the area of another square is
A= 15²
= 15×15
= 225 in.²
So the area of L shaped figure is 225-81= 144 in.²

Question 6.
Boxes of Clancy’s Energy Bars are rectangular prisms. How many lateral faces does each box have?

Answer: 4

Explanation: As Lateral faces are not included in the bases, so rectangular prism has 4.

Share and Show – Page No. 611

Use a net to find the surface area.

Question 1.

_______ ft²

Answer: 24 ft²

Explanation: The area of each face is 2 ft×2 ft= 4 ft and the number of faces is 6, so surface area is 6×4= 24 ft²

Question 2.

Answer: 432 cm²

Explanation:
The area of face A is 16×6= 96 cm²
The area of face B is 16×8= 128 cm²
The area of face C and D is 1/2 × 6×8= 24 cm²
The area of face E is 16×10= 160 cm²
The surface 96+128+2×24+160= 432 cm²

Question 3.

_______ in.²

Answer: 155.5 in.²

Explanation:

The area of face A and E is  8 ½ × 3½
= 17/2 × 7/2
= 119/4
= 29.75 in.²
The area of face B and F is 8 ½×4
= 17 ½ × 4
= 34 in.²
The area of face C and D is 3 ½×4
7/2 × 4= 14 in.²
The surface area is 2×29.75+2×34+2×14
= 59.5+68+28
= 155.5 in.²

On Your Own

Use a net to find the surface area.

Question 4.

_______ m²

Answer:

Explanation:

The area of face A and E is 8×3= 24 m²
The area of face B and F is 8×5= 40 m²
The area of face C and D is 3×5= 15 m²
The surface area is 2×24+2×40+2×15
= 48+80+30
= 158 m²

Question 5.

_______ \(\frac{□}{□}\) in.²

Answer:

Explanation:

The area of each face is 7 1/2 × 7 1/2
= 15/2 × 15/2
= 225/4 in.²
The no.of faces are 6 and the surface area is 6× 225/4
= 675/4
= 337 1/2 in.²

Question 6.
Attend to Precision Calculate the surface area of the cube in Exercise 5 using the formula S = 6s². Show your work.

Answer: 337 1/2 in.²

Explanation: As S= s²
= 6(7 1/2)²
= 6(15/2)²
= 6(225/4)
= 675/2
= 337 1/2 in.²

Unlock the Problem – Page No. 612

Question 7.
The Vehicle Assembly Building at Kennedy Space Center is a rectangular prism. It is 218 m long, 158 m wide, and 160 m tall. There are four 139 m tall doors in the building, averaging 29 m in width. What is the building’s outside surface area when the doors are open?
a. Draw each face of the building, not including the floor.

Answer:

Question 7.
b. What are the dimensions of the 4 walls?

Answer: The 2 walls measure 218 m ×160 m and 2 walls measure by 158 m×160 m.

Question 7.
c. What are the dimensions of the roof?

Answer: The dimensions of the roof are 218 m×158 m.

Question 7.
d. Find the building’s surface area (not including the floor) when the doors are closed.
_______ m²

Answer: 1,54,764 m²

Explanation:
The area of two walls is 218×160= 34,880 m²
The area of the other two walls is 158×160= 25,280 m²
The area of the roof 158×218= 34,444 m²
The surface area is 2× 34,880+ 2× 25,280+ 34,444
= 69,760+ 50,560+ 34,444
= 1,54,764 m²

Question 7.
e. Find the area of the four doors.
_______ m²

Answer: 16,124 m²

Explanation: Area of a door is 139×29 = 4031 m²
And the area of 4 doors is 4×4031= 16,124 m²

Question 7.
f. Find the building’s surface area (not including the floor) when the doors are open.
_______ m²

Answer: 1,38,640 m²

Explanation: The building’s surface area (not including the floor) when the doors are open is
1,54,764 – 16,124= 1,38,640 m²

Question 8.
A rectangular prism is 1 \(\frac{1}{2}\) ft long, \(\frac{2}{3}\) ft wide, and \(\frac{5}{6}\) ft high. What is the surface area of the prism in square inches?
_______ in.²

Answer: 808 in.²

Explanation: The area of two faces is 1 1/2× 5/6
= 3/2 × 5/6
= 5/4 cm²
The area of two faces is 2/3 × 5/6
= 5/9 ft²
The area of two faces is 1 1/2× 2/3
= 3/2 × 2/3
= 1 ft²
The surface area of the prism is 2(wl+hl+hw)
= 2(5/4 + 5/9 + 1)
= 2( 1.25+0.55+1)
= 2.5+1.1+2
= 5.61 ft²
As 1 square foot = 144 square inches
so 5.61×144 = 807.84
= 808 in.²

Question 9.
A gift box is a rectangular prism. The box measures 8 inches by 10 inches by 3 inches. What is its surface area?
_______ in.²

Answer: 268 in.²

Explanation:
The area of face A and Face E is 8×10= 80 in.²
The area of face B and Face F is 8×3= 24 in.²
The area of face C and Face D is 10×3= 30 in.²
The surface area is 2×80+2×24+2×30
= 160+48+60
= 268 in.²

Surface Area of Prisms – Page No. 613

Use a net to find the surface area.

Question 1.

_______ cm²

Answer: 104 cm²

Explanation: Surface area= 2(wl+hl+hw)
= 2(6×5+2×5+2×6)
= 2(30+10+12)
= 2(52)
= 104 cm²

Question 2.

_______ in.²

Answer: 118 in.²

Explanation: Surface area= 2(wl+hl+hw)
= 2(3.5×4+6×4+6×3.5)
= 2(59)
= 118 in.²

Question 3.

_______ ft²

Answer: 486 ft²

Explanation: Surface area= 2(wl+hl+hw)
= 2(9×9+9×9+9×9)
= 2(81+81+81)
= 2(243)
= 486 ft²

Question 4.

_______ cm²

Answer: 336 cm^2.

Explanation: Area = 1/2 bh
= 1/2 (6)(8)
= 3×8
= 24.
As there are 2 triangles, so 2×24= 48.
Surface Area= (wl+hl+hw)
= (6×12+8×12+12×10)
= 228
Total Surface area = 228+48
= 336 cm²

Problem Solving

Question 5.
A shoe box measures 15 in. by 7 in. by 4 \(\frac{1}{2}\) in. What is the surface area of the box?
_______ in.²

Answer: 408 in.²

Explanation:
The area of two faces is 15×7= 105 in.²
The area of two faces is 15× 4 1/2
= 15 × 9/2
= 15 × 4.5
= 67.5 in.²
The area of two faces is 7× 4 1/2
= 7× 9/2
= 7× 4.5
= 31.5 in.²
The surface area is 2×105+ 2×67.5+ 2×31.5
= 210+ 135+ 63
= 408 in.²

Question 6.
Vivian is working with a styrofoam cube for art class. The length of one side is 5 inches. How much surface area does Vivian have to work with?
_______ in.²

Answer: 150 in.²

Explanation:
The area of each face is 5×5= 25 in.²
The number of faces that styrofoam cube has is 6
So the surface area is 6×25= 150 in.²

Question 7.
Explain why a two-dimensional net is useful for finding the surface area of a three-dimensional figure.

Answer: Two-dimensional net is useful because by using a two-dimensional net you can calculate the surface area of each face and add them up to find the surface area of the three-dimensional figure.

Lesson Check – Page No. 614

Question 1.
What is the surface area of a cubic box that contains a baseball that has a diameter of 3 inches?
_______ in.²

Answer: 54 in.²

Explanation:
The area of each face is 3×3= 9 in.²
The number of faces for a cubic box is 6 in.²
The surface area of box that contains a baseball is 6×9= 54 in.²

Question 2.
A piece of wood used for construction is 2 inches by 4 inches by 24 inches. What is the surface area of the wood?
_______ in.²

Answer: 304 in.²

Explanation:
The area of two faces is 4×2= 8 in.²
The area of two faces is 2×24= 48 in.²
The area of two faces is 24×4= 96 in.²
So the surface area is 2×8+ 2×48+ 2×96
= 16+96+192
= 304 in.²

Spiral Review

Question 3.
Detergent costs $4 per box. Kendra graphs the equation that gives the cost y of buying x boxes of detergent. What is the equation?

Answer: Y= 4X.

Explanation: The total price Y and the price is equal to 4 × X, and X is the number of boxes that Kendra buys. So the equation is Y=4X.

Question 4.
A trapezoid with bases that measure 8 inches and 11 inches has a height of 3 inches. What is the area of the trapezoid?
_______ in.²

Answer: 28.5 in.²

Explanation:
Area of a trapezoid is 1/2(b1+b2)h
= 1/2(8+11)3
= 1/2(19)3
= 1/2 (57)
= 28.5 in.²

Question 5.
City Park is a right triangle with a base of 40 yd and a height of 25 yd. On a map, the park has a base of 40 in. and a height of 25 in. What is the ratio of the area of the triangle on the map to the area of City Park?

Answer: 1296:1.

Explanation:
Area= 1/2 bh
= 1/2 (40)(25)
= (20)(25)
= 500 yd²
So area of city park is 500 yd²
Area= 1/2 bh
= 1/2 (40)(25)
= (20)(25)
= 500 in²
So area on the map is 500 in
as 1 yd²= 1296 in²
So 500 in² = 500×1296
= 648,000
So, the ratio of the area of the triangle on the map to the area of City Park is 648,000:500
= 1296:1.

Question 6.
What is the surface area of the prism shown by the net?

Answer: 72 square units.

Explanation:
The area of two faces is 18 squares
The area of two faces is 6 squares
The area of two faces is 12 squares
So the surface area is 2×18+ 2×6+ 2×12
= 72 square units.

Share and Show – Page No. 617

Question 1.
Use a net to find the surface area of the square pyramid.

_______ cm²

Answer: 105 cm²

Explanation:

Area of the base 5×5= 25 ,
and area of one face is 1/2 × 5 × 8
= 5× 4
= 20 cm²
The surface area of a pyramid is 25+ 4×20
= 25+80
= 105 cm²

Question 2.
A triangular pyramid has a base with an area of 43 cm² and lateral faces with bases of 10 cm and heights of 8.6 cm. What is the surface area of the pyramid?
_______ cm²

Answer: 172 cm²

Explanation:
The area of one face is 1/2×10×8.6
= 5×8.6
= 43 cm²
The surface area of the pyramid is 43+3×43
= 43+ 129
= 172 cm²

Question 3.
A square pyramid has a base with a side length of 3 ft and lateral faces with heights of 2 ft. What is the lateral area of the pyramid?
_______ ft²

Answer: 12 ft²

Explanation:
The area of one face is 1/2×3×2= 3 ft²
The lateral area of the pyramid is 4×3= 12 ft²

On Your Own

Use a net to find the surface area of the square pyramid.

Question 4.

_______ ft²

Answer: 208 ft²

Explanation:

The area of the base is 8×8= 64
The area of one face is 1/2 ×8×9
= 36 ft²
The surface area of the pyramid is 64+4×36
= 64+144
= 208 ft²

Question 5.

_______ cm²

Answer: 220 cm²

Explanation:

The area of base is 10×10= 100
The area of one place is 1/2×10×6
= 10×3
= 30
The surface area of the pyramid is 100+4×30
= 100+120
= 220 cm²

Question 6.

_______ in.²

Answer: 264 in.²

Explanation:

The area of the base is 8×8= 64
The area of one face is 1/2×8×12.5
= 4×12.5
= 50 in.²
The surface area of the pyramid is 64+ 4×50
= 64+200
= 264 in.²

Question 7.
The Pyramid Arena is located in Memphis, Tennessee. It is in the shape of a square pyramid, and the lateral faces are made almost completely of glass. The base has a side length of about 600 ft and the lateral faces have a height of about 440 ft. What is the total area of the glass in the Pyramid Arena?
_______ ft²

Answer: 5,28,000 ft²

Explanation:
The area of one face is 1/2×600×440= 1,32,000 ft²
The surface of tha lateral faces is 4× 1,32,000= 5,28,000 ft²
So, the total area of the glass in the arena is 5,28,000 ft²

Problem Solving + Applications – Page No. 618

Use the table for 8–9.

Question 8.
The Great Pyramids are located near Cairo, Egypt. They are all square pyramids, and their dimensions are shown in the table. What is the lateral area of the Pyramid of Cheops?
_______ m²

Answer: 82,800 m²

Explanation:
The area of one face is 1/2×230×180
= 230×90
= 20,700 m²
The lateral area of the pyramid of Cheops is 4×20,700= 82,800 m²

Question 9.
What is the difference between the surface areas of the Pyramid of Khafre and the Pyramid of Menkaure?
_______ m²

Answer: 93,338 m²

Explanation:
The area of the base is 215×215= 46,225
The area of one face is 1/2×215×174
= 215× 87
18,705 m²
The surface area of Pyramid Khafre is= 46,225+4×18,705
= 46,225+ 74820
= 121,045 m²
The area of the base 103×103= 10,609
The area of one face is 1/2×103×83
= 8549÷2
= 4274.4 m²
The surface area of the Pyramid of Menkaure is 10,609+4×4274.5
= 10,609+ 17,098
= 27,707 m²

The difference between the surface areas of the Pyramid of Khafre and the Pyramid of Menkaure
= 121,405-27,707
= 93,338 m²

Question 10.
Write an expression for the surface area of the square pyramid shown.

Answer: 6x+9 ft^2.

Explanation: The expression for the surface area of the square pyramid is 6x+9 ft^2.

Question 11.
Make Arguments A square pyramid has a base with a side length of 4 cm and triangular faces with a height of 7 cm. Esther calculated the surface area as (4 × 4) + 4(4 × 7) = 128 cm². Explain Esther’s error and find the correct surface area

Answer: 72 cm².

Explanation: Esther didn’t apply the formula correctly, she forgot to include 1/2 in the calculated surface area.
The correct surface area is (4×4)+4(1/2 ×4×7)
= 16+4(14)
= 16+56
= 72 cm².

Question 12.
Jose says the lateral area of the square pyramid is 260 in.². Do you agree or disagree with Jose? Use numbers and words to support your answer.

Answer: 160 in.²

Explanation: No, I disagree with Jose as he found surface area instead of the lateral area, so the lateral area is
4×1/2×10×8
= 2×10×8
= 160 in.²

Surface Area of Pyramids – Page No. 619

Use a net to find the surface area of the square pyramid.

Question 1.

_______ mm²

Answer: 95 mm²

Explanation:

The area of the base is 5×5= 25 mm²
The area of one face is 1/2×5×7
= 35/2
= 17.5 mm²
The surface area is 25+4×17.5
= 25+4×17.5
= 25+70
= 95 mm²

Question 2.

_______ cm²

Answer: 612 cm²

Explanation:

The area of the base is 18×18= 324 cm²
The area of one face is 1/2×18×8
= 18×4
=  72 cm²
The surface area is 324+4×72
= 25+4×17.5
= 25+70
= 612 cm²

Question 3.

_______ yd²

Answer: 51.25 yd²

Explanation:

The area of the base is 2.5×2.5= 6.25  mm²
The area of one face is 1/2×2.5×9
= 22.5/2
= 11.25 yd²
The surface area is 25+4×17.5
= 6.25+4×11.25
= 6.25+45
= 51.25 yd²

Question 4.

_______ in.²

Answer: 180 in²

Explanation:

The area of the base is 10×10= 100 in²
The area of one face is 1/2×4×10
= 2×10
= 20 in²
The surface area is 100+4×20
= 100+4×20
= 100+80
= 180 in²

Problem Solving

Question 5.
Cho is building a sandcastle in the shape of a triangular pyramid. The area of the base is 7 square feet. Each side of the base has a length of 4 feet and the height of each face is 2 feet. What is the surface area of the pyramid?
_______ ft²

Answer: 19 ft²

Explanation:
The area of one face is 1/2×4×2= 4 ft²
The surface area of the triangular pyramid is 7+3×4
= 7+12
= 19 ft²

Question 6.
The top of a skyscraper is shaped like a square pyramid. Each side of the base has a length of 60 meters and the height of each triangle is 20 meters. What is the lateral area of the pyramid?
_______ m²

Answer: 2400 m²

Explanation:
The area of the one face is 1/2×60×20
= 600 m²
The lateral area of the pyramid is 4×600= 2400 m²

Question 7.
Write and solve a problem finding the lateral area of an object shaped like a square pyramid.

Answer: Mary has a triangular pyramid with a base of 10cm and a height of 15cm. What is the lateral area of the pyramid?

Explanation:
The area of one face is 1/2×10×15
= 5×15
= 75 cm²
The lateral area of the triangular pyramid is 3×75
= 225 cm²

Lesson Check – Page No. 620

Question 1.
A square pyramid has a base with a side length of 12 in. Each face has a height of 7 in. What is the surface area of the pyramid?
_______ in.²

Answer: 312 in.²

Explanation:
The area of the base is 12×12= 144 in.²
The area of one face is 1/2×12×7
= 6×7
= 42 in.²
The surface area of the square pyramid is 144+4×42
= 144+ 168
= 312 in.²

Question 2.
The faces of a triangular pyramid have a base of 5 cm and a height of 11 cm. What is the lateral area of the pyramid?
_______ cm²

Answer: 82.5 cm²

Explanation:
The area of one face is 1/2×5×11
= 55/2
= 27.5 cm²
The lateral area of the triangular pyramid is 3×27.5= 82.5 cm²

Spiral Review

Question 3.
What is the linear equation represented by the graph?

Answer: y=x+1.

Explanation: As the figure represents that every y value is 1 more than the corresponding x value, so the linear equation is y=x+1.

Question 4.
A regular octagon has sides measuring about 4 cm. If the octagon is divided into 8 congruent triangles, each has a height of 5 cm. What is the area of the octagon?
_______ cm²

Answer:

Explanation:
Area is 1/2bh
= 1/2× 4×5
= 2×5
= 10 cm²
So the area of each triangle is 10 cm²
and the area of the octagon is 8×10= 80 cm²

Question 5.
Carly draws quadrilateral JKLM with vertices J(−3, 3), K(3, 3), L(2, −1), and M(−2, −1). What is the best way to classify the quadrilateral?

Answer: It is a Trapezoid.

Explanation: It is a Trapezoid.

Question 6.
A rectangular prism has the dimensions 8 feet by 3 feet by 5 feet. What is the surface area of the prism?
_______ ft²

Answer: 158 ft²

Explanation:
The area of the two faces of the rectangular prism is 8×3= 24 ft²
The area of the two faces of the rectangular prism is 8×5= 40 ft²
The area of the two faces of the rectangular prism is 3×5= 15 ft²
The surface area of the rectangular prism is 2×24+2×40+2×15
= 48+80+30
= 158 ft²

Mid-Chapter Checkpoint – Vocabulary – Page No. 621

Choose the best term from the box to complete the sentence.

Question 1.
_____ is the sum of the areas of all the faces, or surfaces, of a solid figure.

Answer: Surface area is the sum of the areas of all the faces, or surfaces, of a solid figure.

Question 2.
A three-dimensional figure having length, width, and height is called a(n) _____.

Answer: A three-dimensional figure having length, width, and height is called a(n) solid figure.

Question 3.
The _____ of a solid figure is the sum of the areas of its lateral faces.

Answer: The lateral area of a solid figure is the sum of the areas of its lateral faces.

Concepts and Skills

Question 4.
Identify and draw a net for the solid figure.

Answer: Triangular prism

Explanation:

Question 5.
Use a net to find the lateral area of the square pyramid.

_______ in.²

Answer: 216 in.²

Explanation:

The area of one face is 1/2×9×12
= 9×6
= 54 in.²
The lateral area of the square pyramid is 4×54= 216 in.²

Question 6.
Use a net to find the surface area of the prism.

_______ cm²

Answer: 310 cm²

Explanation:

The area of face A and E is 10×5= 50 cm²
The area of face B and F is 10×7= 70 cm²
The area of face C and D is 7×5= 35 cm²
The surface area of the prism is 2×50+2×70+2×35
= 100+140+70
= 310 cm²

Page No. 622

Question 7.
A machine cuts nets from flat pieces of cardboard. The nets can be folded into triangular pyramids used as pieces in a board game. What shapes appear in the net? How many of each shape are there?

Answer: 4 triangles.

Explanation: There are 4 triangles.

Question 8.
Fran’s filing cabinet is 6 feet tall, 1 \(\frac{1}{3}\) feet wide, and 3 feet deep. She plans to paint all sides except the bottom of the cabinet. Find the area of the sides she intends to paint.
_______ ft²

Answer: 56 ft²

Explanation:
The two lateral face area is 6×1 1/3
= 6× 4/3
= 2×4
= 8 ft²
The area of the other two lateral faces is 6×3= 18
The area of the top and bottom is 3× 1 1/3
= 3× 4/3
= 4 ft²
The area of the sides she intends to paint is 2×8+2×18+4
= 16+36+4
= 56 ft²

Question 9.
A triangular pyramid has lateral faces with bases of 6 meters and heights of 9 meters. The area of the base of the pyramid is 15.6 square meters. What is the surface area of the pyramid?

Answer: 96.6 m²

Explanation:
The area of one face is 1/2× 6× 9
= 3×9
= 27 m²
The surface area of the triangular pyramid is 15.6+3×27
= 15.6+ 81
= 96.6 m²

Question 10.
What is the surface area of a storage box that measures 15 centimeters by 12 centimeters by 10 centimeters?
_______ cm²

Answer: 900 cm²

Explanation:
The area of two faces is 15×12= 180 cm²
The area of another two faces is 15×10= 150 cm²
The area of the other two faces is 10×12= 120 cm²
So surface area of the storage box is 2×180+2×150+2×120 cm²
= 360+300+240
= 900 cm²

Question 11.
A small refrigerator is a cube with a side length of 16 inches. Use the formula S = 6s² to find the surface area of the cube.
_______ in.²

Answer: 1,536 in.²

Explanation:
Area = s²
= 6×(16)²
= 6× 256
= 1,536 in.²

Share and Show – Page No. 625

Question 1.
A prism is filled with 38 cubes with a side length of \(\frac{1}{2}\) unit. What is the volume of the prism in cubic units?
_______ \(\frac{□}{□}\) cubic units

Answer: 4.75 cubic units

Explanation:
The volume of the cube is S³
The volume of a cube with S= (1/2)³
= 1/2×1/2×1/2
= 1/8
= 0.125 cubic units
As there are 38 cubes so 38×0.125= 4.75 cubic units.

Question 2.
A prism is filled with 58 cubes with a side length of \(\frac{1}{2}\) unit. What is the volume of the prism in cubic units?
_______ \(\frac{□}{□}\) cubic units

Answer: 7.25 cubic units.

Explanation:
The volume of the cube is S³
The volume of a cube with S= (1/2)³
= 1/2×1/2×1/2
= 1/8
= 0.125 cubic units
As there are 58 cubes so 58×0.125= 7.25 cubic units.

Find the volume of the rectangular prism.

Question 3.

_______ cubic units

Answer: 33 cubic units.

Explanation:
The volume of the rectangular prism is= Width×Height×Length
= 5 1/2 ×3×2
= 11/2 ×3×2
= 33 cubic units.

Question 4.

_______ \(\frac{□}{□}\) cubic units

Answer: 91 1/8 cubic units.

Explanation:
The volume of the rectangular prism is= Width×Height×Length
= 4 1/2 ×4 1/2×4 1/2
= 9/2 ×9/2×9/2
= 729/8
= 91 1/8 cubic units.

Question 5.
Theodore wants to put three flowering plants in his window box. The window box is shaped like a rectangular prism that is 30.5 in. long, 6 in. wide, and 6 in. deep. The three plants need a total of 1,200 in.³ of potting soil to grow well. Is the box large enough? Explain.

Answer: No, the box is not large enough as the three plants need a total of 1,200 in.³ and here volume is 1,098 in.³

Explanation:
Volume= Width×Height×Length
= 30.5×6×6
= 1,098 in.³

Question 6.
Explain how use the formula V = l × w × h to verify that a cube with a side length of \(\frac{1}{2}\) unit has a volume of \(\frac{1}{8}\) of a cubic unit.

Answer: 1/8 cubic units

Explanation:
As length, width and height is 1/2′ so
Volume = Width×Height×Length
= 1/2 × 1/2 × 1/2
= 1/8 cubic units

Problem Solving + Applications – Page No. 626

Use the diagram for 7–10.

Question 7.
Karyn is using a set of building blocks shaped like rectangular prisms to make a model. The three types of blocks she has are shown at right. What is the volume of an A block? (Do not include the pegs on top.)
\(\frac{□}{□}\) cubic units

Answer: 1/2 cubic units

Explanation: Volume = Width×Height×Length
= 1× 1/2 ×1
= 1/2 cubic units

Question 8.
How many A blocks would you need to take up the same amount of space as a C block?
_______ A blocks

Answer: No of blocks required to take up the same amount of space as a C block is 4 A blocks.

Explanation: Volume = Width×Height×Length
= 1×2×1
= 2 cubic unit
No of blocks required to take up the same amount of space as a C block is 1/2 ÷2
= 2×2
= 4 A blocks

Question 9.
Karyn puts a B block, two C blocks, and three A blocks together. What is the total volume of these blocks?
_______ \(\frac{□}{□}\) cubic units

Answer: 6 1/2 cubic units

Explanation: The volume of A block is
Volume = Width×Height×Length
= 1×1 ×1/2
= 1/2 cubic units.
As Karyn puts three A blocks together, so 3× 1/2= 3/2 cubic units.
The volume of B block is
Volume = Width×Height×Length
= 1×1 × 1
= 1 cubic units.
As Karyn puts only one B, so 1 cubic unit.
The volume of C block is
Volume = Width×Height×Length
= 2×1×1
= 2 cubic units.
As Karyn puts two C blocks together, so 2× 2= 4 cubic units.
So, the total volume of these blocks is 3/2 + 1+ 4
= 3/2+5
= 13/2
= 6 1/2 cubic units

Question 10.
Karyn uses the blocks to make a prism that is 2 units long, 3 units wide, and 1 \(\frac{1}{2}\) units high. The prism is made of two C blocks, two B blocks, and some A blocks. What is the total volume of A blocks used?
_______ cubic units

Answer: 3 cubic units.

Explanation:
Volume = Width×Height×Length
= 2×3×1 1/2
= 2×3× 3/2
= 9 cubic units.
The total volume of A block used is 9-(2×2)-(2×1)
= 9- 4- 2
= 9-6
= 3 cubic units.

Question 11.
Verify the Reasoning of Others Jo says that you can use V = l × w × h or V = h × w × l to find the volume of a rectangular prism. Does Jo’s statement make sense? Explain.

Answer: Yes

Explanation: Yes, Jo’s statement makes sense because by the commutative property we can change the order of the variables of length, width, height and both will produce the same result.

Question 12.
A box measures 5 units by 3 units by 2 \(\frac{1}{2}\) units. For numbers 12a–12b, select True or False for the statement.
12a. The greatest number of cubes with a side length of \(\frac{1}{2}\) unit that can be packed inside the box is 300.
12b. The volume of the box is 37 \(\frac{1}{2}\) cubic units.
12a. __________
12b. __________

Answer:
12a True.
12b True.

Explanation: The volume of the cube is S³
The volume of a cube with S= (1/2)³
= 1/2×1/2×1/2
= 1/8 cubic units
As there are 300 cubes so 300× 1/8= 75/2
= 37 1/2 cubic units.

Fractions and Volume – Page No. 627

Find the volume of the rectangular prism.

Question 1.

_______ \(\frac{□}{□}\) cubic units

Answer: 6 3/4 cubic units

Explanation: Volume = Width×Height×Length
= 3× 1 1/2× 1 1/2
= 3× 3/2 × 3/2
= 27/4
= 6 3/4 cubic units

Question 2.

_______ \(\frac{□}{□}\) cubic units

Answer: 22 1/2 cubic units

Explanation: Volume = Width×Height×Length
= 5×1× 4 1/2
= 5× 9/2
= 45/2
= 22 1/2 cubic units

Question 3.

_______ \(\frac{□}{□}\) cubic units

Answer: 16 1/2 cubic units.

Explanation: Volume = Width×Height×Length
= 5 1/2× 1 1/2× 2
= 11/2×3/2×2
= 33/2
= 16 1/2 cubic units.

Question 4.

_______ \(\frac{□}{□}\) cubic units

Answer: 28 1/8 cubic units.

Explanation: Volume = Width×Height×Length
= 2 1/2× 2 1/2 × 4 1/2
= 5/2 × 5/2 × 9/2
= 225/8
= 28 1/8 cubic units.

Problem Solving

Question 5.
Miguel is pouring liquid into a container that is 4 \(\frac{1}{2}\) inches long by 3 \(\frac{1}{2}\) inches wide by 2 inches high. How many cubic inches of liquid will fit in the container?
_______ \(\frac{□}{□}\) in.³

Answer: 31 1/2 cubic units

Explanation: Volume = Width×Height×Length
= 4 1/2 × 3 1/2 ×2
= 9/2 × 7/2 × 2
= 63/2
= 31 1/2 cubic units

Question 6.
A shipping crate is shaped like a rectangular prism. It is 5 \(\frac{1}{2}\) feet long by 3 feet wide by 3 feet high. What is the volume of the crate?
_______ \(\frac{□}{□}\) ft³

Answer: 49 1/2 ft³

Explanation: Volume = Width×Height×Length
= 5 1/2 × 3 × 3
= 11/2 ×9
= 99/2
= 49 1/2 ft³

Question 7.
How many cubes with a side length of \(\frac{1}{4}\) unit would it take to make a unit cube? Explain how you determined your answer.

Answer: There will be 4×4×4= 64 cubes and 1/4 unit in the unit cube.

Explanation:
As the unit cube has a 1 unit length, 1 unit wide, and 1 unit height
So length 4 cubes = 4× 1/4= 1 unit
width 4 cubes = 4× 1/4= 1 unit
height 4 cubes = 4× 1/4= 1 unit
So there will be 4×4×4= 64 cubes and 1/4 unit in the unit cube.

Lesson Check – Page No. 628

Question 1.
A rectangular prism is 4 units by 2 \(\frac{1}{2}\) units by 1 \(\frac{1}{2}\) units. How many cubes with a side length of \(\frac{1}{2}\) unit will completely fill the prism?

Answer: 120 cubes

Explanation:
No of cubes with a side length of 1/2 unit is
Length 8 cubes= 8× 1/2= 4 units
Width 5 cubes= 5× 1/2= 5/2= 2 1/2 units
Height 3 cubes= 3× 1/2= 3/2= 1 1/2 units
So there are 8×5×3= 120 cubes in the prism.

Question 2.
A rectangular prism is filled with 196 cubes with \(\frac{1}{2}\)-unit side lengths. What is the volume of the prism in cubic units?
_______ \(\frac{□}{□}\) cubic units

Answer: 24 1/2 cubic units.

Explanation: As it takes 8 cubes with a side length of 1/2 to form a unit cube, so the volume of the prism in the cubic units is 196÷8= 24 1/2 cubic units.

Spiral Review

Question 3.
A parallelogram-shaped piece of stained glass has a base measuring 2 \(\frac{1}{2}\) inches and a height of 1 \(\frac{1}{4}\) inches. What is the area of the piece of stained glass?
_______ \(\frac{□}{□}\) in.²

Answer: 3 1/8 in.²

Explanation: Area of a parallelogram = base×height
= 2 1/2 × 1 1/4
= 5/2 × 5/4
= 25/8
= 3 1/8 in.²

Question 4.
A flag for the sports club is a rectangle measuring 20 inches by 32 inches. Within the rectangle is a yellow square with a side length of 6 inches. What is the area of the flag that is not part of the yellow square?
_______ in.²

Answer: 604 in.²

Explanation: Area of a flag= Length×width
= 20×32
= 640 in.²
Area of the yellow square= S²
= 6
= 36 in.²
So the area of the flag that is not a part of the yellow square is 640-36= 604 in.²

Question 5.
What is the surface area of the rectangular prism shown by the net?

_______ square units

Answer: 80 square units

Explanation:
Area of two faces is 12 squares
Area of other two faces is 16 squares
Area of another two faces is 12 squares
So the surface area is 2×12+2×16+2×12
= 24+32+24
= 80 square units

Question 6.
What is the surface area of the square pyramid?

_______ cm²

Answer: 161 cm²

Explanation: The area of the base is 7×7= 49 cm²
And the area of one face is 1/2 × 7× 8
= 7×4
= 28 cm²
The surface area of the square pyramid is 49+4×28
= 49+112
= 161 cm²

Share and Show – Page No. 631

Find the volume.

Question 1.

_______ \(\frac{□}{□}\) in.³

Answer: 3,937 1/2 in.³

Explanation: Volume= Length× wide× heght
= 10 1/2 ×15 × 25
= 11/2 × 15 × 25
= 4,125/2
= 3,937 1/2 in.³

Question 2.

_______ \(\frac{□}{□}\) in.³

Answer: 27/512 in.³

Explanation: Volume= Length× wide× height
=3/8 ×3/8 × 3/8
= 27/512 in.³

On Your Own

Find the volume of the prism.

Question 3.

_______ \(\frac{□}{□}\) in.³

Answer: 690 5/8in.³

Explanation: Volume= Length× wide× height
= 8 1/2 × 6 1/2 × 12 1/2
= 17/2 × 13/2× 25/2
= 5525/2
= 690 5/8in.³

Question 4.

_______ \(\frac{□}{□}\) in.³

Answer: 125/4096 in.³

Explanation: Volume= Length× wide× height
= 5/16 ×5/16 × 5/16
= 125/4096 in.³

Question 5.

_______ yd³

Answer: 20 yd³

Explanation:
Area= 3 1/3 yd²
So Area= wide×height
3 1/3= w × 1 1/3
10/3= w× 4/3
w= 10/3 × 3/4
w= 5/2
w= 2.5 yd
Volume= Length×width×height
= 6× 2.5× 1 1/3
= 6×2.5× 4/3
= 2×2.5×4
= 20 yd³

Question 6.
Wayne’s gym locker is a rectangular prism with a width and height of 14 \(\frac{1}{2}\) inches. The length is 8 inches greater than the width. What is the volume of the locker?
_______ \(\frac{□}{□}\) in.³

Answer: 4,730 5/8 in.³

Explanation: As length is 8 inches greater than width, so 14 1/2+ 8
= 29/2+8
= 45/2
= 22 1/2 in
Then volume= Length×width×height
= 22 1/2 × 14 1/2 × 14 1/2
= 45/2× 29/2× 29/2
= 37845/8
= 4,730 5/8 in.³

Question 7.
Abraham has a toy box that is in the shape of a rectangular prism.

The volume is _____.
_______ \(\frac{□}{□}\) ft³

Answer: 33 3/4 ft³

Explanation: Volume of rectangular prism is= Length×width×height
= 4 1/2× 2 1/2× 3
= 9/2 × 5/2× 3
= 135/3
= 33 3/4 ft³

Aquariums – Page No. 632

Large public aquariums like the Tennessee Aquarium in Chattanooga have a wide variety of freshwater and saltwater fish species from around the world. The fish are kept in tanks of various sizes.
The table shows information about several tanks in the aquarium. Each tank is a rectangular prism.

Find the length of Tank 1.
V = l w h
52,500 = l × 30 × 35
\(\frac{52,500}{1,050}\) = l
50 = l
So, the length of Tank 1 is 50 cm.

Solve.

Question 8.
Find the width of Tank 2 and the height of Tank 3.

Answer: Width of Tank 2= 8m, Height of the Tank 3= 10 m

Explanation:
The volume of Tank 2= 384 m³
so V= LWH
384=  12×W×4
W= 384/48
W= 8 m
So the width of Tank 2= 8m
The volume of Tank 3= 2160 m
So V= LWH
2160= 18×12×H
H= 2160/216
H= 10 m
So the height of the Tank 3= 10 m

Question 9.
To keep the fish healthy, there should be the correct ratio of water to fish in the tank. One recommended ratio is 9 L of water for every 2 fish. Find the volume of Tank 4. Then use the equivalencies 1 cm³ = 1 mL and 1,000 mL = 1 L to find how many fish can be safely kept in Tank 4.

Answer: 35 Fishes

Explanation:
Volume of Tank 4 = LWH
= 72×55×40
= 1,58,400 cm³
As 1 cm³ = 1 mL and 1,000 mL = 1 L
1,58,400 cm³ = 1,58,400 mL and 1,58,400 mL = 158.4 L
So tank can keep safely (158.4÷ 9)×2
= (17.6)× 2 = 35.2
= 35 Fishes

Question 10.
Use Reasoning Give another set of dimensions for a tank that would have the same volume as Tank 2. Explain how you found your answer.

Answer: Another set of dimensions for a tank that would have the same volume as Tank 2 is 8m by 8m by 6m.
So when we multiply the product will be 384

Volume of Rectangular Prisms – Page No. 633

Find the volume.

Question 1.

_______ \(\frac{□}{□}\) m³

Answer: 150 5/16 m³

Explanation: Volume= Length×width×height
= 5× 3 1/4× 9 1/4
= 5× 13/4 × 37/4
= 2405/16
= 150 5/16 m³

Question 2.

_______ \(\frac{□}{□}\) in.³

Answer: 27 1/2 in.³

Explanation: Volume= Length×width×height
= 5 1/2 × 2 1/2 × 2
= 11/2 × 5/2 × 2
= 55/2
= 27 1/2 in.³

Question 3.

_______ \(\frac{□}{□}\) mm³

Answer: 91 1/8 mm³

Explanation: Volume= Length×width×height
= 4 1/2 × 4 1/2 × 4 1/2
= 9/2 × 9/2 × 9/2
= 729/8
= 91 1/8 mm³

Question 4.

_______ \(\frac{□}{□}\) ft³

Answer: 112 1/2 ft³

Explanation: Volume= Length×width×height
= 7 1/2 × 2 1/2 × 6
= 15/2 × 5/2 × 6
= 225/2
= 112 1/2 ft³

Question 5.

_______ m³

Answer: 36 m³

Explanation:
The area of shaded face is Length × width= 8 m²
Volume of the prism= Length×width×height
= 8 × 4 1/2
= 8 × 9/2
= 4 × 9
= 36 m³

Question 6.

_______ \(\frac{□}{□}\) ft³

Answer: 30 3/8 ft³

Explanation: Volume of the prism= Length×width×height
= 2 1/4 × 6 × 2 1/4
= 9/4 × 6 × 9/4
= 243/8
= 30 3/8 ft³

Problem Solving

Question 7.
A cereal box is a rectangular prism that is 8 inches long and 2 \(\frac{1}{2}\) inches wide. The volume of the box is 200 in.³. What is the height of the box?
_______ in.

Answer: H= 10 in

Explanation: As volume = 200 in.³. So
V= LWH
200= 8 × 2 1/2 × H
200= 8 × 5/2 × H
200= 20 × H
H= 10 in

Question 8.
A stack of paper is 8 \(\frac{1}{2}\) in. long by 11 in. wide by 4 in. high. What is the volume of the stack of paper?
_______ in.³

Answer: 374 in.³

Explanation: The volume of the stack of paper= LWH
= 8 1/2 × 11 × 4
= 17/2 × 11 × 4
= 374 in.³

Question 9.
Explain how you can find the side length of a rectangular prism if you are given the volume and the two other measurements. Does this process change if one of the measurements includes a fraction?

Answer: We can find the side length of a rectangular prism if you are given the volume and the two other measurements by dividing the value of the volume by the product of the values of width and height of the prism. And the process doesn’t change if one of the measurements include a fraction.

Lesson Check – Page No. 634

Question 1.
A kitchen sink is a rectangular prism with a length of 19 \(\frac{7}{8}\) inches, a width of 14 \(\frac{3}{4}\) inches, and height of 10 inches. Estimate the volume of the sink.

Answer: 3,000 in.³

Explanation: Length = 19 7/8 as the number was close to 20 and width 14 3/4 which is close to 15 and height is 10
So Volume= LBH
= 20 × 15 × 10
= 3,000 in.³

Question 2.
A storage container is a rectangular prism that is 65 centimeters long and 40 centimeters wide. The volume of the container is 62,400 cubic centimeters. What is the height of the container?

Answer: H= 24 cm

Explanation: Volume of container= LBH
Volume= 62,400 cubic centimeters
62,400 = 65× 40 × H
62,400 = 2600 × H
H= 62,400/ 2600
H= 24 cm

Spiral Review

Question 3.
Carrie started at the southeast corner of Franklin Park, walked north 240 yards, turned and walked west 80 yards, and then turned and walked diagonally back to where she started. What is the area of the triangle enclosed by the path she walked?
_______ yd²

Answer: 9,600 yd²

Explanation:
Area of triangle= 1/2 bh
= 1/2 × 240 × 80
= 240 × 40
= 9,600 yd²

Question 4.
The dimensions of a rectangular garage are 100 times the dimensions of a floor plan of the garage. The area of the floor plan is 8 square inches. What is the area of the garage?

Answer: 80,000 in²

Explanation: As 1 in²= 10,000 in², so area of the floor plan 8 in
= 8×10000
= 80,000 in²

Question 5.
Shiloh wants to create a paper-mâché box shaped like a rectangular prism. If the box will be 4 inches by 5 inches by 8 inches, how much paper does she need to cover the box?

Answer: 184 in²

Explanation: Area of the rectangular prism= 2(wl+hl+hw)
= 2(4×5 + 5×8 + 8×4)
= 2(20+40+32)
= 2(92)
= 184 in²

Question 6.
A box is filled with 220 cubes with a side length of \(\frac{1}{2}\) unit. What is the volume of the box in cubic units?
_______ \(\frac{□}{□}\) cubic units

Answer: 27.5 cubic units.

Explanation: The volume of a cube side is (1/2)³ = 1/8
So 220 cubes= 220× 1/8
= 27.5 cubic units.

Share and Show – Page No. 637

Question 1.
An aquarium tank in the shape of a rectangular prism is 60 cm long, 30 cm wide, and 24 cm high. The top of the tank is open, and the glass used to make the tank is 1 cm thick. How much water can the tank hold?
_______ cm³

Answer: So tank can hold 37,352 cm³

Explanation: As Volume= LBH
Let’s find the inner dimensions of the tank, so 60-2 × 30-2 × 24-1
= 58×28×23
= 37,352 cm³

Question 2.
What if, to provide greater strength, the glass bottom were increased to a thickness of 4 cm? How much less water would the tank hold?
_______ cm³

Answer: 4,872 cm³

Explanation: As the glass bottom was increased to a thickness of 4 cm, 60-2 × 30-2 × 24-4
= 58×28×20
= 32,480 cm³
So the tank can hold 37,352- 32,480= 4,872 cm³

Question 3.
An aquarium tank in the shape of a rectangular prism is 40 cm long, 26 cm wide, and 24 cm high. If the top of the tank is open, how much tinting is needed to cover the glass on the tank? Identify the measure you used to solve the problem.
_______ cm³

Answer: 4,208 cm³  tinting needed to cover the glass on the tank.

Explanation:
The lateral area of the two faces is 26×24= 624 cm²
The lateral area of the other two faces is 40×24= 960 cm²
And the area of the top and bottom is 40×26= 1040 cm²
So the surface area of the tank without the top is 2×624 + 2×960 + 1040
= 1,248+1,920+1,040
= 4,208 cm³

Question 4.
The Louvre Museum in Paris, France, has a square pyramid made of glass in its central courtyard. The four triangular faces of the pyramid have bases of 35 meters and heights of 27.8 meters. What is the area of glass used for the four triangular faces of the pyramid?

Answer: 1946 m²

Explanation: The area of one face is 1/2 × 35 × 27.8= 486.5 m²
And the area of glass used for the four triangular faces of the pyramid is 4×486.5= 1946 m²

On Your Own – Page No. 638

Question 5.
A rectangular prism-shaped block of wood measures 3 m by 1 \(\frac{1}{2}\) m by 1 \(\frac{1}{2}\) m. How much of the block must a carpenter carve away to obtain a prism that measures 2 m by \(\frac{1}{2}\) m by \(\frac{1}{2}\) m?
_______ \(\frac{□}{□}\) m³

Answer: 6 1/4 m³

Explanation: The volume of the original block= LWH
= 3 × 1 1/2 × 1 1/2
= 3× 3/2 × 3/2
= 27/4
= 6 3/4 m²
And volume of carpenter carve is 2× 1/2 × 1/2
= 1/2 m²
So, the carpenter must carve 27/4 – 1/2
= 25/2
= 6 1/4 m³

Question 6.
The carpenter (Problem 5) varnished the outside of the smaller piece of wood, all except for the bottom, which measures \(\frac{1}{2}\) m by \(\frac{1}{2}\) m. Varnish costs $2.00 per square meter. What was the cost of varnishing the wood?
$ _______

Answer: $8.50

Explanation: The area of two lateral faces are 2×1/2= 1 m²
The area of the other two lateral faces are 2×1/2= 1 m²
The area of the top and bottom is 1/2×1/2= 1/4 m²
And the surface area is 2×1 + 2×1 + 1/4
= 2+2+1/4
= 17/4
= 4.25 m²
And the cost of vanishing the wood is $2.00× 4.25= $8.50

Question 7.
A wax candle is in the shape of a cube with a side length of 2 \(\frac{1}{2}\) in. What volume of wax is needed to make the candle?
_______ \(\frac{□}{□}\) in.³

Answer:

Explanation: The Volume of wax is needed to make the candle is= LWH
= 2 1/2 × 2 1/2 × 2 1/2
= 5/2 × 5/2 × 5/2
= 125/8
= 15 5/8 in.³

Question 8.
Describe A rectangular prism-shaped box measures 6 cm by 5 cm by 4 cm. A cube-shaped box has a side length of 2 cm. How many of the cube-shaped boxes will fit into the rectangular prismshaped box? Describe how you found your answer.

Answer: 12 cube-shaped boxes

Explanation: As 6 small boxes can fit on the base i.e 6 cm by 5 cm, as height is 4cm there can be a second layer of 6 small boxes. So, there will be a total of 12 cube-shaped boxes and will fit into a rectangular prism-shaped box

Question 9.
Justin is covering the outside of an open shoe box with colorful paper for a class project. The shoe box is 30 cm long, 20 cm wide, and 13 cm high. How many square centimeters of paper are needed to cover the outside of the open shoe box? Explain your strategy
_______ cm²

Answer: 1,900 cm²

Explanation:
The area of the two lateral faces of the shoebox is 20×13= 260 cm²
The area of another two lateral faces of the shoebox is 30×13= 390 cm²
The area of the top and bottom is 30×20= 600 cm²
So, the surface area of the shoebox without the top is 2×260 + 2× 390 + 600
= 520+780+600
= 1,900 cm²