Category: Envision Math

Go through the enVision Math Common Core Grade 3 Answer Key Topic 8 Use Strategies and Properties to Add and Subtract regularly and improve your accuracy in solving questions.

enVision Math Common Core 2nd Grade Answers Key Topic 8 Use Strategies and Properties to Add and Subtract

Essential Question: How can sums and differences be estimated and found mentally?

EnVision STEM Project: Traits and the Environment
Do Research Use the Internet or other sources to find out how the environment can influence plants or animals. Describe a trait in an animal or plant that can change due to the environment.
Journal: Write a Report Include what you found. Also in your report:

• Make a table that includes the plant or animal, the trait, and changes in the environment. Record any related data about the environment, such as temperature or rainfall.
• Include information about why the trait is useful.
• Write and solve addition problems using your data. Use estimation to check for reasonableness.

Review What You Know

Vocabulary
Choose the best term from the box. Write it on the blank.

• difference
• equation
• number line
• sum

Question 1.
The amount that is left after you subtract is the ___.
Answer: The amount that is left after you subtract is the difference.

Question 2.
A line that shows numbers in order from left to right is a(n) ___.
Answer:  A line that shows numbers in order from left to right is a number line

Question 3.
The total when you add is the ___.
Answer: The total when you add is the sum.

Question 4.
Both sides of a(n) ___ are equal.
Answer: Both sides of an equation are equal.

Addition and Subtraction Strategies
Find the sum or difference. Show your work.

Question 5.
32 + 58
Answer:
32 + 58
(30 + 2) + (50 +8) = 90

Question 6.
27 + 46
Answer:
27 + 46 = 73
(20 + 7) + (40 + 6) = 73

Question 7.
73 – 52
Answer:
73 – 52 = 21
(70 + 3) – ( 50 + 2) = 21

Question 8.
63 +16
Answer:
63 + 16 = 79
(60 + 3) + (10 + 6) = 79

Question 9.
88 – 28
Answer:
88 – 28 = 60
(80 + 8) – (20 + 8) = 60

Question 10.
76 – 49
Answer:
76 – 49 = 27
(70 + 6) – (40 + 9) = 27

Numerical Expressions

Question 11.
Atif puts 45 rocks in a display box. He has 54 rocks in all. Which expression can be used to find how many rocks are not in the display box?
A. 45 + 54
B. 45 + 45
C. 54 – 45
D. 54 – 54
Answer:
Option C is the correct answer because
Total rocks = 54
Atif puts in a display box = 45
Then total rocks – Atif puts in a display box = rocks are not in a display box.

Counting Money no

Question 12.
Tony has the coins shown at the right. Does he have enough money to buy a toy car that costs 86¢? Explain.

Answer:
No, because the sum of all cents is not enough to buy a toy car.

Pick a Project

PROJECT 8A
How much citrus is grown in Florida?
Project: Plan a Citrus Grove

PROJECT 8B
Would you like to travel across the country?
Project: Create and Perform Skits

PROJECT 8C
How can you add and subtract large numbers without a calculator?
Project: Make a Mental Math Game

PROJECT 8D
How many people live in our country?
Project: Design a Class Census and Give an Estimation Test

Lesson 8.1 Addition Properties

Activity

Solve & Share

Olivia arranges cups of buttons on three trays. She records the number of buttons on each cup. Which tray has the most buttons? Use place-value blocks or drawings to help solve the problem.

Are you making the same calculations more than once? How can you use structure to help solve the problem?

Look Back! Olivia pours all the buttons on Tray A into a bowl. She then divides the buttons equally into 8 cups. How many buttons are in each cup? Explain.

Visual Learning Bridge

Essential Question
What Are Some Properties of Addition?

A.
You can use properties of addition to join groups.

Parentheses show what to do first.

Associative (Grouping) Property of Addition: You can group addends in any way and the sum will be the same.

В.
Commutative (Order) Property of Addition: You can add numbers in any order and the sum will be the same.

57 + 35 = 35 + 57

C.
Identity (Zero) Property of Addition: The sum of zero and any number is that same number.

39 + 0 = 39

Convince Me!
Use Structure Pick one of the properties above. Explain how you can use a number line to show an example of that property.

Guided Practice

Do You Understand?

Question 1.
Ralph says he can rewrite (4 + 5) + 21 as 9 + 21. Do you agree? Why or why not?
Answer:
Yes I  agree
because 4 + 5 = 9
So, In place of 4 + 5, we can write has 9

Question 2.
What property is shown with this equation? How do you know?
65 + (18 + 38) = (18 + 38) + 65
Answer:
This equation is in the form of Associative property. This property allows us to change the groupings of addition or multiplication and keep the same value.

Do You Know How?

In 3 and 4, identify each property.

Question 3.
4 + (15 + 26) = (4 + 15) + 26
Answer:
The equation is in the form of Associative property.

Associative property allows us to change groupings of addition or multiplication and keeps the same value.

Question 4.
17 + 0 = 17
Answer:
This equation is in the form of the Identity property of addition.

The identity property of 1 says that any number multiplied by 1 keeps its identity.

In 5-7, write each missing number.

Question 5.
__ + 90 = 90
Answer:
0 + 90 = 90

Question 6.
42 + 23 = 23 + __
Answer:
42 + 23 = 23 + 42

Question 7.
(2 + __) + 36 = 2 + (23 + 36).
Answer:
(2 + 23) + 36 = 2 + (23 + 36)

Independent Practice

In 8-11, identify each property.

Question 8.
19 + 13 = 13 + 19
Answer:
The equation is in the form of
a + b = b + a
It is a commutative property.

Commutative property states that the change in the order of numbers in the addition or multiplication operation does not change the sum or the product.

Question 9.
18 + 0 = 18
Answer:
It is a identity property.

The identity property of 1 says that any number multiplied by 1 keeps its identity.

Question 10.
16 + (14 + 13) = (16 + 14) + 13
Answer:
The equation is in the form of Associative property.This allows us to change groupings of addition or multiplication and keep the same value.

Question 11.
(39 + 12) + 8 = (12 + 39) + 8
Answer:
The equation is in the form of the commutative property.

Commutative property states that the change in the order of numbers in an addition or multiplication and keeps the same value.

In 12-19, write each missing number.

Question 12.
25 + 62 = __ + 25
Answer:
25 + 62 = 62 + 25
Thus the missing number is 62.

Question 13.
(22 + 32) + 25 = ___ + (22 + 32)
Answer:
(22 + 32) + 25 = 25 + (22 + 32)
Thus the missing number is 25.

Question 14.
23 + __ + 11 = 23 + 11
Answer:
23 + 0 + 11 = 23 + 11
Thus the missing number 0.

Question 15.
10 + (45 + 13) = (__ + 45) + 13
Answer:
10 + ( 45 + 13 ) = (10 + 45) + 13
Thus the missing number is 10.

Question 16.
(__ + 0) + 14 = 7 + 14
Answer:
(7 + 0) + 14 =  7 + 14
Thus the missing number is 7.

Question 17.
(12 + 2) + 20 = ___+ 20
Answer:
(12 + 2) + 20 = 14 + 20
Thus the missing number is 14.

Question 18.
34 + (2 + 28) = (___ + 28) + 34
Answer:
34 + (2 + 28) = ( 2 + 28 ) + 34
Thus the missing number is 2.

Question 19.
(50 + 30) + __ = 50 + (30 + 20)
Answer:
(50 + 30) + 20 = 50 + (30 + 20)
Thus the missing number is 20.

Problem Solving

Question 20.
Make Sense and Persevere Gino packs his blue and green pencils into boxes. He puts 8 pencils in each box. How many boxes does Gino use?

Answer:
Blue pencils = 23
Green pencils =17
Gino puts pencils in each box = 8
Blue pencils + Green pencils = 40
Therefore 40 ÷ 5 = 5
Gino uses 5 boxes.

Question 21.
Group the addends below in a different way to get the same sum. Write the new equation.
(42 + 14) + 6 = 62
Answer:
New equation is  (6+ 14) + 42 = 62

Question 22.
Vocabulary How is the Commutative Property of Addition like the Commutative Property of Multiplication?
Answer:
The commutative property states that the numbers on which we operate can be moved from their position without making any difference to the answer.
The property holds for addition and multiplication.

Question 23.
enVision® STEM A lionfish has 13 spines on its back, 2 more spines near its stomach, and 3 more near its tail. Using a property of add isition, write two different equations to find how many spines a lionfish has. What property did you use?
Answer:
Lionfish has spines on its back = 13
More spines near it’s stomach = 2
More near it’s tail = 3
To find the total spines on lionfish we are using two different properties of addition.
1. Associative property of addition
(13 + 2) + 3 = 13 + (2 + 3)
2. Commutative property of addition
13 + 2 + 3 = 3 + 2 + 13
Total spines on lionfish = 18

Question 24.
Higher Order Thinking Barry says he can subtract numbers in any order and the difference will stay the same. Is Barry correct? Give an example to support your answer.
Answer: No
3 – 2 = 1
2 – 3 = -1
By this example, we can say that the subtraction of numbers in any order will not be the same.

Assessment Practice

Question 25.
Use place value to find the sum of 33 +42 + 17.
A. 89
B. 90
C. 91
D. 92
Answer:
D is the answer because
33 + 42 + 17 = 92 

Question 26.
Use properties of operations to find the sum of 22 + 30 + 28.
A. 80
B. 70
C. 60
D. 50
Answer:
To find the sum of 22 + 30 + 28 we are using commutative property.
We write the as
22 + 28 + 30; 30 + 28 +  30,
From this equation, we can calculate 22,28 in an easy way.

Lesson 8.2 Algebra: Addition Patterns

Activity

Solve & Share

Shade three sums that are next to each other on the addition table. Add the first and third sums you shaded. Find a pattern using that total and the second sum you shaded. How are the total and the second sum related? Is this true for other sets of three sums next to each other?

You can look for relationships in the addition table. The numbers in the shaded column and shaded row are addends. The other numbers are the sums.

Look Back! Explain how you can test to see if the relationship among the three sums that are next to each other is a pattern.

Essential Question How Can You Find Addition Patterns?

Visual Learning Bridge

A.
Helen found the sum of the purple numbers in the red square. Then she found the sum of the green numbers. The sums form a pattern. Find the sums and describe the pattern.

You can use a variety of strategies to find the sums!

B.
Use the Associative Property.
44 + 48 = 44 + (2 + 46)
= (44 + 2) + 46
= 46 + 46

C.
Use mental math.
44 +48 = (44 + 2) + (48 – 2)
= 46 +46
44 +48 = 46 + 46

The sum of the purple numbers is equal to the sum of the green numbers. That’s a pattern!

D.
Use the Commutative and Associative Properties.
44 + 48 = (10 + 34) + (12 + 36)
46 + 46 = (12 + 34) + (10 + 36)
Use the properties to rearrange the addends.
(10 + 34) + (12 + 36) = (10 + 34) + (12 + 36)
The sum of the purple numbers is 92.
The sum of the green numbers is 92.
The sums are double the middle number in the red square. That is another pattern.

Convince Me!
Generalize The red square above is 3 squares tall by 3 squares wide. Sebastian says there are other size squares in the addition table that have patterns. Describe a different-size square and its patterns.

Guided Practice

Do You Understand?

Question 1.
Are the sums of any two sets of diagonal corner numbers in a 3-by-3 square in a standard addition table always equal? Explain.
Answer: Yes

Do You Know How?

Question 2.
Look at the addition table in Box A on the previous page. Why do the numbers going down to the right on the diagonal increase by 2? Explain.
Answer: In the box A the numbers from left to right is sequence numbers so, in the diagonal box the numbers are going from down to right.

Independent Practice

In 3 and 4, use the table at the right.

Question 3.
Look at the sums that are shaded the same color. Describe a pattern shown by these pairs of sums. Explain why this pattern is true.

Answer:

Question 4.
Find other pairs of sums with a similar pattern. Shade them on the table. Explain why you chose those sums.

In 5 and 6, use the table at the right.

Question 5.
Shade the table to show a pattern you see. Describe your pattern.

Answer:

Question 6.
Explain why your pattern is true.
Answer:

Problem Solving

Question 7.
Look for Relationships Greg drew a rectangle on the addition table at the right. He colored the corners. Find the sum of the green corners. Find the sum of the orange corners. What pattern do you notice?

Answer:

Question 8.
Draw another rectangle on the addition table. See if Greg’s pattern is true for this rectangle.
Answer:

Question 9.
Explain why Greg’s pattern works.
Answer:

Question 10.
Which multiplication fact does the number line show? Write a related division fact.

Answer:
multiple of 5.

Question 11.
Higher Order Thinking Pierre made an addition table. He skip counted by 3 for the addends. Find and describe a pattern in Pierre’s table.

Answer:

Assessment Practice

Question 12.
Look at the shaded cells  the addition table below.

Which pattern and property of operations are shown in the shaded cells?
A. Each orange sum is equal to zero plus the other addend; The Identity Property of Addition
B. Each green sum is 10 greater than one of its addends; The Identity Property of Addition
C. Each green sum is ten greater than the sum before; The Associative Property of Addition
D. There are no patterns or properties.
Answer:

Lesson 8.3 Mental Math: Addition

Activity

Solve & Share

A school store sold 436 pencils last week and 7 packages that each had 4 pencils today. Use mental math to find how many pencils were sold in all. Explain how you found your answer.

You can use structure by examining the quantities in the problem.

Look Back! What is another way you can find the sum of 436 pencils plus 7 packages of 4 pencils each using mental math?

Visual Learning Bridge

Essential Question
How Can You Add with Question Mental Math?

A.
Dr. Gomez recorded the number of Northern Right Whales, Atlantic Spotted Dolphins, and Western Atlantic Harbor Seals she saw off the coast of Florida in two different years. How many whales did Dr. Gomez see during the two years?

You can use an open number line, mental math strategies, and properties of operations to solve this problem.

B.
One Way Find 325 + 114. Use the adding on strategy.

Start at 325. Break apart 114.
Add 100 to 325.
Add 10 to 425.
Add 4 to 435.
325 + 114 = 439
Dr. Gomez saw 439 whales.

C.
Another Way
Find 325 + 114. Use the make ten strategy. Break apart
114. 114 = 5 + 100 + 9
Add 5 to 325 to make ten.
325 + 5 = 330
Then, add 100.
330 + 100 = 430
Break 114 apart to find a number that makes a ten when added to 325.

Finally, add 9.
430 + 9 = 439
325 + 114 = 439
Dr. Gomez saw 439 whales.

Convince Me!
Model with Math Show two ways to find the total number of dolphins seen.

Guided Practice

Do You Understand?

Question 1.
Compare the One Way and on the previous page. How are they the same? How are they different?

Question 2.
Use mental math to find how many animals Dr. Gomez saw during Year 2. Show your work.
Answer:
Dr.Gomez saw whales in second year = 114
Dr.Gomez saw Dolphins in second year = 171
Dr. Gomez saw seals in second year = 212
Total number of animals that Dr.Gomez saw during second year is
114 + 171 +212
100 + 10 + 4 + 100 + 70 + 1 + 200 + 10 + 2
400 + 90 + 5 + 2
400 + 90 + 7
400 + 97 = 497

Do You Know How?

Question 3.
Use the make ten strategy to add 738 + 126.
126 = 2 + 24 + 100
738 + ___ = 740
740 + __ = 764
764 + __ = 864
So, 738 + 126 = ___.
Answer:
126 = 2 + 24 + 100
738 + 2 = 740
740 + 24 = 764
764 + 100 = 864
So, 738 + 126 = 864

Question 4.
Use the adding  strategy to add 325 + 212.
212 = 200 + 10 + 2
325 + 200 = ___
525 + 10 = ___
___ + 2 = 537
So, 325 + 212 = __.

Answer:
212 = 200 + 10 + 2
325 + 200 = 525
525 + 10 = 535
535 + 2 = 537
So, 325 + 212 = 537

Independent Practice

In 5-12, find each sum using mental math or an open number line.

Question 5.
252 +44
Answer:
To find the sum here we are using mental math.
252 + 44
200 + 50 + 2 + 40 + 4
200 + 90 + 6
200 + 96 = 296

Question 6.
236 + 243
Answer:
To find the sum here we are using mental math.
236 + 243
200 + 30 + 6 + 200 +40 + 3
400 + 70 + 9
400 + 79 = 479

Question 7.
651 + 150
Answer:
To find the sum here we are using mental math.
651 + 150
600+ 50 + 1 +100 + 50
700 + 100 + 1
600 + 1 = 601

Question 8.
378 + 542
Answer:
To find the sum here we are using mental math.
378 + 542
300 + 70 + 8 + 500 + 40 + 2
800 + 110 + 10
800 + 120 = 920

Question 9.
473 + 198
Answer:
To find the sum here we are using mental math.
473 + 198
400 + 70 + 3 + 100 + 90 + 8
500 + 160 + 11
500 + 171 = 671

Question 10.
319 + 339
Answer:
To find the sum here we are using mental math.
319 + 339
300 + 10 + 9 + 300 + 30 + 9
600 + 40 + 18
600 + 58 = 658

Question 11.
208 + 511
Answer:
To find the sum here we are using mental math.
208 + 511
200 + 8 + 500 + 10 + 10 + 1
700 + 18 + 10 +1
710 + 19 = 729

Question 12.
523 + 169
Answer:
To find the sum here we are using mental math.
523 + 169
500 + 20 + 3 + 100 + 60 + 9
600 + 80 + 12
600 + 92 = 692

Problem Solving

Question 13.
Higher Order Thinking Maxine earns $8 each hour that she works as a cashier. She starts with $233. Today she cashiers for 6 hours. How much does she have at the end of the day? Explain how you found your answer.
Answer:
Given,
Maxine earns $8 each hour that she works as a cashier.
She starts with $233. Today she cashiers for 6 hours.
6 × $8 = $48
$233 + $48 = $281
Therefore Maxine have $281 at the end of the day.

Question 14.
Lauren sorted the 4 solids into 2 groups. Use mathematical terms to explain how she sorted the solids.

Answer:

Question 15.
The Rodriguez family drives 229 miles on Friday and 172 miles on Saturday. Explain how you can use the adding on strategy to find the total number of miles the Rodriguez family drives.
Answer:
Rodriguez family drives on Friday = 229 miles
Rodriguez family drives on Saturday = 172
Total number of miles that Rodriguez family drives = 229 + 172
200 + 20 + 9 + 100 + 70 + 2
300 + 90 + 11
300 + 101 = 401

Question 16.

Critique Reasoning Bill added 438 + 107. He recorded his reasoning below. Critique Bill’s reasoning. Are there any errors? If so, explain the errors.
Find 438 + 107.
I’ll think of 7 as 2 + 5.
438 + 2 = 440
440 + 7 = 447
447 + 100 = 547
So, 438 + 107 is 547.
Answer:
Above equation has an error, In place of adding 7 with 440  we want to add 5 then it is correct.

Assessment Practice

Question 17.
Find 153 + 121. Break apart 121 by place value, and then use the adding on strategy. Select numbers to complete the equations.

Answer:

Question 18.
Find 123 + 176. Break apart 176 by place value, and then use the adding on strategy. Select numbers to complete the equations.

Answer:

Lesson 8.4 Mental Math: Subtraction

Activity

Solve & Share

Peyton wants to buy one item that costs $425 and another item that costs $210. If she gets the discount shown on the sign below, what is the sale price? Explain how you can use mental math to find your answer.

Even when you use mental math, you can still show your work! You can construct arguments using mental math.

Look Back! What is another way you can use mental math to solve the problem?

Visual Learning Bridge

Essential Question
How Can You Subtract with Mental Math?

A.
A store is having a sale on jackets. A jacket is on sale for $197 less than the original price. What is the sale price?

You can use mental math and the relationship between addition and subtraction to solve this problem.

The difference is the answer when subtracting two numbers.

B.
One Way
Count Back on the Number Line Find 352 – 197.
To subtract 197 on an open number line, you can subtract 200, and then add 3.
352 – 200 = 152
152 + 3 = 155

So, 352 – 197 = 155. The sale price is $155.

C.
Another Way
Count Up on the Number Line
Find 352 – 197.
To find 352 – 197, you can think addition: 197 + ? = 352
197 + 3 = 200
200 + 100 = 300
300 + 52 = 352
Addition and subtraction are inverse operations.

3+ 100 + 52 = 155
197 + 155 = 352, so 352 – 197 = 155.
The sale price is $155.

Convince Me! Construct Arguments Which of the two ways above would you use to solve 762 – 252? Explain.

Another Example!
You can make a simpler problem to find 352 – 197.
Add 3 to both numbers.
352 + 3 = 355 and 197 + 3 = 200.
Then you have 355 – 200.
So, 355 – 200 = 155 and 352 – 197 = 155.

If you get stuck using one strategy, another strategy may be easier!

Guided Practice

Do You Understand?

Question 1.
In the One Way example on the previous page, why do you add 3 to 152 instead of subtracting 3 from 152?
Answer: We have to add or subtract the same numbers for making the correct answer.

Question 2.
Suppose a computer costs $573. If you buy it today, it costs $498. What is the discount? Show your work.
Answer:
Given,
Suppose a computer costs $573. If you buy it today, it costs $498.
$573 – $498 = $92
Thus the discount is $92.

Do You Know How?

In 3-6, solve using mental math.

Question 3.
846 – 18
848 – 20 =___
Answer:
846 – 18
848 – 20
800 + 40 + 8 – 20
800 + 48 – 20
848 – 20 = 828

Question 4.
534 – 99
535 – 100 = __
Answer:
534 – 99
535 – 100
500 + 30 + 5 – 100
500 + 35 – 100
535 – 100 = 435

Question 5.

873 – 216
877 – 220 = ___
Answer:
873 – 216
877 – 220
800 + 70 + 7 – 200 + 20
800 + 77 – 220
877 – 220 = 657

Question 6.
782 – 347
785 – 350 = ___
Answer:
782 – 347
785 – 350
700 + 80 + 5 – 300 + 50
700 + 85 – 350
785 – 350 = 435

Question 7.
Find 400 – 138 using the make ten strategy
138 + __ = 140
140 + ___ = 200
200 + __ = 400
___ + __ + ___ = ___
Answer:
138 + 2 = 140
140 + 60 = 200
200 + 200 = 400
2 + 60 + 200 = 262

Independent Practice

In 8-15, use an open number line or the think addition strategy to find each difference.

Question 8.
128 – 19
Answer: 109

Question 9.
887 – 18
Answer: 869

Question 10.
339 – 117
Answer: 222

Question 11.
468 – 224
Answer: 244

Question 12.
784 – 515
Answer: 269

Question 13.
354 – 297
Answer: 57

Question 14.
853 – 339
Answer: 514

Question 15.
638 – 372
Answer: 266

Problem Solving

Question 16.
Sarah has $350. How much money will she have after buying the computer at the sale price?

Answer:
Given,
Sarah has $350.
$350 – $58 = $292

Question 17.
Model with Math Jessica has an array with 9 columns. There are 36 counters in the array. How many rows does her array have? Show how to represent the problem and find the answer.
Answer:
Given,
Jessica has an array with 9 columns.
There are 36 counters in the array.
9 × 4 = 36
Thus there are 4 rows and 9 columns.

Question 18.
Of the students at Paul’s school, 270 are girls and 298 are boys. There are 354 students at Alice’s school. How many more students are there at Paul’s school than at Alice’s school?
Answer:
Girls at Paul’s school = 270
Boys at Paul’s school = 298
Students at Alice’s school = 354
Students at Paul’s school = 568
568 – 354 = 214
Paul’s school has 214 students more than Alice’s school students.

Question 19.
Higher Order Thinking To find 357 – 216, Tom added 4 to each number and then subtracted. Saul added 3 to each number and then subtracted. Will both ways work to find the correct answer? Explain.
Answer:
To find 357 – 216, Tom added 4 to each number and then subtracted.
357+4 = 361
216 + 4 = 220
361 – 220 = 141
Saul added 3 to each number and then subtracted.
357+3 = 360
216 + 3 = 219
360 – 219 = 141

Assessment Practice

Question 20.
Use the relationship between addition and subtraction to find 233 – 112. Select numbers from the box to complete the work on the open number line and the equations.

Answer:

Lesson 8.5 Round Whole Numbers

Activity

Solve & Share
Think about ways to find numbers that tell about how much or about how many. Derek has 277 stickers. What number can you use to describe about how many stickers Derek has? Explain how you decided.

Think about whether you need to be precise.

Look Back! Derek gets 3 more packages of 10 stickers. About how many stickers does Derek have now?

Essential Question
How Can You Round Numbers?

Visual Learning Bridge

A.
About how many rocks does Tito have? Round 394 to the nearest ten.

Place value is the value of the place a digit has in a number. Think about the place value of the digits in 394.

When you round to the nearest ten, you are finding the closest multiple of ten for a given number.

B.
You can use place-value understanding and a number line to round to the nearest ten.

394 is closer to 390 than 400, so 394 rounds to 390.
Tito has about 390 rocks.

C.
About how many rocks does Donna have? Round 350 to the nearest hundred.

If a number is halfway between, round to the greater number
350 is halfway between 300 and 400, so 350 rounds to 400.
Donna has about 400 rocks.

Convince Me! Make Sense and Persevere Susan says, “I am thinking of a number that has a four in the hundreds place and a two in the ones place. When you round it to the nearest hundred, it is 500.” What number could Susan be thinking of? What other numbers could be Susan’s number?

Another Example
About how many rocks does Carl have? Round 345 to the nearest ten and hundred.
Round to the nearest ten.
345 is halfway between 340 and 350, so 345 rounds to 350.

Round to the nearest hundred.
345 is closer to 300 than 400, so 345 rounds to 300.

Guided Practice

Do You Understand?

Question 1.
What number is halfway between 200 and 300?
Answer:
The halfway between 200 and 300 is 250.

Question 2.
Sheri rounds 678 to 680. What place does she round to?
Answer:

Here the digits in the one’s place is  8 in 678, now round to the nearest multiple of 10 which is greater than the number.
Given,
Sheri rounds 678 to 680.
The nearest ten to 678 is 680.

Question 3.
Tito adds one more rock to his collection on the previous page. About how many rocks does he have now, rounded to the nearest ten? Rounded to the nearest hundred? Explain.
Answer:

Do You Know How?

In 4-6, round to the nearest ten.

Question 4.

Answer: 517 to the nearest ten is 520.

Question 5.
149
Answer:

Here the digits in the one’s place of 149 is 9 , now round to the nearest multiple of 10 which is greater than the number.
149 to the nearest ten is 150.

Question 6.
732
Answer:

Here the digits in the one’s place of 732 is 2, now round to the nearest multiple of 10

732 to the nearest ten is 730.

In 7-9, round to the nearest hundred.

Question 7.

Answer:

When rounding to the nearest hundred, look at the ten’s digit. If the numbers is 0,1,2,3,4 you will round to previous hundred.

640 to the nearest hundred is 600.

Question 8.
305
Answer:

When rounding to the nearest hundred, look at the ten’s digit. If the numbers is 0,1,2,3,4 you will round to the previous hundred.

305 to the nearest hundred is 300.

Question 9.
166
Answer:

When rounding to the nearest hundred , look at the ten’s digit.If the numbers is 5,6,7,8,9 you will round to the next hundred.

166 to the nearest hundred is 200.

Independent Practice

In 10-12, round to the nearest ten.

Question 10.
88
Answer:

Here the digits in the one’s place of 88 is 8 , now round to the nearest multiple of 10 which is greater than the number.
88 to the nearest ten is 90.

Question 11.
531
Answer:

Here the digits in the one’s place of 531 is 1, now round to the nearest multiple of 10 which is less than the number.
531 to the nearest ten is 530.

Question 12.
855
Answer:

Here the digits in the one’s place of 855 is 5, now round to the nearest multiple of 10 which is greater than the number.
855 to the nearest ten is 860.

In 13-15, round to the nearest hundred.

Question 13.
428
Answer:

When rounding to the nearest hundred, look at the ten’s digit of the number.If the numbers are 0,1,2,3,4, you will round down to the previous hundred.

428 to the nearest hundred is 400.

Question 14.
699
Answer:

When rounding to the nearest hundred look at the ten’s digit of the given number.If the digits are 5,7,8,9 you will round to the previous hundred.

699 to the nearest hundred is 700.

Question 15.
750
Answer: when rounding to the nearest hundred, look at the ten’s digit of the number. If the digits are 5,6,7,8,9 you will round to the next hundred.

750 to the nearest hundred is 700. or 800.

Problem Solving

Question 16.
The Leaning Tower of Pisa in Italy has 294 steps. To the nearest ten, about how many steps are there? To the nearest hundred, about how many steps are there?

Answer:
Given,
The Leaning Tower of Pisa in Italy has 294 steps.
294 to the nearest ten is 290.
294 to the nearest hundred is 300.
Thus there are about 300 steps.

Question 17.
Critique Reasoning Zoe says 247 rounded to the nearest hundred is 300 because 247 rounds to 250 and 250 rounds to 300. Is Zoe correct? Explain.
Answer:
Zoe says 247 rounded to the nearest hundred is 300 because 247 rounds to 250 and 250 rounds to 300.
247 rounded to the nearest hundred is 200.
So, Zoe is not correct.

Question 18.
Use the number line to show a number that rounds to 200 when it is rounded to the nearest ten.

Answer:

Question 19.
Name the least number of coins you can use to show $0.47. What are the coins?
Answer:

Question 20.
Suppose you are rounding to the nearest hundred. What is the greatest number that rounds to 600? What is the least number that rounds to 600?
Answer:
The greatest number that round 600 to the nearest 100 would be 600 + 50 – 1 = 649. It won’t be 650 because of 650 rounds to 700.

Question 21.
Higher Order Thinking A 3-digit number has the digits 2,5, and 7. To the nearest hundred, it rounds to 800. What is the number? Show how you found the answer.
Answer:
For the number to be rounded off to 800, it needs to be either 750 or above or 850 and below.
With the given numbers, we can only have the following combinations:
257~300
275~300
527~500
572~600
725~700
752~800
And hence, 752 is the only combination of numbers that fits.

Question 22.
Emil says, “I am thinking of a number that is greater than 142, rounds to 100 when rounded to the nearest hundred, and has a 5 in the ones place.” What is Emil’s number?

What else can you try if you get stuck?

Answer: 542

Assessment Practice

Question 23.
Select all the numbers that will equal $100 when rounded to the nearest hundred.
$10
$110
$ 89
$150
$91
Answer:
$110, $89, $91 are rounded to the nearest hundred.

Question 24.
Select all the numbers that will equal 70 when rounded to the nearest ten.
62
75
72
83
73
Answer:
The number 70 rounded to the nearest ten is 75, 72, 73.

Lesson 8.6 Estimate Sums

Activity

Solve & Share

Look at the table below. Is the mass of a female and male sun bear together more or less than the mass of one female black bear? Without finding an exact answer, explain how you can decide.

You can use symbols, numbers, and words to be precise in your explanation.

Look Back! Why is an exact answer not needed to solve the problem?

Essential Question
How Can You Estimate Sums?

Visual Learning Bridge

A.
Do the two pandas together weigh more than 500 pounds?
Estimate 255 + 329.

You can estimate to find about how much the two pandas weigh.

B.
One Way
Round to the nearest hundred.

255 + 329 is about 600.
600 > 500
The pandas together weigh more than 500 pounds.

C.
Another Way
Use compatible numbers.

Compatible numbers are numbers that are close to the addends, but easy to add mentally

255 + 329 is about 575 and 575 > 500.
The total weight is more than 500 pounds.

Convince Me! Be Precise Sandy said, “Just look at the numbers. 200 and 300 is 500. The pandas weigh over 500 pounds because one panda weighs 255 pounds and the other weighs 329 pounds.” What do you think she means? Use numbers, words, or symbols to explain.

Another Example!
Suppose one panda ate 166 pounds of bamboo in a week and another ate 158 pounds. About how many pounds of bamboo did the two pandas eat?

You can estimate 166 + 158 by rounding to the nearest ten.

The pandas ate about 330 pounds of bamboo in a week.

Guided Practice

Do You Understand?

Question 1.
Two addends are rounded to greater numbers. Is the estimate greater than or less than the actual sum?
Answer:

Question 2.
Mary and Todd estimate 143 + 286. They have different answers. Can they both be correct? Explain why or why not.
Answer:
Mary and Todd has only one answer that is 143 + 286 = 429

Do You Know How?

Round to the nearest ten to estimate.

Question 3.
218 + 466 ___+___= ___
Answer:

Here the digit in the one’s place of 218 is 8, now round to the nearest multiple of 10 which is greater than the ten.
218 round to the nearest ten to estimate is 220.

The digit in the one’s place of 466 is 6, now round to the nearest multiple of ten which is greater than the number.
466 round to the nearest ten to estimate is 470
220 + 470 = 690

Question 4.
108 + 223 ___ + ___ = ____
Answer:

Here the digit in the one’s place of 108 is 8, now round to the nearest multiple of 10 which is greater than the number.
108 round to the nearest ten to estimate is 110.

The digit in the one’s place of 223 is 3, now round to the nearest multiple of 10 which is less than the number.
223 round to the nearest ten to estimate is 220.
110 + 220 = 330

Round to the nearest hundred to estimate.

Question 5.
514 + 258 ___ + ___ = ____
Answer:

Here the digit in the one’s place of 514 is 4, now round to the nearest multiple of 10 which is less than the number.
514 round to the nearest hundred to estimate is 500.

The digit in the one’s place of 258 is round to the nearest ten which is greater than the number.
258 round to the nearest hundred to estimate is 300.
500 + 300 = 800

Question 6.
198 + 426 ___ + ___ = ____
Answer:

Here the digit in the one’s place of 198 is 8,now round to the nearest multiple of ten which is greater than the number.
198 round to the nearest hundred to estimate is 200.

The digit in the one’s place of 426 is 6, now round to the nearest multiple of 10 which is greater than the number.
426 round to the nearest hundred to estimate is 400.
200 + 400 = 600

Independent Practice

In 7-10, round to the nearest ten to estimate.

Question 7.
138 + 435
Answer:

Here the digit in the one’s place of 138 is 8, now round to the nearest multiple of ten which is greater than the number.
138 round to the nearest ten to estimate is 140.

The digit in the one’s place of 435 is 5 now round to the nearest multiple of ten which is greater than the number.
435 round to the nearest ten to estimate is 440.
140 + 440 = 580

Question 8.
563 + 289
Answer:

Here the digit in the one’s place of 563 is 5 now round to the nearest multiple of ten which is less than the number.
563 round to the nearest ten to estimate is 560.

The digit in the one’s place of 289 is 9 now round to the nearest multiple of ten which is greater than the number.
289 round to the nearest ten to estimate is 290
560 + 290 = 850

Question 9
644 + 172
Answer:

Here the digit in the one’s place of 644 is 4 now round to the nearest multiple of ten which is less than the number.
644 round to the nearest ten to estimate is 640.

The digit in the one’s place of 172 is 2, now round to the nearest multiple of ten which is less than the number.
172 round to the nearest ten to estimate is 170
640 + 170 = 810

Question 10.
376 + 295
Answer:

Here the digit in the one’s place of the number 376 is 6 now round to the nearest ten which is greater than the number.
376 round to the nearest ten to estimate is 380.

The digit in the one’s place of 295 is 5 now round to the nearest multiple of ten which is greater than the number.
295 round to the nearest ten to estimate is 300
380 + 300 = 680

In 11-14, round to the nearest hundred to estimate.

Question 11.
403 + 179
Answer:

When rounding to the nearest hundred look at the ten’s digit of the number which the number is 0,1,2,3,4 you will round to the previous hundred.
403 round to the nearest hundred to estimate is 400.

Here the ten’s place of the 179 is 7 you will round to the next hundred.
179 round to the nearest hundred to estimate is 200.
400 + 200 = 600

Question 12.
462 + 251
Answer:

Here the ten’s place of the number is 462 is 6 you will round to the previous hundred.

462 round to the nearest hundred to estimate is 500.

Here the ten’s place of the number 251 is 5 you will round to the next hundred.
251 round to the nearest hundred to estimate is 300.
500 + 300 = 800

Question 13.
274 + 443
Answer:

Here the digit in the ten’s place of 274 is 7 you will round to the next hundred.
274 round to the nearest hundred to estimate is 300.

The digit in the ten’s of 443 is  4 you will round to the previous hundred.
443 round to the nearest hundred to estimate is 400.
300 + 400 = 700

Question 14.
539 + 399
Answer:

The digit in the ten’s place of 539 is 3 you will round to the previous hundred.
539 round to the nearest hundred to estimate is 500.

The digit in the ten’s place of 399 is 9 you will round to the next hundred.
399 round to the nearest hundred to estimate is 400.
500 + 400 = 900

In 15-18, use compatible numbers to estimate.

Question 15.
175 + 126
Answer:

Compatible numbers are the pair of numbers that are easy to add, subtract, multiply or division mentally.

Here 170 and 130 are the compatible numbers to the 175 and 126.

Question 16.
167 + 27
Answer:

Compatible numbers are the pair of numbers that are easy to add, subtract, multiply or division mentally.

Here 170 and 30 are the compatible number for 167 and 27.

Question 17.
108 + 379
Answer:

Compatible numbers are the pair of numbers that are easy to add, subtract, multiply or division mentally.

Here 110 and 380 are the compatible numbers for 108 and 379.

Question 18.
145 + 394
Answer:

Compatible numbers are the pair of numbers that are easy to add, subtract, multiply or division mentally.

Here 150 and 390 are the compatible numbers for 145 and 394.

Problem Solving

Use the table to answer 19 and 20.

Question 19.
Ms. Tyler drove from Albany to Boston, and then from Boston to Baltimore. To the nearest ten miles, about how many miles did she drive in all?
Answer:

Question 20.

Ms. Tyler drove from Boston to New York City and back again. To the nearest hundred miles, about how many miles did she drive?

Answer:

Given,
Ms. Tyler drives from Boston to New York City  = 211
Ms.Tyler drives back to nearest hundred miles to Boston  = 211 – 100
= 111
How many miles did she drive  equals to 211 + 111 = 322

Question 21.
Reasoning Jen has $236. Dan has $289. Do Jen and Dan have more than $600 in all? Estimate to solve. Explain.
Answer:
Jen has $236
Don has $286
Jen and Don has 236 + 289 = 525
No, They did have more than $600

Question 22.
Ralph has 75¢. How much more money does he need to buy a pencil for 90¢? Complete the diagram.

Answer:
Ralph has 75 cents.
He need to buy a pencil for 90 cents.
Ralph need to buy a pencil = 90 cents – 75 cents =15 cents

Question 23.
Higher Order Thinking Susan drove 247 miles on Wednesday morning. Then she drove 119 miles on Wednesday afternoon. On Thursday, Susan drove 326 miles. About how far did Susan drive in all? Explain the method you used to estimate.

Remember that you learned different estimation strategies.

Assessment Practice

Question 24.
Round to the nearest 10 to estimate the sums.

Answer:
273 round to the nearest ten to estimate is 270
616 round to the nearest ten to estimate is 620
270 + 620 = 890
542 round to the nearest ten to estimate is 540
338 round to the nearest ten to estimate is 340
540 + 340 = 880
435 round to the nearest ten to estimate is 440
441 round to the nearest ten to estimate is 440
440 – 440 = 0

Question 25.
Round to the nearest 100 to estimate the sums.

Answer:
173 round to the nearest hundred to estimate is 200
139 round to the nearest hundred to estimate is 100
200 – 100 = 100
155 round to the nearest hundred to estimate is 200
177 round to the nearest hundred to estimate is 200
200 – 200 = 0
289 round to the nearest hundred to estimate is 300
18 round to the nearest hundred to estimate is 0
300 – 0 = 300

Lesson 8.7 Estimate Differences

Activity

Solve & Share

Sara collected 220 cans on Monday, 80 cans on Tuesday, and 7 cartons with 8 cans each on Wednesday to recycle. Pierre collected 112 cans. About how many more cans did Sara collect than Pierre?

You can make sense and persevere. What is a good strategy for solving this problem?

Look Back! Which strategy gives an estimate that is closest to the exact answer? How did you decide?

Essential Question
How Can You Estimate Differences?

A.
All of the tickets for a concert were sold. So far, 126 people have arrived at the concert. About how many people who have tickets have not arrived?
Estimate 493 – 126 by rounding.

You can estimate to find about how many.

B.
One Way
Round each number to the nearest hundred and subtract.

About 400 people have not yet arrived.

C.
Another Way
Round each number to the nearest ten and subtract.

About 360 people have not yet arrived.

Convince Me! Model with Math Suppose 179 people have arrived at the concert. Estimate how many people have not arrived.

Another Example!
You can use compatible numbers to estimate differences.

375 and 150 are compatible numbers for 372 and 149.

Guided Practice

Do You Understand?

Question 1.
Does rounding to the nearest ten or nearest hundred give an estimate closer to the exact answer for 295 – 153?

Answer:

Here the digit of the one’s place of 295 is 5 now round to the nearest multiple of ten which is greater than the number.
295 round to the nearest ten to estimate is 290

The digit in the one’s place of 153 is3 now round to the nearest multiple of ten which is less than the number.
153 round to the nearest ten to estimate is 140
290 – 140 =150

Question 2.
A theater sells 408 tickets. Two hundred seventy-three people arrive. About how many more people are expected to arrive? Use compatible numbers. Show your work.

Answer:

Given,

A theater sells tickets = 408

People arrive to theater = 253

Expected people to arrive = 408 – 253 = 155

The compatible numbers for 408 and 253 is 410 and 250

410  – 250 = 150

Do You Know How?

In 3-6, estimate. Use rounding or compatible numbers. Tell how you made each estimate.

Question 3.
321 – 182
Answer:

Compatible numbers are pair of numbers that are easy to add, subtract, multiply and division mentally.

The compatible numbers for 321 and 182 is 320 and 180.

Question 4.
655 – 189
Answer:

Compatible numbers are pair of numbers that are easy to add, subtract, multiply or division mentally.

The compatible numbers for 655 and 189 is 660 and 190

Question 5.
763 – 471
Answer:

Compatible numbers are pair of numbers that are easy to add, subtract multiply, and division mentally.

The compatible numbers for 763 and 471 are 760 and 470.

Question 6.
816 – 297
Answer:

Compatible numbers are pair of numbers that are easy to add, subtract, multiply and division mentally.

The compatible numbers for 816 and 297 are 820 and 300.

Independent Practice

In 7-10, round to the nearest hundred to estimate.

Question 7.
286 – 189
Answer:

When rounding to the nearest hundred look at the ten’s place.The ten’s place of 286 is 8 now round to the next hundred.
286 round to the nearest hundred to estimate is 300.

The ten’s place of the number 189 is 8 now round to the next hundred
189 round to the nearest hundred to estimate is 200.
300 – 200 = 100

Question 8.
461 – 216

Answer:

When rounding to the nearest hundred look at the ten’s place.The ten’s place of the number 461 is 6 now round to the next hundred.

461 round to the nearest hundred to estimate is 500.

The ten’s place of the number 216 is 1 now round to the  previous hundred.
216 round to the nearest hundred to estimate is 200.
500 – 200 = 300

Question 9.
891 – 686
Answer:

When rounding to the nearest hundred look at the ten’s place of the number 891 is 9 now round to the next hundred.
891 round to the nearest hundred to estimate is 900.

The ten’s place of the number 686 is 8 now round to the next hundred.
686 round to the nearest hundred to estimate is 700.
900 – 700 = 200

Question 10.
724 – 175
Answer:

When rounding to the nearest hundred look at the ten’s place of the number 724 is 2 now round to the previous hundred.
724 round to the nearest hundred to estimate is 600.

The ten’s place of the number 175 is 7 now round to the next hundred.
175 round to the nearest hundred to estimate is 200
600 – 200 = 400

In 11-14, round to the nearest ten to estimate.

Question 11.
766 – 492
Answer:

Here the digit of the one’s place of the number 766 is 6 now round to the nearest ten which is greater than the number.
766 round to the nearest ten to estimate is 770.

The one’s place of the number 492 is 2 now round to the nearest ten which is less than the number.
492 round to the nearest ten to estimate is 500.
770 – 500 = 270

Question 12.
649 – 487
Answer:

Here the digit of the one’s place of the number 649 is 9 now round to the nearest hundred which is greater than the number.
649 round to the nearest ten to estimate is 650.

The one’s place of the number 487 is 7 now round to the nearest hundred which is greater than the number.
487 round to the nearest ten to estimate is 490.
650 – 490 = 169

Question 13.
241 – 117
Answer:

Here the digit in the one’s place of the number 241 is 1 now round to the nearest hundred which is less than the number.
241 round to the nearest ten to estimate is 240.

The one’s place of the number 117 is 7 now round to the nearest hundred which is greater than the number.
117 round to the nearest ten to estimate is 120.
240 – 120 = 120

Question 14.
994 – 679
Answer:

Here the digit of the one’s place of the number 994 is 4 now round to the nearest hundred which is less than the number.
994 round to the nearest ten to estimate is 1000.

The digit in the one’s place of the number 679 is 9 which is greater than the number.
679 round to the nearest ten to estimate is 680.
1000 – 680 = 320

In 15-18, use compatible numbers to estimate.

Question 15.
760 – 265
Answer:

Compatible numbers are pair of numbers that are easy to add, subtract, multiply and division mentally.

The compatible numbers for 760 and 265 are 760 and 270.

Question 16.
355 – 177
Answer:

Compatible numbers are the pair of numbers that are easy to add, subtract, multiply and division mentally

Compatible numbers for 355 and 177 is 360 and 180.

Question 17.
481 – 105
Answer:

Compatible numbers are the pair of numbers that are easy to add, subtract, multiply and division mentally.

The compatible numbers for 481 and 105 is 480 and 110.

Question 18.
794 – 556
Answer:

Compatible numbers are the pair of numbers that are easy to add, subtract, multiply and division mentally.

The compatible numbers for 794 and 556 are 800 and 560.

Problem Solving

Question 19.
The Grand Concert Hall sold 100 more tickets on Sunday than on Friday. On what day did it sell about 150 tickets more than it sold on Sunday?

Use estimation strategies to help find your answers.

Question 20.
Model with Math Find the total number of tickets sold on Thursday and Friday. Explain what math you used.
Answer:
Number of tickets sold on Thursday = 323
Number of tickets sold on Friday = 251
Total number of tickets sold on Thursday and Friday equal to
321 + 251
300 + 20 + 1 + 200 + 50 +1
500 + 70 + 2
500 + 72 = 572

Question 21.
Algebra Anna and Joe write reports for their science class. Anna’s report is 827 words long. Joe’s report is 679 words long. Round each report length to the nearest ten and estimate how many more words Anna’s report is. Then write an equation that shows exactly how many more words Anna’s report is.
Answer:
Anna report is 827 long words.
Joe’s report is 679 long words.
827 round to the nearest ten to estimate is 830
679 round to the nearest ten to estimate is 680
830 – 680 = 150
Anna report is 150  long words more than Joe’s long report

Question 22.
Higher Order Thinking One week Mrs. Runyan earned $486, and the next week she earned $254. If Mrs. Runyan’s goal was to earn $545, by about how much did she exceed her goal? Show how you used estimation to find your answer.
Answers:
Mrs Runyan earned in one week = $486
Mrs Runyan earned in next week = $254
Total she earned in two weeks = $ 486 + $255
=$740
But she goal to earn is $545
She exceed her goal = $740 – $545
She exceed her goal is $ 195

Question 23.
About how many inches longer was a Brachiosaurus than a Tyrannosaurus rex?

Answer:
Brachiosaurus length = 972 inches.
Tyrannosaurus length = 468 inches.
972 – 468 = 504
Brachiosaurus length is 504 longer than Tyrannosaurus length.

Assessment Practice

Question 24.
Estimate 753 – 221 by rounding each number to the nearest ten.
A. 540
B. 530
C. 520
D. 510
Answer:
753 round to the nearest ten to estimate is 750.
221 round to the nearest ten to estimate is 220
750 – 220 = 530
Thus the correct answer is option B.

Question 25.
Estimate 812 – 369 by rounding each number to the nearest hundred.
A. 500
B. 400
C. 300
D. 200
Answer:
812 round to the nearest hundred to estimate is 810.
369 round to the nearest hundred to estimate is 370.
810 – 370 = 440
440 to the nearest hundred is 400.
Thus the correct answer is option B.

Lesson 8.8 Model with Math

Activity

Solve & Share

A pond has 458 rosy red minnows, 212 white cloud minnows, and 277 goldfish. How many more minnows than goldfish live in the pond?

Thinking Habits
Be a good thinker! These questions can help you.
• How can I use math I know to help solve this problem?
• How can I use pictures, objects, or an equation to represent the problem?
• How can I use numbers, words, and symbols to solve the problem?

Look Back! Model with Math Explain what math you used to solve this problem.

Essential Question How Can You Model With math?

A.
David has $583 to spend on soccer uniforms. He buys this soccer jersey and 2 soccer shorts. How much money does David spend?

What math do I need to use to solve the problem?

I need to show what I know and then choose the needed operations.

B. How can I model with math? I can

• apply math I know to solve the problem.
• use a bar diagram and equations to represent the problem.
• use an unknown to represent the number I am trying to find.

Here’s my thinking…

C.
I will use a bar diagram and an equation.
The hidden question is: How much does David spend on shorts?

$35 + $35 = ?
$35 + $35 = $70. The shorts cost $70.
So I need to find the total including the jersey.

$70+ $109 = ?
$70 + $109 = $179. David spends $179.

Convince Me! Model with Math How does the bar diagram help you model with math?

Guided Practice

Model with Math
Harris’s office building has 126 windows. Morgan’s bank has 146 windows. Devon’s bank has 110 windows. How many more windows do the banks have altogether than the office building?

One way you can model  math is by using bar diagrams to represent each step of a two-step problem.

Question 1.
What is the hidden question you need to answer before you can solve the problem?
Answer:
How many windows do the banks have in all?

Question 2.
Solve the problem. Complete the bar diagrams. Show the equations you used.
Answer:
Morgan’s bank has 146 windows.
Devon’s bank has 110 windows.
Morgan’s and Devon’s office has 146 + 110 = 256 windows
Another office windows = 256 -126=130

Independent Practice

Model with Math
Regina’s bakery made 304 pies in January. Her bakery made 34 fewer pies in February. How many pies did her bakery make in both months?

Question 3.
What is the hidden question you need to answer before you can solve the problem?

Answer:
How many pies did the bakery make in February?

Question 4.
Solve the problem. Complete the bar diagrams. Show the equations you used.
Answer:

304 – 34 = ?
304 – 34 = 270
304 + 270 = ?
304 + 270 = 574

Question 5.
How would your equations change if the bakery made 34 more pies in February than in January?
Answer:
I would add 304 + 34 = ? instead of subtracting 304 – 34 = ? to find the number of pies baked in February.

Problem Solving

Performance Task

Skyscraper Heights
The Empire State Building in New York is 159 meters taller than the Republic Plaza in Denver. The John Hancock Building in Chicago is 122 meters taller than the Republic Plaza. The Empire State Building is 712 miles away from the Hancock Building. The Hancock Building is 920 miles away from the Republic Plaza. Manuel wants to know how tall the Hancock Building is. Answer Exercises 6-9 to solve the problem.

Answer:

Question 6.
Critique Reasoning Tom said to solve the problem, you should add 159 to the height of the Empire State Building. Do you agree? Explain why or why not.
Answer:

Question 7.
Model with Math What is the hidden question you need to solve in this problem? How can you represent the hidden question?
Answer:

Question 8.
Model with Math Solve the problem. Show the equations you used.
Answer:

Question 9.
Use Appropriate Tools Which tool would you use to represent and explain how to solve the problem: counters, cubes, or place-value blocks? Explain.

Model with math means you apply math you have learned to solve problems.

Answer:

Topic 8 Fluency Practice

Activity

Find a Match
Work with a partner. Point to a clue. Read the clue. Look below the clues to find a match. Write the clue letter in the box next to the match. Find a match for every clue. Once you find the matches, write the fact families for each of the facts.

Clues
A The missing number is 9.

B The missing number is 10.

C The missing number is 3.

D The missing number is 6.

E The missing number is 7.

F The missing number is 4.

G The missing number is 8.

H The missing number is 5.

Answer:

Topic 8 Vocabulary Review

Glossary

Understand Vocabulary

Circle the property of addition shown in the following examples.

Word List
• Associative Property of Addition
• Commutative Property of Addition
• compatible numbers
• estimate
• Identity Property of Addition
• inverse operations
• mental math
• place value
• round

Question 1.
17 + 14 = 14 + 17
Associative Property Commutative Property Identity Property
Answer: Associative property.

Question 2.
93 + 0 = 93
Associative Property Commutative Property Identity Property
Answer: Identity property.

Question 3.
8 + (5 + 9) = (8 + 5) + 9
Associative Property Commutative Property Identity Property
Answer: Commutative property.

Question 4.
65 + 0 = 0 + 65
Associative Property Commutative Property Identity Property
Answer: Identity property.

Choose the best term from the box. Write it in the blank.

Question 5.
You ___ when you use the nearest multiple of ten or a hundred.
Answer: You Estimate sum and differences in the number when you use the nearest multiple of ten or a hundred.

Question 6.
Addition and subtraction are ___
Answer:
Addition and subtraction are the arithmetic operations.

Question 7.
___ is the value given to the place a digit has in a number.
Answer: Place value is the value given to the place a digit has in a number.

Question 8.
Numbers that are easy to compute mentally are ___
Answer: Numbers that are easy to compute mentally are mental math.

Question 9.
You do not need pencil and paper when using ___
Answer: You do not need pencil and paper when using mental math.

Use Vocabulary in Writing

Question 10.
Jim found that 123 + 284 is about 400. Explain what Jim did. Use at least 3 terms from the Word List in your answer.
Answer: Jim used rounding numbers to add the given numbers.

Topic 8 Vocabulary Reteaching

Set A
pages 289-292
You can use properties of addition to help solve addition problems.
The Commutative Property of Addition
12 + = 15 + 12
You can order addends in any way, and the sum will be the same.
12 + 15 = 15 + 12

The Associative Property of Addition
3 + (7 + 8) = (3 + ) + 8
You can group addends in any way, and the sum will be the same.
3 + (7 + 8) = (3 + 7) + 8
The Identity Property of Addition
30 + = 30
The sum of any number and zero is that same number.
30 + 0 = 30

Remember that both sides of the equal sign must have the same value.

In 1-6, write each missing number.

Question 1.
18 + __ = 18
Answer:
18 + 0 = 18

Question 2.
14 + (16 + 15) = (___ +16) + 15
Answer:
14 + ( 16 + 15) = ( 14 + 16) + 15

Question 3.
___ + 13 = 13 + 17
Answer:
17 + 13 = 13 + 17

Question 4.
28 + (__ + 22) = 28 + (22 + 25)
Answer:
28 + (25 + 22) = 28 + ( 22 + 25 )

Question 5.
62 + 21 + 0 = 62 + ___
Answer:
62 + 21 + 0 = 62 + 21

Question 6.
___ + (26 + 78) = (31 + 26) + 78
Answer:
31 + (26 + 78) = (31 + 26) + 78

Question 7.
Use 78 and 34 to write an equation that shows the Commutative Property of Addition.

Answer:
78 + 34 = 34 + 78

Set B
pages 293-296
You can find patterns using an addition table.

The green sums increase by 2 down the column and are even numbers.
The yellow sums increase by 2 down the column and are odd numbers.
Use examples to make generalizations!

Remember that properties can help you understand patterns.

Question 1.
Find the doubles-plus-2 facts. What pattern do you notice about the sums?
Answer:

Question 2.
Explain why your pattern works.
Answer:

Set C
pages 297-300
Use mental math to find 374 + 238.
Break apart 238: 200 + 30+ 8.
Add hundreds, tens, and ones.
374 + 200 = 574
574 + 30 = 604
604 + 8 = 612
So, 374 + 238 = 612.

Remember that you can break apart addends when finding sums mentally.

Question 1.
302 + 56
Answer:
Using mental math
300 + 2 + 50 + 6
300 + 50 + 8
300 + 58 = 358

Question 2.
463 + 418
Answer:
Using mental math
463 + 418
400 + 60 + 3 + 400 + 10 + 8
800 + 70 + 11
800 + 81 = 881

Question 3.
222 + 725
Answer:
Using mental math
222 + 725
200 + 20 + 2 + 700 + 20 + 5
900 + 40 + 7
900 + 47 = 947

Question 4.
689 + 115
Answer:
Using mental math
689 + 115
600  + 80 + 9 + 100 + 10 + 5
700 + 90 + 14
700 + 104 = 804

Set D
pages 301-304
Think addition to find 400 – 168.
Count on.
168 + 2 = 170
170 + 30 = 200
200 + 200 = 400
2 +30 + 200 = 232
So, 400 – 168 = 232.

Remember that you can count on when subtracting mentally.

Question 1.
523 – 163
Answer:
Using mental math
523 – 163
500 + 20 + 3 – 100 + 60 + 3
400 – 40 = 360

Question 2.
847 – 372
Answer:
Using mental math
847 – 372
800 + 40 + 7 – 300 + 70 + 2
500 – 30 – 5
500 – 25 = 575

Question 3.
768 – 259
Answer:
Using mental math
768 – 259
700 + 60 + 8 – 200 + 50 + 9
500 + 10 – 1
500 + 9 = 509

Question 4.
282 – 125
Answer:
Using mental math
282 – 125
200 + 80 + 2 – 100 + 20 + 5
100 + 60 – 3
100 + 57 = 157

Set E
pages 305-308
You can use a number line to round.

Nearest ten: 437 rounds to 440.

Nearest hundred: 437 rounds to 400.
Think about place value when you round.

Remember that if a number is halfway between, round to the greater number.

Question 1.
Round 374 to the nearest ten and the nearest hundred.
Answer:
374 round to the nearest ten to estimate is 370.
374 round to the nearest hundred to estimate is 400.

Question 2.
Round 848 to the nearest ten and the nearest hundred.
Answer:
842 round to the nearest ten to estimate is 850.
842 round to the nearest hundred to estimate is 800.

Question 3.
Mark’s family traveled 565 miles. Rounded to the nearest ten, about how many miles did they travel?
Answer:
Mark’s family traveled 565 miles
565 round to the nearest ten to estimate is 570.
Mark’s family traveled 570 miles.
Question 4.
Sara collected 345 shells. Rounded to the nearest hundred, about how many shells did she collect?
Answer:

Set F
pages 309-312

Estimate 478 + 112.
Round each addend to the nearest ten.

Round each addend to the nearest hundred.

Use compatible numbers.

Remember that compatible numbers are numbers close to the actual numbers and are easier to add mentally.

Round to the nearest hundred.

Question 1.
367 + 319
Answer:

Compatible numbers are the pair of numbers that are easy to add, subtract, multiply and division mentally.

Compatible numbers for 367 and 319 is 400 and 300.

Question 2.
737 + 127
Answer:

Compatible numbers are the pair of numbers that are easy to add, subtract, multiply and division mentally,

The compatible numbers for 737 and 127 is 700 and 100

Round to the nearest ten.

Question 3.
298 + 542
Answer:

Compatible numbers are the pair of numbers that are easy to add, subtract, multiply and division mentally.

Compatible numbers for 298 and 542 is 300 and 540.

Question 4.
459 + 85
Answer:

Compatible numbers are the pair of numbers that are easy to add, subtract, multiply and division mentally.

The compatible numbers for 459 and 85 is 460 and 90

Use compatible numbers.

Question 5.
372 + 173
Answer:

Compatible numbers are the pair of numbers that are easy to add, subtract, multiply and division mentally.

The compatible numbers for 372 and 173 is 370 and 170

Question 6.
208 + 164
Answer:

Compatible numbers are the pair of numbers that are easy to add, subtract, multiply and division mentally.

The compatible numbers for 208 and 164 is 210 and 160

Question 7.
Will rounding to the nearest ten or the nearest hundred give a closer estimate of 314 + 247? Explain your answer.
Answer:
314 + 247 = 561
314 round to the nearest ten to estimate is 310
247 round to the nearest ten to estimate is 250
310  +250 = 560

Set G
pages 313-316
Estimate 486 – 177.
Round each number to the nearest hundred.

Round to the nearest hundred.

Use compatible numbers.

Remember that an estimate is close to the actual answer.

Question 1.
527 – 341
Answer:

Question 2.
872 – 184
Answer:

Round to the nearest ten.

Question 3.
387 – 298
Answer:

Here the digit in the one’s place of the number 387 is 7 now round to the nearest ten which is greater than the number.

387 round to the nearest ten is 390

The digit in the one’s place of the number 298 is 8 now round to the nearest ten which is greater than the number.

287 round to the nearest ten is 290

390 – 290 = 100

Question 4.
659 – 271
Answer:

Here the digit in the one’s place of the number 659 is 9 now round to the nearest ten which is greater than the number.

659 round to the nearest ten is 660

The digit in the one’s place of the number 271 is 1 which is less than the number.

271 is round to the nearest ten to 270

660 – 270 = 390

Use compatible numbers.

Question 5.
472 – 228
Answer:

Compatible numbers are the pair of numbers that are easy to add, subtract, multiply and division mentally.

The compatible numbers for 472 and 228 is 470 and 230

Question 6.
911 – 347
Answer:

Compatible numbers are the pair of numbers that are easy to add, subtract, multiply and division mentally.

Compatible numbers for 911 and 347 is 910 and 350

Question 7.
Will rounding to the nearest ten or the nearest hundred give a closer estimate of 848 – 231? Explain your answer.
Answer:
848 -231 = 617
848 round to the nearest hundred to estimate is 800
231 round to the nearest hundred to estimate is 200
800 – 200 = 600

Set H
pages 317-320
Think about these questions to help you model with math.

Thinking Habits
• How can I use math I know to help solve this problem?
• How can I use pictures, objects, or an equation to represent the problem?
• How can I use numbers, words, and symbols to solve the problem?

Remember to apply the math you know to solve problems.
Elena has $265. She buys a jacket that costs $107 and a sweater that costs $69. How much money does Elena have left?

Question 1.
What is the hidden question you need to answer before you can solve the problem?
Answer:

Question 2.
Solve the problem. Draw bar diagrams to represent the problem. Show the equations you used.
Answer:

Toni read 131 pages on Monday, 56 pages on Tuesday, and some pages on Wednesday. She read 289 pages in all. How many pages did Toni read on Wednesday?

Ans:

Toni read pages on Monday = 131

Toni read pages on Tuesday = 56

Toni read pages on Wednesday = 289 – 131 + 56 = 102

Question 1.
What is the hidden question you need to answer before you can solve the problem?
Answer:

Question 2.
Solve the problem. Use equations to represent your work.
Answer:

Topic 8 Assessment Practice

Question 1.
What is the sum of 243, 132, and 157?
Answer:
243 + 132 + 157 = 532

Question 2.
Find a reasonable estimate for the sum of 171 and 69. Select all that apply.
175 + 70 = 245
100 + 60 = 160
170 + 70 = 240
175 + 75 = 250
130 + 70 = 200
Answer:

For the sum of 171 and 69 is we will appy 175 + 70 = 245 in the equation we are using the  compatible numbers

170 + 70 = 240 in this equation we are using the nearest ten.

Question 3.
Subtract 382 – 148 mentally. Which of the following should you do first to find the difference?
A. Add 2 to 148 and add 2 to 382.
B. Add 2 to 148 and subtract 2 from to 382.
C. Subtract 8 from 148 and subtract 2 from 382.
D. Subtract 12 from 382 and add 12 to 148.
Answer:
To find the difference first to subtract 12 from 382 and add 12 to 148.

Question 4.
Estimate the difference of 765 and 333. Explain your estimate.
Answer:
The difference of 765 and 333 is
765 – 333 = 432
Our estimate is to Subtract the lowest value from the highest value.

Question 5.
Look at the sums in the shaded column. Look at the addends. What pattern do you see? Explain why this pattern is always true.

Answer:

Question 6.
Use mental math to add 332 and 154. Which of these shows how to break apart the numbers into hundreds, tens, and ones?
A. Break 332 into 320 + 12. Break 154 into 125 + 29.
B. Break 332 into 100+ 230 + 2. Break 154 into 100 + 52 + 2.
C. Break 332 into 300 + 16 + 16. Break 154 into 100 + 27 + 27.
D. Break 332 into 300 + 30+ 2. Break 154 into 100 + 50 + 4.
Answer:
D is the correct answer .

Question 7.
Use mental math to find 634 – 528.
Answer:
634 – 528
600 + 30 + 4 – 500 + 20 + 8
100 +10 – 4
110 – 4 = 106

Question 8.
Select all the equations that are true.
32 + 56 + 10 = 10 + 56 + 32
(49 + 28) + 5 = 49 + (28 + 5)
56 + 890 = 890 + 56
82 + 0 = 82
45+ 27 = 27 + 35
Answer:

Yes all the equations are true they are in the associative property, commutative property and identity property.

Question 9.
Look at the table.

Write the letters A, B, C, and D above the number line to show each number rounded to the nearest ten.

Answer:

Question 10.
Use the subtraction work shown.

Which strategy shows a way to check the work using inverse operations?
A. Subtract 57 from 169.
B. Add 57 and 112.
C. Add 112 and 169.
D. Add 169 and 57.
Answer:
Inverse means reverse operation
Inverse operation is the Add 169 and 57

Question 11.
Estimate the sum of 405 and 385 by rounding to the nearest hundred. Explain your estimate.
Answer:
The sum of 405 + 385 = 790

When rounding to the nearest hundred look at the ten’s place of the number. The ten’s place of the 790 is 9 now round to the next hundred.
790 round to the nearest hundred to estimate is 800.

Question 12.
What addition equation can be used to check the answer for 456 – 342 = 114? Draw a bar diagram to show how the numbers in this problem are related.
Answer:

Question 13.
Find the sum of 350, 62, and 199.
A. Draw a bar diagram that represents the problem.
B. What is the first step you would do to solve this problem using mental math?
Answer:

The sum of 350 + 62 + 199 = 611

Question 14.
Subtract 341 – 97 mentally. First add 3 to 97 to get 100. What should the next step be? What is the difference?
A. Add 6 to 341. The difference is 247.
B. Add 200 to 100. The difference is 100.
C. Subtract 3 from 341. The difference is 238.
D. Add 3 to 341. The difference is 244.
Answer: The next step is Subtract 3 from 341. The difference is 238.

Question 15.
Round each number on the left to the nearest hundred. Select the appropriate answer.

Answer
668 round to the nearest hundred to estimate is 700.
404 round to the nearest hundred to estimate is 400.
649 round to the nearest hundred to estimate is 600.
489 round to the nearest hundred to estimate is 500.

Question 16.
Explain how to use mental math to find 620 – 278.
Answer:
Using the mental math
620 – 278
600 + 20 – 200 + 70 + 8
600 + 20 – 200 + 78
600 + 20 – 278
620 – 278 = 332

Question 17.
How can you check the answer for 693 – 231 = 462?
A. Subtract 639 – 300 = 339.
B. Subtract 462 – 231 = 231.
C. Add 693 + 231 = 924.
D. Add 462 + 231 = 693.
Answer:
The correct answer is D.
693 – 231 = 462
462 + 231 = 693
Question 18.
Consider the sum of 123, 201, and 387.
A. Estimate the sum by rounding to the nearest ten. Explain your estimate.
B. What is the difference between the exact sum of 123, 201, and 387 and your estimate? Explain your answer.
Answer:
The sum of 123 + 201 + 387 = 711
A. 711 round to the nearest ten to estimate is 710
B. The difference between the exact sum and the estimate sum is 711 – 710 is 1.

Question 19.
Consider the equation 360 = __ + 84.
Use a bar diagram to represent the equation. Then solve for the unknown value.
Answer:
360 = 276 + 84

Topic 8 Performance Task

Vacation Trip
Mia is planning a vacation in Orlando, FL.
The Mia’s Route table shows her route and the miles she will drive.

Use the Mia’s Route table to answer Questions 1-3.

Question 1.
Round each distance to the nearest ten to show about how many miles Mia will drive on each part of her trip.

Memphis, TN, to Birmingham, AL:

Birmingham, AL, to Gainesville, FL:

Gainesville, FL, to Orlando, FL:
Answer:
Mia’s drive from Memphis,TN, to Birmingham, AL is 237 miles.
237 is round to the nearest ten is to estimate is 240.
Mia’s drives from Memphis, TN, to Birmingham, AL is 240 miles.
Mia’s drive from Birmingham, AL to Gainesville, FL is 422
422 round to the nearest ten to estimate is 420
Mia’s drive from Birmingham, AL to Gainesville, FL is 420 miles.
Mia’s drive from Gainesville, FL to Orlando, FL is 183 miles.
183 round to the nearest ten to estimate is 180.
Mia’s drive from Gainesville, FL to Orlando, FL is 180 miles.

Question 2.
Use mental math to find the actual number of miles Mia will drive to reach Gainesville. Show your work.
Answer:
Mia’s started a drive from Memphis, TN to Birmingham, AL is 237 miles.
From Birmingham, AL to Gainesville, FL is 422 miles.
Mia’s drive to reach Gainesville is 237 + 422
By using mental math
237 +422
200 + 30 + 7 + 400 + 20 + 2
600 + 50 + 9
600 + 59 = 659

Question 3.
Mia says, “Birmingham is 185 miles closer to Memphis than to Gainesville.” Her brother says, “No, it is 175 miles closer.” Use mental math to decide who is correct. Show your work.
Answer:
Birmingham  to Memphis is 237 miles.
Birmingham to Gainesville is 422 miles.
Mia’s says the correct
422 – 237
400 + 20 + 2 – 200 + 30 + 7
200 – 10 – 5
200 – 15 = 185

Mia has to book a hotel and buy theme park tickets. The Hotel Prices and Theme Park Prices tables show the total prices for Mia’s stay. The Mia’s Options list shows two plans that Mia can choose from.

Question 4.
Mia has $600 to spend on a hotel and tickets.
Part A

Which option does Mia have enough money for? Explain using estimation.
Part B
Create a new option for Mia. Fill out the table with a hotel and a theme park. Explain why Mia has enough money for this plan.

Question 5.
One theme park has a special offer. For each ticket Mia buys, she gets another ticket free. Shade the squares in the table at the right to show this pattern. Explain why the pattern is true.

Answer:

Go through the enVision Math Common Core Grade 8 Answer Key Topic 7 Understand And Apply The Pythagorean Theorem regularly and improve your accuracy in solving questions.

enVision Math Common Core 8th Grade Answers Key Topic 7 Understand And Apply The Pythagorean Theorem

Topic Essential Question
How can you use the Pythagorean Theorem to solve problems?
Answer:
The Pythagorean Theorem is used to calculate the steepness of slopes of hills or mountains. A surveyor looks through a telescope toward a measuring stick a fixed distance away, so that the telescope’s line of sight and the measuring stick form a right angle.

3-ACT MATH OOO

Go with the Flow
You may have noticed that when you double the base and the height of a triangle, the area is more than doubled. The same is true for doubling the sides of a square or the radius of a circle. So what is the relationship? Think about this during the 3-Act Mathematical Modeling lesson.

Topic 7 enVision STEM Project

Did You Know?
Over two billion people will face water shortages by 2050 according to a 2015 United Nations Environment Program report.

Rainwater can be collected and stored for use in irrigation, industrial uses, flushing toilets, washing clothes and cars, or it can be purified for use as everyday drinking water.
This alternative water source reduces the use of fresh water from reservoirs and wells.

Using water wisely saves money on water and energy bills and extends the life of supply and wastewater facilities.

Roofs of buildings or large tarps are used to collect rainwater.
A rainwater collection system for a building roof that measures 28 feet by 40 feet can provide 700 gallons of water-enough water to support two people for a year—from a rainfall of 1.0 inch.

Even a 5 foot by 7-foot tarp can collect 2 gallons of water from a rainfall total of only 0.1 in.

The rainwater harvesting market is expected to grow 5% from 2016 to 2020.

Your Task: Rainy Days

Rainwater collection is an inexpensive way to save water in areas where it is scarce. One inch of rain falling on a square roof with an area of 100 ft² collects 62 gallons of water that weighs over 500 pounds. You and your classmates will research the necessary components of a rainwater collection system. Then you will use what you know about right triangles to design a slanted roof system that will be used to collect rainwater.
Answer:
It is given that
Rainwater collection is an inexpensive way to save water in areas where it is scarce. One inch of rain falling on a square roof with an area of 100 ft² collects 62 gallons of water that weighs over 500 pounds
Now,
The necessary components of a rainwater collection system are:
A) Catchments B) Coarse mesh C) Gutters D) Conduits E) First-flushing F) Filter G) Storage facility H) Recharge Structures

Topic 7 GET READY!

Review What You Know!

Vocabulary
Choose the best term from the box to complete each definition.
cube root
diagonal
isosceles triangle
perimeter
right triangle
square root

Question 1.
The __________ of a number is a factor that when multiplied by itself gives the number.
Answer:
We know that,
The “Square root” of a number is a factor that when multiplied by itself gives the number
Hence, from the above,
We can conclude that the best term to complete the given definition is a “Square root”

Question 2.
A _________ is a line segment that connects two vertices of a polygon and is not the side.
Answer:
We know that,
A “Diagonal” is a line segment that connects two vertices of a polygon and is not the side
Hence, from the above,
We can conclude that the best term to complete the given definition is a “Diagonal”

Question 3.
The _________ of a figure is the distance around it.
Answer:
We know that,
The “Perimeter” of a figure is the distance around it
Hence, from the above,
We can conclude that the best term to complete the given definition is the “Perimeter”

Question 4.
A ___________ is a triangle with one right angle.
Answer:
We know that,
A “Right triangle” is a triangle with one right angle
Hence, from the above,
We can conclude that the best term to complete the given definition is a “Right angle”

Simplify Expressions with Exponents

Simplify the expression.
Question 5.
3² + 4²
Answer:
The given expression is: 3² + 4²
So,
3² + 4²
= (3 × 3) + (4 × 4)
= 9 + 16
= 25

Question 6.
2² + 5²
Answer:
The given expression is: 2² + 5²
So,
2² + 5²
= (2 × 2) + (5 × 5)
= 4 + 25
= 29

Question 7.
10² – 8²
Answer:
The given expression is: 10² – 8²
So,
10² – 8²
= (10 × 10) – (8 × 8)
= 100 – 64
= 36

Square Roots

Determine the square root.
Question 8.
\(\sqrt {81}\)
Answer:
The given expression is: \(\sqrt{81}\)
Hence,
\(\sqrt{81}\) = 9

Question 9.
\(\sqrt {144}\)
Answer:
The given expression is: \(\sqrt{144}\)
Hence,
\(\sqrt{144}\) = 12

Question 10.
\(\sqrt {225}\)
Answer:
The given expression is: \(\sqrt{225}\)
Hence,
\(\sqrt{225}\) = 15

Distance on a Coordinate Plane

Determine the distance between the two points.
Question 11.

Answer:
The given graph is:

From the given graph,
The given points are: (2, 5), (7, 5)
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
We know that,
Distance between 2 points = √(x₂ – x₁)² + (y₂ – y₁)²
= √(7 – 2)² + (5 – 5)²
= \(\sqrt{5²}\)
= 5 units
Hence, from the above,
We can conclude that the distance between the given points is: 5 units

Question 12.

Answer:
The given graph is:

From the given graph,
The given points are: (3, 2), (3, 9)
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
We know that,
Distance between 2 points =√(x₂ – x₁)² + (y₂ – y₁)²
= √(3 – 3)² + (9 – 2)²
= \(\sqrt{7²}\)
= 7 units
Hence, from the above,
We can conclude that the distance between the given points is: 7 units

Language Development

Complete the word map using key terms, examples, or illustrations related to the Pythagorean Theorem and its Converse.


Answer:

Topic 7 PICK A PROJECT

PROJECT 7A
Where would you like to bike ride in your neighborhood?
PROJECT: PLAN A METRIC CENTURY RIDE

PROJECT 7B
What designs have you seen on kites?
PROJECT: BUILD A KITE

PROJECT 7C
What buildings in your community have unusual shapes as part of their structure or design?
PROJECT: MAKE A SCRAPBOOK

PROJECT 7D
What geometric designs have you noticed on your clothes?
PROJECT: DESIGN A FABRIC TEMPLATE

3-ACT MATH

3-Act Mathematical Modeling: Go with the Flow

АСТ 1
Question 1.
After watching the video, what is the first question that comes to mind?
Answer:

Question 2.
Write the Main Question you will answer.
Answer:

Question 3.
Make a prediction to answer this Main Question.
_________ % will fit in the third square.
Answer:

Question 4.
Construct Arguments Explain how you arrived at your prediction.

Answer:

ACT 2
Question 5.
What information in this situation would be helpful to know? How would you use that information?
Answer:

Question 6.
Use Appropriate Tools What tools can you use to solve the problem? Explain how you would use them strategically.
Answer:

Question 7.
Model With Math
Represent the situation using mathematics. Use your representation to answer the Main Question.
Answer:

Question 8.
What is your answer to the Main Question? Does it differ from your prediction? Explain.

Answer:

АСТ 3
Question 9.
Write the answer you saw in the video.
Answer:

Question 10.
Reasoning Does your answer match the answer in the video? If not, what are some reasons that would explain the difference?
Answer:

Question 11.
Make Sense and Persevere Would you change your model now that you know the answer? Explain.

Answer:

Act 3
Extension
Reflect
Question 12.
Model with Math
Explain how you used a mathematical model to represent the situation. How did the model help you answer the Main Question?

Answer:

Question 13.
Reason Abstractly How did you represent the situation using symbols? How did you use those symbols to solve the problem?

Answer:

SEQUEL
Question 14.
Construct Arguments Explain why you can use an area formula when the problem involves comparing volumes.

Answer:

Lesson 7.1 Understand the Pythagorean Theorem

Explain It!
Kelly drew a right triangle on graph paper. Kelly says that the sum of the areas of squares with side lengths a and b is the same as the area of a square with side length c.

I can… use the Pythagorean Theorem to find unknown sides of triangles.

A. Do you agree with Kelly? Explain.
Answer:
It is given that
Kelly drew a right triangle on graph paper. Kelly says that the sum of the areas of squares with side lengths a and b is the same as the area of a square with side length c.
We know that,
According to the Pythagorean theorem,
Hypotenuse² = Side length 1² + Side length 2²
So,
From the given graph,
We can observe that
Side length 1 is: a
Side length 2 is: b
The hypotenuse is: c
So,
c² = a² + b²
Hence, from the above,
We can conclude that we can agree with Kelly

B. Sam drew a different right triangle with side lengths a = 5, b = 12, and c = 13. Is the relationship Kelly described true for Sam’s right triangle? Explain.
Answer:
It is given that
Sam drew a different right triangle with side lengths a = 5, b = 12, and c = 13
Now,
From part (a),
The relation according to Kelly is:
c² = a² + b²
So,
13² = 12² + 5²
169 = 144 + 25
169 = 169
Hence, from the above,
We can conclude that the relationship Kelly described is true for Sam’s right-angled triangle

Focus on math practices
Generalize Kelly draws another right triangle. What would you expect to be the relationship between the areas of the squares drawn on each side of the triangle? Explain.
Answer:
It is given that Kelly draws another right triangle
Hence,
If in a triangle, the square on one of the sides equals the sum of the squares on the remaining two sides of the triangle, then the angle contained by the remaining two sides of the triangle is right.”

Essential Question
How does the Pythagorean Theorem relate to the side lengths of a right triangle?
Answer:
The Pythagorean equation relates the sides of a right triangle in a simple way so that if the lengths of any two sides are known the length of the third side can be found. Another corollary of the theorem is that in any right triangle, the hypotenuse is greater than any one of the other sides but less than their sum.

Try It!

A right triangle has side lengths 15 centimeters, 25 centimeters, and 20 centimeters. How can you use the Pythagorean Theorem to write an equation that describes how the side lengths are related?
a² + b² = c²
_______² + ________² = _________²
_________ + _________ = ________
Answer:
It is given that
A right triangle has side lengths 15 centimeters, 25 centimeters, and 20 centimeters.
We know that,
According to the Pythagorean Theorem,
The hypotenuse has the greatest length in the right triangle
Now,
Let the hypotenuse be c
Let the other two sides be a and b
So,
From the given information,
c = 25 centimeters, a = 15 centimeters,and b = 20 centimeters
So,
According to the Pythagorean Theorem,
25² = 15² + 20²
625 = 225 + 400
625 = 625
Hence, from the above,
We can conclude that we proved how the Pythagorean Theorem relates to the lengths of the right triangle

Convince Me!
How do you know that the geometric proof of the Pythagorean Theorem shown above can be applied to all right triangles?
Answer:
It can be proven using the law of cosines or as follows: Let ABC be a triangle with side lengths a, b, and c, with a² + b² = c².  Therefore, the angle between the side of lengths a and b in the original triangle is a right angle. This proof of the converse makes use of the Pythagorean theorem itself.

Try It!

A right triangle has a hypotenuse length of 32 meters. It has one leg with a length of 18 meters. What is the length of the other leg? Express your answer as a square root.
Answer:
It is given that
A right triangle has a hypotenuse length of 32 meters. It has one leg with a length of 18 meters.
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the hypotenuse
a and b are the legs
Now,
Let the other leg be x
So,
32² = 18² + x²
x² = 32² – 18²
x² = 1024 – 324
x² = 700
x = \(\sqrt{700}\)
Hence, from the above,
We can conclude that the length of the other leg is: \(\sqrt{700}\)

KEY CONCEPT

The Pythagorean Theorem is an equation that relates the side lengths of a right triangle, a² + b² = c², where a and b are the legs of a right triangle and c is the hypotenuse.

Do You Understand?
Question 1.
Essential Question How does the Pythagorean Theorem relate to the side lengths of a right triangle?
Answer:
The Pythagorean equation relates the sides of a right triangle in a simple way so that if the lengths of any two sides are known the length of the third side can be found. Another corollary of the theorem is that in any right triangle, the hypotenuse is greater than any one of the other sides but less than their sum.

Question 2.
Use Structure A side of each of the three squares forms a side of a right triangle.
Would any three squares form the sides of a right triangle? Explain.

Answer:
It is given that
A side of each of the three squares forms a side of a right triangle.
Now,
We know that,
The length of all the sides in a square is equal
Now,
From the given figure,
We can observe that
Each side of a square from the three squares form a right triangle
Hence, from the above,
We can conclude that the three squares form the sides of a right triangle

Question 3.
Construct Arguments Xavier said the missing length is about 18.5 units. Without calculating, how can you tell that Xavier solved incorrectly?

Answer:
It is given that
Xavier said the missing length is about 18.5 units
Now,
We know that,
According to the Pythagorean Theorem,
The length of the hypotenuse is the greatest
Now,
The given right triangle is:

So,
According to the Pythagorean Theorem,
The length of the missing side should be greater than 21 and 26
Hence, from the above,
We can conclude that Xavier calculated incorrectly

Do You Know How?
Question 4.
A right triangle has leg lengths of 4 inches and 5 inches. What is the length of the hypotenuse? Write the answer as a square root and round to the nearest tenth of an inch.
Answer:
It is given that
A right triangle has leg lengths of 4 inches and 5 inches
Now,
We know that,
According to the Pythagorean theorem,
c² = a²+ b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
c² = 4² + 5²
c² = 16 + 25
c² = 41
c = \(\sqrt{41}\)
c = 6.4 inches
Hence, from the above,
We can conclude that the length of the hypotenuse is: 6.4 inches

Question 5.
Find the missing side length to the nearest tenth of afoot.

Answer:
The given right triangle is:

Now,
We know that,
According to the Pythagorean theorem,
c² = a²+ b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
14² = 8² + b²
b²= 14² – 8²
b² = 196 – 64
b² = 132
b = \(\sqrt{132}\)
b = 11.5 feet
Hence, from the above,
We can conclude that the length of the missing side is: 11.5 feet

Question 6.
Find the missing side length to the nearest tenth of a millimeter.

Answer:
The given right triangle is:

Now,
We know that,
According to the Pythagorean theorem,
c² = a²+ b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
c² = (3.7)² + (7.5)²
c² = 13.69 + 56.25
c² = 69.94
c = \(\sqrt{69.94}\)
c = 8.4 mm
Hence, from the above,
We can conclude that the length of the missing side is: 8.4 mm

Practice & Problem Solving

Leveled Practice In 7 and 8, find the missing side length of each triangle.
Question 7.

The length of the hypotenuse is ________ units.
Answer:

Question 8.

The length of leg b is about ________ inches.
Answer:

Question 9.
What is the length of the hypotenuse of the triangle when x = 15? Round your answer to the nearest tenth of a unit.

Answer:
The given right angle is:

Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
c² = (3x)² + (4x + 4)²
Now,
When x = 15,
c² = (3 × 15)² + (4 × 15 + 4)²
c² = 45² + 64²
c² = 2,025 + 4,096
c² = 6,121
c = \(\sqrt{6,121}\)
c = 78.2 units
Hence, from the above,
We can conclude that the length of the hypotenuse when x= 15 is: 78.2 units

Question 10.
What is the length of the missing side rounded to the nearest tenth of a centimeter?

Answer:
The given right triangle is:

Now,
We know that,
According to the Pythagorean Theorem,
c² = a²+ b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
a² = (12.9)² + (15.3)²
a² = 166.41 + 234.09
a² = 400.5
a = \(\sqrt{400.5}\)
a = 20 cm
Hence, from the above,
We can conclude that the length of the side a is: 20 cm

Question 11.
Use the Pythagorean Theorem to find the unknown side length of the right triangle.

Answer:
The given right triangle is:

Now,
We know that,
According to the Pythagorean Theorem,
c² = a²+ b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
c² = (10)² + (24)²
c² = 100 + 576
c² = 676
c = \(\sqrt{676}\)
c = 26 m
Hence, from the above,
We can conclude that the length of the side a is: 26 m

Question 12.
What is the length of the unknown leg of the right triangle rounded to the nearest tenth of afoot?

Answer:
The given right triangle is:

Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
9² = 2² + b²
b² = 9² – 2²
b² = 81 – 4
b² = 77
b = \(\sqrt{77}\)
b = 8.8 foot
Hence, from the above,
We can conclude that the length of the unknown leg is: 8.8 foot

Question 13.
A student is asked to find the length of the hypotenuse of a right triangle. The length of one leg is 32 centimeters, and the length of the other leg is 26 centimeters. The student incorrectly says that the length of the hypotenuse is 7.6 centimeters.
a. Find the length of the hypotenuse of the right triangle to the nearest tenth of a centimeter.
Answer:
It is given that
A student is asked to find the length of the hypotenuse of a right triangle. The length of one leg is 32 centimeters, and the length of the other leg is 26 centimeters. The student incorrectly says that the length of the hypotenuse is 7.6 centimeters.
Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
c² = 32² + 26²
c² = 1,024 + 676
c² = 1,700
c = \(\sqrt{1,700}\)
c = 41.2 centimeters
Hence, from the above,
We can conclude that the length of the hypotenuse of a right triangle is: 41.2 centimeters

b. What mistake might the student have made?
Answer:
It is given that
The student incorrectly says that the length of the hypotenuse is 7.6 centimeters.
But,
From part (a),
The length of the hypotenuse is: 41.2 centimeters
Hence, from the above,
We can conclude that the mistake the student might make is the misinterpretation of the length of the hypotenuse

Question 14.
Find the length of the unknown leg of the right triangle.

Answer:
The given right triangle is:

Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
(37.25)² = (12.25)² + b²
b² = (37.25)² – (12.25)²
b² = 1,387.56 – 150.06
b² = 1,237.5
b = \(\sqrt{1,237.5}\)
b = 35.17 units
Hence, from the above,
We can conclude that the length of the unknown leg is: 35.17 units

Question 15.
Higher-Order Thinking A right triangle has side lengths 12 centimeters and 14 centimeters. Name two possible side lengths for the third side, and explain how you solved for each.
Answer:
It is given that
A right triangle has side lengths of
12 centimeters and 14 centimeters.
Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
Now,
Let the length of the third side be x
So,
The possible lengths of the third side may be:
x < 12 centimeters and x > 14 centimeters
Hence, from the above,
We can conclude that the two possible side lengths for the third side are:
x < 12 centimeters and x > 14 centimeters

Assessment Practice
Question 16.
Which right triangle has a hypotenuse that is about 39 feet long?

Answer:
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
For Options A and D:
A) c² = 30² + 15²                     D) c² = 30² + 14²
= 33.54 ft                                 = 33.10 ft
For Options B and C:
B) c² = 36² + 12²                      C) c² = 36² + 15²
= 37.94 ft                                    = 39 ft
Hence, from the above,
We can conclude that Option C matches the given situation

Question 17.
Which right triangle does NOT have an unknown leg length of about 33 cm?

Answer:
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the length of the hypotenuse
a and b are the lengths of the legs
So,
For Options A and D:
A) b² = 35² – 11²                     D) b² = 34² – 9²
= 33.22 cm                              = 32.78 cm
For Options B and C:
B) b² = 30² – 10²                      C) b² = 35² – 12²
= 28.28 cm                              = 32.87 cm
Hence, from the above,
We can conclude that Option B matches the given situation

Lesson 7.2 Understand the Converse of the Pythagorean Theorem

Solve & Discuss It!
Kayla has some straws that she will use for an art project. She wants to glue three of the straws onto a sheet of paper, without overlapping, to make the outline of a right triangle. Which three straws could Kayla use to make a right triangle? Explain.

I can… use the Converse of the Pythagorean Theorem to identify right triangles.
Answer:
It is given that
Kayla has some straws that she will use for an art project. She wants to glue three of the straws onto a sheet of paper, without overlapping, to make the outline of a right triangle.
Now,
We know that,
The converse of the Pythagorean Theorem states that if the square of the third side of a triangle is equivalent to the sum of its two shorter sides, then it must be a right triangle i.e.,
If c² = a²+ b², then the given triangle is a right triangle
So,
From the given straws,
we can observe that
The straws numbered 3, 4, and 5 can be glued to make the outline of a right triangle
The straws numbered 5, 12, and 13 can be glued to make the outline of a right triangle
Hence, from the above,
We can conclude that there are 2 pairs of straws i.e., (3, 4, 5) and (12, 5, 13) to make the outline of a right triangle

Look for Relationships
How could you use the Pythagorean Theorem to determine whether the lengths form a right triangle?
Answer:
According to the converse of the Pythagorean Theorem,
If the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.

Focus on math practices
Use Structure Could Kayla use the straws that form a right triangle to make a triangle that is not a right triangle? Explain.
Answer:
From the given straws,
We can observe that the pair (6, 7, 4) can’t form a right triangle but (3, 4, 5) can form a right triangle
Hence, from the above,
We can conclude that Kayla can use the straws that form a right triangle to make a triangle that is not a right triangle

Essential Question
How can you determine whether a triangle is a right triangle?
Answer:
We can determine the triangle is a right triangle by using the converse of the Pythagorean Theorem
Hence,
According to the converse of the Pythagorean Theorem,
If the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.

Try It!

A triangle has side lengths 4 inches, 5 inches, and 7 inches. Is the triangle a right triangle?

Is a² + b² equal to c²? _______
Is the triangle a right triangle? _________
Answer:
It is given that
A triangle has side lengths 4 inches, 5 inches, and 7 inches.
Now,
We know that,
According to the converse of the Pythagorean Theorem,
If the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.
So,
c² = a² + b²
Where,
c is the hypotenuse that has the longest side length
a and b are the lengths of the legs
So,
7² = 4² + 5²
49 = 16 + 25
49 = 41
So,
49 ≠ 41
Hence, from the above,
We can conclude that
c² ≠ a²+ b²
The given triangle is not a right triangle

Convince Me!
Explain the proof of the Converse of the Pythagorean Theorem in your own words.
Answer:
The converse of the Pythagorean Theorem is:
If the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.

Try It!

A triangle has side lengths 10 feet, \(\sqrt {205}\) feet, and \(\sqrt {105}\) feet. Is this a right triangle? Explain.
Answer:
It is given that
A triangle has side lengths 10 feet, \(\sqrt {205}\) feet, and \(\sqrt {105}\) feet.
Now,
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b², then
The given triangle is a right triangle
We know that,
c is the length of the hypotenuse that has the longest side in a right triangle
So,
(\(\sqrt{205}\))² =(\(\sqrt{105}\))² + 10²
205 = 105 + 100
205 = 205
So,
c² = a² + b²
Hence, from the above,
We can conclude that the triangle with the given side lengths is a right triangle

Try It!

A triangle is inside a trapezoid. Is the triangle a right triangle? Explain.

Answer:
It is given that a triangle is inside a trapezoid
Now,
The given trapezoid is:

From the given trapezoid,
The sides of the triangle are: 17 in., 15 in., \(\sqrt{514}\) in.
Now,
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²,
then, the given triangle is a right triangle
So,
(\(\sqrt{514}\))² = 17² + 15²
514 = 289 + 225
514 = 514
So,
c² = a² + b²
Hence, from the above,
We can conclude that the triangle that is in the trapezoid is a right triangle

KEY CONCEPT

The Converse of the Pythagorean Theorem states that if the sum of the squares of the lengths of two sides of a triangle is equal to the square of the length of the third side, the triangle is a right triangle.

Do You Understand?
Question 1.
Essential Question How can you determine whether a triangle is a right triangle?
Answer:
We can determine the triangle is a right triangle by using the converse of the Pythagorean Theorem
Hence,
According to the converse of the Pythagorean Theorem,
If the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.

Question 2.
Construct Arguments A triangle has side lengths of 3 centimeters, 5 centimeters, and 4 centimeters. Abe used the Converse of the Pythagorean Theorem to determine whether it is a right triangle.
3² + 5² \(\underline{\underline{?}}\) 4²
9 + 25 \(\underline{\underline{?}}\) 16
34 ≠ 16
Abe concluded that it is not a right triangle. Is Abe correct? Explain.
Answer:
It is given that
A triangle has side lengths of 3 centimeters, 5 centimeters, and 4 centimeters
Now,
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²,
then, the triangle is  a right triangle
We know that,
The hypotenuse has the length of the longest side
So,
5²= 3² + 4²
25 = 9 + 16
25 = 25
So,
c² = a² + b²
Hence, from the above,
We can conclude that the given triangle is a right triangle and Abe is not correct

Question 3.
Use Structure When you are given three side lengths for a triangle, how do you know which length to substitute for a, b, or c in the Pythagorean Theorem?
Answer:
First, we will find the squares of all lengths, then we will check which two squares of sides are equal to the square of the third side as per Pythagoras theorem. Hence the two sides would according be a and b and the third side will become c

Do You Know How?
Question 4.
Is the triangle a right triangle? Explain.

Answer:
The given triangle is:

We know that,
According to the converse of the Pythagorean Theorem,
If c² = a² + b²,
then, the triangle is a right triangle
So,
10² = 6² + 8²
100 = 36 + 64
100 = 100
So,
c² = a² + b²
Hence, from the above,
We can conclude that the given triangle is a right triangle

Question 5.
Is the triangle a right triangle? Explain.

Answer:
The given triangle is:

We know that,
According to the converse of the Pythagorean Theorem,
If c² = a² + b²,
then, the triangle is a right triangle
So,
8² = (\(\sqrt{26}\))² + (\(\sqrt{28}\))²
64 = 26 + 28
64 = 54
So,
c² ≠ a² + b²
Hence, from the above,
We can conclude that the given triangle is not a right triangle

Question 6.
Is the purple triangle a right triangle? Explain.

Answer:
The given triangle is:

We know that,
According to the converse of the Pythagorean Theorem,
If c² = a² + b²,
then, the triangle is a right triangle
So,
(20.8)² = 14² + (15.5)²
432.64 = 196 + 240.25
432.64 = 436.25
So,
c² ≠ a² + b²
Hence, from the above,
We can conclude that the purple triangle is not a right triangle

Practice & Problem Solving

Leveled Practice In 7 and 8, determine whether each triangle is a right triangle.
Question 7.

Is the triangle a right triangle? ________
Answer:

Hence, from the above,
We can conclude that the given triangle is not a right triangle

Question 8.

Is the triangle a right triangle? _________
Answer:

Hence, from the above,
We can conclude that the given triangle is a right triangle

Question 9.
Can the sides of a right triangle have lengths 5, 15, and \(\sqrt {250}\)? Explain.
Answer:
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²,
then, the triangle is a right triangle
The hypotenuse has the longest side length
So,
(\(\sqrt{250}\))² = 5² + 15²
250 = 25 + 225
250 = 250
Hence, from the above,
We can conclude that the sides of a right triangle have lengths 5, 15, and \(\sqrt {250}\)

Question 10.
Is ∆PQR a right triangle? Explain.

Answer:
The given triangle is:

We know that,
According to the converse of the Pythagorean Theorem,
If c² = a² + b²,
then, the triangle is a right triangle
So,
(6.25)² = (3.75)² + 5²
39.0625 = 14.0625 + 25
39.0625 = 39.0625
So,
c² = a² + b²
Hence, from the above,
We can conclude that ΔPQR is a right triangle

Question 11.
The green triangle is set inside a rectangle. Is the green triangle a right triangle? Explain.

Answer:
It is given that
The green triangle is set inside a rectangle
Now,
The given figure is:

We know that,
According to the converse of the Pythagorean Theorem,
If c² = a² + b²,
then, the triangle is a right triangle
So,
21² = (\(\sqrt{282}\))² + (\(\sqrt{159}\))²
441 = 282 + 159
441 = 441
So,
c² = a² + b²
Hence, from the above,
We can conclude that the given green triangle is a right triangle

Question 12.
The side lengths of three triangles are shown. Which of the triangles are right triangles?

Answer:
It is given that
The side lengths of the three triangles are shown in the table
Now,
The given table is:

Now,
We know that,
According to the converse of the Pythagorean Theorem,
If c² = a² + b²,
then, the triangle is a right triangle
So
For Triangle 1,
(\(\frac{5}{7}\))² = (\(\frac{3}{7}\))² + (\(\frac{4}{7}\))²
25 = 16 + 9
25 = 25
For Triangle 2,
15² = 8² + 8²
225 = 64 + 64
225 = 128
For Triangle 3,
(\(\frac{13}{17}\))² = (\(\frac{12}{17}\))² + (\(\frac{5}{17}\))²
169 = 144 + 25
169 = 169
So,
The condition
c² = a² + b²
is true for the Triangles 1 and 3
Hence, from the above,
We can conclude that Triangle 1 and Triangle 3 is a right triangle

Question 13.
Construct Arguments Three students draw triangles with the side lengths shown. All three say that their triangle is a right triangle. Which students are incorrect? What mistake might they have made?
Student 1: 22, 33, 55
Student 2: 44, 33, 77
Student 3: 33, 44, 55
Answer:
It is given that
Three students draw triangles with the side lengths shown. All three say that their triangle is a right triangle.
Now,
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²,
then, the triangle is a right triangle
Now,
For student 1,
55² = 33² + 22²
3,025 = 1,089 + 484
3,025 = 1,573
For student 2,
77²= 44² + 33²
5,929 = 1,936 + 1,089
5,929 = 3,025
For student 3,
55² = 44² + 33²
3,025 = 1,936 + 1,089
3,025 = 3,025
So,
The condition
c² = a² + b²
is false for the side lengths of the triangles that are drawn by students 1 and 2
Hence, from the above,
We can conclude that student 1 and student 2 are incorrect

Question 14.
Model with Math
∆JKL is an isosceles triangle. Is KM the height of ∆JKL? Explain.

Answer:
It is given that
ΔJKL is an isosceles triangle
Now,
To find whether KM is the height of ΔJKL,
Find out whether ΔKLM is a right triangle or not
Now,
The given triangle is:

Now,
We know that,
According to the converse of the Pythagorean Theorem,
c² = a²+ b²
So,
From ΔKLM,
(\(\sqrt{340}\))² = 13² + 14²
340 = 169 + 196
340 = 365
So,
c² ≠ a² + b²
So,
ΔKLM is not a right triangle
Hence, from the above,
We can conclude that KM is not the height of ΔJKL

Question 15.
Higher-Order Thinking The side lengths of three triangles are given.
Triangle 1: \(\sqrt{229}\) units, \(\sqrt{225}\) units, 22 units
Triangle 2: \(\sqrt{11 \frac{1}{3}}\) units, \(\sqrt{13 \frac{2}{3}}\) units, 5 units
Triangle 3: 16 units, 17 units, \(\sqrt{545}\) units
a. Which lengths represent the side lengths of a right triangle?
Answer:
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²
then, the triangle is a right triangle
Now,
For Triangle 1,
(\(\sqrt{229}\))² = (\(\sqrt{225}\))² + 22²
229 = 225 + 484
229 = 709
For Triangle 2,
(\(\sqrt{\frac{41}{3}}\))² + (\(\sqrt{\frac{34}{3}}\))² = 5²
\(\frac{41}{3}\) + \(\frac{34}{3}\) = 25
25 = 25
For Triangle 3,
(\(\sqrt{545}\))² = 16² + 17²
545 = 256 + 289
545 = 545
So,
The condition
c² = a²+ b²
is true for triangle 2 and triangle 3
Hence, from the above,
We can conclude that Triangle 2 and Triangle 3 represent the side lengths of a triangle

b. For any triangles that are not right triangles, use two of the sides to make a right triangle.
Answer:

Assessment Practice
Question 16.
Which shaded triangle is a right triangle? Explain.

Answer:
The given figure is:

Now,
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a²+ b²,
then the triangle is a right triangle
Now,
For ΔABC,
144² = 63² + (\(\sqrt{9}\))²
20,736 = 3,969 + 9
20,736 = 3,978
For ΔXYZ,
(\(\sqrt{144}\))² = (\(\sqrt{63}\))² + 9²
144 = 63 + 81
144 = 144
So,
The condition c² = a²+ b² is true for the shaded triangle XYZ
Hence, from the above,
We can conclude that the shaded triangle ΔXYZ is a right triangle

Question 17.
Which triangle is a right triangle?

A. Triangle I only
B. Triangle II only
C. Triangle I and Triangle II
D. Neither Triangle I nor Triangle II
Answer:
The given triangles are:

Now,
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²,
then the triangle is a right triangle
Now,
For Triangle I,
52² = 40² + 48²
2,704 = 1,600 + 2,304
2,704 = 3,904
For Triangle II,
65² = 60² + 25²
4,225 = 3,600 + 625
4,225 = 4,225
Hence, from the above,
We can conclude that Triangle II only is a right triangle

Topic 7 MID-TOPIC CHECKPOINT

Question 1.
Vocabulary How are the hypotenuse and the legs of a right triangle related? Lesson 7-1
Answer:
The relation between the sides and angles of a right triangle is the basis for trigonometry. The side opposite the right angle is called the hypotenuse. The sides adjacent to the right angle are called legs

Question 2.
Given that ∆QPR has side lengths of 12.5 centimeters, 30 centimeters, and 32.5 centimeters, proves ∆QPR is a right triangle. Lesson 7-2
Answer:
It is given that
∆QPR has side lengths of 12.5 centimeters, 30 centimeters, and 32.5 centimeters
Now,
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²,
then, the triangle is a right triangle
So,
(32.5)² = (12.5)² + 30²
1,056.25 = 156.25 + 900
1,056.25 = 1,056.25
So,
The condition
c² = a² + b²
is true for the given side lengths of a triangle
Hence, from the above,
We can conclude that ΔQPR is a right triangle

Question 3.
Ella said that if she knows the lengths of just two sides of any triangle, then she can find the length of the third side by using the Pythagorean Theorem. Is Ella correct? Explain. Lesson 7-1
Answer:
Ella said that if she knows the lengths of just two sides of any triangle, then she can find the length of the third side by using the Pythagorean Theorem.
Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the hypotenuse
a and b are the legs
Now,
If we know a and b, then we can find c
If we know b and c, then we can find a
If we know a and c, then we can find b
Hence, from the above,
We can conclude that Ella is correct

Question 4.
Find the unknown side length. Round to the nearest tenth. Lesson 7-1

Answer:
The given triangle is:

From the given triangle,
We can observe that it is a right triangle
So,
Now,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the hypotenuse
a and b are the legs
Now,
8² = 4² + b²
64 = 16 + b²
b² = 64 – 16
b² = 48
b = \(\sqrt{48}\)
b = 6.9 cm
Hence, from the above,
We can conclude that the length of the unknown side is: 6.9 cm

Question 5.
The height of a shed is 6 m. A ladder leans against the shed with its base 4.5 m away, and its top just reaching the roof. What is the length of the ladder? Lesson 7-1

Answer:
It is given that
The height of a shed is 6 m. A ladder leans against the shed with its base 4.5 m away, and its top just reaching the roof.
Now,
From the given figure,
We can observe that the ladder looks like the hypotenuse of a right triangle for the shed
Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the hypotenuse
a and b are the legs
In this situation,
c is the length of the ladder
a is the height of the shed
b is the length of the base
So,
c² = 6² + (4.5)²
c² = 36 + 20.25
c² = 56.25
c = \(\sqrt{56.25}\)
c = 7.5 m
Hence, from the above,
We can conclude that the length of the ladder is: 7.5 m

Question 6.
Select all the sets of lengths that could represent the sides of a right triangle. Lesson 7-2
☐ 5 cm, 10 cm, 15 cm
☐ 7 in., 14 in., 25 in.
☐ 13 m, 84 m, 85 m
☐ 5 ft, 11 ft, 12 ft
☐ 6ft, 9 ft, \(\sqrt {117}\) ft
Answer:
Let the given options be named as A, B, C, D, and E
Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the hypotenuse and has the longest length
a and b are the legs
So,
A)
15² = 5² + 10²
225 = 25 + 100
225 ≠ 125
B)
25² = 7² + 14²
625 = 49 + 196
625 ≠ 245
C)
85² = 84² + 13²
7,225 = 7,056 + 169
7,225 = 7,225
D)
12² = 11² + 5²
144 = 121 + 25
144 ≠ 146
E)
(\(\sqrt{117}\))² = 6² + 9²
117 = 36 + 81
17 = 117
Hence, from the above,
We can conclude that the side lengths present in options C and E represent the side lengths of a right triangle

Topic 7 MID-TOPIC PERFORMANCE TASK

Javier is standing near a palm tree. He holds an electronic tape measure near his eyes and finds the three distances shown.

PART A
Javier says that he can now use the Pythagorean Theorem to find the height of the tree. Explain. Use vocabulary terms in your explanation.
Answer:
It is given that
Javier says that he can now use the Pythagorean Theorem to find the height of the tree
Now,
According to the Pythagorean Theorem,
In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“. The sides of this triangle have been named Perpendicular (The height of the tree), Base, and Hypotenuse.

PART B
Find the height of the tree. Round to the nearest tenth. Show your work.

Answer:
From the given figure,
We can observe that
To find the height of the tree, we have to find the perpendicular distances of the two right triangles
Now,
Let x be the perpendicular height of the first right triangle
Let y be the perpendicular height of the second right triangle
So,
The height of the tree = x + y
Now,
The side lengths of the first right triangle are: 25 ft, 7 ft, x ft
Now,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the hypotenuse
a and b are the side lengths
So,
25² = 7² + x²
x² = 625 – 49
x² = 576
x = \(\sqrt{576}\)
x = 24 ft
Now,
The side lengths of the second right triangle are: 9 ft, 7ft, y ft
So,
9² = 7² + y²
y² = 81 – 49
y² = 32
y = \(\sqrt{32}\)
y = 5.6 ft
So,
The height of the tree = 24 + 5.6
= 29.6 ft
Hence, from the above,
We can conclude that the height of the tree to the nearest tenth is: 29.6 ft

PART C
Javier moves backward so that his horizontal distance from the palm tree is 3 feet greater. Will the distance from his eyes to the top of the tree also be 3 feet greater? Explain.
Answer:
Yes it will be greater, he is moving back 3 feet so what you are doing is taking the leg (a²) and multiplying it by 3. Once you do, you see triangle 1 has double and so did triangle 2. Triangle 2 was originally 5.6 (rounded to 6) then went up to 26.1. Triangle 1 was originally 24 and went up to 74.7 (or 75)
Step-by-step explanation:
For Triangle 1 (when multiplied by 3):
We know that,
a² + b² = c²
7² + b² = 27²
49 + b² = 729
b² = 680
b  = \(\sqrt{680}\)
b = 26.1 ft
For Triangle 2:
We know that,
a² + b² = c²
7² + b² =75²
b² = 5576
b = \(\sqrt{5,576}\)
b = 74.7 ft

PART D
Could Javier change his horizontal distance from the tree so that the distance from his eyes to the top of the tree is only 20 feet? Explain.
Answer:
Yes, Javier can change his horizontal distance from the tree so that the distance from his eyes to the top of the tree is only 20 feet by moving forward 5 ft

Lesson 7.3 Apply the Pythagorean Theorem to Solve Problems

Solve & Discuss It!
Carlos is giving his friend in another state a new umbrella as a gift. He wants to ship the umbrella in a box he already has. Which box can Carlos use to ship the umbrella? Explain.

I can… use the Pythagorean Theorem to solve problems.
Answer:
It is given that
Carlos is giving his friend in another state a new umbrella as a gift. He wants to ship the umbrella in a box he already has.
Now,
From the given figure,
We can observe that the umbrella is 37.5 inches long.
Now,
If you observe, each box doesn’t have 37 inches wide, however, they are tridimensional figures, which means they have a certain height.
So,
The smaller box is too small to fit the umbrella.
But,
The medium box is perfect because it has 27 inches wide and 27 inches in height, which is enough to fit the umbrella. These are the best dimensions to send the umbrella.
Hence, from the above,
We can conclude that the box that can be used by Carlos to ship the umbrella has the dimensions 27 in × 14 in × 27 in

Make Sense and Persevere
How will the umbrella fit inside any of the boxes?
Answer:
We know that,
The box is a 3-d figure
So,
To fit the whole umbrella in the box, we have to put it in a diagonal manner i.e., like the hypotenuse of a right triangle

Focus on math practices
Construct Arguments Tim says that the diagonal of any of the boxes will always be longer than the sides. Is Tim correct? Explain.
Answer:
We know that,
If we consider a square or any 2-d figure or any 3-d figure that consists of 1 right angle,
then, the diagonal will divide a figure into 2 right triangles
We know that,
We can apply the Pythagorean Theorem for any right triangle
We know that,
In a right triangle,
The hypotenuse is the longest side
We know that,
The hypotenuse in a right triangle is considered as a diagonal in a figure that consists of 1 right angle
Hence, from the above,
We can conclude that Tim is correct

Essential Question
What types of problems can be solved using the Pythagorean Theorem?
Answer:
The Pythagorean theorem is a way of relating the leg lengths of a right triangle to the length of the hypotenuse, which is the side opposite the right angle. Even though it is written in these terms, it can be used to find any of the sides as long as you know the lengths of the other two sides

Try It!

What is the length of the diagonal, d, of a rectangle with length 19 feet and width 17 feet?
leg² + leg² = hypotenuse²
______ ² + _______² = d²
______ + _______ = d²
_______ = d²
________ ≈ d
Answer:
It is given that a rectangle has a length of 19 feet and a width of 17 feet
Now,
We know that,
In a rectangle,
If a diagonal is drawn, then it divides the rectangle into 2 right angles
So,
According to the Pythagorean Theorem,
d² = a² + b²
Where,
d is the diagonal or hypotenuse
a and b are the lengths of the legs
So,
d²= 19² + 17²
d² = 361 + 289
d² = 650
d = \(\sqrt{650}\)
d = 25.4 ft
Hence, from the above,
We can conclude that the length of the diagonal is: 25.4 ft

Convince Me!
If the rectangle were a square, would the process of finding the length of the diagonal change? Explain.
Answer:
We know that,
For any figure i.e., either 2-d figure or 3-d figure with 1 right angle, the diagonal will divide that figure into 2 right angles
We know that,
We will use the Pythagorean Theorem to find the length of any unknown side in the right triangle
Hence, from the above,
We can conclude that even if the rectangle were a square, the process of finding the length of the diagonal will not change

Try It!

A company wants to rent a tent that has a height of at least 10 feet for an outdoor show. Should they rent the tent shown at the right? Explain.

Answer:
It is given that
A company wants to rent a tent that has a height of at least 10 feet for an outdoor show.
Now,
The given figure is:

Now,
To find whether the tent should be rented or not,
We have to find the value of h
We can also observe that the triangle that contains the value of h is a right triangle
Now,
The base of the right triangle = \(\frac{24}{2}\)
= 12 ft
Now,
The side lengths of the right triangle are: h, 15 ft, 12 ft
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
So,
15² = 12² + h²
h² = 225 – 144
h² = 81
h = \(\sqrt{81}\)
h = 9 ft
But,
The given height of the tent is: 10 ft
So,
9 ft < 10 ft
Hence, from the above,
We can conclude that the company should not rent the tent

KEY CONCEPT

You can use the Pythagorean Theorem and its converse to solve problems involving right triangles.

Do You Understand?
Question 1.
Essential Question What types of problems can be solved using the Pythagorean Theorem?
Answer:
The Pythagorean theorem is a way of relating the leg lengths of a right triangle to the length of the hypotenuse, which is the side opposite the right angle. Even though it is written in these terms, it can be used to find any of the sides as long as you know the lengths of the other two sides

Question 2.
Look for Structure How is using the Pythagorean Theorem in a rectangular prism similar to using it in a rectangle?
Answer:
We know that,
The rectangular prism and the rectangle have at least 1 right angle
We know that,
If a 3-d or 2-d figure has 1 right angle, then the diagonal of that figure divides the figure into the right triangles
So,
If we have the right triangle, then we can use the Pythagorean Theorem irrespective of the overall shape of the figure

Question 3.
Construct Arguments Glen found the length of the hypotenuse of a right triangle using \(\sqrt{a^{2}+b^{2}}\). Gigi used \(\sqrt{(a+b)^{2}}\). Who is correct? Explain.
Answer:
It is given that
Glen found the length of the hypotenuse of a right triangle using \(\sqrt{a^{2}+b^{2}}\). Gigi used \(\sqrt{(a+b)^{2}}\).
Now,
We know that,
We can use the Pythagorean Theorem only for the right triangles
The condition for a triangle to become the right triangle is:
Hypotenuse² = Side² + Side²
c² = a² + b²
Hence, from the above,
We can conclude that Glen is correct

Do You Know How?
Question 4.
You are painting the roof of a boathouse. You are going to place the base of a ladder 12 feet from the boathouse. How long does the ladder need to be to reach the roof of the boathouse?

Answer:
It is given that
You are painting the roof of a boathouse. You are going to place the base of a ladder 12 feet from the boathouse.
Now,
From the figure,
We can observe that the roof of a boathouse, ladder, and the base form a right triangle
So,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the height of the boathouse
a is the length of the base
b is the length of the ladder
So,
35² = 12² + b²
b² = 1,225 – 144
b² = 1,081
b = \(\sqrt{1,081}\)
b = 32.9 ft
Hence, from the above,
We can conclude that the length of the ladder that is needed to reach the roof of the boathouse is: 32.9 ft

Question 5.
A box-shaped like a right rectangular prism measures 5 centimeters by 3 centimeters by 2 centimeters. What is the length of the interior diagonal of the prism to the nearest hundredth?
Answer:
It is given that
A box-shaped like a right rectangular prism measures 5 centimeters by 3 centimeters by 2 centimeters.
So,
The dimensions of a rectangular prism is: 5 cm × 3 cm × 2 cm
So,
The length of the rectangular prism is: 5 cm
The width of the rectangular prism is: 3 cm
The height of the rectangular prism is: 2 cm
Now,
We know that,
The length of the diagonal of the prism = \(\sqrt{Length^{2} + Width^{2} + Height^{2}}\)
So,
The length of the diagonal of the rectangular prism = \(\sqrt{5^{2} + 3^{2} + 2^{2}}\)
= \(\sqrt{25 + 9 + 4}\)
= \(\sqrt{38}\)
= 6.16 cm
Hence, from the above,
We can conclude that the length of the interior diagonal of the rectangular prism is: 6.16 cm

Question 6.
A wall 12 feet long makes a corner with a wall that is 14 feet long. The other ends of the walls are about 18.44 feet apart. Do the walls form a right angle? Explain.

Answer:
It is given that
A wall 12 feet long makes a corner with a wall that is 14 feet long. The other ends of the walls are about 18.44 feet apart
Now,
The given figure is:

From the given figure,
We can observe that the given situation forms the right triangle
Now,
We know that,
According to the Pythagorean Therem,
c² = a² + b²
So,
c² = 12² + 14²
c² = 144 + 196
c² = 340
c = \(\sqrt{340}\)
c = 18.44 feet
Hence, from the above,
We can conclude that the walls form a right angle

Practice & Problem Solving

Leveled Practice In 7 and 8, use the Pythagorean Theorem to solve.
Question 7.
You are going to use an inclined plane to lift a heavy object to the top of a shelving unit with a height of 6 feet. The base of the inclined plane is 16 feet from the shelving unit. What is the length of the inclined plane? Round to the nearest tenth of a foot.

The length of the inclined plane is about ________ feet.
Answer:

Question 8.
Find the missing lengths in the rectangular prism.

Answer:

Question 9.
A stainless steel patio heater is shaped like a square pyramid. The length of one side of the base is 19.8 inches. The slant height is 92.8 inches. What is the height of the heater? Round to the nearest tenth of an inch.
Answer:
It is given that
A stainless steel patio heater is shaped like a square pyramid. The length of one side of the base is 19.8 inches. The slant height is 92.8 inches
Now,
We know that,
The slant height is nothing but a length of the diagonal
Now,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the slant height
a is the length of the base of the steel patio heater
b is the height of the heater
So,
(92.8)² = (19.8)² + b²
b² = 8,611.84 – 392.04
c² = 8,219.8
c = \(\sqrt{8,219.8}\)
c = 90.6 inches
Hence, from the above,
We can conclude that the height of the heater is: 90.6 inches

Question 10.
Reasoning What is the measurement of the longest line segment in a right rectangular prism that is 16 centimeters long, 9 centimeters wide, and 7 centimeters tall? Round to the nearest tenth of a centimeter.
Answer:
It is given that
A right rectangular prism is 16 centimeters long, 9 centimeters wide, and 7 centimeters tall
So,
The dimensions of a right rectangular prism are: 16 cm × 9 cm × 7 cm
So,
The length of the right rectangular prism is: 16 cm
The width of the right rectangular prism is: 9 cm
The height of the right rectangular prism is: 7 cm
We know that,
The longest line segment in any 2-d or 3-d figure is a “Diagonal”
Now,
We know that,
The length of the diagonal of the right rectangular prism = \(\sqrt{Length^{2} + Width^{2} + Height^{2}}\)
So,
The length of the diagonal of the right rectangular prism = \(\sqrt{16^{2} + 9^{2} + 7^{2}}\)
= \(\sqrt{256 + 49 + 81}\)
= \(\sqrt{386}\)
= 19.64 cm
Hence, from the above,
We can conclude that the length of the longest line segment in the right rectangular prism is: 19.64 cm

Question 11.
Felipe is making triangles for a stained glass window. He made the design shown but wants to change it. Felipe wants to move the purple triangle to the corner. The purple piece has side lengths of 4.5 inches, 6 inches, and 7 inches. Can the purple piece be moved to the corner? Explain.

Answer:
It is given that
Felipe is making triangles for a stained glass window. He made the design shown as above but wants to change it. Felipe wants to move the purple triangle to the corner. The purple piece has side lengths of 4.5 inches, 6 inches, and 7 inches.
Now,
From the given figure,
We can observe that
If the purple figure is moving to the corner, then one side of the purple figure will become a right angle
So,
Now,
We have to find whether the purple figure will be a right triangle or not if it moves to a corner
Now,
We know that,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²,
then, the triangle is a right triangle
We know that,
The hypotenuse has the longest side length
So,
7² = 4.5² + 6²
49 = 20.25 + 36
49 = 56.25
So,
The condition
c² = a² + b²
is false
Hence, from the above,
We can conclude that the purple figure should not

Question 12.
a. What is the longest poster you could fit in the box? Express your answer to the nearest tenth of an inch.

Answer:
The given figure is:

From the given figure,
We can observe that there will be two longest sides for two pairs of different side lengths
we know that,
The longest side is nothing but a diagonal
Now,
The two pairs of side lengths are: (20, 12, x), and (8, 12, y)
Where,
x and y are the lengths of the longest sides
Now,
We know that
Since the figure consists of the right triangles,
According to the Pythagorean Theorem,
c² = a² + b²
So,
x² = 20² + 12²
x² = 400 + 144
x² = 544
x = \(\sqrt{544}\)
x = 23.32 in.
So,
y² = 12² + 8²
y² = 144 + 64
y² = 208
y = \(\sqrt{208}\)
y = 14.42 in.
Hence, from the above,
We can conclude that the longest poster that you can fit in the box is: 23.32 in.

b. Explain why you can fit only one maximum-length poster in the box, but you can fit multiple 21.5-inch posters in the same box.

Answer:
It is given that
you can fit multiple 21.5-inch posters in the same box.
Now,
From part (a),
We can observe that the length of the longest poster that can fit into the box is: 23.32 in.
So,
21.5 in. < 23.32 in.
Hence, from the above,
We can conclude that since the given size of the posters is less than the maximum length of the poster,
We can fit multiple 21.5-inch posters in the same box instead of 1 poster that is of the maximum length

Question 13.

The corner of a room where two walls meet the floor should be at a right angle. Jeff makes a mark along each wall. One mark is 3 inches from the corner. The other is 4 inches from the corner. How can Jeff use the Pythagorean Theorem to see if the walls form a right angle?
Answer:
It is given that
The corner of a room where two walls meet the floor should be at a right angle. Jeff makes a mark along each wall. One mark is 3 inches from the corner. The other is 4 inches from the corner.
Now,
To see whether the walls form a right angle or not,
We have to see whether the length along the walls is greater than the lengths of the marks from the corners
Now,
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the length along the walls
a is the length of one mark from the corner
b is the length of another mark from the corner
So,
c² = 3² + 4²
c² = 9 + 16
c² = 25
c = \(\sqrt{25}\)
c = 5
So,
From the above value,
We can observe that
c > a and c > b
Hence, from the above,
We can conclude that the walls form a right angle

Question 14.
Higher-Order Thinking It is recommended that a ramp have at least 6 feet of horizontal distance for every 1 foot of vertical rise along an incline. The ramp shown has a vertical rise of 2 feet. Does the ramp show match the recommended specifications? Explain.

Answer:
It is given that
It is recommended that a ramp have at least 6 feet of horizontal distance for every 1 foot of vertical rise along an incline. The ramp shown has a vertical rise of 2 feet
Now,
6 feet of horizontal distance for every 1 foot,
The ramp shown has a vertical rise of 5 feet.
So,
The rate of change (m)  ≤ \(\frac{1 foot}{6 feet}\)
Convert both to a single unit of inches
We know that,
1 foot = 12 inches
So,
m ≤ \(\frac{12 inches}{72 inches}\)
Divide by 4 into both sides
m ≤ \(\frac{4 inches}{18 inches}\)
Now,
For a ramp that has a vertical distance of 2 feet and the same horizontal distance
m = \(\frac{12 inches}{24 inches}\)
m = \(\frac{4 inches}{6 inches}\)
So,
The rate of change when the vertical distance is 6 feet > The rate of change when the vertical distance is 2 feet
Hence, from the above,
We can conclude that the ramp shown above matched the given specifications

Assessment Practice

Question 15.
A machine in a factory cuts out triangular sheets of metal. Which of the triangles are right triangles? Select all that apply.

☐ Triangle 1
☐ Triangle 2
☐ Triangle 3
☐ Triangle 4
Answer:
It is given that
A machine in a factory cuts out triangular sheets of metal.
Now,
According to the converse of the Pythagorean Theorem,
If
c² = a² + b²
then, the triangle is a right triangle
Now,
For Triangle 1,
(\(\sqrt{505}\))² = 12² + 19²
505 = 505
For Triangle 2,
(\(\sqrt{467}\))² = 16² + 19²
467 ≠ 617
For Triangle 3,
(\(\sqrt{596}\))² = 14² + 20²
596 = 596
For Triangle 4,
(\(\sqrt{421}\))² = 11² + 23²
421 ≠ 650
Hence, from the above,
We can conclude that Triangle 1 and Triangle 3 are the right triangles

Question 16.
What is the length b, in feet, of the rectangular plot of land shown?

Answer:
It is given that the given figure is a rectangular plot of land
We know that,
In a rectangle,
The opposite sides are equal and the angles are all 90°
So,
A diagonal forms 2 right triangles in a rectangle
Now,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the length of the diagonal
a is the width
b is the length
Now,
325² = 300² + b²
b²= 1,05,625 – 90,000
b² = 15,625
b = \(\sqrt{15,625}\)
b = 125 ft
Hence, from the above,
We can conclude that the length of b is: 125 ft

Lesson 7.4 Find Distance in the Coordinate

Explore It!
Thomas and Jim are outside the haunted castle ride and want to get to the clown tent in time for the next show.

I can… use the Pythagorean Theorem to find the distance between two points in the coordinate plane.

A. How can you represent the starred locations on a coordinate plane?

Answer:
It is given that
Thomas and Jim are outside the haunted castle ride and want to get to the clown tent in time for the next show.
Now,
The given route of the castle is:

Now,
From the given figure,
To represent the route in the coordinate plane,
The given scale is:
X-axis: 1 cm = 500 feet
Y-axis: 1 cm = 500 feet
Now,
From the given figure,
The coordinates of the starred locations can be:
The coordinates of the haunted house are: (500, 1,500),
The coordinates of the clown tent are: (2,000, 500)
Hence,
The representation of the starred locations in the coordinate plane is:

B. Jim says that the marked yellow paths show the shortest path to the tent. Write an expression to represent this and find the distance Jim walks from the haunted mansion to the clown tent.
Answer:
From part (a),
We know that,
The coordinates of the haunted house are: (500, 1,500),
The coordinates of the clown tent are: (2,000, 500)
Now,
We know that
The linear equation in the slope-intercept form is:
y = mx + b
Where,
m is the slope
b is the initial value (or) y-intercept
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
Now,
We know that,
Slope (m) = y₂ – y₁ / x₂ – x₁
So,
Slope (m) = \(\frac{500 – 1,500}{2,000 – 500}\)
Slope (m) = –\(\frac{1,000}{1,500}\)
Slope (m) = –\(\frac{2}{3}\)
So,
The equation in the slope-intercept form is:
y = –\(\frac{2}{3}\)x + b
So,
3y = -2x + 3b
Now,
Substitute (500, 1,500) or (2,000, 500) in the above equation
So,
1,500 = -4000 + 3b
3b = 5,500
So,
The equation that represents the shortest path to the tent is:
3y = -2x + 5,500
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the haunted mansion and the clown tent = \(\sqrt{(2,000 – 500)^{2} + (500 – 1,500)^{2}}\)
= \(\sqrt{1,500^{2} + 1,000^{2}}\)
= \(\sqrt{2,250,000 + 1,000,000}\)
= 1,802.77 feet
Hence, from the above,
We can conclude that
The equation that represents the shortest path to the clown tent is:
3y = -2x + 5,500
The distance between the haunted mansion and the clown tent is: 1,802.77 feet

Focus on math practices
Construct Arguments Why is the distance between two nonhorizontal and nonvertical points always greater than the horizontal or vertical distance?
Answer:
Let us consider a coordinate plane
Now,
When we draw either a horizontal line or the vertical line,
We can observe that the length will be constant
But,
When we draw non-vertical and non-horizontal lines,
We can observe that the lengths are unknown and not constant
Hence, from the above,
We can conclude that the distance between two nonhorizontal and nonvertical points always greater than the horizontal or vertical distance

Essential Question
How can you use the Pythagorean Theorem to find the distance between two points?
Answer:
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the hypotenuse
a and b are the side lengths or lengths of the legs
Now,
Graphically,
The terms of the Pythagorean Theorem can be expressed as:
c is the distance between two points
a and b are the points
So,
c = \(\sqrt{a^{2} + b^{2}}\)

Try It!

What is the distance between points A and B?
The distance between point A and point B is about ________ units.

Answer:
The given graph is:

The representation of the graph in the coordinate plane is:

From the given graph,
The coordinates of A are: (2, 3)
The coordinates of B are: (4, 1)
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points A and B = \(\sqrt{(4 – 2)^{2} + (1 – 3)^{2}}\)
= \(\sqrt{2^{2} + 2^{2}}\)
= \(\sqrt{4 + 4}\)
= 2.83 units
Hence, from the above,
We can conclude that the distance between points A and B is about 2.83 units

Convince Me!
Why do you need to use the Pythagorean Theorem to find the distance between points A and B?
Answer:
The representation of points A and B in the coordinate plane are:

Now,
When we observe the graph,
We can see that A and B can form a right triangle
Now,
We know that,
The Pythagorean Theorem is only applicable to the right triangles
So,
According to the Pythagorean Theorem,
c² = a² + b²
c = \(\sqrt{a^{2} + b^{2}}\)
where,
c is the distance between points A and B
A and B are the given points

Try It!
Find the perimeter of ∆ABC with vertices (2, 5), (5, -1), and (2, -1).
Answer:
It is given that
∆ABC with vertices (2, 5), (5, -1), and (2, -1)
Now,
The names of the vertices are:
A (2, 5), B (5, -1), and C (2, -1)
We know that,
The perimeter of a triangle is the sum of all the side lengths of a triangle
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
Now,
We know that,
In ∆ABC,
AB and BC are the side lengths
Ac is the hypotenuse
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points A and B = \(\sqrt{(5 – 2)^{2} + (-1 – 5)^{2}}\)
= \(\sqrt{3^{2} + 6^{2}}\)
= \(\sqrt{9 + 36}\)
= 6.70 units
The distance between the points B and C (BC) = \(\sqrt{(2 – 5)^{2} + (-1 + 1)^{2}}\)
= \(\sqrt{3^{2} + 0^{2}}\)
= \(\sqrt{9 + 0}\)
= 3 units
The distance between the points A and C (AC) = \(\sqrt{(2 – 2)^{2} + (-1 – 5)^{2}}\)
= \(\sqrt{0^{2} + 6^{2}}\)
= \(\sqrt{0 + 36}\)
= 6 units
So,
The perimeter of ∆ABC = AB + BC + AC
= 6 + 3 + 6.70
= 15.7 units
Hence, from the above,
We can conclude that the perimeter of ∆ABC is about 15.7 units

Try It!

What are the coordinates, to the nearest tenth, of the third vertex in an isosceles triangle that has one side length of 2 and two side lengths of 5, with vertices at (1, 0) and (1, 2)? The third vertex is in the first quadrant.
Answer:
It is given that
An isosceles triangle that has one side length of 2 and two side lengths of 5, with vertices at (1, 0) and (1, 2)
Now,
Let the third vertex be (x, y)
Now,
The given vertices are:
A (1, 0), B (1, 2), and C (x, y)
It is also given that
BC = 2 units, and AC = 5 units
We know that,
An isosceles triangle has any 2 equal side lengths
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points A and B = \(\sqrt{(2 – 0)^{2} + (1 – 1)^{2}}\)
= \(\sqrt{2^{2} + 0^{2}}\)
= \(\sqrt{4 + 0}\)
= 4 units
The distance between the points B and C = \(\sqrt{(x – 1)^{2} + (y – 2)^{2}}\)
Squaring on both sides
So,
BC² = (x – 1)² + (y – 2)²
The distance between the points A and C = \(\sqrt{(x – 1)^{2} + (y – 0)^{2}}\)
Squaring on both sides
So,
BC² = (x – 1)² + y²
So,
(x – 1)² + (y – 2)² = 4 —– (1)
(x – 1)² + y² = 25 —— (2)
So,
From eq (1) and eq (2),
25 – y² + (y – 2)² = 4
-y² + y² – 4y + 4 = -21
-4y = -25
y = \(\frac{25}{4}\)
So,
(x – 1)² = |25 – (\(\frac{25}{4}\))²|
x = \(\frac{19}{4}\)
Hence, from the above,
We can conclude that the third vertex is: (\(\frac{19}{4}\), \(\frac{25}{4}\))

KEY CONCEPT

You can use the Pythagorean Theorem to find the distance between any two points, P and Q, on the coordinate plane.

Do You Understand?
Question 1.
Essential Question How can you use the Pythagorean Theorem to find the distance between two points?
Answer:
We know that,
According to the Pythagorean Theorem,
c² = a² + b²
Where,
c is the hypotenuse
a and b are the side lengths or lengths of the legs
Now,
Graphically,
The terms of the Pythagorean Theorem can be expressed as:
c is the distance between two points
a and b are the points
So,
c = \(\sqrt{a^{2} + b^{2}}\)

Question 2.
Model with Math
Can you use a right triangle to represent the distance between any two points on the coordinate plane? Explain.
Answer:
Derived from the Pythagorean Theorem, the distance formula is used to find the distance between two points in the plane. We know that,
According to the Pythagorean Theorem,
a²+b²=c²
where,
a and b are the lengths of the legs adjacent to the right angle
c is the length of the hypotenuse.

Question 3.
Generalize How does the fact that the points are on opposite sides of the y-axis affect the process of finding the distance between the two points?
Answer:
The fact that the points are on opposite sides of the y-axis affects the process of finding the distance between the two points because  We need to find the distance between the two points by adding the distances from each of them to the y-axis.

Do You Know How?
In 4-6, use the coordinate plane below.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The given points are:
C (1, 2), D (2, -1), and E (-2, 1)

Question 4.
Find the distance between points C and D. Round to the nearest hundredth.
Answer:
Compare the points C and D with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points C and D = \(\sqrt{(2 – 1)^{2} + (-1 – 2)^{2}}\)
= \(\sqrt{1^{2} + 3^{2}}\)
= \(\sqrt{1 + 9}\)
= 3.16 units
Hence, from the above,
We can conclude that the distance between points C and D is: 3.16 units

Question 5.
Find the perimeter of ∆CDE.
Answer:
We know that,
The “Perimeter” is defined as the sum of all the side lengths
So,
The perimeter of ∆CDE = CD + DE + CE
So,
The distance between the points C and D = \(\sqrt{(2 – 1)^{2} + (-1 – 2)^{2}}\)
= \(\sqrt{1^{2} + 3^{2}}\)
= \(\sqrt{1 + 9}\)
= 3.16 units
The distance between the points D and E = \(\sqrt{(-2 – 2)^{2} + (1 + 1)^{2}}\)
= \(\sqrt{4^{2} + 2^{2}}\)
= \(\sqrt{16 + 4}\)
= 4.47 units
The distance between the points C and E = \(\sqrt{(-2 – 1)^{2} + (1 – 2)^{2}}\)
= \(\sqrt{3^{2} + 1^{2}}\)
= \(\sqrt{1 + 9}\)
= 3.16 units
So,
The perimeter of ∆CDE = 3.16 + 4.47 + 3.16
= 10.79 units
Hence, from the above,
We can conclude that the perimeter of ∆CDE is: 10.79 units

Question 6.
Point B is plotted on the coordinate plane above the x-axis. ∆BDE is equilateral. What are the coordinates of point B to the nearest hundredth?
Answer:
It is given that
Point B is plotted on the coordinate plane above the x-axis. ∆BDE is equilateral.
Now,
Let the unknown vertex be B (x, y)
So,
The given points are:
B (x, y), D (2, -1), and E (-2, 1)
It is given that ΔBDE is equilateral
So,
BD = DE = EB
BD² = DE² = EB²
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points B and D = \(\sqrt{(x – 2)^{2} + (y – 1)^{2}}\)
Squaring on both sides
So,
BD² = (x – 2)² + (y + 1)²
The distance between the points D and E = \(\sqrt{(1 + 1)^{2} + (-2 – 2)^{2}}\)
= \(\sqrt{2^{2} + 4^{2}}\)
= \(\sqrt{4 + 16}\)
= 4.47 units
The distance between the points E and B = \(\sqrt{(x + 2)^{2} + (y – 1)^{2}}\)
Squaring on both sides
So,
EB² = (x + 2)² + (y – 1)²
Now,
(x – 2)² + (y + 1)² = 4.47 —- (1)
(x + 2)² + (y – 1)² = (x – 2)² + (y + 1)²
x² + 2x + 4 + y² + 1 – 2y = x² + 4 – 4x + y² + 1 + 2y
2x + 4 + 1 – 2y = 4 – 4x + 1 + 2y
6x + 5 = 4y + 5
6x = 4y
3x = 2y
x = \(\frac{2}{3}\)y
Now,
From eq (1),
x² + 4 – 4x + y² + 1 + 2y = 4.47
x² + y² -4x + 2y = -0.47
(\(\frac{2}{3}\)y)² + y² – 4 (\(\frac{2}{3}\))y + 2y = -0.47
4y² + 15y + 4.23 = 0
So,
y = -0.30 (or) y = -3.44
So,
x = \(\frac{2}{3}\) (-0.30) (or) x = \(\frac{2}{3}\) (-3.44)
x = -0.2 (or) x = -2.29
Hence, from the above,
We can conclude that the coordinates of point B are: (-0.2, -0.30) or (-2.29, -3.44)

Practice & Problem Solving

Question 7.
Leveled Practice Use the Pythagorean Theorem to find the distance between points P and Q.
Label the length, in units, of each leg of the right triangle.

The distance between point P and point Q is __________ units.
Answer:
From the given coordinate plane,
There are only 2 vertices
Let the third vertex be R(x, y) and the coordinates of R can be found from the coordinate plane
Now,
From the given coordinate plane,
The vertices are:
P (3, 2), Q (9, 10), and R (9, 2)
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points Q and R = \(\sqrt{(9 – 9)^{2} + (10 – 2)^{2}}\)
= \(\sqrt{0^{2} + 8^{2}}\)
= \(\sqrt{0 + 64}\)
= 8 units
The distance between the points P and R = \(\sqrt{(9 – 3)^{2} + (2 – 2)^{2}}\)
= \(\sqrt{0^{2} + 6^{2}}\)
= \(\sqrt{0 + 36}\)
= 6 units
Now,
From the given coordinate plane,
We can observe that P, Q, R form the right triangle
So,
According to the Pythagorean Theorem,
PQ² = QR² + PR²
PQ² = 8² + 6²
PQ² = 64 + 36
PQ² = 100
PQ = \(\sqrt{100}\)
PQ = 10 units
Hence, from the above,
We can conclude that the length of PQ is: 10 units

Question 8.
Find the perimeter of triangle QPR. Round to the nearest hundredth.
Answer:
From the coordinate plane,
The vertices of ΔPQR are:
P (-5, -2), Q (2, -2), and R (-1, 3)
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points P and Q = \(\sqrt{(2 + 5)^{2} + (2 – 2)^{2}}\)
= \(\sqrt{0^{2} + 7^{2}}\)
= \(\sqrt{0 + 49}\)
= 7 units
The distance between the points Q and R = \(\sqrt{(-1 – 2)^{2} + (3 + 2)^{2}}\)
= \(\sqrt{3^{2} + 5^{2}}\)
= \(\sqrt{9 + 25}\)
= 5.83 units
The distance between the points P and R = \(\sqrt{(1 – 5)^{2} + (-3 – 2)^{2}}\)
= \(\sqrt{4^{2} + 5^{2}}\)
= \(\sqrt{16 + 25}\)
= 6.40 units
So,
The perimeter of ΔPQR = PQ + QR + PR
= 7 + 5.83 + 6.40
= 19.23 units
Hence, from the above,
We can conclude that the perimeter of ΔPQR is: 19.23 units

Question 9.
Determine whether the triangle is equilateral, isosceles, or scalene.
Answer:
We know that,
On the basis of the side lengths,
Scalene Triangle – All the side lengths are different
Equilateral Triangle – All the side lengths are the same
Isosceles Triangle – Any two of the side lengths are the same
So,
From Exercise 8,
We can observe that all the side lengths are different
Hence, from the above,
We can conclude that ΔPQR is a scalene Triangle

Question 10.
You walk along the outside of a park starting at point P. Then you take a shortcut represented by \(\overline{P Q}\) on the graph.

a. What is the length of the shortcut in meters? Round to the nearest tenth of a meter.
Answer:
It is given that
You walk along the outside of a park starting at point P. Then you take a shortcut represented by \(\overline{P Q}\) on the graph.
Now,
From the given figure,
We can observe that
The vertices are:
P (0, 0), Q (40, 85), and R (40, 0)
The shortest path is represented as PQ
Now,
Compare the points P and Q with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points P and Q = \(\sqrt{(40 – 0)^{2} + (85 – 0)^{2}}\)
= \(\sqrt{40^{2} + 85^{2}}\)
= \(\sqrt{1,600 + 7,225}\)
= 93.9 m
Hence, from the above,
We can conclude that the length of the shortest path is: 93.9 meters

b. What is the total length of your walk in the park? Round to the nearest tenth of a meter.
Answer:
We know that,
The total length is nothing but the “Perimeter”
So,
The perimeter of the given triangle = PQ + QR + PR
Now,
The distance between the points P and Q = \(\sqrt{(40 – 0)^{2} + (85 – 0)^{2}}\)
= \(\sqrt{40^{2} + 85^{2}}\)
= \(\sqrt{1,600 + 7,225}\)
= 93.9 m
The distance between the points Q and R = \(\sqrt{(40 – 40)^{2} + (85 – 0)^{2}}\)
= \(\sqrt{0^{2} + 85^{2}}\)
= \(\sqrt{0 + 7,225}\)
= 85 m
The distance between the points P and R = \(\sqrt{(40 – 0)^{2} + (0 – 0)^{2}}\)
= \(\sqrt{0^{2} + 40^{2}}\)
= \(\sqrt{0 + 1,600}\)
= 40 m
So,
The total length of your walk in the park = PQ + QR + PR
= 93.9 + 85 + 40
= 218.9 meters
Hence, from the above,
We can conclude that the total length of your walk in the park is: 218.9 meters

Question 11.
Suppose a park is located 3.6 miles east of your home. The library is 4.8 miles north of the park. What is the shortest distance between your home and the library?
Answer:
It is given that
A park is located 3.6 miles east of your home. The library is 4.8 miles north of the park
So,
The representation of the above situation is:

Now,
According to the Pythagorean Theorem,
(The shortest distance between home and library)² = (The distance from home to park)² + (he distance from park to library)²
(The shortest distance between home and library)² = 3.6² + 4.8²
(The shortest distance between home and library)² = 12.96 + 23.04
The shortest distance between home and library = 6
Hence, from the above,
We can conclude that the shortest distance between home and library is: 6 miles

Question 12.
Use Structure Point B has coordinates (2, 1). The x-coordinate of point A is -10. The distance between point A and point B is 15 units. What are the possible coordinates of point A?
Answer:
It is given that
Point B has coordinates (2, 1). The x-coordinate of point A is -10. The distance between point A and point B is 15 units
Now,
Let the coordinates of A be:
(x, y) = (-10, y)
Now,
Compare the points A and B with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points A and B = \(\sqrt{(-10 – 2)^{2} + (y – 1)^{2}}\)
15 = \(\sqrt{12^{2} + (y – 1)^{2}}\)
15 = \(\sqrt{144 + (y – 1)^{2}}\)
Now,
Squaring on both sides
So,
144 + (y – 1)² = 225
(y  1)² = 225 – 144
(y – 1)² = 81
y – 1 = \(\sqrt{81}\)
y – 1 = 9 (or) y – 1 = -9
y = 9 + 1 (or) y = -9 + 1
y = 10 (or) y = -8
Hence, from the above,
We can conclude that the possible coordinates of A are: (-10, 10), and (-10, -8)

Question 13.
Higher-Order Thinking ∆EFG and ∆HIJ have the same perimeter and side lengths. The coordinates are E(6, 2), F(9, 2), G(8, 7), H(0, 0), and I(0, 3). What are the possible coordinates of point J?
Answer:
It is given that
∆EFG and ∆HIJ have the same perimeter and side lengths. The coordinates are E(6, 2), F(9, 2), G(8, 7), H(0, 0), and I(0, 3)
So,
According to the side lengths,
EF = HI, FG = IJ, and GE = JH
So,
EF² = HI², FG²= JI², and GE² = JH²
Now,
Let the unknown vertex be J (x, y)
Now,
Compare the given points with (x₁, y₁), (x₂, y₂)
Now,
We know that,
The distance between two points = √(x₂–  x₁) + (y₂ – y₁)²
So,
The distance between the points E and F = \(\sqrt{(9 – 6)^{2} + (2 – 2)^{2}}\)
= \(\sqrt{3^{2} + 0^{2}}\)
= \(\sqrt{9 + 0}\)
= 3 units
The distance between the points F and G = \(\sqrt{(8 – 9)^{2} + (7 – 2)^{2}}\)
= \(\sqrt{1^{2} + 5^{2}}\)
= \(\sqrt{1 + 25}\)
= 5.09 units
The distance between the points G and E = \(\sqrt{(8 – 6)^{2} + (7 – 2)^{2}}\)
= \(\sqrt{2^{2} + 5^{2}}\)
= \(\sqrt{4 + 25}\)
= 5.38 units
Now,
The distance between the points H and I = \(\sqrt{(0 – 0)^{2} + (3 – 0)^{2}}\)
= \(\sqrt{3^{2} + 0^{2}}\)
= \(\sqrt{9 + 0}\)
= 3 units
The distance between the points I and J = \(\sqrt{(x – 0)^{2} + (y – 3)^{2}}\)
5.09 = \(\sqrt{x^{2} + (y – 3)^{2}}\)
Squaring on both sides
So,
x² + (y – 3)² = 25.90 units
The distance between the points J and H = \(\sqrt{(x – 0)^{2} + (y – 0)^{2}}\)
5.38 = \(\sqrt{x^{2} + y^{2}}\)
Squaring on both sides
So,
x² + y² = 28.94 units
Now,
28.94 – y² + (y – 3)² = 25.90
y² + 9 – 6y – y² = 25.90 – 28.94
9 – 6y = -3.04
-6y = -3.04 – 9
6y = 12.04
y = 2
Now,
Substitute the value of y in eq 2
x² + 4 = 28.94
x² = 24.94
x = 4.99 (or) x = -4.99
Hence, from the baove,
We can conclude that the missing vertex is: J (4.99, 2) or J (-4.99, 2)