Go Math Grade 8 Answer Key PDF | Chapterwise Grade 8 HMH Go Math Solution Key

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HMH Go Math 8th Grade Answer Key

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Grade 8 HMH Go Math – Answer Keys

Grade 8 McGraw Hill Glencoe – Answer Keys

  • Chapter 1: Real Numbers
  • Chapter 2: Equations in One Variable
  • Chapter 3: Equations in Two Variables
  • Chapter 4: Functions
  • Chapter 5: Triangles and the Pythagorean Theorem
  • Chapter 6: Transformations
  • Chapter 7: Congruence and Similarity
  • Chapter 8: Volume and Surface Area
  • Chapter 9: Scatter Plots and Data Analysis

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Solutions to Go Math | Middle School Grade 7 Answer Key PDF Download for All Chapters

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Go Math Answer Key for Grade 7: Mastering in Maths subject is very important than other subjects. It is more demanding in real-time situations. Every kid should practice and solve all chapters grade 7 questions covered in Go Math Books. Go Math Middle School Grade 7 Answer Key is the perfect solution for getting basic & fundamental maths concepts knowledge. Find out the correct & updated Go Math Grade 7 Answer Key and understand the topics covered in Go Math Textbooks.

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Go Math Grade 7 Answer Key Free Pdf Download

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Grade 7 HMH Go Math – Answer Keys

Grade 7 McGraw Hill Glencoe – Answer Keys

  • Chapter 1: Ratios and Proportional Reasoning
  • Chapter 2: Percents
  • Chapter 3: Integers
  • Chapter 4: Rational Number
  • Chapter 5: Expressions
  • Chapter 6: Equations and Inequalities
  • Chapter 7: Geometric Figures
  • Chapter 8: Measure Figures
  • Chapter 9: Probability
  • Chapter 10: Statistics
  • Chapter 11: Statistical Measures
  • Chapter 12: Statistical Display

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Go Math Grade 2 Answer Key | Download Primary School Second Grade HMH Go Math Answers

Go Math Grade 2 Answer Key

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HMH Go Math Second Grade Answers provided are given by subject experts and meets the Common Core State Standards 2019 Curriculum. Primary School Go Math 2nd Grade Answer Key includes Questions from Exercises, Chapter Tests, Assessments, Review Tests, Cumulative Practice, etc. Make the most out of the quick resources and grasp the fundamentals of maths that might be useful for you in the mere future.

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Learn the concepts of 2nd Grade Math in a fun and engaging way. HMH Go Math Grade 2 Answer Key is curated by subject experts keeping in mind students’ level of understanding. Keep your kid on the right track and help them become proficient in maths. Access the Chapterwise Go Math Second Grade Answer Key in PDF Format via quick links available. Just click on the respective chapter you feel like preparing and improve on the areas you feel difficult.

  • Vocabulary Reader: Whales
  • Chapter 1
  • Chapter 2
  • Vocabulary Reader: All About Animals
  • Chapter 3
  • Chapter 4
  • Chapter 5
  • Chapter 6
  • Vocabulary Reader: Making a Kite
  • Chapter 7
  • Chapter 8
  • Chapter 9
  • Chapter 10
  • Vocabulary Reader: A Farmer’s Job
  • Chapter 11
  • Picture Glossary

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Go Math Primary School Grade 4 Answer Key PDF for All Chapters | Download Go Math Solutions Key for Grade 4

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Go Math Answer Key for Grade 4: Clearing all math exams can be tough for students who are pursuing 4th grade but with Go Math Grade 4 Answer Key it can be easy. Because this solutions key is prepared by our highly-experienced subject experts after ample research and easy to understand the concepts too. So, students are implied to answer all the questions while practicing and cross-check the solutions through HMH Go Math 4th Grade Answer Key.

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Common Core Grade 4 HMH Go Math – Answer Keys

Grade 4 Homework Practice FL.

Common Core – Grade 4 – Practice Book

Grade 4 Homework FL. – Answer Keys

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Go Math Grade 1 Answer Key | HMH Go Math First Grade Solutions PDF Free Download

Go Math Grade 1 Answer Key

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  • Vocabulary Reader: Animals in Our World
  • Chapter 1
  • Chapter 2
  • Chapter 3
  • Chapter 4
  • Chapter 5
  • Vocabulary Reader: Around the Neighborhood
  • Chapter 6
  • Chapter 7
  • Chapter 8
  • Vocabulary Reader: All Kinds of Weather
  • Chapter 9
  • Chapter 10
  • Vocabulary Reader: On the Move
  • Chapter 11
  • Chapter 12
  • Picture Glossary

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Students who are in need of Go Math Grade 1 Homework or Assignments help can access the Grade 1 Go Math Answer Key to resolve their queries. The major benefits of referring to HMH Go Math 1st Grade Answers are as follows

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Solutions Key of Go Math Grade 5 Textbook | Download HMH Go Math 5th Grade Answer Key Pdf

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Go Math Answer Key for Grade 5: Parents who are concentrating on their kid’s studies must try this Go math 5th Grade Answer Key. It is the most important preparation material for finishing the homework and efficient preparation. You have to practice with this Go math answer key for grade 5 and clear all your queries and score high marks in the exam.

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All Chapters Go Math Grade 5 Answer Key Pdf Free Download

Compiled list of Go Math 5th Standard chapterwise Answer Key pdf improves your skills and performance in the standard tests. A clear-cut explanation of concepts makes students to grasp the topics clearly and score well in the exams. It also develops your interest in mathematical concepts and helps students to stand out from the crowd. However, Go Math Primary School Grade 5 Answer Key pdf can be helpful during your assessments and you can attempt the test with utmost confidence.

Grade 5 HMH Go Math – NEW

Grade 5 Math Common Core Tests

  • Test 1 Session 1 (page 2)
  • Test 2 Session 1 (page 3)
  • Test 3 Session 1 (page 4)
  • Test 3 Session 2 (page 5)
  • Test 4 Session 1 (page 6)
  • Test 4 Session 2 (page 7)

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Download Go Math Answer Key for Grades K-8 | HMH Go Math Solution Key for Grades Kindergarten, 1, 2, 3, 4, 5, 6, 7, 8

Go Math Answer Key: HMH Go Math Solutions Key provided helps students to have learning targets and achieve success at chapter and lesson level and makes learning visible. Give your kid the right amount of knowledge he needs as a part of your preparation by taking the help of our HMH Go Math Answer Key for Grades K-8. Resolve all your queries and assess your preparation standard using the Common Core Go Math Solution Key.

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Go Math Grade K Answer Key | Download HMH Go Math Kindergarten Solution Key

Go Math Grade K Answer Key

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  • Vocabulary Reader: Fall Festival!
  • Chapter 1
  • Chapter 2
  • Chapter 3
  • Chapter 4
  • Chapter 5
  • Chapter 6
  • Chapter 7
  • Chapter 8
  • Vocabulary Reader: School Fun
  • Chapter 9
  • Chapter 10
  • Vocabulary Reader: Plants All Around
  • Chapter 11
  • Chapter 12
  • Picture Glossary

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Primary School Go Math Grade 3 Answer Key | HMH Go Math 3rd Grade Solution Key

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3rd Grade Go Math Answer Key All Chapters

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Grade 3 HMH Go Math – Answer Keys

Grade 3 HMH Go Math – Extra Practice Questions and Answers

Grade 3 HMH Go Math – Answer Keys

  • Chapter 1: Addition and Subtraction within 1,000 Assessment Test
  • Chapter 2: Represent and Interpret Data Assessment Test
  • Chapter 3: Understand Multiplication Assessment Test
  • Chapter 4: Multiplication Facts and Strategies Assessment Test
  • Chapter 5: Use Multiplication Facts Assessment Test
  • Chapter 6: Understand Division Assessment Test
  • Chapter 7: Division Facts and Strategies Assessment Test
  • Chapter 8: Understand Fractions Assessment Test
  • Chapter 9: Compare Fractions Assessment Test
  • Chapter 10: Time, Length, Liquid Volume, and Mass Assessment Test
  • Chapter 11: Perimeter and Area Assessment Test
  • Chapter 12:Two-Dimensional Shapes Assessment Test

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Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem

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Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem

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Chapter 12- Lesson 1: 

Chapter 12- Lesson 2: 

Chapter 12- Lesson 3: 

Guided Practice – The Pythagorean Theorem – Page No. 378

Question 1.
Find the length of the missing side of the triangle
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Lesson 1: The Pythagorean Theorem img 1
a2 + b2 = c2 → 242 + ? = c2 → ? = c2
The length of the hypotenuse is _____ feet.
_____ feet

Answer: The length of the hypotenuse is 26 feet.

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

Question 2.
Mr. Woo wants to ship a fishing rod that is 42 inches long to his son. He has a box with the dimensions shown.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Lesson 1: The Pythagorean Theorem img 2
a. Find the square of the length of the diagonal across the bottom of the box.
________ inches

Answer: 1700 inches.

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

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

Answer: 42.42 inches.

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

ESSENTIAL QUESTION CHECK-IN

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

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

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

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

Question 4.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Lesson 1: The Pythagorean Theorem img 3
________ cm

Answer: 8.9 cm.

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

Question 5.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Lesson 1: The Pythagorean Theorem img 4
________ in.

Answer: 11.5 in.

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

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

Answer: 75.4 cm

Explanation: Let’s consider the diagonal of the TV screen as C = 152cm, length as A = 132 cm, and height of the screen as B.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem
As C2 = A2 + B2
   1522 = 1322 + B2
23,104 = 17,424 + B2
B2 = 23,104 – 17,424
B = √5680 => 75.365
So the height of the screen B = 75.4cm

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

Answer: 14.1in.

Explanation:
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem

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

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

Answer: 22.6 ft.

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

Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem

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

Question 9.
What is the longest flagpole (in whole feet) that could be shipped in a box that measures 2 ft by 2 ft by 12 ft?
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Lesson 1: The Pythagorean Theorem img 5
________ ft

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

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

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

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

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

Explanation: From the above details we will get a diagram as shown below.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem

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

Question 11.
Justify Reasoning A tree struck by lightning broke at a point 12 ft above the ground as shown. What was the height of the tree to the nearest tenth of a foot? Explain your reasoning.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Lesson 1: The Pythagorean Theorem img 6
________ ft

Answer: The total height of the tree was 52.8ft

Explanation:

Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem
By using the Pythagorean Theorem
a2 + b= c2
122 + 392 = c2
144 + 1521 = c2
1665 = c2
We are told to round the length of the hypotenuse to the nearest tenth of a foot, therefore: c = 40.8ft.
Therefore, the total height of the tree was:
height = a+c
height = 12 +40.8
height = 52.8 feet

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

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

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

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

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

Answer: c = x√ 2

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

Question 14.
Persevere in Problem Solving A square hamburger is centered on a circular bun. Both the bun and the burger have an area of 16 square inches.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Lesson 1: The Pythagorean Theorem img 7
a. How far, to the nearest hundredth of an inch, does each corner of the burger stick out from the bun? Explain.
________ in

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

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

 

Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem

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

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

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

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

Explanation:

Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem

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

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

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

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

Question 1.
Lashandra used grid paper to construct the triangle shown.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Lesson 2: Converse of the Pythagorean Theorem img 8
a. What are the lengths of the sides of Lashandra’s triangle?
_______ units, _______ units, _______ units,

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

Question 1.
b. Use the converse of the Pythagorean Theorem to determine whether the triangle is a right triangle.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Lesson 2: Converse of the Pythagorean Theorem img 9
The triangle that Lashandra constructed is / is not a right triangle.
_______ a right triangle

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

Explanation:
Verifying with Pythagorean formula a2 + b= c2
82 + 62 = 102
64 + 36 =100
100 = 100.

Question 2.
A triangle has side lengths 9 cm, 12 cm, and 16 cm. Tell whether the triangle is a right triangle.
Let a = _____, b = _____, and c = ______.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Lesson 2: Converse of the Pythagorean Theorem img 10
By the converse of the Pythagorean Theorem, the triangle is / is not a right triangle.
_______ a right triangle

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

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

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

Answer: It is a right-angled triangle.

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

ESSENTIAL QUESTION CHECK-IN

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 18.
History In ancient Egypt, surveyors made right angles by stretching a rope with evenly spaced knots as shown. Explain why the rope forms a right angle.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Lesson 2: Converse of the Pythagorean Theorem img 11

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

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

Converse of the Pythagorean Theorem – Page No. 386

Question 19.
Justify Reasoning Yoshi has two identical triangular boards as shown. Can he use these two boards to form a rectangle? Explain.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Lesson 2: Converse of the Pythagorean Theorem img 12

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

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

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

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

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

FOCUS ON HIGHER ORDER THINKING

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

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

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

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

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

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

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

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

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

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

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

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

Question 1.
Approximate the length of the hypotenuse of the right triangle to the nearest tenth using a calculator.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Lesson 3: Distance Between Two Points img 13
_______ units

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

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

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

Answer: Distance between points on the coordinate plane is 13

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

Question 3.
A plane leaves an airport and flies due north. Two minutes later, a second plane leaves the same airport flying due east. The flight plan shows the coordinates of the two planes 10 minutes later. The distances in the graph are measured in miles. Use the Pythagorean Theorem to find the distance shown between the two planes.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Lesson 3: Distance Between Two Points img 14
_______ miles

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

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

ESSENTIAL QUESTION CHECK-IN

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

Answer:

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

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

Using the Pythagorean Theorem:

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

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

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

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

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

Question 5.
A metal worker traced a triangular piece of sheet metal on a coordinate plane, as shown. The units represent inches. What is the length of the longest side of the metal triangle? Approximate the length to the nearest tenth of an inch using a calculator. Check that your answer is reasonable.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Lesson 3: Distance Between Two Points img 15
_______ in.

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

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

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

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

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

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

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

Question 7.
The coordinates of the vertices of a rectangle are given by R(- 3, – 4), E(- 3, 4), C (4, 4), and T (4, – 4). Plot these points on the coordinate plane at the right and connect them to draw the rectangle. Then connect points E and T to form diagonal \(\overline { ET } \).
a. Use the Pythagorean Theorem to find the exact length of \(\overline { ET } \).
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Lesson 3: Distance Between Two Points img 16

Answer: The diagonal ET is about 10.63 units long.

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

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

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

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

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

Answer: Ships P and Q are farthest apart

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

Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem
d= √( x2 – x1)2 + √( y2 – y1)2
The distance d1 between the two points P(-2,5) and Q(-7,-5) is:
d1 = √( xQ – xP)2 + √( yQ – yP)2
= √(-7 – (-2))2 + √(- 5 – 5)2
= √(-5)2 + √(-10)2
= √(25+100) => √125 = 11.18

The distance d2 between the two points Q(-7,-5) and R(2,-3) is:
d3 = √( xR – xQ)2 + √( yR – yQ)2
  = √(2 – (-7))2 + √(- 3 – 5)2
= √(9)2 + √(2)2
= √(81+4) => √85 = 9.22

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

Distance Between Two Points – Page No. 392

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

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

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

Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem

Question 10.
Justify Reasoning The graph shows the location of a motion detector that has a maximum range of 34 feet. A peacock at point P displays its tail feathers. Will the motion detector sense this motion? Explain.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Lesson 3: Distance Between Two Points img 17

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

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

FOCUS ON HIGHER ORDER THINKING

Question 11.
Persevere in Problem Solving One leg of an isosceles right triangle has endpoints (1, 1) and (6, 1). The other leg passes through the point (6, 2). Draw the triangle on the coordinate plane. Then show how you can use the Distance Formula to find the length of the hypotenuse. Round your answer to the nearest tenth.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Lesson 3: Distance Between Two Points img 18

Answer: 7.1 units.

Explanation:

Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem

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

Question 12.
Represent Real-World Problems The figure shows a representation of a football field. The units represent yards. A sports analyst marks the locations of the football from where it was thrown (point A) and where it was caught (point B). Explain how you can use the Pythagorean Theorem to find the distance the ball was thrown. Then find the distance.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Lesson 3: Distance Between Two Points img 19
_______ yards

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

Explanation:
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem
To find the distance between points A and B, we draw segment AB and label its length d. Then we draw vertical segment AC and Horizontal segment CB. We label the lengths of these segments a and b. triangle ACB is a right triangle with hypotenuse AB.
Since AC is vertical segment, its length, a, is the difference between its y-coordinates. Therefore, a = 26 – 14 = 12 units.
Since CB is horizontal segment, its length b is the difference between its x-coordinates. Therefore, b = 75 – 40 = 35units.
We use the Pythagorean Theorem to find d, the length of segment AB.
d2 = a2 + b2
d2 = 122 + 352
d2 = 144 + 1225
d2 = 1369 => d = √1369 => 37
The distance between point A and B is 37 yards

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

12.1 The Pythagorean Theorem

Find the length of the missing side.

Question 1.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Model Quiz img 20
________ meters

Answer: Length of missing side is 28m

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

Question 2.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Model Quiz img 21
________ ft

Answer: Length of missing side is 34ft

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

12.2 Converse of the Pythagorean Theorem

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

Question 3.
11, 60, 61
____________

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

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

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

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

Question 5.
15, 35, 38
____________

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

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

Question 6.
28, 45, 53
____________

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

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

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

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

12.3 Distance Between Two Points

Find the distance between the given points. Round to the nearest tenth.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Model Quiz img 22

Question 8.
A and B
________ units

Answer: Distance between A and B is 6.7 units

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

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

Question 9.
B and C
________ units

Answer: Distance between B and C is 7.07 units

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

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

Question 10.
A and C
________ units

Answer: Distance between A and C is 6.403 units

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

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

ESSENTIAL QUESTION

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

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

Selected Response – Mixed Review – Page No. 394

Question 1.
What is the missing length of the side?
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Mixed Review img 23
A. 9 ft
B. 30 ft
C. 39 ft
D. 120 ft

Answer: C

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

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

Answer: D

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

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

Answer: C

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

Question 4.
To the nearest tenth, what is the distance between point F and point G?
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Mixed Review img 24
Options:
A. 4.5 units
B. 5.0 units
C. 7.3 units
D. 20 units

Answer: A.

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

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

Answer: D

Explanation:
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem
By Pythagorean theorem
a2+b2=c 2
532+282= C2
2,809+784= C2
C2 = 9,409-5,184
C2 = 3,593
C= √3,593
C= 59.94 feet
=60 feet.

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

Answer: D.

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

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

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

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

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

Answer: A

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

Mini-Task

Question 8.
A fallen tree is shown on the coordinate grid below. Each unit represents 1 meter.
Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem Mixed Review img 25
a. What is the distance from A to B?
_______ meters

Answer: 13.34  m.

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

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

Answer: 16.3 m.

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

Final Words

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

Go Math Grade 8 Answer Key Chapter 1 Real Numbers

go-math-grade-8-chapter-1-real-numbers-answer-key

The Solutions for Go Math Grade 8 Answer Key Chapter 1 Real Numbers are given in detail here. Get the step by step explanations for all the question in Go Math Grade 8 Chapter 1 Real Numbers Answer Key and start your practice today. You can find the best ways to learn maths by using Go Math Grade 8 Answer Key. So, Download Go Math Grade 8 Chapter 1 Real Numbers Solution Key and make use of the given resources.

Go Math Grade 8 Chapter 1 Real Numbers Answer Key

It is essential for the students to choose the best material to practice the questions. Because by practicing only you can score the highest marks in the exams. Download HMH Go Math Grade 8 Answer Key Chapter 1 Real Numbers PDF for free. Quick learning is possible with our Go Math Grade 8 Chapter 1 Real Numbers Answer Key. Find a better way to make your learning simple by clicking on the below-provided links.

Lesson 1: Rational and Irrational Numbers

Lesson 2: Sets of real Numbers

Lesson 3: Ordering Real Numbers

Model Quiz

Mixed Review

Guided Practice – Rational and Irrational Numbers – Page No. 12

Write each fraction or mixed number as a decimal.

Question 1.
\(\frac{2}{5}\) =

Answer:
0.4

Explanation:
\(\frac{2}{5}\) = \(\frac{2 × 2}{5 × 2}\) = \(\frac{4}{10}\) = 0.4

Question 2.
\(\frac{8}{9}\) =

Answer:
0.88

Explanation:
\(\frac{8}{9}\) = \(\frac{8 × 10}{9 × 10}\) = \(\frac{80}{9 × 10}\) = \(\frac{8.88}{10}\) = 0.88

Question 3.
3 \(\frac{3}{4}\) =

Answer:
3.75

Explanation:
3 \(\frac{3}{4}\) =\(\frac{15}{4}\) = 3.75

Question 4.
\(\frac{7}{10}\) =

Answer:
0.7

Explanation:
\(\frac{7}{10}\) = 0.7

Question 5.
2 \(\frac{3}{8}\) =

Answer:
2.375

Explanation:
2 \(\frac{3}{8}\) = \(\frac{19}{8}\) = 2.375

Question 6.
\(\frac{5}{6}\) =

Answer:
0.833

Explanation:
\(\frac{5}{6}\) = \(\frac{5 × 10}{6 × 10}\) = \(\frac{50}{6 × 10}\) = \(\frac{8.33}{10}\) = 0.833

Write each decimal as a fraction or mixed number in simplest form

Question 7.
0.675
\(\frac{□}{□}\)

Answer:
\(\frac{27}{40}\)

Explanation:
\(\frac{0.675 × 1000}{1 × 1000}\) = \(\frac{675}{1000}\) = \(\frac{675/25}{1000/25}\) = \(\frac{27}{40}\)

Question 8.
5.6
______ \(\frac{□}{□}\)

Answer:
5 \(\frac{3}{5}\)

Explanation:
\(\frac{5.6 × 10}{10}\) = \(\frac{56}{10}\) = 5 \(\frac{6}{10}\) = 5 \(\frac{6/2}{10/2}\) = 5 \(\frac{3}{5}\)

Question 9.
0.44
\(\frac{□}{□}\)

Answer:
\(\frac{11}{25}\)

Explanation:
\(\frac{0.44 × 100}{1 × 100}\) = \(\frac{44}{100}\) = \(\frac{44/4}{100/4}\) = \(\frac{11}{25}\)

Question 10.
0.\(\bar{4}\)
\(\frac{□}{□}\)

Answer:
\(\frac{4}{9}\)

Explanation:
Let x = 0.\(\bar{4}\)
Now, 10x = 4.\(\bar{4}\)
10x – x = 4.\(\bar{4}\) – 0.\(\bar{4}\)
9x = 4
x = \(\frac{4}{9}\)

Question 11.
0.\(\overline { 26 } \)
\(\frac{□}{□}\)

Answer:
\(\frac{26}{99}\)

Explanation:
Let x = 0.\(\overline {26}\)
Now, 100x = 26.\(\overline{26}\)
100x – x = 26.\(\overline{26}\) – 0.\(\overline {26}\)
99x = 26
x = \(\frac{26}{99}\)

Question 12.
0.\(\overline { 325 } \)
\(\frac{□}{□}\)

Answer:
\(\frac{325}{999}\)

Explanation:
Let x = 0.\(\overline {325}\)
Now, 1000x = 325.\(\overline{325}\)
1000x – x = 325.\(\overline{325}\) – 0.\(\overline {325}\)
999x = 325
x = \(\frac{325}{999}\)

Solve each equation for x

Question 13.
x2 = 144
± ______

Answer:
x=±12

Explanation:
x2 = 144
Taking square roots on both the sides
x2=±144
x = ±12

Question 14.
x2 = \(\frac{25}{289}\)
± \(\frac{□}{□}\)

Answer:
x = ±\(\frac{5}{17}\)

Explanation:
x2 = \(\frac{25}{289}\)
Taking square roots on both the sides
x2=±√\(\frac{25}{289}\)
x = ±\(\frac{5}{17}\)

Question 15.
x3 = 216
______

Answer:
x = 6

Explanation:
x3 = 216
Taking cube roots on both the sides
3x3= 3√216
x = 6

Approximate each irrational number to two decimal places without a calculator.

Question 16.
\(\sqrt { 5 } \) ≈ ______

Answer:
2.236

Explanation:
x = \(\sqrt { 5 } \)
The 5 is in between 4 and 6
Take square root of each year
√4 < √5 < √6
2 < √5 < 3
√5 = 2.2
(2.2)² = 4.84
(2.25)² = 5.06
(2.5)³ = 5.29
A good estimate for √5 is 2.25

Question 17.
\(\sqrt { 3 } \) ≈ ______

Answer:
1.75

Explanation:
\(\sqrt { 3 } \)
1 < 3 < 4
√1 < √3 < √4
1 < √3 < 2
√3 = 1.6
(1.65)² = 2.72
(1.7)² = 2.89
(1.75)² = 3.06
A good estimate for √3 is 1.75

Question 18.
\(\sqrt { 10 } \) ≈ ______

Answer:
3.15

Explanation:
\(\sqrt { 10 } \)
9 < 10 < 16
√9 < √10 < √16
3 < √10 < 4
√10 = 3.1
(3.1)² = 9.61
(3.15)² = 9.92
(3.2)² = 10.24
A good estimate for √10 is 3.15

Question 19.
What is the difference between rational and irrational numbers?
Type below:
_____________

Answer:

Rational number can be expressed as a ration of two integers such as 5/2
Irrational number cannot be expressed as a ratio of two integers such as √13

Explanation:
A rational number is a number that can be express as the ratio of two integers. A number that cannot be expressed that way is irrational.

1.1 Independent Practice – Rational and Irrational Numbers – Page No. 13

Question 20.
A \(\frac{7}{16}\)-inch-long bolt is used in a machine. What is the length of the bolt written as a decimal?
______ -inch-long

Answer:
0.4375 inch

Explanation:
The length of the bolt is \(\frac{7}{16}\)-inch
Let, x = \(\frac{7}{16}\)
Multiplying by 125 on both nominator and denominator
x = \(\frac{7×125}{16×125}\) = \(\frac{875}{2000}\) =\(\frac{437.5}{1000}\) = 0.4375

Question 21.
The weight of an object on the moon is \(\frac{1}{6}\) its weight on Earth. Write \(\frac{1}{6}\) as a decimal.
______

Answer:
0.1666

Explanation:
The weight of the object on the moon is \(\frac{1}{6}\)
Let, x = \(\frac{1}{6}\)
Multiplying by 100 on both nominator and denominator
x = \(\frac{1×100}{6×100}\) = \(\frac{16.6}{100}\) =0.166

Question 22.
The distance to the nearest gas station is 2 \(\frac{4}{5}\) kilometers. What is this distance written as a decimal?
______

Answer:
2.8

Explanation:
The distance of the nearest gas station is 2 \(\frac{4}{5}\)
Let, x = 2 \(\frac{4}{5}\)
Multiplying by 100 on both nominator and denominator
x = 2 \(\frac{4×100}{5×100}\) = \(\frac{80}{100}\) =0.8

Question 23.
A baseball pitcher has pitched 98 \(\frac{2}{3}\) innings. What is the number of innings written as a decimal?
______

Answer:
98.6

Explanation:
A baseball pitcher has pitched 98 \(\frac{2}{3}\) innings.
98 \(\frac{2}{3}\) = 98 + 2/3
= (294/3) + (2/3)
296/3
98.6

Question 24.
A heartbeat takes 0.8 second. How many seconds is this written as a fraction?
\(\frac{□}{□}\)

Answer:
\(\frac{4}{5}\)

Explanation:
A heartbeat takes 0.8 seconds.
0.8
There are 8 tenths.
8/10 = 4/5

Question 25.
There are 26.2 miles in a marathon. Write the number of miles using a fraction.
\(\frac{□}{□}\)

Answer:
26\(\frac{1}{5}\)

Explanation:
There are 26.2 miles in a marathon.
26.2 miles
262/10
131/5
26 1/5 miles

Question 26.
The average score on a biology test was 72.\(\bar{1}\). Write the average score using a fraction.
\(\frac{□}{□}\)

Answer:
80 \(\frac{1}{9}\)

Explanation:
The average score on a biology test was 72.\(\bar{1}\).
72.\(\bar{1}\)
Let x = 72.\(\bar{1}\)
10x = 10(72.\(\bar{1}\))
10x = 721.1
-x = -0.1
9x = 721
x = 721/9
x = 80 1/9

Question 27.
The metal in a penny is worth about 0.505 cent. How many cents is this written as a fraction?
\(\frac{□}{□}\)

Answer:
\(\frac{101}{200}\)

Explanation:
The metal in a penny is worth about 0.505 cent.
0.505 cent
505 thousandths
505/1000
101/200 cents

Question 28.
Multistep An artist wants to frame a square painting with an area of 400 square inches. She wants to know the length of the wood trim that is needed to go around the painting.
Go Math Grade 8 Answer Key Chapter 1 Real Numbers Lesson 1: Rational and Irrational Numbers img 1
a. If x is the length of one side of the painting, what equation can you set up to find the length of a side?
x2 = ______

Answer:
x² = 400

Explanation:
The area of a square is the square of its equal side, x
x² = 400

Question 28.
b. Solve the equation you wrote in part a. How many solutions does the equation have?
x = ± ______

Answer:
x = ± 20

Explanation:
Take the square root on both sides. Solve
x = ± 20

Question 28.
c. Do all of the solutions that you found in part b make sense in the context of the problem? Explain.
Type below:
_____________

Answer:
No. Both values of x do not make sense.

Explanation:
The length cannot be negative, hence negative value does not make sense.
No. Both values of x do not make sense.

Question 28.
d. What is the length of the wood trim needed to go around the painting?
P = ______ inches

Answer:
Length P = 20 + 2y

Rational and Irrational Numbers – Page No. 14

Question 29.
Analyze Relationships To find \(\sqrt { 15 } \), Beau found 32 = 9 and 42 = 16. He said that since 15 is between 9 and 16, \(\sqrt { 15 } \) must be between 3 and 4. He thinks a good estimate for \(\sqrt { 15 } \) is \(\frac { 3+4 }{ 2 } \) = 3.5. Is Beau’s estimate high, low, or correct? Explain.
_____________

Answer:
3.85

Explanation:
15 is closer to 16
√15 is closer to √16
Beau’s estimate is low.
(3.8)² = 14.44
(3.85)² = 14.82
(3.9)² = 15.21
√15 is 3.85

Question 30.
Justify Reasoning What is a good estimate for the solution to the equation x3 = 95? How did you come up with your estimate?
x ≈ ______

Answer:
x ≈  4.55

Explanation:
3√x = 95
x = 3√95
64 < 95 < 125
Take the cube root of each number
3√64 < 3√95  < 3√125
4 < 3√95 < 5
3√95 = 4.6
(4.5)³ = 91.125
(4.55)³ = 94.20
(4.6)³ = 97.336
3√95 = 4.55

Question 31.
The volume of a sphere is 36π ft3. What is the radius of the sphere? Use the formula V = \(\frac { 4 }{ 3 } \)πr3 to find your answer.
Go Math Grade 8 Answer Key Chapter 1 Real Numbers Lesson 1: Rational and Irrational Numbers img 2
r = ______

Answer:
r = 3

Explanation:
V = 4/3 πr³
36π = 4/3 πr³
r³ = 36π/π . 3/4
r³ = 27
r = 3√27
r = 3

FOCUS ON HIGHER ORDER THINKING

Question 32.
Draw Conclusions Can you find the cube root of a negative number? If so, is it positive or negative? Explain your reasoning.
_____________

Answer:
Yes

Explanation:
Yes. The cube root of a negative number would be negative. Because the product of three negative signs is always negative.

Question 33.
Make a Conjecture Evaluate and compare the following expressions.
\(\sqrt { \frac { 4 }{ 25 } } \) and \(\frac { \sqrt { 4 } }{ \sqrt { 25 } } \) \(\sqrt { \frac { 16 }{ 81 } } \) and \(\frac { \sqrt { 16 } }{ \sqrt { 81 } } \) \(\sqrt { \frac { 36 }{ 49 } } \) and\(\frac { \sqrt { 36 } }{ \sqrt { 49 } } \)
Use your results to make a conjecture about a division rule for square roots. Since division is multiplication by the reciprocal, make a conjecture about a multiplication rule for square roots.
Expressions are: _____________

Answer:
Evaluating and comparing
√4/25 = 2/5
√16/81 = 4/9
√36/49 = 6/7
Conjecture about a division rule for square roots
√a/√b = √(a/b)
Conjecture about a multiplication rule for square roots
√a × √b

Question 34.
Persevere in Problem Solving
The difference between the solutions to the equation x2 = a is 30. What is a? Show that your answer is correct.
_____

Answer:
30

Explanation:
x2 = a
x = ±√a
√a – (-√a) = 30
√a + √a = 30
2√a = 30
√a = 15
a = 225
x2 = 225
x = ±225
x = ±15
15 – (-15) = 15 + 15 = 30

Guided Practice – Sets of real Numbers – Page No. 18

Write all names that apply to each number.

Question 1.
\(\frac{7}{8}\)
Options:
a. Real Numbers
b. Rational Numbers
c. Integers
d. Whole Numbers
e. Irrational Numbers

Answer:
a. Real Numbers
b. Rational Numbers

Question 2.
\(\sqrt { 36 } \)
Options:
a. Real Numbers
b. Rational Numbers
c. Integers
d. Whole Numbers
e. Irrational Numbers

Answer:
a. Real Numbers
b. Rational Numbers
c. Integers
d. Whole Numbers

Explanation:
\(\sqrt { 36 } \) = 6

Question 3.
\(\sqrt { 24 } \)
Options:
a. Real Numbers
b. Rational Numbers
c. Integers
d. Whole Numbers
e. Irrational Numbers

Answer:
a. Real Numbers
e. Irrational Numbers

Question 4.
0.75
Options:
a. Real Numbers
b. Rational Numbers
c. Integers
d. Whole Numbers
e. Irrational Numbers

Answer:
a. Real Numbers
b. Rational Numbers

Question 5.
0
Options:
a. Real Numbers
b. Rational Numbers
c. Integers
d. Whole Numbers
e. Irrational Numbers

Answer:
a. Real Numbers
b. Rational Numbers
c. Integers
d. Whole Numbers

Question 6.
−\(\sqrt { 100 } \)
Options:
a. Real Numbers
b. Rational Numbers
c. Integers
d. Whole Numbers
e. Irrational Numbers

Answer:
a. Real Numbers
b. Rational Numbers
c. Integers

Explanation:
−\(\sqrt { 100 } \) = – 10

Question 7.
5.\(\overline { 45 } \)
Options:
a. Real Numbers
b. Rational Numbers
c. Integers
d. Whole Numbers
e. Irrational Numbers

Answer:
a. Real Numbers
b. Rational Numbers

Question 8.
−\(\frac{18}{6}\)
Options:
a. Real Numbers
b. Rational Numbers
c. Integers
d. Whole Numbers
e. Irrational Numbers

Answer:
a. Real Numbers
b. Rational Numbers
c. Integers

Explanation:
−\(\frac{18}{6}\) = -3

Tell whether the given statement is true or false. Explain your choice.

Question 9.
All whole numbers are rational numbers.
i. True
ii. False

Answer:
i. True

Explanation:
All whole numbers are rational numbers.
Whole numbers are a subset of the set of rational numbers and can be written as ratio of the whole number to 1.

Question 10.
No irrational numbers are whole numbers.
i. True
ii. False

Answer:
i. True

Explanation:
True. Whole numbers are ration numbers.

Identify the set of numbers that best describes each situation. Explain your choice.

Question 11.
the change in the value of an account when given to the nearest dollar
Options:
a. Real Numbers
b. Rational Numbers
c. Integer Numbers
d. Whole Numbers
e. Irrational Numbers

Answer:
c. Integer Numbers

Explanation:
The change can be a whole dollar amount and can be positive, negative or zero.

Question 12.
the markings on a standard ruler
Go Math Grade 8 Answer Key Chapter 1 Real Numbers Lesson 2: Sets of real Numbers img 3
Options:
a. Real Numbers
b. Rational Numbers
c. Integer Numbers
d. Whole Numbers
e. Irrational Numbers

Answer:
b. Rational Numbers

Explanation:
The ruler is marked every 1/16t inch.

ESSENTIAL QUESTION CHECK-IN

Question 13.
What are some ways to describe the relationships between sets of numbers?

Answer:
There are two ways that we have been using until now to describe the relationships between sets of numbers

  • Using a scheme or a diagram as the one on page 15.
  • Verbal description, for example, “All irrational numbers are real numbers.”

1.2 Independent Practice – Sets of real Numbers – Page No. 19

Write all names that apply to each number. Then place the numbers in the correct location on the Venn diagram.
Go Math Grade 8 Answer Key Chapter 1 Real Numbers Lesson 2: Sets of real Numbers img 4

Question 14.
\(\sqrt { 9 } \)
Options:
a. Real Numbers
b. Rational Numbers
c. Integer Numbers
d. Whole Numbers
e. Irrational Numbers

Answer:
a. Real Numbers
b. Rational Numbers
c. Integer Numbers
d. Whole Numbers

Explanation:
\(\sqrt { 9 } \) = 3

Question 15.
257
Options:
a. Real Numbers
b. Rational Numbers
c. Integer Numbers
d. Whole Numbers
e. Irrational Numbers

Answer:
a. Real Numbers
b. Rational Numbers
c. Integer Numbers
d. Whole Numbers

Question 16.
\(\sqrt { 50 } \)
Options:
a. Real Numbers
b. Rational Numbers
c. Integer Numbers
d. Whole Numbers
e. Irrational Numbers

Answer:
a. Real Numbers
e. Irrational Numbers

Question 17.
8 \(\frac{1}{2}\)
Options:
a. Real Numbers
b. Rational Numbers
c. Integer Numbers
d. Whole Numbers
e. Irrational Numbers

Answer:
a. Real Numbers
b. Rational Numbers

Question 18.
16.6
Options:
a. Real Numbers
b. Rational Numbers
c. Integer Numbers
d. Whole Numbers
e. Irrational Numbers

Answer:
a. Real Numbers
b. Rational Numbers

Question 19.
\(\sqrt { 16 } \)
Options:
a. Real Numbers
b. Rational Numbers
c. Integer Numbers
d. Whole Numbers
e. Irrational Numbers

Answer:
a. Real Numbers
b. Rational Numbers
c. Integer Numbers
d. Whole Numbers

Explanation:
\(\sqrt { 16 } \) = 4

Identify the set of numbers that best describes each situation. Explain your choice.

Question 20.
the height of an airplane as it descends to an airport runway
Options:
a. Real Numbers
b. Rational Numbers
c. Integer Numbers
d. Whole Numbers
e. Irrational Numbers

Answer:
d. Whole Numbers

Explanation:
Whole. The height of an airplane as it descents to an airport runway is a whole number greater than 0

Question 21.
the score with respect to par of several golfers: 2, – 3, 5, 0, – 1
Options:
a. Real Numbers
b. Rational Numbers
c. Integer Numbers
d. Whole Numbers
e. Irrational Numbers

Answer:
c. Integer Numbers

Explanation:
Integers. The scores are counting numbers, their opposites, and zero.

Question 22.
Critique Reasoning Ronald states that the number \(\frac{1}{11}\) is not rational because, when converted into a decimal, it does not terminate. Nathaniel says it is rational because it is a fraction. Which boy is correct? Explain.
i. Ronald
ii. Nathaniel

Answer:
ii. Nathaniel

Explanation:
Nathaniel is correct.
A fraction is a rational real number, even if it is not a terminating decimal.

Sets of real Numbers – Page No. 20

Question 23.
Critique Reasoning The circumference of a circular region is shown. What type of number best describes the diameter of the circle? Explain your answer.
Go Math Grade 8 Answer Key Chapter 1 Real Numbers Lesson 2: Sets of real Numbers img 5
Options:
a. Real Numbers
b. Rational Numbers
c. Irrational Numbers
d. Integers
e. Whole Numbers

Answer:
e. Whole Numbers

Explanation:
Circumference of the circle
A = 2πr
π = 2πr
Diameter is twice the radius
2r = 1
Whole

Question 24.
Critical Thinking A number is not an integer. What type of number can it be?
Options:
a. Real Numbers
b. Rational Numbers
c. Integers
d. Whole Numbers
e. Irrational Numbers

Answer:
b. Rational Numbers
e. Irrational Numbers

Question 25.
A grocery store has a shelf with half-gallon containers of milk. What type of number best represents the total number of gallons?
Options:
a. Real Numbers
b. Rational Numbers
c. Integers
d. Whole Numbers
e. Irrational Numbers

Answer:
b. Rational Numbers

FOCUS ON HIGHER ORDER THINKING

Question 26.
Explain the Error Katie said, “Negative numbers are integers.” What was her error?
Type below:
_______________

Answer:
Her error is that she stated that all negative numbers are integers. Some negative numbers are integers such as -4 but some are not such an -0.8

Question 27.
Justify Reasoning Can you ever use a calculator to determine if a number is rational or irrational? Explain.
Type below:
_______________

Answer:
Not always.

Explanation:
Not always.
If the calculator shows a terminating decimal, the number is rational but otherwise, it is not possible as you can only see a few digits.

Question 28.
Draw Conclusions The decimal 0.\(\bar{3}\) represents \(\frac{1}{3}\). What type of number best describes 0.\(\bar{9}\) , which is 3 × 0.\(\bar{3}\)? Explain.
Type below:
_______________

Answer:
1

Explanation:
let x = 0.9999999
10x = 9.99999999
10x = 9 + 0.999999999
10x = 9 + x
9x = 9
x=1.

Question 29.
Communicate Mathematical Ideas Irrational numbers can never be precisely represented in decimal form. Why is this?

Answer:
Because irrational numbers are nonrepeating, otherwise they could be represented as a fraction. Although a potential counter-example to this claim is that some irrational numbers can only be represented in decimal form, for example, 0.1234567891011121314151617…, 0.24681012141618202224…, 0.101101110111101111101111110… are all irrational numbers.

Guided Practice – Ordering Real Numbers – Page No. 24

Compare. Write <, >, or =.

Question 1.
\(\sqrt { 3 } \) + 2 ________ \(\sqrt { 3 } \) + 3

Answer:
\(\sqrt { 3 } \) + 2 < \(\sqrt { 3 } \) + 3

Explanation:
\(\sqrt { 3 } \) is between 1 and 2
\(\sqrt { 3 } \) + 2 is between 3 and 4
\(\sqrt { 3 } \) + 3 is between 4 and 5
\(\sqrt { 3 } \) + 2 < \(\sqrt { 3 } \) + 3

Question 2.
\(\sqrt { 11 } \) + 15 _______ \(\sqrt { 8 } \) + 15

Answer:
\(\sqrt { 11 } \) + 15 > \(\sqrt { 8 } \) + 15

Explanation:
\(\sqrt { 11 } \) is between 3 and 4
\(\sqrt { 8 } \) is between 2 and 3
\(\sqrt { 11 } \) + 15 is between 18 and 19
\(\sqrt { 8 } \) + 15 is between 17 and 18
\(\sqrt { 11 } \) + 15 > \(\sqrt { 8 } \) + 15

Question 3.
\(\sqrt { 6 } \) + 5 _______ 6 + \(\sqrt { 5 } \)

Answer:
\(\sqrt { 6 } \) + 5 < 6 + \(\sqrt { 5 } \)

Explanation:
\(\sqrt { 6 } \) is between 2 and 3
\(\sqrt { 5 } \) is between 2 and 3
\(\sqrt { 6 } \) is between 7 and 8
\(\sqrt { 5 } \) is between 8 and 9
\(\sqrt { 6 } \) + 5 < 6 + \(\sqrt { 5 } \)

Question 4.
\(\sqrt { 9 } \) + 3 _______ 9 + \(\sqrt { 3 } \)

Answer:
\(\sqrt { 9 } \) + 3 < 9 + \(\sqrt { 3 } \)

Explanation:
\(\sqrt { 9 } \) + 3
9 + \(\sqrt { 3 } \)
\(\sqrt { 3 } \) is between 1 and 2
\(\sqrt { 9 } \) + 3 = 3 + 3 = 6
9 + \(\sqrt { 3 } \) is between 10 and 11
\(\sqrt { 9 } \) + 3 < 9 + \(\sqrt { 3 } \)

Question 5.
\(\sqrt { 17 } \) – 3 _______ -2 + \(\sqrt { 5 } \)

Answer:
\(\sqrt { 17 } \) – 3 > -2 + \(\sqrt { 5 } \)

Explanation:
\(\sqrt { 17 } \) is between 4 and 5
\(\sqrt { 5 } \) is between 2 and 3
\(\sqrt { 17 } \) – 3 is between 1 and 2
-2 + \(\sqrt { 5 } \) is between 0 and 1
\(\sqrt { 17 } \) – 3 > -2 + \(\sqrt { 5 } \)

Question 6.
10 – \(\sqrt { 8 } \) _______ 12 – \(\sqrt { 2 } \)

Answer:
10 – \(\sqrt { 8 } \) < 12 – \(\sqrt { 2 } \)

Explanation:
\(\sqrt { 8 } \) is between 2 and 3
\(\sqrt { 2 } \) is between 1 and 2
10 – \(\sqrt { 8 } \) is between 8 and 7
12 – \(\sqrt { 2 } \) is between 11 and 10
10 – \(\sqrt { 8 } \) < 12 – \(\sqrt { 2 } \)

Question 7.
\(\sqrt { 7 } \) + 2 _______ \(\sqrt { 10 } \) – 1

Answer:
\(\sqrt { 7 } \) + 2 > \(\sqrt { 10 } \) – 1

Explanation:
\(\sqrt { 7 } \) is between 2 and 3
\(\sqrt { 10 } \) is between 3 and 4
\(\sqrt { 7 } \) + 2 is between 4 and 5
\(\sqrt { 10 } \) – 1 is between 2 and 3
\(\sqrt { 7 } \) + 2 > \(\sqrt { 10 } \) – 1

Question 8.
\(\sqrt { 17 } \) + 3 _______ 3 + \(\sqrt { 11 } \)

Answer:
\(\sqrt { 17 } \) + 3 > 3 + \(\sqrt { 11 } \)

Explanation:
\(\sqrt { 17 } \) is between 4 and 5
\(\sqrt { 11 } \) is between 3 and 4
\(\sqrt { 17 } \) + 3 is between 7 and 8
3 + \(\sqrt { 11 } \) is between 6 and 7
\(\sqrt { 17 } \) + 3 > 3 + \(\sqrt { 11 } \)

Question 9.
Order \(\sqrt { 3 } \), 2 π, and 1.5 from least to greatest. Then graph them on the number line.
\(\sqrt { 3 } \) is between _________ and _____________ , so \(\sqrt { 3 } \) ≈ ____________.
π ≈ 3.14, so 2 π ≈ _______________.
Go Math Grade 8 Answer Key Chapter 1 Real Numbers Lesson 3: Ordering Real Numbers img 6
From least to greatest, the numbers are ______________, _____________________ ,_________________.
Type below:
___________

Answer:
1.5, \(\sqrt { 3 } \), 2 π

Explanation:
\(\sqrt { 3 } \) is between 1.7 and 1.75
π = 3.14; 2 π = 6.28
grade 8 chapter 1 image 1
1.5, \(\sqrt { 3 } \), 2 π

Question 10.
Four people have found the perimeter of a forest using different methods. Their results are given in the table. Order their calculations from greatest to least.
Go Math Grade 8 Answer Key Chapter 1 Real Numbers Lesson 3: Ordering Real Numbers img 7
Type below:
___________

Answer:
\(\sqrt { 17 } \) – 2, 1+ π/2, 2.5, 12/5

Explanation:
\(\sqrt { 17 } \) – 2
\(\sqrt { 17 } \) is between 4 and 5
Since, 17 is closer to 16, the estimated value is 4.1
1+ π/2
1 + (3.14/2) = 2.57
12/5 = 2.4
2.5
\(\sqrt { 17 } \) – 2, 1+ π/2, 2.5, 12/5

ESSENTIAL QUESTION CHECK-IN

Question 11.
Explain how to order a set of real numbers.
Type below:
___________

Answer:
Evaluate the given numbers and write in decimal form. Plot on number line and arrange the numbers accordingly.

Independent Practice – Ordering Real Numbers – Page No. 25

Order the numbers from least to greatest.

Question 12.
\(\sqrt { 7 } \), 2, \(\frac { \sqrt { 8 } }{ 2 } \)
Type below:
____________

Answer:
\(\frac { \sqrt { 8 } }{ 2 } \), 2, \(\sqrt { 7 } \)

Explanation:
\(\sqrt { 7 } \), 2, \(\frac { \sqrt { 8 } }{ 2 } \)
\(\sqrt { 7 } \) is between 2 and 3
Since 7 is closer to 9, (2.65)² = 7.02, hence the estimated value is 2.65
\(\frac { \sqrt { 8 } }{ 2 } \)
\(\sqrt { 8 } \) is between 2 and 3
Since 8 is closer to 9, (2.85)² = 8.12, hence the estimated value is 2.85
2.85/2 = 1.43
grade 8 chapter 1 image 3
\(\frac { \sqrt { 8 } }{ 2 } \), 2, \(\sqrt { 7 } \)

Question 13.
\(\sqrt { 10 } \), π, 3.5
Type below:
____________

Answer:
π, \(\sqrt { 10 } \), 3.5

Explanation:
\(\sqrt { 10 } \), π, 3.5
\(\sqrt { 10 } \) is between 3 and 4
Since, 10 is closer to 9, (3.15)² = 9.92, hence the estimated value is 3.15
π = 3.14
3.5
grade 8 chapter 1 image 4
π, \(\sqrt { 10 } \), 3.5

Question 14.
\(\sqrt { 220 } \), −10, \(\sqrt { 100 } \), 11.5
Type below:
____________

Answer:
-10, √100, 11.5, √220

Explanation:
\(\sqrt { 220 } \), −10, \(\sqrt { 100 } \), 11.5
196 < 220 < 225
√196 < √220 < √225
14 < √220 < 15
√220 = 14.5
√100 = 10
grade 8 chapter 1 image 5
-10, √100, 11.5, √220

Question 15.
\(\sqrt { 8 } \), −3.75, 3, \(\frac{9}{4}\)
Type below:
____________

Answer:
−3.75, \(\frac{9}{4}\), \(\sqrt { 8 } \)

Explanation:
\(\sqrt { 8 } \), −3.75, 3, \(\frac{9}{4}\)
\(\sqrt { 8 } \) is between 2 and 3
Since, 8 is closer to 9, (2.85)² = 8.12, hence the estimated value is 2.85
-3.75 = 3
9/4 = 2.25
grade 8 chapter 1 image 6
−3.75, \(\frac{9}{4}\), \(\sqrt { 8 } \)

Question 16.
Your sister is considering two different shapes for her garden. One is a square with side lengths of 3.5 meters, and the other is a circle with a diameter of 4 meters.
a. Find the area of the square.
_______ m2

Answer:
(3.5)² = 12.25

Explanation:
Area of the square = x²
Area = (3.5)² = 12.25

Question 16.
b. Find the area of the circle.
_______ m2

Answer:
π(2)² = 12.56

Explanation:
Area of the circle = πr² where r = d/2 = 4/2 = 2
Area = π(2)² = 12.56

Question 16.
c. Compare your answers from parts a and b. Which garden would give your sister the most space to plant?
___________

Answer:
12.25 < 12.56
The circle will give more space

Question 17.
Winnie measured the length of her father’s ranch four times and got four different distances. Her measurements are shown in the table.
a. To estimate the actual length, Winnie first approximated each distance to the nearest hundredth. Then she averaged the four numbers. Using a calculator, find Winnie’s estimate.
Go Math Grade 8 Answer Key Chapter 1 Real Numbers Lesson 3: Ordering Real Numbers img 8
______

Answer:
7.4815

Explanation:
\(\sqrt { 60 } \) = 7.75
58/8 = 7.25
7.3333
7 3/5 = 7.60
Average = (7.75 + 7.25 + 7.33 + 7.60)/4 = 7.4815

Question 17.
b. Winnie’s father estimated the distance across his ranch to be \(\sqrt { 56 } \) km. How does this distance compare to Winnie’s estimate?
____________

Answer:
They are nearly identical

Explanation:
\(\sqrt { 56 } \) = 7.4833
They are nearly identical

Give an example of each type of number.

Question 18.
a real number between \(\sqrt { 13 } \) and \(\sqrt { 14 } \)
Type below:
____________

Answer:
A real number between \(\sqrt { 13 } \) and \(\sqrt { 14 } \)
Example: 3.7

Explanation:
\(\sqrt { 13 } \) = 3.61
\(\sqrt { 13 } \) = 3.74
A real number between \(\sqrt { 13 } \) and \(\sqrt { 14 } \)
Example: 3.7

Question 19.
an irrational number between 5 and 7
Type below:
____________

Answer:
An irrational number between 5 and 7
Example: \(\sqrt { 29 } \)

Explanation:
5² = 25 and 7² = 49
An irrational number between 5 and 7
Example: \(\sqrt { 29 } \)

Ordering Real Numbers – Page No. 26

Question 20.
A teacher asks his students to write the numbers shown in order from least to greatest. Paul thinks the numbers are already in order. Sandra thinks the order should be reversed. Who is right?
Go Math Grade 8 Answer Key Chapter 1 Real Numbers Lesson 3: Ordering Real Numbers img 9
_____________

Answer:
Neither are correct

Explanation:
\(\sqrt { 115 } \), 115/11, 10.5624
\(\sqrt { 115 } \) is between 10 and 11
Since, 115 is closer to 121, (10.7)² = 114.5, hence the estimated value is 10.7
115/11 = 10.4545
10.5624
Neither are correct

Question 21.
Math History
There is a famous irrational number called Euler’s number, symbolized with an e. Like π, its decimal form never ends or repeats. The first few digits of e are 2.7182818284.
a. Between which two square roots of integers could you find this number?
Type below:
_____________

Answer:
The square of e lies between 7 and 8
2.718281828
(2.72)² = 7.3984
Hence, it lies between \(\sqrt { 7 } \) = 2.65 and \(\sqrt { 8 } \) = 2.82

Question 21.
b. Between which two square roots of integers can you find π?
Type below:
_____________

Answer:
3.142
(3.14)² = 9.8596
Hence. it lies between \(\sqrt { 9 } \) = 3 and \(\sqrt { 10 } \) = 3.16

H.O.T.

FOCUS ON HIGHER ORDER THINKING

Question 22.
Analyze Relationships
There are several approximations used for π, including 3.14 and \(\frac{22}{7}\). π is approximately 3.14159265358979 . . .
a. Label π and the two approximations on the number line.
Go Math Grade 8 Answer Key Chapter 1 Real Numbers Lesson 3: Ordering Real Numbers img 10
Type below:
_____________

Answer:
grade 8 chapter 1 image 7

Question 22.
b. Which of the two approximations is a better estimate for π? Explain.
Type below:
_____________

Answer:
As we can see from the number line, 22/7 is closer to π, so we can conclude that 22/7 is a better estimation for π.

Question 22.
c. Find a whole number x so that the ratio \(\frac{x}{113}\) is a better estimate for π than the two given approximations.
Type below:
_____________

Answer:
355/113 is a better estimation for π, because 355/113 = 3.14159292035 = 3.14159265358979 = π

Question 23.
Communicate Mathematical Ideas
What is the fewest number of distinct points that must be graphed on a number line, in order to represent natural numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers? Explain.
_______ points

Answer:
2 points

Explanation:
There need to be plotting of at least 2 points because a rational number can never be equal to an irrational number. So let’s say 5 points are the same among six but the 6th will be different as there both rational numbers and irrational numbers included.

Question 24.
Critique Reasoning
Jill says that 12.\(\bar{6}\) is less than 12.63. Explain her error.
Type below:
_____________

Answer:
12.\(\bar{6}\) = 12.666
12.\(\bar{6}\) > 12.63

1.1 Rational and Irrational Numbers – Model Quiz – Page No. 27

Write each fraction as a decimal or each decimal as a fraction.

Question 1.
\(\frac{7}{20}\)
_______

Answer:
0.35

Explanation:
\(\frac{7}{20}\) = 0.35

Question 2.
1.\(\overline { 27} \)
______ \(\frac{□}{□}\)

Answer:
1\(\frac{28}{99}\)

Explanation:
1.\(\overline { 27} \)
x = 1.\(\overline { 27} \)
100x = 100(1.\(\overline { 27} \))
100x = 127(\(\overline { 27} \))
x = .\(\overline { 27} \)
99x = 127
x = 127/99
x = 1 28/99

Question 3.
1 \(\frac{7}{8}\)
______

Answer:
1.875

Explanation:
1 \(\frac{7}{8}\)
1 + 7/8
8/8 + 7/8
15/8 = 1.875

Solve each equation for x.

Question 4.
x2 = 81
± ______

Answer:
± 9

Explanation:
x2 = 81
x = ± 81
x = ± 9

Question 5.
x3 = 343
______

Answer:
x = 7

Explanation:
x3 = 343
x = 7

Question 6.
x2 = \(\frac{1}{100}\)
± \(\frac{□}{□}\)

Answer:
± \(\frac{1}{10}\)

Explanation:
x2 = \(\frac{1}{100}\)
x = ± \(\frac{1}{10}\)

Question 7.
A square patio has an area of 200 square feet. How long is each side of the patio to the nearest 0.05?
______ feet

Answer:
14.15 feet

Explanation:
The area of a square is found by multiplying the side of the square by itself. Therefore, to find the side of the square, we have to take the square root of the area.
Let’s denote with A the area of the patio and with s each side of the square.
We have:
A = 200
A = s.s
s = \(\sqrt { A } \) = \(\sqrt { 200 } \)
Following the steps as in “Explore Activity” on page 9, we can make an estimation for the irrational number:
196 < 200 < 225
\(\sqrt { 196 } \) < \(\sqrt { 200 } \) < \(\sqrt { 225 } \)
14 < \(\sqrt { 200 } \) < 15
We see that 200 is much closer to 196 than to 225, therefore the square root of it should be between 14 and 14.5. To make a better estimation, we pick some numbers between 14 and 14.5 and calculate their squares:
(14.1)² = 198.81
(14.2)² = 201.64
14.1 < \(\sqrt { 200 } \) < 14.2
\(\sqrt { 200 } \) = 14.15
We see that 200 is much closer to 14.1 than to 14.2, therefore the square root of it should be between 14.1 and 14.15. If we round to the nearest 0.05, we have:
s = 14.15

1.2 Sets of Real Numbers

Write all names that apply to each number.

Question 8.
\(\frac { 121 }{ \sqrt { 121 } }\)
Type below:
___________

Answer:
Rational, whole, integer, real numbers

Explanation:
\(\frac { 121 }{ \sqrt { 121 } }\)
121/11 = 11

Question 9.
\(\frac{π}{2}\)
Type below:
___________

Answer:
Irrational, real numbers

Question 10.
Tell whether the statement “All integers are rational numbers” is true or false. Explain your choice.
___________

Answer:
True

Explanation:
“All integers are rational numbers” is true, because every integer can be expressed as a fraction with a denominator equal to 1. The set of integer A a subset of rational numbers.

1.3 Ordering Real Numbers

Compare. Write <, >, or =.

Question 11.
\(\sqrt { 8 }\) + 3 _______ 8 + \(\sqrt { 3 }\)

Answer:
\(\sqrt { 8 }\) + 3 < 8 + \(\sqrt { 3 }\)

Explanation:
4 < 8 < 9
\(\sqrt { 4 }\) < \(\sqrt { 8 }\) < \(\sqrt { 9 }\)
2 < \(\sqrt { 8 }\) < 3
1 < 3 < 4
\(\sqrt { 1 }\) < \(\sqrt { 3 }\) < \(\sqrt { 4 }\)
1 < \(\sqrt { 3 }\) < 2
\(\sqrt { 8 }\) + 3 is between 5 and 6
8 + \(\sqrt { 3 }\) is between 9 and 10
\(\sqrt { 8 }\) + 3 < 8 + \(\sqrt { 3 }\)

Question 12.
\(\sqrt { 5 }\) + 11 _______ 5 + \(\sqrt { 11 }\)

Answer:
\(\sqrt { 5 }\) + 11 > 5 + \(\sqrt { 11 }\)

Explanation:
\(\sqrt { 5 }\) lies in between 2 and 3
\(\sqrt { 11 }\) lies in between 3 and 4
\(\sqrt { 5 }\) + 11 lies in between 13 and 14
5 + \(\sqrt { 11 }\) lies in between 8 and 9
\(\sqrt { 5 }\) + 11 > 5 + \(\sqrt { 11 }\)

Order the numbers from least to greatest.

Question 13.
\(\sqrt { 99 }\), π2, 9.\(\bar { 8 }\)
Type below:
_______________

Answer:
π2, 9.\(\bar { 8 }\), \(\sqrt { 99 }\)

Explanation:
\(\sqrt { 99 }\), π2, 9.\(\bar { 8 }\)
99 lies between 9² and 10²
99 is closer to 100, hence \(\sqrt { 99 }\) is closer to 10
(9.9)² = 98.01
(9.95)² = 99.0025
(10)² = 100
\(\sqrt { 99 }\) = 9.95
π² = 9.86
9.88888 = 9.89
grade 8 chapter 1 image 8
π2, 9.\(\bar { 8 }\), \(\sqrt { 99 }\)

Question 14.
\(\sqrt { \frac { 1 }{ 25 } } \), \(\frac{1}{4}\), 0.\(\bar { 2 }\)
Type below:
____________

Answer:
\(\sqrt { \frac { 1 }{ 25 } } \), 0.\(\bar { 2 }\), \(\frac{1}{4}\)

Explanation:
\(\sqrt { \frac { 1 }{ 25 } } \), \(\frac{1}{4}\), 0.\(\bar { 2 }\)
\(\sqrt { \frac { 1 }{ 25 } } \) = 1/5 = 0.2
1/4 = 0.25
0.\(\bar { 2 }\) = 0.222 = 0.22
grade 8 chapter 1 image 9
\(\sqrt { \frac { 1 }{ 25 } } \), 0.\(\bar { 2 }\), \(\frac{1}{4}\)

Essential Question

Question 15.
How are real numbers used to describe real-world situations?
Type below:
_______________

Answer:
In real-world situations, we use real numbers to count or make measurements. They can be seen as a convention for us to quantify things around, for example, the distance, the temperature, the height, etc.

Selected Response – Mixed Review – Page No. 28

Question 1.
The square root of a number is 9. What is the other square root?
Options:
a. -9
b. -3
c. 3
d. 81

Answer:
a. -9

Explanation:
We know that every positive number has two square roots, one positive and one negative. We are given the principal square root (9), so the other square root would be its negative (-9). To prove that, we square both numbers and we compare the results:
9 • 9 = 81
(-9). (-9)= 81

Question 2.
A square acre of land is 4,840 square yards. Between which two integers is the length of one side?
Options:
a. between 24 and 25 yards
b. between 69 and 70 yards
c. between 242 and 243 yards
d. between 695 and 696 yards

Answer:
b. between 69 and 70 yards

Explanation:
The area of a square is found by multiplying the side of the square by itself. Therefore, to Bud the side of the square, we have to take the square root of the area.
Let’s denote with A the area of the land and with each side of the square. We have:
A = 4840
A = s . s
A = s²
s = √A = √4840
Following the steps as in °Explore Activity on page 9, we can make an estimation for the irrational number:
4761 < 4840 < 4900
\(\sqrt { 4761 }\) < \(\sqrt { 4840 }\) < \(\sqrt { 4900 }\)
69 < \(\sqrt { 4840 }\) < 70
Each side of the land is between 69 and 70 yards.

Question 3.
Which of the following is an integer but not a whole number?
Options:
a. -9.6
b. -4
c. 0
d. 3.7

Answer:
b. -4

Explanation:
Whole numbers are not negative
-4 is an integer but not a whole number

Question 4.
Which statement is false?
Options:
a. No integers are irrational numbers.
b. All whole numbers are integers.
c. No real numbers are irrational numbers.
d. All integers greater than 0 are whole numbers.

Answer:
c. No real numbers are irrational numbers.

Explanation:
Rational and irrational numbers are real numbers.

Question 5.
Which set of numbers best describes the displayed weights on a digital scale that shows each weight to the nearest half pound?
Options:
a. whole numbers
b. rational numbers
c. real numbers
d. integers

Answer:
b. rational numbers

Explanation:
The scale weighs nearest to 1/2 pound.

Question 6.
Which of the following is not true?
Options:
a. π2 < 2π + 4
b. 3π > 9
c. \(\sqrt { 27 }\) + 3 > 172
d. 5 – \(\sqrt { 24 }\) < 1

Answer:
c. \(\sqrt { 27 }\) + 3 > 172

Explanation:
a. π2 < 2π + 4
(3.14)² < 2(3.14) + 4
9.86 < 10.28
True
b. 3π > 9
9.42 > 9
True
c. \(\sqrt { 27 }\) + 3 > 172
5.2 + 3 > 8.5
8.2 > 8.5
False
d. 5 – \(\sqrt { 24 }\) < 1
5 – 4.90 < 1
0.1 < 1
True

Question 7.
Which number is between \(\sqrt { 21 }\) and \(\frac{3π}{2}\) ?
Options:
a. \(\frac{14}{3}\)
b. 2 \(\sqrt { 6 }\)
c. 5
d. π + 1

Answer:

Explanation:
a. \(\sqrt { 21 }\) and \(\frac{3π}{2}\)
\(\sqrt { 21 }\) = 4.58
\(\frac{3π}{2}\) = 4.71
14/3 = 4.67
b. 2\(\sqrt { 6 }\) = 4.90
c. 5
d. π + 1 = 3.14 + 1 = 4.14

Question 8.
What number is shown on the graph?
Go Math Grade 8 Answer Key Chapter 1 Real Numbers Mixed Review img 11
Options:
a. π+3
b. \(\sqrt { 4 }\) + 2.5
c. \(\sqrt { 20 }\) + 2
d. 6.\(\overline { 14 } \)

Answer:
c. \(\sqrt { 20 }\) + 2

Explanation:
6.48
a. π+3 = 3.14 + 3 = 6.14
b. \(\sqrt { 4 }\) + 2.5 = 2 + 2.5 = 4.5
c. \(\sqrt { 20 }\) + 2 = 4.47 + 2 = 6.47
d. 6.\(\overline { 14 } \) = 6.1414

Question 9.
Which is in order from least to greatest?
Options:
a. 3.3, \(\frac{10}{3}\), π, \(\frac{11}{4}\)
b. \(\frac{10}{3}\), 3.3, \(\frac{11}{4}\), π
c. π, \(\frac{10}{3}\), \(\frac{11}{4}\), 3.3
d. \(\frac{11}{4}\), π, 3.3, \(\frac{10}{3}\)

Answer:
d. \(\frac{11}{4}\), π, 3.3, \(\frac{10}{3}\)

Explanation:
10/3 = 3.3333333
11/4 = 2.75
grade 8 chapter 1 image 10

Mini-Task

Question 10.
The volume of a cube is given by V = x3, where x is the length of an edge of the cube. The area of a square is given by A = x2, where x is the length of a side of the square. A given cube has a volume of 1728 cubic inches.
a. Find the length of an edge.
______ inches

Answer:
12 inches

Explanation:
V = x3
A = x2
1728 = x3
x = 12
The length of an edge = 12 in

Question 10.
b. Find the area of one side of the cube.
______ in2

Answer:
144 in2

Explanation:
A = (12)² = 144
Area of the side of the cube = 144 in2

Question 10.
c. Find the surface area of the cube.
______ in2

Answer:
864 in2

Explanation:
SA = 6 (12)² = 864
The surface area of the cube = 864 in2

Question 10.
d. What is the surface area in square feet?
______ ft2

Answer:
6 ft2

Explanation:
SA = 864/144 = 6
The surface area of the cube = 6 ft2

Conclusion:

We hope the details prevailed in this Grade 8 Go Math Answer Key Chapter 1 Real Numbers is helpful for you guys. Make use of the above links and try to solve all the problems. This HMH Go Math Grade 8 Answer Key also helps to complete the homework within the time without any mistakes. Stick to our ccssmathanswers.com site to get the pdf links of all the chapters. Feel free to post your comments in the below box. We will try to clarify your doubts as early as possible. All the Best Guys!!!

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

go-math-grade-6-chapter-11-surface-area-and-volume-answer-key

Go Math solutions for Class 6 Maths Provide detailed explanations for all the questions provided in the HMH Go Math. We provide topic wise Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume to help the students clear their doubts by offering an understanding of concepts in depth. You can practice different types of questions in Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume. 

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

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

Lesson 1: Three-Dimensional Figures and Nets

Lesson 2: Investigate • Explore Surface Area Using Nets

Lesson 3: Algebra • Surface Area of Prisms

Lesson 4: Algebra • Surface Area of Pyramids

 Mid-Chapter Checkpoint

Lesson 5: Investigate • Fractions and Volume

Lesson 6: Algebra • Volume of Rectangular Prisms

Lesson 7: Problem Solving • Geometric Measurements

Chapter 11 Review/Test

Share and Show – Page No. 599

Identify and draw a net for the solid figure.

Question 1.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 1

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

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

Question 2.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 2
Answer: Cube or Rectangular prism.

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

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

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

Question 3.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 3

Answer: Triangular pyramid.

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

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

Question 4.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 4
Answer: Cube

Explanation: The net has six squares.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume

On Your Own

Identify and draw a net for the solid figure.

Question 5.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 5

Answer: Triangular prism.

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

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

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

Answer:  Rectangular Prism.

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

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

Problem Solving + Applications – Page No. 600

Solve.

Question 7.
The lateral faces and bases of crystals of the mineral galena are congruent squares. Identify the shape of a galena crystal.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 7
Answer: Cube

Explanation: The shape of the galena is Cube.

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

Question 8.
Rhianon draws the net below and labels each square. Can Rhianon fold her net into a cube that has letters A through G on its faces? Explain.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 8

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

Question 9.
Describe A diamond crystal is shown. Describe the figure in terms of the solid figures you have seen in this lesson.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 9

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

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

Answer:
The bases are Triangle
The lateral faces are Rectangle

Three-Dimensional Figures and Nets – Page No. 601

Identify and draw a net for the solid figure.

Question 1.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 10

Answer: Rectangular Prism

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

Question 2.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 11

Answer: Cube, Rectangular prism

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

Question 3.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 12

Answer: Square Pyramid

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

Question 4.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 13

Answer: Triangular Prism

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

Problem Solving

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

Answer: 4

Explanation: As triangled pyramids have four faces.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume

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

Answer: 36

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

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

Answer:

Explanation:

Lesson Check – Page No. 602

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

Answer: 5

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

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

Answer: Triangular Prism

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

Spiral Review

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

Answer: 3.2 miles

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

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

Answer: Y= 240- 30X.

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

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

Answer: 351 mm2

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

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

Answer: Parallelogram

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

Share and Show – Page No. 605

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

Question 1.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 14

Answer:

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

Find the surface area of the rectangular prism.

Question 2.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 15

Answer: 222 cm2

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

Question 3.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 16

Answer:

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

Question 4.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 17

Answer: 350 cm2

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

Problem Solving + Applications

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

Answer: 1700 cm2

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

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

Answer:

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

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

Answer: 416 cm2

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

What’s the Error? – Page No. 606

Question 8.
Emilio is designing the packaging for a new MP3 player. The box for the MP3 player is 5 cm by 3 cm by 2 cm. Emilio needs to find the surface area of the box.
Look at how Emilio solved the problem. Find his error.
STEP 1: Draw a net.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 18
STEP 2: Find the areas of all the faces and add them.
Face A: 3 × 2 = 6 cm2.
Face B: 3 × 5 = 15 cm2.
Face C: 3 × 2 = 6 cm2.
Face D: 3 × 5 = 15 cm2.
Face E: 3 × 5 = 15 cm2.
Face F: 3 × 5 = 15 cm2.
The surface area is 6 + 15 + 6 + 15 + 15 + 15 = 72 cm2.
Correct the error. Find the surface area of the prism.

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

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

Question 9.
For numbers 9a–9d, select True or False for each statement.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 19
9a. The area of face A is 10 cm2.
9b. The area of face B is 10 cm2.
9c. The area of face C is 40 cm2.
9d. The surface area of the prism is 66 cm2.

9a. The area of face A is 10 cm2.

Answer: True

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

9b. The area of face B is 10 cm2.

Answer: False

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

9c. The area of face C is 40 cm2.

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

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

Answer: 160 cm2.

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

Explore Surface Area Using Nets – Page No. 607

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

Question 1.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 20
_______ square units

Answer: 52 square units.

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

Question 2.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 21
_______ square units

Answer: 112 square units.

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

Question 3.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 22

Answer: 102 mm2

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

Question 4.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 23
_______ in.2

Answer: 58 in.2

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

Question 5.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 24
_______ ft2

Answer: 77 ft2

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

Problem Solving

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

Answer: 248 in.2

Explanation: Surface area of a cereal box is 2(wl+hl+hw)
= 2(2×6+14×6+14×2)
= 2(12+84+28)
= 2(124)
= 248 in.2
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume

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

Answer: 97 in.2

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

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

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

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

Lesson Check – Page No. 608

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

Answer: 412 cm2.

Explanation: Surface area is 2(wl+hl+hw)
= 2(10×7+8×7+8×10)
= 2(70+56+80)
= 2(206)
= 412 cm2.

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

Answer: 62 in.2

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

Spiral Review

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

Answer: X<477.

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

Question 4.
The table shows the number of mixed CDs y that Jason makes in x hours.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 25
Which equation describes the pattern in the table?

Answer: y= 5x

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

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

Answer: 144 in.2

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

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

Answer: 4

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

Share and Show – Page No. 611

Use a net to find the surface area.

Question 1.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 26
_______ ft2

Answer: 24 ft2

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

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

Question 2.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 27

Answer: 432 cm2

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

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

Question 3.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 28
_______ in.2

Answer: 155.5 in.2

Explanation:

Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume
The area of face A and E is  8 ½ × 3½
= 17/2 × 7/2
= 119/4
= 29.75 in.2
The area of face B and F is 8 ½×4
= 17 ½ × 4
= 34 in.2
The area of face C and D is 3 ½×4
7/2 × 4= 14 in.2
The surface area is 2×29.75+2×34+2×14
= 59.5+68+28
= 155.5 in.2

On Your Own

Use a net to find the surface area.

Question 4.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 29
_______ m2

Answer:

Explanation:
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume
The area of face A and E is 8×3= 24 m2
The area of face B and F is 8×5= 40 m2
The area of face C and D is 3×5= 15 m2
The surface area is 2×24+2×40+2×15
= 48+80+30
= 158 m2

Question 5.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 30
_______ \(\frac{□}{□}\) in.2

Answer:

Explanation:

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

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

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

Answer: 337 1/2 in.2

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

Unlock the Problem – Page No. 612

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

Answer:

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

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

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

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

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

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

Answer: 1,54,764 m2

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

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

Answer: 16,124 m2

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

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

Answer: 1,38,640 m2

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

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

Answer: 808 in.2

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

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

Answer: 268 in.2

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

Surface Area of Prisms – Page No. 613

Use a net to find the surface area.

Question 1.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 31
_______ cm2

Answer: 104 cm2

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

Question 2.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 32
_______ in.2

Answer: 118 in.2

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

Question 3.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 33
_______ ft2

Answer: 486 ft2

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

Question 4.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 34
_______ cm2

Answer: 336 cm2.

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

Problem Solving

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

Answer: 408 in.2

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

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

Answer: 150 in.2

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

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

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

Lesson Check – Page No. 614

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

Answer: 54 in.2

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

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

Answer: 304 in.2

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

Spiral Review

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

Answer: Y= 4X.

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

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

Answer: 28.5 in.2

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

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

Answer: 1296:1.

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

Question 6.
What is the surface area of the prism shown by the net?
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 35
Answer: 72 square units.

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

Share and Show – Page No. 617

Question 1.
Use a net to find the surface area of the square pyramid.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 36
_______ cm2

Answer: 105 cm2

Explanation:

Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume
Area of the base 5×5= 25 ,
and area of one face is 1/2 × 5 × 8
= 5× 4
= 20 cm2
The surface area of a pyramid is 25+ 4×20
= 25+80
= 105 cm2

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

Answer: 172 cm2

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

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

Answer: 12 ft2

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

On Your Own

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

Question 4.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 37
_______ ft2

Answer: 208 ft2

Explanation:
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume
The area of the base is 8×8= 64
The area of one face is 1/2 ×8×9
= 36 ft2
The surface area of the pyramid is 64+4×36
= 64+144
= 208 ft2

Question 5.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 38
_______ cm2

Answer: 220 cm2

Explanation:

Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume
The area of base is 10×10= 100
The area of one place is 1/2×10×6
= 10×3
= 30
The surface area of the pyramid is 100+4×30
= 100+120
= 220 cm2

Question 6.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 39
_______ in.2

Answer: 264 in.2

Explanation:

Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume
The area of the base is 8×8= 64
The area of one face is 1/2×8×12.5
= 4×12.5
= 50 in.2
The surface area of the pyramid is 64+ 4×50
= 64+200
= 264 in.2

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

Answer: 5,28,000 ft2

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

Problem Solving + Applications – Page No. 618

Use the table for 8–9.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 40

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

Answer: 82,800 m2

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

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

Answer: 93,338 m2

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

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

Question 10.
Write an expression for the surface area of the square pyramid shown.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 41

Answer: 6x+9 ft2.

Explanation: The expression for the surface area of the square pyramid is 6x+9 ft2.

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

Answer: 72 cm2.

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

Question 12.
Jose says the lateral area of the square pyramid is 260 in.2. Do you agree or disagree with Jose? Use numbers and words to support your answer.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 42

Answer: 160 in.2

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

Surface Area of Pyramids – Page No. 619

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

Question 1.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 43
_______ mm2

Answer: 95 mm2

Explanation:

Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume
The area of the base is 5×5= 25 mm2
The area of one face is 1/2×5×7
= 35/2
= 17.5 mm2
The surface area is 25+4×17.5
= 25+4×17.5
= 25+70
= 95 mm2

Question 2.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 44
_______ cm2

Answer: 612 cm2

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

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

Question 3.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 45
_______ yd2

Answer: 51.25 yd2

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

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

Question 4.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 46
_______ in.2

Answer: 180 in2

Explanation:

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

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

Problem Solving

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

Answer: 19 ft2

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

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

Answer: 2400 m2

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

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

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

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

Lesson Check – Page No. 620

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

Answer: 312 in.2

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

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

Answer: 82.5 cm2

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

Spiral Review

Question 3.
What is the linear equation represented by the graph?
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 47

Answer: y=x+1.

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

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

Answer:

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

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

Answer: It is a Trapezoid.

Explanation: It is a Trapezoid.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume

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

Answer: 158 ft2

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

Mid-Chapter Checkpoint – Vocabulary – Page No. 621

Choose the best term from the box to complete the sentence.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 48

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

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

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

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

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

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

Concepts and Skills

Question 4.
Identify and draw a net for the solid figure.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 49

Answer: Triangular prism

Explanation:

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

Question 5.
Use a net to find the lateral area of the square pyramid.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 50
_______ in.2

Answer: 216 in.2

Explanation:

Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume
The area of one face is 1/2×9×12
= 9×6
= 54 in.2
The lateral area of the square pyramid is 4×54= 216 in.2

Question 6.
Use a net to find the surface area of the prism.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 51
_______ cm2

Answer: 310 cm2

Explanation:

Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume
The area of face A and E is 10×5= 50 cm2
The area of face B and F is 10×7= 70 cm2
The area of face C and D is 7×5= 35 cm2
The surface area of the prism is 2×50+2×70+2×35
= 100+140+70
= 310 cm2

Page No. 622

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

Answer: 4 triangles.

Explanation: There are 4 triangles.

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

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

Answer: 56 ft2

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

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

Answer: 96.6 m2

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

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

Answer: 900 cm2

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

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

Answer: 1,536 in.2

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

Share and Show – Page No. 625

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

Answer: 4.75 cubic units

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

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

Answer: 7.25 cubic units.

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

Find the volume of the rectangular prism.

Question 3.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 52
_______ cubic units

Answer: 33 cubic units.

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

Question 4.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 53
_______ \(\frac{□}{□}\) cubic units

Answer: 91 1/8 cubic units.

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

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

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

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

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

Answer: 1/8 cubic units

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

Problem Solving + Applications – Page No. 626

Use the diagram for 7–10.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 54

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

Answer: 1/2 cubic units

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

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

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

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

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

Answer: 6 1/2 cubic units

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

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

Answer: 3 cubic units.

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

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

Answer: Yes

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

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

Answer:
12a True.
12b True.

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

Fractions and Volume – Page No. 627

Find the volume of the rectangular prism.

Question 1.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 55
_______ \(\frac{□}{□}\) cubic units

Answer: 6 3/4 cubic units

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

Question 2.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 56
_______ \(\frac{□}{□}\) cubic units

Answer: 22 1/2 cubic units

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

Question 3.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 57
_______ \(\frac{□}{□}\) cubic units

Answer: 16 1/2 cubic units.

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

Question 4.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 58
_______ \(\frac{□}{□}\) cubic units

Answer: 28 1/8 cubic units.

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

Problem Solving

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

Answer: 31 1/2 cubic units

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

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

Answer: 49 1/2 ft3

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

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

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

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

Lesson Check – Page No. 628

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

Answer: 120 cubes

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

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

Answer: 24 1/2 cubic units.

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

Spiral Review

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

Answer: 3 1/8 in.2

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

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

Answer: 604 in.2

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

Question 5.
What is the surface area of the rectangular prism shown by the net?
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 59
_______ square units

Answer: 80 square units

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

Question 6.
What is the surface area of the square pyramid?
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 60
_______ cm2

Answer: 161 cm2

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

Share and Show – Page No. 631

Find the volume.

Question 1.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 61
_______ \(\frac{□}{□}\) in.3

Answer: 3,937 1/2 in.3

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

Question 2.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 62
_______ \(\frac{□}{□}\) in.3

Answer: 27/512 in.3

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

On Your Own

Find the volume of the prism.

Question 3.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 63
_______ \(\frac{□}{□}\) in.3

Answer: 690 5/8in.3

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

Question 4.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 64
_______ \(\frac{□}{□}\) in.3

Answer: 125/4096 in.3

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

Question 5.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 65
_______ yd3

Answer: 20 yd3

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

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

Answer: 4,730 5/8 in.3

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

Question 7.
Abraham has a toy box that is in the shape of a rectangular prism.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 66
The volume is _____.
_______ \(\frac{□}{□}\) ft3

Answer: 33 3/4 ft3

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

Aquariums – Page No. 632

Large public aquariums like the Tennessee Aquarium in Chattanooga have a wide variety of freshwater and saltwater fish species from around the world. The fish are kept in tanks of various sizes.
The table shows information about several tanks in the aquarium. Each tank is a rectangular prism.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 67
Find the length of Tank 1.
V = l w h
52,500 = l × 30 × 35
\(\frac{52,500}{1,050}\) = l
50 = l
So, the length of Tank 1 is 50 cm.

Solve.

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

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

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

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

Answer: 35 Fishes

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

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

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

Volume of Rectangular Prisms – Page No. 633

Find the volume.

Question 1.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 68
_______ \(\frac{□}{□}\) m3

Answer: 150 5/16 m3

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

Question 2.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 69
_______ \(\frac{□}{□}\) in.3

Answer: 27 1/2 in.3

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

Question 3.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 70
_______ \(\frac{□}{□}\) mm3

Answer: 91 1/8 mm3

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

Question 4.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 71
_______ \(\frac{□}{□}\) ft3

Answer: 112 1/2 ft3

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

Question 5.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 72
_______ m3

Answer: 36 m3

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

Question 6.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 73
_______ \(\frac{□}{□}\) ft3

Answer: 30 3/8 ft3

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

Problem Solving

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

Answer: H= 10 in

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

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

Answer: 374 in.3

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

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

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

Lesson Check – Page No. 634

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

Answer: 3,000 in.3

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

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

Answer: H= 24 cm

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

Spiral Review

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

Answer: 9,600 yd2

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

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

Answer: 80,000 in2

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

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

Answer: 184 in2

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

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

Answer: 27.5 cubic units.

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

Share and Show – Page No. 637

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

Answer: So tank can hold 37,352 cm3

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

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

Answer: 4,872 cm3

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

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

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

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

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

Answer: 1946 m2

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

On Your Own – Page No. 638

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

Answer: 6 1/4 m3

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

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

Answer: $8.50

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

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

Answer:

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

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

Answer: 12 cube-shaped boxes

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

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

Answer: 1,900 cm2

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

Problem Solving Geometric Measurements – Page No. 639

Read each problem and solve.

Question 1.
The outside of an aquarium tank is 50 cm long, 50 cm wide, and 30 cm high. It is open at the top. The glass used to make the tank is 1 cm thick. How much water can the tank hold?
_______ cm3

Answer: So water tank can hold 66,816 cm3

Explanation: The volume of inner dimensions of the aquarium is 50-2 × 50-2 × 30-1
= 48×48×29
= 66,816 cm3
So water tank can hold 66,816 cm3

Question 2.
Arnie keeps his pet snake in an open-topped glass cage. The outside of the cage is 73 cm long, 60 cm wide, and 38 cm high. The glass used to make the cage is 0.5 cm thick. What is the inside volume of the cage?
_______ cm3

Answer: The volume of the cage is 1,59,300 cm3

Explanation: The volume of inner dimensions is 73-1 × 60-1 × 38-0.5
= 72×59×37.5
= 1,59,300 cm3
So, the volume of the cage is 1,59,300 cm3

Question 3.
A display number cube measures 20 in. on a side. The sides are numbered 1–6. The odd-numbered sides are covered in blue fabric and the even-numbered sides are covered in red fabric. How much red fabric was used?
_______ in.2

Answer: 1200 in.2

Explanation: The area of each side of a cube is 20×20= 400 in.2, as there are 3 even-numbered sides on the cube. So there will be
3×400= 1200 in.2

Question 4.
The caps on the tops of staircase posts are shaped like square pyramids. The side length of the base of each cap is 4 inches. The height of the face of each cap is 5 inches. What is the surface area of the caps for two posts?
_______ in.2

Answer: 112 in.2

Explanation: The area of the base is 4×4= 16 in.2
The area of one face is 1/2×5×4= 10 in.2
The surface area of one cap is 16+4×10
= 16+40
= 56 in.2
And the surface area of the caps for two posts is 2×56= 112 in.2

Question 5.
A water irrigation tank is shaped like a cube and has a side length of 2 \(\frac{1}{2}\) feet. How many cubic feet of water are needed to completely fill the tank?
_______ \(\frac{□}{□}\) ft3

Answer: 15 5/8 ft3

Explanation: Volume= LWH
= 2 1/2 × 2 1/2 × 2 1/2
= 5/2 × 5/2 × 5/2
= 125/8
= 15 5/8 ft3

Question 6.
Write and solve a problem for which you use part of the formula for the surface area of a triangular prism.

Answer: In a triangular prism, the triangular end has a base of 5cm and the height is 8 cm. The length of each side is 4 cm and the height of the prism is 10 cm. What is the lateral area of this triangular prism?

Explanation: The area of two triangular faces is 1/2 × 5 × 8
= 5×4
= 20 cm2
The area of two rectangular faces is 4×10= 40 cm2
The lateral area is 2×20+2×40
= 40+80
= 120 cm2

Lesson Check – Page No. 640

Question 1.
Maria wants to know how much wax she will need to fill a candle mold shaped like a rectangular prism. What measure should she find?

Answer: Maria needs to find the volume of the mold.

Question 2.
The outside of a closed glass display case measures 22 inches by 15 inches by \(\frac{1}{2}\) inches. The glass is 12 inch thick. How much air is contained in the case?
_______ in.3

Answer: 3381 in.3

Explanation: The inner dimensions are 22-1× 15-1 × 12- 1/2
= 21 ×14×23/2
= 3381 in.3

Spiral Review

Question 3.
A trapezoid with bases that measure 5 centimeters and 7 centimeters has a height of 4.5 centimeters. What is the area of the trapezoid?
_______ cm2

Answer: 27 cm2

Explanation: Area of trapezoid= 1/2 ×(7+5)×4.5
= 6×4.5
= 27 cm2

Question 4.
Sierra has plotted two vertices of a rectangle at (3, 2) and (8, 2). What is the length of the side of the rectangle?
_______ units

Answer: 5 units.

Explanation: The length of the side of the rectangle is 8-3= 5 units.

Question 5.
What is the surface area of the square pyramid?
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 74
_______ m2

Answer: 104 m2

Explanation: The area of the base 4×4= 16
The area of the one face is 1/2 × 4 × 11
= 2×11
= 22 m2
The surface area of the square pyramid is 16+4×22
= 16+88
= 104 m2

Question 6.
A shipping company has a rule that all packages must be rectangular prisms with a volume of no more than 9 cubic feet. What is the maximum measure for the height of a box that has a width of 1.5 feet and a length of 3 feet?
_______ feet

Answer: 2 feet.

Explanation: As given volume = 9 cubic feet
So 1.5×3×H < 9
4.5×H < 9
H< 9/4.5
and H<2
So maximum measure for the height of the box is 2 feet.

Chapter 11 Review/Test – Page No. 641

Question 1.
Elaine makes a rectangular pyramid from paper.
The base is a _____. The lateral faces are _____.
The base is a ___________ .
The lateral faces are ___________ .

Answer:
The base is a rectangle.
The lateral faces are triangles.

Question 2.
Darrell paints all sides except the bottom of the box shown below.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 75
Select the expressions that show how to find the surface area that Darrell painted. Mark all that apply.
Options:
a. 240 + 240 + 180 + 180 + 300 + 300
b. 2(20 × 12) + 2(15 × 12) + (20 × 15)
c. (20 × 12) + (20 × 12) + (15 × 12) + (15 × 12) + (20 × 15)
d. 20 × 15 × 12

Answer: b,c

Explanation: The expressions that show how to find the surface area is 2(20 × 12) + 2(15 × 12) + (20 × 15), (20 × 12) + (20 × 12) + (15 × 12) + (15 × 12) + (20 × 15)

Question 3.
A prism is filled with 44 cubes with \(\frac{1}{2}\)-unit side lengths. What is the volume of the prism in cubic units?
_______ \(\frac{□}{□}\) cubic unit

Answer:

Explanation:
The volume of a cube with S= (1/2)3
= 1/2×1/2×1/2
= 1/8
= 0.125 cubic units
As there are 44 cubes so 44×0.125=5.5 cubic units.

Question 4.
A triangular pyramid has a base with an area of 11.3 square meters, and lateral faces with bases of 5.1 meters and heights of 9 meters. Write an expression that can be used to find the surface area of the triangular pyramid.

Answer: 11.3+ 3 × 1/2+ 5.1×9

Explanation: The expression that can be used to find the surface area of the triangular pyramid is 11.3+ 3 × 1/2+ 5.1×9

Page No. 642

Question 5.
Jeremy makes a paperweight for his mother in the shape of a square pyramid. The base of the pyramid has a side length of 4 centimeters, and the lateral faces have heights of 5 centimeters. After he finishes, he realizes that the paperweight is too small and decides to make another one. To make the second pyramid, he doubles the length of the base in the first pyramid.
For numbers 5a–5c, choose Yes or No to indicate whether the statement is correct.
5a. The surface area of the second pyramid is 144 cm2.
5b. The surface area doubled from the first pyramid to the second pyramid.
5c. The lateral area doubled from the first pyramid to the second pyramid.
5a. ___________
5b. ___________
5c. ___________

Answer:
5a. True.
5b. False
5c. True.

Explanation:
The area of the base is 4×4= 16 cm2.
The area of one face is 1/2×4×5
= 2×5
= 10 cm2.
The surface area of the First pyramid is 16+ 4×10
= 16+40
= 56 cm2.
The area of the base is 8×8= 64
The area of one face is 1/2×8×5
= 4×5
= 20 cm2.
The surface area od the second pyramid is 64+ 4×20
= 64+80
= 144 cm2.

Question 6.
Identify the figure shown and find its surface area. Explain how you found your answer.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 76

Answer: 369 in2

Explanation:
The area of the base is 9×9= 81 in2
The area of one face is 1//2 × 16× 9
= 8×9
= 72 in2
The surface area of a square pyramid is 81+ 4× 72
= 81+ 288
= 369 in2

Question 7.
Dominique has a box of sewing buttons that is in the shape of a rectangular prism.
The volume of the box is 2 \(\frac{1}{2}\) in. × 3 \(\frac{1}{2}\) in. × _____ = _____.

Answer: 17.5 in3

Explanation: The volume of the box is 2 1/2 × 3 1/2 × 2
= 5/2 × 7/2 × 2
= 5/2 × 7
= 35/2
= 17.5 in3

Page No. 643

Question 8.
Emily has a decorative box that is shaped like a cube with a height of 5 inches. What is the surface area of the box?
_______ in.2

Answer: 150 in.2

Explanation: Surface area of the box is 6 a2
So 6 × 52
= 6×5×52
= 150 in.2

Question 9.
Albert recently purchased a fish tank for his home. Match each question with the geometric measure that would be most appropriate for each scenario.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 77

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

Question 10.
Select the expressions that show the volume of the rectangular prism. Mark all that apply.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 78Options:
a. 2(2 units × 2 \(\frac{1}{2 }\) units) + 2(2 units × \(\frac{1}{2}\) unit) + 2(\(\frac{1}{2}\) unit × 2 \(\frac{1}{2}\) units)
b. 2(2 units × \(\frac{1}{2}\) unit) + 4(2 units × 2 \(\frac{1}{2}\) units)
c. 2 units × \(\frac{1}{2}\) unit × 2 \(\frac{1}{2}\) units
d. 2.5 cubic units

Answer: c, d

Explanation: 2 units ×1/2 unit × 2 1/2 units and 2.5 cubic units

Page No. 644

Question 11.
For numbers 11a–11d, select True or False for the statement.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 79
11a. The area of face A is 8 square units.
11b. The area of face B is 10 square units.
11c. The area of face C is 8 square units.
11d. The surface area of the prism is 56 square units.
11a. ___________
11b. ___________
11c. ___________
11d. ___________

Answer:
11a. True.
11b. True.
11c. False.
11d. False.

Explanation:
The area of the face A is 4×2= 8 square units
The area of the face B is 5×2= 10 square units
The area of the face C is 5×4= 20 square units
So the surface area is 2×8+2×10+2×20
= 16+20+40
= 76 square units

Question 12.
Stella received a package in the shape of a rectangular prism. The box has a length of 2 \(\frac{1}{2}\) feet, a width of 1 \(\frac{1}{2}\) feet, and a height of 4 feet.
Part A
Stella wants to cover the box with wrapping paper. How much paper will she need? Explain how you found your answer

Answer: 39.5 ft2

Explanation:
The area of two lateral faces is 4 × 2 1/2
= 4 × 5/2
= 2×5
= 10 ft2
The area of another two lateral faces is 4 × 1 1/2
= 4 × 3/2
= 2×3
= 6 ft2
The area of the top and bottom is 2 1/2 × 1 1/2
= 5/2 × 3/2
= 15/4
= 3 3/4 ft2
So Stella need 2×10+ 2×6 + 2 × 15/4
= 20+ 12+15/2
= 20+12+7.5
= 39.5 ft2

Question 12.
Part B
Can the box hold 16 cubic feet of packing peanuts? Explain how you know

Answer: The box cannot hold 16 cubic feet of the packing peanuts

Explanation: Volume = LWH
= 2 1/2 ×1 1/2 × 4
= 5/2 × 3/2 ×4
= 5×3
= 15 ft3
So the box cannot hold 16 cubic feet of the packing peanuts.

Page No. 645

Question 13.
A box measures 6 units by \(\frac{1}{2}\) unit by 2 \(\frac{1}{2}\) units.
For numbers 13a–13b, select True or False for the statement.
13a. The greatest number of cubes with a side length of \(\frac{1}{2}\) unit that can be packed inside the box is 60.
13b. The volume of the box is 7 \(\frac{1}{2}\) cubic units.
13a. ___________
13b. ___________

Answer:
13a. True
13b. True.

Explanation:
Length is 12 × 1/2= 6 units
Width is 1× 1/2= 1/2 units
Height is 5× 1/2= 5/2 units
So, the greatest number of cubes with a side length of 1/2 unit that can be packed inside the box is 12×1×5= 60
The volume of the cube is S3
The volume of a cube with S= (1/2)3
= 1/2×1/2×1/2
= 1/8
= 0.125 cubic units
As there are 60 cubes so 60×0.125= 7.5cubic units.

Question 14.
Bella says the lateral area of the square pyramid is 1,224 in.2. Do you agree or disagree with Bella? Use numbers and words to support your answer. If you disagree with Bella, find the correct answer.

Answer: 900 in2

Explanation:
Area= 4× 1/2 bh
= 4× 1/2 × 18 × 25
= 2× 18 × 25
=  900 in2
So lateral area is 900 in2, so I disagree

Question 15.
Lourdes is decorating a toy box for her sister. She will use self-adhesive paper to cover all of the exterior sides except for the bottom of the box. The toy box is 4 feet long, 3 feet wide, and 2 feet high. How many square feet of adhesive paper will Lourdes use to cover the box?
_______ ft2

Answer: 40 ft2

Explanation:
The area of two lateral faces is 4×2= 8 ft2
The area of another two lateral faces is 3×2= 6 ft2
The area of the top and bottom is 4×3= 12 ft2
So Lourdes uses to cover the box is 2×8 + 2×6 + 12
= 16+12+12
= 40 ft2

Question 16.
Gary wants to build a shed shaped like a rectangular prism in his backyard. He goes to the store and looks at several different options. The table shows the dimensions and volumes of four different sheds. Use the formula V = l × w × h to complete the table.
Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 80

Answer:
Length of shed 1= 12 ft
Width of shed 2= 12 ft
Height of shed 3= 6 ft
Volume of shed 4= 1200 ft3

Explanation: Volume= LWH
Volume of shed1= 960 ft
So 960= L×10×8
960= 80×L
L= 960/80
L= 12 ft
Volume of shed2= 2160 ft
So 2160= 18×W×10
960= 180×W
W= 2160/180
W= 12 ft
Volume of shed3= 288 ft
So 288= 12×4×H
288= 48×H
H= 288/48
W= 6 ft
Volume of shed2= 10×12×10
So V= 10×12×10
V= 1200 ft3

Page No. 646

Question 17.
Tina cut open a cube-shaped microwave box to see the net. How many square faces does this box have?
_______ square faces

Answer: The box has 6 square faces.

Question 18.
Charles is painting a treasure box in the shape of a rectangular prism.
Which nets can be used to represent Charles’ treasure box? Mark all that apply.
Options:
a. Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 81
b. Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 82
c. Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 83
d. Go Math Grade 6 Answer Key Chapter 11 Surface Area and Volume img 84

Answer: a and b can be used to represent Charle’s treasure box.

Question 19.
Julianna is lining the inside of a basket with fabric. The basket is in the shape of a rectangular prism that is 29 cm long, 19 cm wide, and 10 cm high. How much fabric is needed to line the inside of the basket if the basket does not have a top? Explain your strategy.
_______ cm2

Answer: 1511 cm2

Explanation: The surface area= 2(WL+HL+HW)
The surface area of the entire basket= 2(19×29)+2(10×29)+2(10×19)
= 2(551)+2(290)+2(190)
= 1102+580+380
= 2,062 cm2
The surface area of the top is 29×19= 551
So Julianna needs 2062-551= 1511 cm2

Conclusion

Click on the related links and begin your preparation. Make your preparation perfect by practicing the problems a number of times. We wish the info provided in the Go Math 6th Standard Answer Key Chapter 11 Surface Area and Volume is satisfactory for all of you. Keep in touch with us to get the latest updates regarding the HMH Go Math Grade 6 Answer Key.

Go Math Grade 6 Answer Key Chapter 5 Model Percents

go-math-grade-6-chapter-5-model-percents-answer-key

Go Math Grade 6 Answer Key Chapter 5 Model Percents Pdf is available here. So, the pupils who are in search of the solutions of Chapter 5 Model Percents can get them on this page along with images. Relate the questions in real-time and make your practice best. Students who are preparing for exams must have the best material. Our team will provide step by step explanations for all the questions on Go Math Grade 6 Answer Key.

Go Math Grade 6 Chapter 5 Model Percents Answer Key

Make yourself comfortable by using HMH Go math Grade 6 Answer Key Chapter 5 Model Percents. So, make use of the resources of Go Math Answer Key to score good marks in the exams. Test your skills by solving the problems given at the end of the chapter. Just click on the links and start solving the problems.

Lesson 1: Investigate • Model Percents

Lesson 2: Write Percents as Fractions and Decimals

Lesson 3: Write Fractions and Decimals as Percents

Mid-Chapter Checkpoint

Lesson 4: Percent of a Quantity

Lesson 5: Problem Solving • Percents

Lesson 6: Find the Whole from a Percent

Chapter 5 Review/Test

Share and Show – Page No. 271

Write a ratio and a percent to represent the shaded part.

Question 1.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 1
Type below:
_____________

Answer:
53% and \(\frac{53}{100}\)

Explanation:
53 squares are shaded out of 100.
So, 53% and 35/100 are the answers.

Question 2.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 2
Type below:
_____________

Answer:
1% and \(\frac{100}{100}\)

Explanation:
100 out of 100 squares are shaded
So, So, 1% and 100/100 are the answers.

Question 3.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 3
Type below:
_____________

Answer:
40% and \(\frac{40}{100}\)

Explanation:
40 squares are shaded out of 100.
So, 40% and 40/100 are the answers.

Model the percent and write it as a ratio.

Question 4.
30%
\(\frac{□}{□}\)

Answer:
grade 6 chapter 5 image 1

Explanation:
30% is 30 out of 100
30 out of 100 squares is 30/100
30% = \(\frac{30}{100}\)

Question 5.
5%
\(\frac{□}{□}\)

Answer:
grade 6 chapter 5 image 2

Explanation:
5% is 5 out of 100
5 out of 100 squares is 5/100
5% = \(\frac{5}{100}\)

Question 6.
75%
\(\frac{□}{□}\)

Answer:
grade 6 chapter 5 image 3

Explanation:
75% is 75 out of 100
75 out of 100 squares is 75/100
75% = \(\frac{75}{100}\)

Problem Solving + Applications

Question 7.
Use a Concrete Model Explain how to model 32% on a 10-by-10 grid. How does the model represent the ratio of 32 to 100?
Type below:
_____________

Answer:
grade 6 chapter 5 image 4

Question 8.
A floor has 100 tiles. There are 24 black tiles and 35 brown tiles. The rest of the tiles are white. What percent of the tiles are white?
_______ %

Answer:
41%

Explanation:
A floor has 100 tiles. There are 24 black tiles and 35 brown tiles.
24 + 35 = 59
100 – 59 = 41 tiles are white
41 tiles out of 100 are white tiles

Pose a Problem – Page No. 272

Question 9.
Javier designed a mosaic wall mural using 100 tiles in 3 different colors: yellow, blue, and red. If 64 of the tiles are yellow, what percent of the tiles are either red or blue?
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 4
To find the number of tiles that are either red or blue, count the red and blue squares. Or subtract the number of yellow squares, 64, from the total number of squares, 100.
36 out of 100 tiles are red or blue.
The ratio of red or blue tiles to all tiles is \(\frac{36}{100}\).
So, the percent of the tiles that are either red or blue is 36%.
Write another problem involving a percent that can be solved by using the mosaic wall mural.
Type below:
_____________

Answer:
Sam designed a mosaic wall mural using 100 squares using two colors. She represented the squares with red and blue colors. She has 54 red tiles. What percent of other tiles she can use with blue color?
100 – 54 = 46 blue tiles.

Question 10.
Select the 10-by-10 grids that model 45%. Mark all that apply.
Options:
a. Go Math Grade 6 Answer Key Chapter 5 Model Percents img 5
b. Go Math Grade 6 Answer Key Chapter 5 Model Percents img 6
c. Go Math Grade 6 Answer Key Chapter 5 Model Percents img 7
d. Go Math Grade 6 Answer Key Chapter 5 Model Percents img 8
e. Go Math Grade 6 Answer Key Chapter 5 Model Percents img 9

Answer:
a. Go Math Grade 6 Answer Key Chapter 5 Model Percents img 5
c. Go Math Grade 6 Answer Key Chapter 5 Model Percents img 7
e. Go Math Grade 6 Answer Key Chapter 5 Model Percents img 9

Model Percents – Page No. 273

Write a ratio and a percent to represent the shaded part.

Question 1.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 10
Type below:
_____________

Answer:
31% and \(\frac{31}{100}\)

Explanation:
31 squares are shaded out of 100.
So, 31% and 31/100 are the answers.

Question 2.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 11
Type below:
_____________

Answer:
70% and \(\frac{70}{100}\)

Explanation:
70 squares are shaded out of 100.
So, 70% and 70/100 are the answers.

Question 3.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 12
Type below:
_____________

Answer:
48% and \(\frac{48}{100}\)

Explanation:
48 squares are shaded out of 100.
So, 48% and 48/100 are the answers.

Model the percent and write it as a ratio.

Question 4.
97%
\(\frac{□}{□}\)

Answer:
grade 6 chapter 5 image 5

Explanation:
97% is 97 out of 100
97 out of 100 squares is 97/100
97% = \(\frac{97}{100}\)

Question 5.
24%
\(\frac{□}{□}\)

Answer:
grade 6 chapter 5 image 6

Explanation:
24% is 24 out of 100
24 out of 100 squares is 24/100
24% = \(\frac{24}{100}\)

Question 6.
50%
\(\frac{□}{□}\)

Answer:
grade 6 chapter 5 image 7

Explanation:
50% is 50 out of 100
50 out of 100 squares is 50/100
50% = \(\frac{50}{100}\)

Problem Solving

The table shows the pen colors sold at the school supply store one week. Write the ratio comparing the number of the given color sold to the total number of pens sold. Then shade the grid.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 13

Question 7.
Black
\(\frac{□}{□}\)

Answer:
\(\frac{49}{100}\)
grade 6 chapter 5 image 8

Explanation:
The total number of pens sold = 36 + 49 + 15 = 100
Black : total number of pens sold = 49:100
49 out of 100 squares need to shade the grid

Question 8.
Not Blue
\(\frac{□}{□}\)

Answer:
\(\frac{64}{100}\)
grade 6 chapter 5 image 9

Explanation:
Not Blue = Black + Red = 49 + 15 = 64

Question 9.
Is every percent a ratio? Is every ratio a percent? Explain.
Type below:
_____________

Answer:
Every percent is a ratio but not all ratios are percent. All ratios can be expressed as percents, decimals, or fractions or in ratio form.

Lesson Check – Page No. 274

Question 1.
What percent of the large square is shaded?
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 14
_______ %

Answer:
63%

Explanation:
63 squares are shaded out of 100.
So, 63% and 63/100 are the answers.

Question 2.
Write a ratio to represent the shaded part.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 15
\(\frac{□}{□}\)

Answer:
\(\frac{10}{100}\)

Explanation:
63 squares are shaded out of 100.
63/100 is the answer.

Spiral Review

Question 3.
Write a number that is less than −2 \(\frac{4}{5}\) and greater than −3 \(\frac{1}{5}\).
Type below:
_____________

Answer:
-2.9, -3.0, -3.1

Explanation:
−2 \(\frac{4}{5}\) = -14/5 = -2.8
−3 \(\frac{1}{5}\) = -16/5 = -3.2
-2.9, -3.0, -3.1 are the numbers less than −2 \(\frac{4}{5}\) and greater than −3 \(\frac{1}{5}\)

Question 4.
On a coordinate grid, what is the distance between (2, 4) and (2, –3)?
_______ units

Answer:
7 units

Explanation:
|-3| = 3
4+ 0 = 4; 0 + 3 = 3
4 + 3 = 7

Question 5.
Each week, Diana spends 4 hours playing soccer and 6 hours babysitting. Write a ratio to compare the time Diana spends playing soccer to the time she spends babysitting.
\(\frac{□}{□}\)

Answer:
\(\frac{2}{3}\)

Explanation:
Each week, Diana spends 4 hours playing soccer and 6 hours babysitting.
The ratio to compare the time Diana spends playing soccer to the time she spends babysitting is 4:6 or 4/6 = 2/3

Question 6.
Antwone earns money at a steady rate mowing lawns. The points (1, 25) and (5, 125) appear on a graph of the amount earned versus number of lawns mowed. What are the coordinates of the point on the graph with an x-value of 3?
Type below:
_____________

Answer:
(3, 75)

Explanation:
y2-y1/x2-x1.
Y2 is 125, Y1 is 25, X2 is 5, and X1 is 1.
You then plug the numbers in, 125-25=100. 5-1=4.
Then you divide 100/4, in which you get 25. So you time 25 by 3, getting 75.

Share and Show – Page No. 277

Write the percent as a fraction.

Question 1.
80%
\(\frac{□}{□}\)

Answer:
\(\frac{80}{100}\)

Explanation:
80% is 80 out of 100
80 out of 100 squares is 80/100

Question 2.
150%
______ \(\frac{□}{□}\)

Answer:
1\(\frac{1}{2}\)

Explanation:
150% is 150 out of 100
150 out of 100 squares is 150/100 = 3/2 = 1 1/2

Question 3.
0.2%
\(\frac{□}{□}\)

Answer:
\(\frac{2}{1,000}\)

Explanation:
0.2% is 0.2 out of 100
0.2 out of 100 squares is 0.2/100 = 2/1,000

Write the percent as a decimal.

Question 4.
58%
______

Answer:
0.58

Explanation:
58% is 58 out of 100
58 out of 100 squares is 58/100
58/100 = 0.58

Question 5.
9%
______

Answer:
0.09

Explanation:
9% is 9 out of 100
9 out of 100 squares is 9/100
9/100 = 0.09

On Your Own

Write the percent as a fraction or mixed number.

Question 6.
17%
\(\frac{□}{□}\)

Answer:
\(\frac{17}{100}\)

Explanation:
17% is 17 out of 100
17 out of 100 squares is 17/100

Question 7.
20%
\(\frac{□}{□}\)

Answer:
\(\frac{1}{5}\)

Explanation:
20% is 20 out of 100
20 out of 100 squares is 20/100 = 2/10 = 1/5

Question 8.
125%
______ \(\frac{□}{□}\)

Answer:
1\(\frac{1{4}\)

Explanation:
125% is 125 out of 100
125 out of 100 squares is 125/100 = 1 1/4

Question 9.
355%
______ \(\frac{□}{□}\)

Answer:
3\(\frac{11}{20}\)

Explanation:
355% is 355 out of 100
355 out of 100 squares is 355/100 = 3 11/20

Question 10.
0.1%
\(\frac{□}{□}\)

Answer:
\(\frac{1}{1,000}\)

Explanation:
0.1% is 0.1 out of 100
0.1 out of 100 squares is 0.1/100 = 1/1,000

Question 11.
2.5%
\(\frac{□}{□}\)

Answer:
\(\frac{1}{40}\)

Explanation:
2.5% is 2.5 out of 100
2.5 out of 100 squares is 2.5/100 = 25/1,000 = 1/40

Write the percent as a decimal.

Question 12.
89%
______

Answer:
0.89

Explanation:
89% is 89 out of 100
89 out of 100 squares is 89/100
89/100 = 0.89

Question 13.
30%
______

Answer:
0.3

Explanation:
30% is 30 out of 100
30 out of 100 squares is 30/100
30/100 = 0.3

Question 14.
2%
______

Answer:
0.02

Explanation:
2% is 2 out of 100
2 out of 100 squares is 2/100
2/100 = 0.02

Question 15.
122%
______

Answer:
1.22

Explanation:
122% is 122 out of 100
122 out of 100 squares is 122/100
122/100 = 1.22

Question 16.
3.5%
______

Answer:
0.035

Explanation:
3.5% is 3.5 out of 100
3.5 out of 100 squares is 3.5/100
3.5/100 = 0.035

Question 17.
6.33%
______

Answer:
0.0633

Explanation:
6.33% is 6.33 out of 100
6.33 out of 100 squares is 6.33/100
6.33/100 = 0.0633

Question 18.
Use Reasoning Write <, >, or =.
21.6% ______ \(\frac{1}{5}\)

Answer:
21.6% > \(\frac{1}{5}\)

Explanation:
1/5 × 100/100 = 100/500 = 0.2/100 = 0.2%
21.6% > 0.2%

Question 19.
Georgianne completed 60% of her homework assignment. Write the portion of her homework that she still needs to complete as a fraction.
\(\frac{□}{□}\)

Answer:
\(\frac{2}{5}\)

Explanation:
Georgianne completed 60% of her homework assignment.
60/100
She needs to complete 40% of her homework = 40/100 = 2/5

Problem Solving + Applications – Page No. 278

Use the table for 20 and 21.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 16

Question 20.
What fraction of computer and video game players are 50 years old or more?
\(\frac{□}{□}\)

Answer:
\(\frac{13}{50}\)

Explanation:
computer and video game players,
50 or more are of 26% = 26/100 = 13/50

Question 21.
What fraction of computer and video game players are 18 years old or more?
\(\frac{□}{□}\)

Answer:
\(\frac{49}{100}\)

Explanation:
18 years old or more are of 49% = 49/100

Question 22.
Box A and Box B each contain black tiles and white tiles. They have the same total number of tiles. In Box A, 45% of the tiles are black. In Box B, \(\frac{11}{20}\) of the tiles are white. Compare the number of black tiles in the boxes. Explain your reasoning.
Type below:
_____________

Answer:
In Box A, 45% of the tiles are black.
In Box B, \(\frac{11}{20}\) of the tiles are white.
11/20 = 0.55 = 55/100 = 55%
100 – 55 = 45%
Both Box A and Box B have an equal number of black tiles

Question 23.
Mr. Truong is organizing a summer program for 6th grade students. He surveyed students to find the percent of students interested in each activity. Complete the table by writing each percent as a fraction or decimal.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 17
Type below:
_____________

Answer:
Sports = 48% = 48/100 = 0.48
Cooking = 23% = 23/100
Music = 20% = 20/100
Art = 9% = 9/100 = 0.09

Write Percents as Fractions and Decimals – Page No. 279

Write the percent as a fraction or mixed number.

Question 1.
44%
\(\frac{□}{□}\)

Answer:
\(\frac{11}{25}\)

Explanation:
44% is 44 out of 100
44 out of 100 squares is 44/100 = 11/25

Question 2.
32%
\(\frac{□}{□}\)

Answer:
\(\frac{8}{25}\)

Explanation:
32% is 32 out of 100
32 out of 100 squares is 32/100 = 8/25

Question 3.
116%
______ \(\frac{□}{□}\)

Answer:
1 \(\frac{4}{25}\)

Explanation:
116% is 116 out of 100
116 out of 100 squares is 116/100 = 1 4/25

Question 4.
250%
______ \(\frac{□}{□}\)

Answer:
2\(\frac{1}{2}\)

Explanation:
250% is 250 out of 100
250 out of 100 squares is 250/100 = 2 1/2

Question 5.
0.3%
\(\frac{□}{□}\)

Answer:
\(\frac{3}{1,000}\)

Explanation:
0.3% is 0.3 out of 100
0.3 out of 100 squares is 0.3/100
3/1,000

Question 6.
0.4%
\(\frac{□}{□}\)

Answer:
\(\frac{1}{250}\)

Explanation:
0.4% is 0.4 out of 100
0.4 out of 100 squares is 0.4/100 = 4/1,000 = 1/250

Question 7.
1.5%
\(\frac{□}{□}\)

Answer:
\(\frac{3}{200}\)

Explanation:
1.5% is 1.5 out of 100
1.5 out of 100 squares is 1.5/100 = 15/1,000 = 3/200

Question 8.
12.5%
\(\frac{□}{□}\)

Answer:
\(\frac{1}{8}\)

Explanation:
12.5% is 12.5 out of 100
12.5 out of 100 squares is 12.5/100 = 125/1,000 = 25/200 = 5/40 = 1/8

Write the percent as a decimal.

Question 9.
63%
______

Answer:
0.63

Explanation:
63% is 63 out of 100
63 out of 100 squares is 63/100
63/100 = 0.63

Question 10.
110%
______

Answer:
1.1

Explanation:
110% is 110 out of 100
110 out of 100 squares is 110/100 = 1.1

Question 11.
42.15%
______

Answer:
0.4215

Explanation:
42.15% is 42.15 out of 100
42.15 out of 100 squares is 42.15/100 = 0.4215

Question 12.
0.1%
______

Answer:
0.001

Explanation:
0.1% is 0.1 out of 100
0.1 out of 100 squares is 0.1/100  = 0.001

Problem Solving

Question 13.
An online bookstore sells 0.8% of its books to foreign customers. What fraction of the books are sold to foreign customers?
\(\frac{□}{□}\)

Answer:
\(\frac{1}{125}\)

Explanation:
An online bookstore sells 0.8% of its books to foreign customers.
0.8% = 0.8/100 = 8/1,000 = 1/125

Question 14.
In Mr. Klein’s class, 40% of the students are boys. What decimal represents the portion of the students that are girls?
______

Answer:
0.4

Explanation:
In Mr. Klein’s class, 40% of the students are boys.
40/100 = 0.4

Question 15.
Explain how percents, fractions, and decimals are related. Use a 10-by-10 grid to make a model that supports your explanation.
Type below:
_____________

Answer:
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 1
53 squares are shaded out of 100.
53% or \(\frac{53}{100}\) or 0.53

Lesson Check – Page No. 280

Question 1.
The enrollment at Sonya’s school this year is 109% of last year’s enrollment. What decimal represents this year’s enrollment compared to last year’s?
______

Answer:
1.09 represents this year’s enrollment compared to last year’s

Explanation:
The enrollment at Sonya’s school this year is 109% of last year’s enrollment.
109% = 109/100 = 1.09

Question 2.
An artist’s paint set contains 30% watercolors and 25% acrylics. What fraction represents the portion of the paints that are watercolors or acrylics? Write the fraction in simplest form.
\(\frac{□}{□}\)

Answer:
\(\frac{11}{20}\)

Explanation:
An artist’s paint set contains 30% watercolors and 25% acrylics.
30 + 25 = 55% = 55/100 = 11/20

Spiral Review

Question 3.
Write the numbers in order from least to greatest.
-5.25 1.002 -5.09
Type below:
_____________

Answer:
-5.25, -5.09, 1.002

Question 4.
On a coordinate plane, the vertices of a rectangle are (2, 4), (2, −1), (−5, −1), and ( −5, 4). What is the perimeter of the rectangle?
______ units

Answer:
24 units

Explanation:
(2, 4) to (2, −1) is 4 + 1 = 5
(2, −1) to (−5, −1) is 2 + 5 = 7
5 + 7 + 5 + 7 = 24

Question 5.
The table below shows the widths and lengths, in feet, for different playgrounds. Which playgrounds have equivalent ratios of width to length?
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 18
Type below:
_____________

Answer:
12/20 and 16.5/27.5 are equal

Explanation:
12/20 = 0.6
15/22.5 = 0.666
20/25 = 0.8
16.5/27.5 = 0.6

Question 6.
What percent represents the shaded part?
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 19
_______ %

Answer:
85%

Explanation:
85 squares are shaded out of 100.
85%

Share and Show – Page No. 283

Write the fraction or decimal as a percent.

Question 1.
\(\frac{3}{25}\)
_______ %

Answer:
12%

Explanation:
3/25 ÷ 25/25 = 0.12/1 = 12/100 = 12%

Question 2.
\(\frac{3}{10}\)
_______ %

Answer:
30%

Explanation:
3/10 ÷ 10/10 = 0.3 = 0.3 × 100/100 = 30/100 = 30%

Question 3.
0.717
_______ %

Answer:
71.7%

Explanation:
0.717 = 717/100 = 71.7%

Question 4.
0.02
_______ %

Answer:
2%

Explanation:
0.02 = 2/100 = 2%

On Your Own

Write the number in two other forms ( fraction, decimal, or percent). Write the fraction in simplest form.

Question 5.
0.01
Type below:
_____________

Answer:
1% and \(\frac{1}{100}\)

Explanation:
0.01 as a fraction 1/100
0.01 as percent 1%

Question 6.
\(\frac{13}{40}\)
Type below:
_____________

Answer:
0.325 and 32.5%

Explanation:
\(\frac{13}{40}\) as decimal 0.325
\(\frac{13}{40}\) as percent 32.5/100 = 32.5%

Question 7.
\(\frac{6}{5}\)
Type below:
_____________

Answer:
1.2 and 120%

Explanation:
\(\frac{6}{5}\) as decimal 1.2
\(\frac{6}{5}\) as percent 120/100 = 120%

Question 8.
0.08
Type below:
_____________

Answer:
8% and \(\frac{8}{100}\)

Explanation:
0.08 as a fraction 8/100
0.08 as percent 8%

The table shows the portion of Kim’s class that participates in each sport. Use the table for 9–10.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 20

Question 9.
Do more students take part in soccer or in swimming? Explain your reasoning.
Type below:
_____________

Answer:
Soccer = 1/5 = 0.2
Swimming = 0.09
0.2 > 0.09
more students take part in Soccer

Question 10.
Explain What percent of Kim’s class participates in one of the sports listed? Explain how you found your answer
_______ %

Answer:
23%

Explanation:
Kim’s class participates in Baseball that is mentioned with 23%

Question 11.
For their reading project, students chose to either complete a character study, or write a book review. \(\frac{1}{5}\) of the students completed a character study, and 0.8 of the students wrote a book review. Joia said that more students wrote a book review than completed a character study. Do you agree with Joia? Use numbers and words to support your answer
Type below:
_____________

Answer:
1/5 = 0.2
0.2 < 0.8
More students completed writing a book review.
I agree with Joia

Sand Sculptures – Page No. 284

Every year, dozens of teams compete in the U.S. Open Sandcastle Competition. Recent winners have included complex sculptures in the shape of flowers, elephants, and racing cars.

Teams that participate in the contest build their sculptures using a mixture of sand and water. Finding the correct ratios of these ingredients is essential for creating a stable sculpture.

The table shows the recipes that three teams used. Which team used the greatest percent of sand in their recipe?
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 21
Convert to percents. Then order from least to greatest.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 22
From least to greatest, the percents are 75%, 84%, 95%.
So, Team B used the greatest percent of sand.
Solve.

Question 12.
Which team used the greatest percent of water in their recipe?
Type below:
_____________

Answer:
Team A used the greatest percent of water in their recipe

Explanation:
Team A, 10/10+30 = 10/40 = 0.25 = 25%
Team B, 1/20 × 5/5 = 5/100 = 5%
Team C, 0.16 = 16%

Question 13.
Some people say that the ideal recipe for sand sculptures contains 88.9% sand. Which team’s recipe is closest to the ideal recipe?
Type below:
_____________

Answer:
Team C

Question 14.
Team D used a recipe that consists of 20 cups of sand, 2 cups of flour, and 3 cups of water. How does the percent of sand in Team D’s recipe compare to that of the other teams?
Type below:
_____________

Answer:
Total number of cups together = 20 + 2+ 3 =25 cups
20/25 × 100 = 80/100 = 80%

Write Fractions and Decimals as Percents – Page No. 285

Write the fraction or decimal as a percent.

Question 1.
\(\frac{7}{20}\)
_______ %

Answer:
35%

Explanation:
7/20 = 0.35 = 35%

Question 2.
\(\frac{3}{50}\)
_______ %

Answer:
6%

Explanation:
3/50 = 0.06 = 6%

Question 3.
\(\frac{1}{25}\)
_______ %

Answer:
4%

Explanation:
1/25 = 0.04 = 4%

Question 4.
\(\frac{5}{5}\)
_______ %

Answer:
0.01%

Explanation:
5/5 = 1 = 0.01%

Question 5.
0.622
_______ %

Answer:
6.22%

Explanation:
0.622 = 6.22/100 = 6.22%

Question 6.
0.303
_______ %

Answer:
3.03%

Explanation:
0.303 = 3.03/100 = 3.03%

Question 7.
0.06
_______ %

Answer:
6%

Explanation:
0.06 = 6/100 = 6%

Question 8.
2.45
_______ %

Answer:
245%

Explanation:
2.45 × 100/100 = 245/100 = 245%

Write the number in two other forms (fraction, decimal, or percent). Write the fraction in simplest form

Question 9.
\(\frac{19}{20}\)
Type below:
_____________

Answer:
0.95 and 95%

Explanation:
\(\frac{19}{20}\) as a decimal 0.95
\(\frac{19}{20}\) as a percentage 95%

Question 10.
\(\frac{9}{16}\)
Type below:
_____________

Answer:
0.5625 and 56.25%

Explanation:
\(\frac{9}{16}\) as a decimal 0.5625
\(\frac{9}{16}\) as a percentage 56.25%

Question 11.
0.4
Type below:
_____________

Answer:
\(\frac{2}{5}\) and 40%

Explanation:
0.4 as a fraction 2/5
0.4 as a percentage 40/100 = 40%

Question 12.
0.22
Type below:
_____________

Answer:
\(\frac{11}{50}\) and 22%

Explanation:
0.22 as a fraction 11/50
0.22 as a percentage 22/100 = 22%

Problem Solving

Question 13.
According to the U.S. Census Bureau, \(\frac{3}{25}\) of all adults in the United States visited a zoo in 2007. What percent of all adults in the United States visited a zoo in 2007?
_______ %

Answer:
12%

Explanation:
According to the U.S. Census Bureau, \(\frac{3}{25}\) of all adults in the United States visited a zoo in 2007.
\(\frac{3}{25}\) = 0.12 = 12%

Question 14.
A bag contains red and blue marbles. Given that \(\frac{17}{20}\) of the marbles are red, what percent of the marbles are blue?
_______ %

Answer:
15%

Explanation:
The total number of marbles = 20
If 17 marbles are red, the remaining 3 marbles out of 20 are blue marbles
3/20 = 0.15 = 15%

Question 15.
Explain two ways to write \(\frac{4}{5}\) as a percent.
Type below:
_____________

Answer:
Decimal =0.8.
Percentage =80%

Explanation:
4/5 = 0.8 = 80/100 = 80%

Lesson Check – Page No. 286

Question 1.
The portion of shoppers at a supermarket who pay by credit card is 0.36. What percent of shoppers at the supermarket do NOT pay by credit card?
_______ %

Answer:
36%

Explanation:
The portion of shoppers at a supermarket who pay by credit card is 0.36.
0.36 = 0.36 × 100/100 = 36/100 = 36%

Question 2.
About \(\frac{23}{40}\) of a lawn is planted with Kentucky bluegrass. What percent of the lawn is planted with Kentucky bluegrass?
_______ %

Answer:
57.5%

Explanation:
About \(\frac{23}{40}\) of a lawn is planted with Kentucky bluegrass.
23/40 = 0.575 = 0.575 × 100/100 = 57.5/100 = 57.5%

Spiral Review

Question 3.
A basket contains 6 peaches and 8 plums. What is the ratio of peaches to total pieces of fruit?
Type below:
_____________

Answer:
6:14

Explanation:
total pieces of fruit 6 + 8 = 14
the ratio of peaches to total pieces of fruit is 6:14

Question 4.
It takes 8 minutes for 3 cars to move through a car wash. At the same rate, how many cars can move through the car wash in 24 minutes?
_______ cars

Answer:
9 cars

Explanation:
It takes 8 minutes for 3 cars to move through a car wash.
3/8 × 24 = 9 cars

Question 5.
A 14-ounce box of cereal sells for $2.10. What is the unit rate?
$ _______ per ounce

Answer:
$0.15 per ounce

Explanation:
$2.10/14 × 14/14 = $0.15 per ounce

Question 6.
A model railroad kit contains curved tracks and straight tracks. Given that 35% of the tracks are curved, what fraction of the tracks are straight? Write the fraction in simplest form.
\(\frac{□}{□}\)

Answer:
\(\frac{7}{20}\)

Explanation:
A model railroad kit contains curved tracks and straight tracks. Given that 35% of the tracks are curved,
35% = 35/100 = 7/20

Vocabulary – Page No. 287

Choose the best term from the box to complete the sentence.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 23

Question 1.
A _____ is a ratio that compares a quantity to 100.
Type below:
_____________

Answer:
percent

Concepts and Skills

Write a ratio and a percent to represent the shaded part.

Question 2.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 24
Type below:
_____________

Answer:
17% and \(\frac{17}{100}\)

Explanation:
17 squares are shaded out of 100.
So, 17% and 17/100 are the answers.

Question 3.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 25
Type below:
_____________

Answer:
60% and \(\frac{60}{100}\)

Explanation:
60 squares are shaded out of 100.
So, 60% and 60/100 are the answers.

Question 4.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 26
Type below:
_____________

Answer:
7% and \(\frac{7}{100}\)

Explanation:
7 squares are shaded out of 100.
So, 7% and 7/100 are the answers.

Question 5.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 27
Type below:
_____________

Answer:
11% and \(\frac{11}{100}\)

Explanation:
11 squares are shaded out of 100.
So, 11% and 11/100 are the answers.

Question 6.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 28
Type below:
_____________

Answer:
82% and \(\frac{82}{100}\)

Explanation:
82 squares are shaded out of 100.
So, 82% and 82/100 are the answers.

Question 7.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 29
Type below:
_____________

Answer:
36% and \(\frac{36}{100}\)

Explanation:
36 squares are shaded out of 100.
So, 36% and 36/100 are the answers.

Write the number in two other forms (fraction, decimal, or percent).

Write the fraction in simplest form.

Question 8.
0.04
Type below:
_____________

Answer:
\(\frac{1}{25}\) and 4%

Explanation:
0.04 as a fraction 4/100 = 1/25
0.04 as a decimal 0.04 × 100/100 = 4/100 = 4%

Question 9.
\(\frac{3}{10}\)
Type below:
_____________

Answer:
0.3 and 30%

Explanation:
\(\frac{3}{10}\) as a decimal 0.3
\(\frac{3}{10}\) as a percentage 0.3 × 100/100 = 30/100 = 30%

Question 10.
1%
Type below:
_____________

Answer:
\(\frac{1}{100}\) and 0.01

Explanation:
1% as a fraction 1/100
1% as a decimal 1/100 = 0.01

Question 11.
1 \(\frac{1}{5}\)
Type below:
_____________

Answer:
1.2 and 120%

Explanation:
1 \(\frac{1}{5}\) as a decimal = 6/5 = 1.2
1 \(\frac{1}{5}\) as a percentage 1.2 × 100/100 = 120/100 = 120%

Question 12.
0.9
Type below:
_____________

Answer:
\(\frac{90}{100}\) and 90%

Explanation:
0.9 as a fraction 0.9 × 100/100 = 90/100 = 90%

Question 13.
0.5%
Type below:
_____________

Answer:
\(\frac{5}{1,000}\) and 0.005

Explanation:
0.5% as a fraction = 0.5/100 = 5/1,000
0.5% as a decimal = 0.5/100 = 0.005

Question 14.
\(\frac{7}{8}\)
Type below:
_____________

Answer:
0.875 and 87.5%

Explanation:
\(\frac{7}{8}\) as a decimal 0.875
\(\frac{7}{8}\) as a percentage 87.5/100 = 87.5%

Question 15.
355%
Type below:
_____________

Answer:
\(\frac{71}{20}\) and 35.5

Explanation:
355% as a decimal 355/100 = 71/20 = 35.5

Page No. 288

Question 16.
About \(\frac{9}{10}\) of the avocados grown in the United States are grown in California. About what percent of the avocados grown in the United States are grown in California?
_______ %

Answer:
90%

Explanation:
About \(\frac{9}{10}\) of the avocados grown in the United States are grown in California.
9/10 × 10/10 = 90/100 = 90%

Question 17.
Morton made 36 out of 48 free throws last season. What percent of his free throws did Morton make?
_______ %

Answer:
75%

Explanation:
Morton made 36 out of 48 free throws last season.
36/48 = 0.75 = 75/100 = 75%

Question 18.
Sarah answered 85% of the trivia questions correctly. What fraction describes this percent?
\(\frac{□}{□}\)

Answer:
\(\frac{17}{20}\)

Explanation:
Sarah answered 85% of the trivia questions correctly.
85% = 85/100 = 17/20

Question 19.
About \(\frac{4}{5}\) of all the orange juice in the world is produced in Brazil. About what percent of all the orange juice in the world is produced in Brazil?
_______ %

Answer:
80%

Explanation:
About \(\frac{4}{5}\) of all the orange juice in the world is produced in Brazil.
4/5 = 0.8 × 100/100 = 80/100 = 80%

Question 20.
If you eat 4 medium strawberries, you get 48% of your daily recommended amount of vitamin C. What fraction of your daily amount of vitamin C do you still need?
\(\frac{□}{□}\)

Answer:
\(\frac{13}{25}\)

Explanation:
If you eat 4 medium strawberries, you get 48% of your daily recommended amount of vitamin C.
48% = 48/100
100 – 48 = 52
52% = 52/100 = 13/25 of your daily amount of vitamin C do you still need

Share and Show – Page No. 290

Find the percent of the quantity.

Question 1.
25% of 320
_______

Answer:
80

Explanation:
Write the percent as a rate per 100
25% = 25/100
25/100 × 320 = 80

Question 2.
80% of 50
_______

Answer:
40

Explanation:
Write the percent as a rate per 100
80% = 80/100
80/100 × 50 = 40

Question 3.
175% of 24
_______

Answer:
42

Explanation:
Write the percent as a rate per 100
175% = 175/100
175/100 × 24 = 42

Question 4.
60% of 210
_______

Answer:
126

Explanation:
Write the percent as a rate per 100
60% = 60/100
60/100 × 210 = 126

Question 5.
A jar contains 125 marbles. Given that 4% of the marbles are green, 60% of the marbles are blue, and the rest are red, how many red marbles are in the jar?
_______ marbles

Answer:
45 marbles

Explanation:
A jar contains 125 marbles.
4% of the marbles are green = 125 × 4/100 = 5
60% of the marbles are blue = 125 × 60/100 = 75
Red Marbles = Total Number of Marbles -[Number of Green Marbles + Number of Blue Marbles]
Red Marbles = 125 – (5 + 75) = 125 – 80 = 45

Question 6.
There are 32 students in Mr. Moreno’s class and 62.5% of the students are girls. How many boys are in the class?
_______ students

Answer:
12 students

Explanation:
There are 32 students in Mr. Moreno’s class
62.5% of the students are girls = 32 × 62.5/100 = 20
boys = 32 – 20 = 12

On Your Own – Page No. 291

Find the percent of the quantity.

Question 7.
60% of 90
_______

Answer:
54

Explanation:
Write the percent as a rate per 100
60% = 60/100
60/100 × 90 = 54

Question 8.
25% of 32.4
_______

Answer:
8.1

Explanation:
Write the percent as a rate per 100
25% = 25/100
25/100 × 32.4 = 8.1

Question 9.
110% of 300
_______

Answer:
330

Explanation:
Write the percent as a rate per 100
110% = 110/100
110/100 × 300 = 330

Question 10.
0.2% of 6500
_______

Answer:
13

Explanation:
Write the percent as a rate per 100
0.2% = 0.2/100
0.2/100 × 6500 = 13

Question 11.
A baker made 60 muffins for a cafe. By noon, 45% of the muffins were sold. How many muffins were sold by noon?
_______ muffins

Answer:
27 muffins

Explanation:
A baker made 60 muffins for a cafe. By noon, 45% of the muffins were sold.
60 × 45%
60 × 45/100 = 27

Question 12.
There are 30 treasures hidden in a castle in a video game. LaToya found 80% of them. How many of the treasures did LaToya find?
_______ treasures

Answer:
24 treasures

Explanation:
There are 30 treasures hidden in a castle in a video game.
LaToya found 80% of them.
30 × 80/100 = 24

Question 13.
A school library has 260 DVDs in its collection. Given that 45% of the DVDs are about science and 40% are about history, how many of the DVDs are about other subjects?
_______ DVDs

Answer:
39 DVDs

Explanation:
A school library has 260 DVDs in its collection.
45% of the DVDs are about science = 260 × 45/100 = 117
40% are about history = 260 × 40/100 = 104
other subjects = 260 – (117 + 104) = 260 – 221 = 39

Question 14.
Mitch planted cabbage, squash, and carrots on his 150-acre farm. He planted half the farm with squash and 22% with carrots. How many acres did he plant with cabbage?
_______ acres

Answer:

Explanation:
Mitch planted cabbage, squash, and carrots on his 150-acre farm.
He planted half the farm with squash 150/2 = 75
22% with carrots = 150 × 22/100 = 33
cabbage = 150 – (75 + 33) = 150 – 108 = 42

Question 15.
45% of 60 _______ 60% of 45

Answer:
45% of 60 = 60% of 45

Explanation:
45% of 60
45/100 × 60 = 27
60% of 45
60/100 × 45 = 27
45% of 60 = 60% of 45

Question 16.
10% of 90 _______ 90% of 100

Answer:
10% of 90 _______ 90% of 100

Explanation:
10% of 90
10/100 × 90 = 9
90% of 100
90/100 × 100 = 90
10% of 90 < 90% of 100

Question 17.
75% of 8 _______ 8% of 7.5

Answer:
75% of 8 > 8% of 7.5

Explanation:
75% of 8
75/100 × 8 = 6
8% of 7.5
8/100 × 7.5 = 0.6
75% of 8 > 8% of 7.5

Question 18.
Sarah had 12 free throw attempts during a game and made at least 75% of the free throws. What is the greatest number of free throws Sarah could have missed during the game?
_______ free throws

Answer:
3 free throws

Explanation:
Sarah had 12 free throw attempts during a game and made at least 75% of the free throws.
So, she missed 25% of the free throws.
12 × 25/100 = 3

Question 19.
Chrissie likes to tip a server in a restaurant a minimum of 20%. She and her friend have a lunch bill that is $18.34. Chrissie says the tip will be $3.30. Her friend says that is not a minimum of 20%. Who is correct? Explain.
Type below:
_____________

Answer:
100% = $18.34
10% = $18.34 / 10 = 1.834
20% = 1.834 × 2 = 3.66800 = $3.70
Her friend is correct because $3.70 is more than $3.30.

Unlock The Problem – Page No. 292

Question 20.
One-third of the juniors in the Linwood High School Marching Band play the trumpet. The band has 50 members and the table shows what percent of the band members are freshmen, sophomores, juniors, and seniors. How many juniors play the trumpet?
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 30
a. What do you need to find?
Type below:
_____________

Answer:
The percent of the band members are freshmen, sophomores, juniors, and seniors. How many juniors play the trumpet

Question 20.
b. How can you use the table to help you solve the problem?
Type below:
_____________

Answer:
percent of the band members that are Juniors: 24%
In 50 members of the band, 50×24/100 = 12 are Juniors. One-third of them play the trumpet, which makes 12×(1/3) = 4 members.

Question 20.
c. What operation can you use to find the number of juniors in the band?
Type below:
_____________

Answer:
percent of the band members that are Juniors: 24%
In 50 members of the band, 50×24/100 = 12 are Juniors.

Explanation:

Question 20.
d. Show the steps you use to solve the problem.
Type below:
_____________

Answer:
percent of the band members that are Juniors: 24%
In 50 members of the band, 50×24/100 = 12 are Juniors. One-third of them play the trumpet, which makes 12×(1/3) = 4 members.

Question 20.
e. Complete the sentences.
The band has _____ members. There are _____ juniors in the band. The number of juniors who play the trumpet is _____.
Type below:
_____________

Answer:
The band has 50 members. There are 12 juniors in the band. The number of juniors who play the trumpet is 4.

Question 21.
Compare. Circle <, >, or =.
a. 25% of 44 Ο 20% of 50
b. 10% of 30 Ο 30% of 100
c. 35% of 60 Ο 60% of 35
25% of 44 _____ 20% of 50
10% of 30 _____ 30% of 100
35% of 60 _____ 60% of 35

Answer:
25% of 44 >  20% of 50
10% of 30 < 30% of 100
35% of 60 = 60% of 35

Explanation:
25% of 44 = 25/100 × 44 = 11
20% of 50 = 20/100 × 50 = 1000/100 = 10
25% of 44  > 20% of 50
10% of 30 = 10/100 × 30 = 3
30% of 100 = 30/100 × 100 = 30
10% of 30 < 30% of 100
35% of 60 = 35/100 × 60 = 21
60% of 35 = 60/100 × 35 = 21
35% of 60 = 60% of 35

Percent of a Quantity – Page No. 293

Find the percent of the quantity.

Question 1.
60% of 140
_____

Answer:
84

Explanation:
60% of 140
60/100 × 140 = 84

Question 2.
55% of 600
_____

Answer:
330

Explanation:
55% of 600
55/100 × 600 = 330

Question 3.
4% of 50
_____

Answer:
2

Explanation:
4% of 50
4/100 × 50 = 2

Question 4.
10% of 2,350
_____

Answer:
235

Explanation:
10% of 2,350
10/100 × 2,350 = 235

Question 5.
160% of 30
_____

Answer:
48

Explanation:
160% of 30
160/100 × 30 = 48

Question 6.
105% of 260
_____

Answer:
273

Explanation:
105% of 260
105/100 × 260 = 273

Question 7.
0.5% of 12
_____

Answer:
0.06

Explanation:
0.5% of 12
0.5/100 × 12 = 0.06

Question 8.
40% of 16.5
_____

Answer:
6.6

Explanation:
40% of 16.5
40/100 × 16.5 =  6.6

Problem Solving

Question 9.
The recommended daily amount of vitamin C for children 9 to 13 years old is 45 mg. A serving of a juice drink contains 60% of the recommended amount. How much vitamin C does the juice drink contain?
_____ mg

Answer:
27 mg

Explanation:
The recommended daily amount of vitamin C for children 9 to 13 years old is 45 mg. A serving of a juice drink contains 60% of the recommended amount.
45% of 60 = 45/100 × 60 = 27

Question 10.
During a 60-minute television program, 25% of the time is used for commercials and 5% of the time is used for the opening and closing credits. How many minutes remain for the program itself?
_____ minutes

Answer:
42 minutes

Explanation:
60 minutes of tv
25% + 5% = 30%
30%= 0.30
60 times 0.30= 18
60-18=42
inly 42 minutes are used for the program itself

Question 11.
Explain two ways you can find 35% of 700.
Type below:
_____________

Answer:
First way
700 : 100 = x : 35
x = 700 × 35 : 100
x = 245
Second way
700 : 100 × 35 =
245

Lesson Check – Page No. 294

Question 1.
A store has a display case with cherry, peach, and grape fruit chews. There are 160 fruit chews in the display case. Given that 25% of the fruit chews are cherry and 40% are peach, how many grape fruit chews are in the display case?
_____ grape fruit chews

Answer:
56 grape fruit chews

Explanation:
A store has a display case with cherry, peach, and grape fruit chews. There are 160 fruit chews in the display case. Given that 25% of the fruit chews are cherry and 40% are peach,
25% + 40% +?% = 100%
65% + ?% = 100%
?% = 35%
.35×160 = 56

Question 2.
Kelly has a ribbon that is 60 inches long. She cuts 40% off the ribbon for an art project. While working on the project, she decides she only needs 75% of the piece she cut off. How many inches of ribbon does Kelly end up using for her project?
_____ inches

Answer:
18 inches

Explanation:
Length of ribbon = 60 inches
Part of ribbon cut off for an art project = 40%
So, the Length of the ribbon remains is given by
40% of 60 = 40/100 × 60 = 24
Part of a piece she only needs from cut off = 75%
so, the Length of ribbon she need end up using in her project is given by
75/100 × 24 = 18

Spiral Review

Question 3.
Three of the following statements are true. Which one is NOT true?
|−12| > 1      |0| > −4      |20| > |−10|        6 < |−3|
Type below:
_____________

Answer:
|−12| > 1
12 > 1; True
|0| > −4
0 > -4; True
|20| > |−10|
20 > 10; True
6 < |−3|
6 < 3; False

Question 4.
Miyuki can type 135 words in 3 minutes. How many words can she expect to type in 8 minutes?
_____ words

Answer:
360 words

Explanation:
Miyuki can type 135 words in 3 minutes.
135/3 = 45
45 × 8 = 360

Question 5.
Which percent represents the model?
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 31
_____ %

Answer:
63%

Explanation:
63 squares are shaded out of 100
63%

Question 6.
About \(\frac{3}{5}\) of the students at Roosevelt Elementary School live within one mile of the school. What percent of students live within one mile of the school?
_____ %

Answer:
60%

Explanation:
About \(\frac{3}{5}\) of the students at Roosevelt Elementary School live within one mile of the school.
3/5 × 100/100 = 60/100 = 60%

Share and Show – Page No. 297

Question 1.
A geologist visits 40 volcanoes in Alaska and California. 15% of the volcanoes are in California. How many volcanoes does the geologist visit in California and how many in Alaska?
Type below:
_____________

Answer:
40 volcanoes = 100% of them
100 – 15% = 85%
Number of volcanoes in California = 15% of 40 volcanoes = 0.15 x 40 = 6
Number of volcanoes in Alaska = 85% of 40 volcanoes 0.85 x 40 = 34

Question 2.
What if 30% of the volcanoes were in California? How many volcanoes would the geologist have visited in California and how many in Alaska?
Type below:
_____________

Answer:
Number of volcanoes in California = 30% of 40 = 30/100 x 40 = 12
Number of volcanoes in Alaska = 70% of 40 = 70/100 x 40 = 28

Question 3.
Ricardo has $25 to spend on school supplies. He spends 72% of the money on a backpack and the rest on a large binder. How much does he spend on the backpack? How much does he spend on the binder?
Type below:
_____________

Answer:
$18 on Backpack $7 on binder.
If you turn the percent into a decimal .72 and multiply .72 by 25 you get 18 which is the cost of the backpack.
subtract 18 from 25 and you get $7 left meaning the binder was $7

Question 4.
Kevin is hiking on a trail that is 4.2 miles long. So far, he has hiked 80% of the total distance. How many more miles does Kevin have to hike in order to complete the trail?
Type below:
_____________

Answer:
0.84 miles

Explanation:
Kevin is hiking on a trail that is 4.2 miles long. So far, he has hiked 80% of the total distance.
80% of 4.2 = 80/100 x 4.2 = 3.36
4.2 – 3.36 = 0.84 miles

On Your Own – Page No. 298

Question 5.
Jordan takes 50% of the cherries from a bowl. Then Mei takes 50% of the remaining cherries. Finally, Greg takes 50% of the remaining cherries. There are 3 cherries left. How many cherries were in the bowl before Jordan arrived?
_____ cherries

Answer:
24 cherries

Explanation:
Let total cherries in a bowl=x
Jordan takes cherries=50% of x = 50x/100
Remaining cherries = x – 50x/100 = x/2
Mei takes cherries=50% of 50x/100 = x/4
remaining cherries= x/2 – x/4 = x/4
Greg takes cherries=50% of x/4 = x/8
remaining cherries = x/4 – x/8 = x/8
Now,remaining cherries in a bowl=3
x/8 =3
x = 8 × 3 = 24

Question 6.
Each week, Tasha saves 65% of the money she earns babysitting and spends the rest. This week she earned $40. How much more money did she save than spend this week?
$ _____

Answer:
Tasha saved $26 and spent $14

Explanation:
Since 65% of 40 is 26, that’s how much Tasha saves. Then do 40 – 26 to get 14, which is how much she spends.
So Tasha saved $26 and spent $14.

Question 7.
An employee at a state park has 53 photos of animals found at the park. She wants to arrange the photos in rows so that every row except the bottom row has the same number of photos. She also wants there to be at least 5 rows. Describe two different ways she can arrange the photos
Type below:
_____________

Answer:
5 rows of 10 photos and last row with 3 photos,
6 rows of 8 photos and last row with 5 photos,
7 rows of 7 photos and last row with 4 photos,
Also, reverse the rows and photos in each row (ex 5 rows 10 photos=10 rows 5 photos) to get another 3 sets.

Question 8.
Explain a Method Maya wants to mark a length of 7 inches on a sheet of paper, but she does not have a ruler. She has pieces of wood that are 4 inches, 5 inches, and 6 inches long. Explain how she can use these pieces to mark a length of 7 inches.
Type below:
_____________

Answer:
Maya can put the 5 and 6-inch pieces together to get 11 inches. She can then subtract the length of the 4-inch piece to get 7 inches.

Question 9.
Pierre’s family is driving 380 miles from San Francisco to Los Angeles. On the first day, they drive 30% of the distance. On the second day, they drive 50% of the distance. On the third day, they drive the remaining distance and arrive in Los Angeles. How many miles did Pierre’s family drive each day? Write the number of miles in the correct box.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 32
Type below:
_____________

Answer:
76 miles

Explanation:
Pierre’s family is driving 380 miles from San Francisco to Los Angeles.
On the first day, they drive 30% of the distance. 380 × 30/100 = 114
On the second day, they drive 50% of the distance. 380 × 50/100 = 190
They traveled 80%.
On the third day, they drive the remaining distance and arrive in Los Angeles.
380 × 20/100 = 76 miles

Problem Solving Percents – Page No. 299

Read each problem and solve.

Question 1.
On Saturday, a souvenir shop had 125 customers. Sixty-four percent of the customers paid with a credit card. The other customers paid with cash. How many customers paid with cash?T
_____ costumers

Answer:
45 costumers

Explanation:
On Saturday, a souvenir shop had 125 customers. Sixty-four percent of the customers paid with a credit card.
125 × 64/100 = 80
100 – 64 = 36
125 × 36/100 = 45

Question 2.
A carpenter has a wooden stick that is 84 centimeters long. She cuts off 25% from the end of the stick. Then she cuts the remaining stick into 6 equal pieces. What is the length of each piece?
_____ cm

Answer:
10 1/2 cm

Explanation:
A carpenter has a wooden stick that is 84 centimeters long. She cuts off 25% from the end of the stick. Then she cuts the remaining stick into 6 equal pieces.
84 × 75/100 = 63
63/6 = 10 1/2

Question 3.
A car dealership has 240 cars in the parking lot and 17.5% of them are red. Of the other 6 colors in the lot, each color has the same number of cars. If one of the colors is black, how many black cars are in the lot?
_____ black cars

Answer:
33 black cars

Explanation:
number of red cars 17.5% × 240 = 42
number of cars of other colors = 240 – 42 = 198
number of black cars 1/6 × 198 = 33

Question 4.
The utilities bill for the Millers’ home in April was $132. Forty-two percent of the bill was for gas, and the rest was for electricity. How much did the Millers pay for gas, and how much did they pay for electricity?
Type below:
_____________

Answer:
Amount of money paid for gas = 132 * (42/100) dollars
= 5544/100 dollars
= 55.44 dollars
Then
The amount of money paid for electricity = (132 – 55.44) dollars
= 76.56 dollars
So the Millers paid 55.44 dollars for gas and 76.56 dollars for electricity in the month of April.

Question 5.
Andy’s total bill for lunch is $20. The cost of the drink is 15% of the total bill and the rest is the cost of the food. What percent of the total bill did Andy’s food cost? What was the cost of his food?
Type below:
_____________

Answer:
$17

Explanation:
Andy paid $20 total for his lunch (100%).
15% is for drink.
Therefore, 100 – 15 = 85% is the percent that was constituted by the food.
85% of 20 is equal to 0.85 × 20 is equal to:
17 × 20/20 = 17
Andy’s food cost $17.

Question 6.
Write a word problem that involves finding the additional amount of money needed to purchase an item, given the cost and the percent of the cost already saved.
Type below:
_____________

Answer:
Each week, Tasha saves 65% of the money she earns babysitting and spends the rest. This week she earned $40. How much more money did she save than spend this week?
Tasha saved $26 and spent $14

Lesson Check – Page No. 300

Question 1.
Milo has a collection of DVDs. Out of 45 DVDs, 40% are comedies and the remaining are action-adventures. How many actionadventure DVDs does Milo own?
_____ DVDs

Answer:
27 DVDs

Explanation:
100%-40%=60%
60/100*45=27
27 DVD’s are action-adventure

Question 2.
Andrea and her partner are writing a 12-page science report. They completed 25% of the report in class and 50% of the remaining pages after school. How many pages do Andrea and her partner still have to write?
_____ pages

Answer:
9 pages

Explanation:
first 50% + 25% = 75%
then you can do 75% of 12
75% = 0.75
of = multiplication
0.75 • 12 which should equal 9
so they have 9 pages left

Spiral Review

Question 3.
What is the absolute value of \(\frac{-4}{25}\)?
\(\frac{□}{□}\)

Answer:
\(\frac{4}{25}\)

Explanation:
|\(\frac{-4}{25}\)| = 4/25

Question 4.
Ricardo graphed a point by starting at the origin and moving 5 units to the left. Then he moved up 2 units. What is the ordered pair for the point he graphed?
Type below:
_____________

Answer:
(-5, 2)

Explanation:
In a coordinate system, the coordinates of the origin are (0, 0).
If he moves 5 units to the left, he is moving in the negative direction along the x-axis, and x takes the value -5.
If he moves up 2 units, he is moving in the positive direction along the y-axis, and y takes the value 2.
The ordered pair (x, y) is (-5, 2).

Question 5.
The population of birds in a sanctuary increases at a steady rate. The graph of the population over time has the points (1, 105) and (3, 315). Name another point on the graph.
Type below:
_____________

Answer:
You could do (2, 210) or (4, 420) or (5, 525)

Question 6.
Alicia’s MP3 player contains 1,260 songs. Given that 35% of the songs are rock songs and 20% of the songs are rap songs, how many of the songs are other types of songs?
_____ songs

Answer:
567 songs

Explanation:
Since 55% of the songs are rock and rap, 45% of the songs are other.
To find 45% of 1260 we multiply by the decimal:
1260 x 0.45 = 567
Therefore 567 of the songs are other.

Share and Show – Page No. 303

Find the unknown value.

Question 1.
9 is 25% of _____.
_____

Answer:
36

Explanation:
25/100 ÷ 25/25 = 1/4
1/4 = 9/s
1/4 × 9/9 = 9/36
the unknown value is 36

Question 2.
14 is 10% of _____.
_____

Answer:
140

Explanation:
10/100 ÷ 10/10 = 1/10
1/10 = 14/s
1/10 × 14/14 = 14/140
the unknown value is 140

Question 3.
3 is 5% of _____.
_____

Answer:
6

Explanation:
5/10 ÷ 5/5 = 1/2
1/2 × 3/3 = 3/6
the unknown value is 6

Question 4.
12 is 60% of _____.
_____

Answer:
20

Explanation:
60/100 ÷ 60/60 = 60/100
60/100 ÷ 5/5 = 12/20
the unknown value is 20

On Your Own

Find the unknown value.

Question 5.
16 is 20% of _____.
_____

Answer:
80

Explanation:
20/100 ÷ 20/20 = 1/5
1/5 × 16/16 = 16/80
the unknown value is 80

Question 6.
42 is 50% of _____.
_____

Answer:
84

Explanation:
50/100 ÷ 50/50 = 1/2
1/2 × 42/42 = 42/84
the unknown value is 84

Question 7.
28 is 40% of _____.
_____

Answer:
70

Explanation:
40/100 ÷ 40/40 = 1/2.5
1/2.5 × 28/28 = 28/70
the unknown value is 70

Question 8.
60 is 75% of _____.
_____

Answer:
80

Explanation:
75/100 ÷ 75/75 = 60/s
60 × 100 = 6000/75 = 80
the unknown value is 80

Question 9.
27 is 30% of _____.
_____

Answer:
90

Explanation:
30/100 ÷ 30/30 = 3/10
3/10 × 9/9 = 27/90
the unknown value is 90

Question 10.
21 is 60% of _____.
_____

Answer:
35

Explanation:
60/100 ÷ 60/60 = 3/5
3/5 × 7/7 = 21/35
the unknown value is 35

Question 11.
12 is 15% of _____.
_____

Answer:
80

Explanation:
15/100 ÷ 15/15 = 3/20
3/20 × 4/4 = 12/80
the unknown value is 80

Solve.

Question 12.
40% of the students in the sixth grade at Andrew’s school participate in sports. If 52 students participate in sports, how many sixth graders are there at Andrew’s school?
_____ students

Answer:
130 students

Explanation:
52/s = 40%
52/s = 40/100
s = 40/100 × 52 = 130

Question 13.
There were 136 students and 34 adults at the concert. If 85% of the seats were filled, how many seats are in the auditorium?
_____ seats

Answer:
80 seats

Explanation:
There are 170 seats filled total. 170 is 85% of 200. There are 200 seats in the auditorium.
If you were to solve for x in the equation 40% = 32/x, you would get x = 80.

Use Reasoning Algebra Find the unknown value.

Question 14.
40% = \(\frac{32}{?}\)
_____

Answer:
80

Explanation:
40/100 = 32/?
40/100 ÷ 40/40 = 2/5
2/5 × 16/16 = 32/80
the unknown value is 80

Question 15.
65% = \(\frac{91}{?}\)
_____

Answer:
140

Explanation:
65/100 = 91/?
65/100 ÷ 65/65 = 13/20
13/20 × 7/7 = 91/140
the unknown value is 140

Question 16.
45% = \(\frac{54}{?}\)
_____

Answer:
120

Explanation:
45/100 ÷ 45/45 = 9/20
9/20 × 6/6 = 54/120

Problem Solving + Applications – Page No. 304

Use the advertisement for 17 and 18.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 33

Question 17.
Corey spent 20% of his savings on a printer at Louie’s Electronics. How much did Corey have in his savings account before he bought the printer?
$ _____

Answer:
$800

Explanation:
(printer cost) = 0.20 * (savings)
(printer cost)/0.20 = (savings)
savings = 5*(printer cost)
Corey’s savings was 5 times that amount.
savings = 5 × 160 = 800

Question 18.
Kai spent 90% of his money on a laptop that cost $423. Does he have enough money left to buy a scanner? Explain.
Type below:
_____________

Answer:
$42.3

Explanation:
He spent 90% of his money. So, he left 10% of money with him.
423 × 10/100 = 42.3 left to buy a scanner

Question 19.
Maurice has completed 17 pages of the research paper he is writing. That is 85% of the required length of the paper. What is the required length of the paper?
_____ pages

Answer:
20 pages

Explanation:
Maurice has completed 17 pages of the research paper he is writing. That is 85% of the required length of the paper.
85%=17 ? what about 100%
100multiplied by 17 divided by 85% =20

Question 20.
Of 250 seventh-grade students, 175 walk to school. What percent of seventh-graders do not walk to school?
_____ %

Answer:
30%

Explanation:
it’s either 30 percent or 70. 70 percent walks to school and 30 percent DO NOT walk to school

Question 21.
What’s the Error? Kate has made 20 free throws in basketball games this year. That is 80% of the free throws she has attempted. To find the total number of free throws she attempted, Kate wrote the equation \(\frac{80}{100}=\frac{?}{20}\). What error did Kate make?
Type below:
_____________

Answer:
20 free throws is 80% of the total attempted
80% to decimal is:
80/100 = 0.8
If total attempted is x, we can say:
20 is 80% (0.8) of x
We can now write an algebraic equation:
20 = 0.8x
We simply solve this for x, that is the number of free throws she attempted:
20 = 0.8x
x = 20/0.8 = 25

Question 22.
Maria spent 36% of her savings to buy a smart phone. The phone cost $90. How much money was in Maria’s savings account before she purchased the phone? Find the unknown value.
$ _____

Answer:
$ 250

Explanation:
let her savings be A
A/Q-
36% of A = $90
36/100 of A = $90
A = 90×100/36
A= $ 250

Find the Whole from a Percent – Page No. 305

Find the Whole from a Percent

Question 1.
9 is 15% of _____.
_____

Answer:
60

Explanation:
15/100 ÷ 15/15 = 3/20
3/20 × 3/3 = 9/60
the unknown value is 60

Question 2.
54 is 75% of _____.
_____

Answer:
72

Explanation:
75/100 ÷ 75/75 = 3/4
3/4 × 18/18 = 54/72
the unknown value is 72

Question 3.
12 is 2% of _____.
_____

Answer:
600

Explanation:
2/100 = 1/50
1/50 × 12/12 = 12/600
the unknown value is 600

Question 4.
18 is 50% of _____.

Answer:
36

Explanation:
50/100 = 1/2
1/2 × 18/18 = 18/36
the unknown value is 36

Question 5.
16 is 40% of _____.
_____

Answer:
40

Explanation:
40/100 = 2/5
2/5 × 8/8 = 16/40
the unknown value is 40

Question 6.
56 is 28% of _____.
_____

Answer:
200

Explanation:
28/100 = 14/50 = 7/25
7/25 × 8/8 = 56/200
the unknown value is 200

Question 7.
5 is 10% of _____.
_____

Answer:
50

Explanation:
10/100 = 1/10
1/10 × 5/5 = 5/50
the unknown value is 50

Question 8.
24 is 16% of _____.
_____

Answer:
150

Explanation:
16/100 = 4/25
4/25 × 6/6 = 24/150
the unknown value is 150

Question 9.
15 is 25% of _____.
_____

Answer:
60

Explanation:
25/100 = 1/4
1/4 × 15/15 = 15/60
the unknown value is 60

Problem Solving

Question 10.
Michaela is hiking on a weekend camping trip. She has walked 6 miles so far. This is 30% of the total distance. What is the total number of miles she will walk?
_____ miles

Answer:
20 miles

Explanation:
Since 6mi=30%,
You should find ten percent.
This is how, divide both sides by 3, and this gives you
2m=10% (2m being 2 miles)
So, to find 100%, you need to multiply both sides by 10
20m=100%
So now, Michaela will walk 20 miles this weekend

Question 11.
A customer placed an order with a bakery for muffins. The baker has completed 37.5% of the order after baking 81 muffins. How many muffins did the customer order?
_____ muffins

Answer:
216 muffins

Explanation:
A customer placed an order with a bakery for muffins. The baker has completed 37.5% of the order after baking 81 muffins.
37.5/100=0.375 and 81/0.375=216
so the answer is 216

Question 12.
Write a question that involves finding what number is 25% of another number. Solve using a double number line and check using equivalent ratios. Compare the methods.
Type below:
_____________

Answer:
25% of 15 = 25/100 × 15 = 375/100 = 3.75

Lesson Check – Page No. 306

Question 1.
Kareem saves his coins in a jar. 30% of the coins are pennies. If there are 24 pennies in the jar, how many coins does Kareem have?
_____ coins

Answer:
80 coins

Explanation:
24=30%
find 100%
24=30%
diivde by 3
8=10%
multiply 10
80=100%
80 coins

Question 2.
A guitar shop has 19 acoustic guitars on display. This is 19% of the total number of guitars. What is the total number of guitars the shop has?
_____ guitars

Answer:
100 guitars

Explanation:
Let’s find out how much 1% is worth first.
19 guitars = 19%
therefore 19 ÷ 19 = [ 1 guitar = 1% ]
The total number of guitars is going to be 100%,
so if 1% × 100 = 100%, then 1 guitar × 100 = 100 guitars total.

Spiral Review

Question 3.
On a coordinate grid, in which quadrant is the point (−5, 4) located?
Type below:
_____________

Answer:
Quadrant II

Explanation:
(-5, 4)
-5 is the negative point of the x coordinate
4 is the positive point of the y coordinate
Quadrant II

Question 4.
A box contains 16 cherry fruit chews, 15 peach fruit chews, and 12 plum fruit chews. Which two flavors are in the ratio 5 to 4?
Type below:
_____________

Answer:
peach fruit chews and plum fruit chews are in the ratio 5 to 4

Explanation:
15 peach fruit chews, and 12 plum fruit chews
15/12 = 5/4

Question 5.
During basketball season, Marisol made \(\frac{19}{25}\) of her free throws. What percent of her free throws did Marisol make?
_____ %

Answer:
76%

Explanation:
During the basketball season, Marisol made \(\frac{19}{25}\) of her free throws.
(19 ÷ 25) × 100 = 76%. Marisol made 76% of her free throws.

Question 6.
Landon is entering the science fair. He has a budget of $115. He has spent 20% of the money on new materials. How much does Landon have left to spend?
$ _____

Answer:
$92

Explanation:
Landon has $92 left because if you divide 115/.20 you get 23 and then you subtract 115-23=92 or $92.

Chapter 5 Review/Test – Page No. 307

Question 1.
What percent is represented by the shaded part?
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 34
Options:
a. 46%
b. 60%
c. 64%
d. 640%

Answer:
c. 64%

Explanation:
64 squares are shaded out of 100.
So, 64% and 64/100 are the answers.

Question 2.
Write a percent to represent the shaded part.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 35
_____ %

Answer:
42%

Explanation:
42 squares are shaded out of 100.
So, 42% and 42/100 are the answers.

Question 3.
Rosa made a mosaic wall mural using 42 black tiles, 35 blue tiles and 23 red tiles. Write a percent to represent the number of red tiles in the mural.
_____ %

Answer:
23%

Explanation:
42+35+23= 100
So plug it in.
23/100
23%
Your answer is 23%.

Question 4.
Model 39%.
Type below:
_____________

Answer:
grade 6 chapter 5 image 10

Explanation:
39 squares out of 100 need to shaded

Page No. 308

Question 5.
For 5a–5d, choose Yes or No to indicate whether the percent and the fraction represent the same amount.
5a. 50% and \(\frac{1}{2}\)
5b. 45% and \(\frac{4}{5}\)
5c. \(\frac{3}{8}\) and 37.5%
5d. \(\frac{2}{10}\) and 210%
5a. _____________
5b. _____________
5c. _____________
5d. _____________

Answer:
5a. Yes
5b. No
5c. Yes
5d. No

Explanation:
1/2 = 0.5 × 100/100 = 50/100 = 50%
4/5 = 0.8 × 100/100 = 80/100 = 80%
3/8 = 0.375 × 100/100 = 37.5/100 = 37.5%
2/10 = 0.2 × 100/100 = 20/100 = 20%

Question 6.
The school orchestra has 25 woodwind instruments, 15 percussion instruments, 30 string instruments, and 30 brass instruments. Select the portion of the instruments that are percussion. Mark all that apply.
Options:
a. 15%
b. 1.5
c. \(\frac{3}{20}\)
d. 0.15

Answer:
a. 15%
c. \(\frac{3}{20}\)
d. 0.15

Explanation:
25 + 15 + 30 + 30 = 100
15 percussion instruments = 15/100 = 15% = 0.15

Question 7.
For a science project, \(\frac{3}{4}\) of the students chose to make a poster and 0.25 of the students wrote a report. Rosa said that more students made a poster than wrote a report. Do you agree with Rosa? Use numbers and words to support your answer
Type below:
_____________

Answer:
Yes, because 3/4 is equal to 0.75 and 0.75 > 0.25
Or 0.25 is equal to 1/4, and 1/4 < 3/4

Question 8.
Select other ways to write 0.875. Mark all that apply.
Options:
a. 875%
b. 87.5%
c. \(\frac{7}{8}\)
d. \(\frac{875}{100}\)

Answer:
c. \(\frac{7}{8}\)

Explanation:
0.875 = 8.75/100 = 8.75%

Page No. 309

Question 9.
There are 88 marbles in a bin and 25% of the marbles are red.
There are _____________ red marbles in the bin.

Answer:
There are 22 red marbles in the bin.

Explanation:
88 × 25% = 88 × 25/100 = 22

Question 10.
Harrison has 30 CDs in his music collection. If 40% of the CDs are country music and 30% are pop music, how many CDs are other types of music?
_____ CDs

Answer:
9 CDs

Explanation:
Harrison has 30 CDs in his music collection. If 40% of the CDs are country music and 30% are pop music,
40 + 30 = 70
100 – 70 = 30%
30 × 30/100 = 9

Question 11.
For numbers 11a–11b, choose <, >, or =.
11a. 30% of 90 Ο 35% of 80
11b. 25% of 16 Ο 20% of 25
30% of 90 _____ 35% of 80
25% of 16 _____ 20% of 25

Answer:
30% of 90 < 35% of 80
25% of 16 < 20% of 25

Explanation:
30% of 90 = 30/100 × 90 = 27
35% of 80 = 35/100 × 80 = 28
30% of 90 < 35% of 80
25% of 16 = 25/100 × 16 = 4
20% of 25 = 20/100 × 25 = 5
25% of 16 < 20% of 25

Question 12.
There were 200 people who voted at the town council meeting. Of these people, 40% voted for building a new basketball court in the park. How many people voted against building the new basketball court? Use numbers and words to explain your answer.
Type below:
_____________

Answer:
There were 200 people who voted at the town council meeting. Of these people, 40% voted for building a new basketball court in the park.
100 – 40% = 60%
200 × 60/100 = 120 people

Page No. 310

Question 13.
James and Sarah went out to lunch. The price of lunch for both of them was $20. They tipped their server 20% of that amount. How much did each person pay if they shared the price of lunch and the tip equally?
$ _____

Answer:
$12

Explanation:
James and Sarah went out to lunch. The price of lunch for both of them was $20. They tipped their server 20% of that amount.
20% of 20 = 20/100 × 20 = 4
20 + 4 = 24
24/2 = 12
$12

Question 14.
A sandwich shop has 30 stores and 60% of the stores are in California. The rest of the stores are in Nevada.
Part A
How many stores are in California and how many are in Nevada?
Type below:
_____________

Answer:
30 × 60/100 = 18 stores in California
30 – 18 = 12 stores in Nevada

Question 14.
Part B
The shop opens 10 new stores. Some are in California, and some are in Nevada. Complete the table.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 36
Type below:
_____________

Answer:
grade 6 chapter 5 image 11

Explanation:
100 – 45 = 55%
55% of 40 = 55/100 × 40 = 22
45% of 40 = 45/100 × 40 = 18

Question 15.
Juanita has saved 35% of the money that she needs to buy a new bicycle. If she has saved $63, how much money does the bicycle cost? Use numbers and words to explain your answer
$ _____

Answer:
$180

Explanation:
Juanita has saved 35% of the money that she needs to buy a new bicycle. If she has saved $63,
35/100 = 7/20
7/20 × 9/9 = 63/180
The bicycle cost is $180

Page No. 311

Question 16.
For 16a–16d, choose Yes or No to indicate whether the statement is correct.
16a. 12 is 20% of 60.
16b. 24 is 50% of 48.
16c. 14 is 75% of 20.
16d. 9 is 30% of 30.
16a. _____________
16b. _____________
16c. _____________
16d. _____________

Answer:
16a. Yes
16b. Yes
16c. No
16d. Yes

Explanation:
20% of 60 = 20/100 × 60 = 12
50% of 48 = 50/100 × 48 = 24
75% of 20 = 75/100 × 20 = 15
30% of 30 = 30/100 × 30 = 9

Question 17.
Heather and her family are going to the grand opening of a new amusement park. There is a special price on tickets this weekend. Tickets cost $56 each. This is 70% of the cost of a regular price ticket
Part A
What is the cost of a regular price ticket? Show your work.
$ _____

Answer:
$80

Explanation:
70/100 = 56/s
s = 56 × 100/70 = 80

Question 17.
Part B
Heather’s mom says that they would save more than $100 if they buy 4 tickets for their family on opening weekend. Do you agree or disagree with Heather’s mom? Use numbers and words to support your answer. If her statement is incorrect, explain the correct way to solve it.
Type below:
_____________

Answer:
80 × 4 = 320
56 × 4 = 224
320 – 224 = 96
$96

Question 18.
Elise said that 0.2 equals 2%. Use words and numbers to explain her mistake.
Type below:
_____________

Answer:
0.2 × 100/100 = 20/100 = 2%

Page No. 312

Question 19.
Write 18% as a fraction.
\(\frac{□}{□}\)

Answer:
\(\frac{9}{50}\)

Explanation:
18% = 18/100 = 9/50

Question 20.
Noah wants to put a variety of fish in his new fish tank. His tank is large enough to hold a maximum of 70 fish.
Part A
Complete the table.
Go Math Grade 6 Answer Key Chapter 5 Model Percents img 37
Type below:
_____________

Answer:
grade 6 chapter 5 image 12

Explanation:
70 × 20/100 = 14
70 × 40/100 = 28
70 × 30/100 = 21

Question 20.
Part B
Has Noah put the maximum number of fish in his tank? Use numbers and words to explain how you know. If he has not put the maximum number of fish in the tank, how many more fish could he put in the tank?
Type below:
_____________

Answer:
No, since 20% + 40% + 30% = 90%, he can add 10% in the tank.

Conclusion:

Test your knowledge by solving the problems from Go Math Grade 6 Answer Key Chapter 5 Model Percents. Get the solutions for Mid Chapter Checkpoint and Review Test along with the exercise problems in Go Math Grade 6 Chapter 5 Model Percents Solution Key. Quick learning and best practice come in a single hand with our Go Math Grade 6 Solution Key Chapter 5 Model Percents @ ccssmathanswers.com

Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations

go-math-grade-8-chapter-8-solving-systems-of-linear-equations-answer-key

Students of Grade 8 can get a detailed explanation for all the questions in Go Math Answer Key Chapter 8 Solving Systems of Linear Equations. In addition to the exercise problems we also provide the solutions for the review test. So, go through all the answers and explanations provided by the math experts in Go Math Grade 8 Chapter 8 Solving Systems of Linear Equations Answer Key. Our aim is to provide easy and simple tricks to solve the problems in Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations.

Download Go Math Grade 8 Chapter 8 Solving Systems of Linear Equations Answer Key Pdf

Students who are interested to secure the highest marks in the exams are suggested to download the Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations pdf. All the solutions are provided in the pdf format as per the list of the chapters provided in the latest edition. Hence refer to Go Math 8th Grade Solution Key to learning the easy way of maths practice. Check the list of the topics covered in Chapter 8 Solving Systems of Linear Equations from the following section.

Lesson 1: Solving Systems of Linear Equations by Graphing

Lesson 2: Solving Systems by Substitution

Lesson 3: Solving Systems by Elimination

Lesson 4: Solving Systems by Elimination with Multiplication

Lesson 5: Solving Solving Special Systems

Model Quiz

Review

Guided Practice – Solving Systems of Linear Equations by Graphing – Page No. 232

Solve each system by graphing.

Question 1.
\(\left\{\begin{array}{l}y=3 x-4 \\y=x+2\end{array}\right.\)
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson1: Solving Systems of Linear Equations by Graphing img 1
Type below:
______________

Answer:
grade 8 chapter 8 image 1

Explanation:
y = 3x – 4
y = x + 2
The solution of thr linear system of equations is the intersection point of the two equations.
(3, 5) is the solution of the system of equations.
If x = 3, y = 3(3) – 4 = 9 – 4 = 5; y = 3 + 2 = 5
5 = 5; True

Question 2.
\(\left\{\begin{array}{l}x-3 y=2 \\-3x+9y=-6\end{array}\right.\)
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson1: Solving Systems of Linear Equations by Graphing img 2
Type below:
______________

Answer:
grade 8 chapter 8 image 2
Infinitely many solutions

Explanation:
x – 3y = 2
-3x + 9y = -6
x – 3y – x = -x + 2
-3y = -x + 2
y = 1/3 . x – 2/3
-3x + 9y + 3x = 3x – 6
9y = 3x – 6
y = 3/9 . x – 6/9
y = 1/3 . x – 2/3
The solution of the linear system of equations is the intersection of the two equations.
Infinitely many solutions

Question 3.
Mrs. Morales wrote a test with 15 questions covering spelling and vocabulary. Spelling questions (x) are worth 5 points and vocabulary questions (y) are worth 10 points. The maximum number of points possible on the test is 100.
a. Write an equation in slope-intercept form to represent the number of questions on the test.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson1: Solving Systems of Linear Equations by Graphing img 3
Type below:
______________

Answer:
y = -x + 15

Explanation:
Mrs. Morales wrote a test with 15 questions covering spelling and vocabulary. Spelling questions (x) are worth 5 points and vocabulary questions (y) are worth 10 points.
x + y = 15
x + y – x = -x + 15
y = -x + 15

Question 3.
b. Write an equation in slope-intercept form to represent the total number of points on the test.
Type below:
______________

Answer:
y = -1/2 . x + 10

Explanation:
The total number of points on test is 100
5x + 10y = 100
5x + 10y – 5x = -5x + 100
10y = -5x + 100
y = -5/10 . x + 100/10
y = -1/2 . x + 10

Question 3.
c. Graph the solutions of both equations.
Type below:
______________

Answer:
grade 8 chapter 8 image 3

Question 3.
d. Use your graph to tell how many of each question type are on the test.
_________ spelling questions
_________ vocabulary questions

Answer:
10 spelling questions
5 vocabulary questions

ESSENTIAL QUESTION CHECK-IN

Question 4.
When you graph a system of linear equations, why does the intersection of the two lines represent the solution of the system?
Type below:
______________

Answer:
To solve a system of linear equations means finding the solutions that satisfy all the equations of that system. When we graph a system of linear equations, the intersection point lies on the line of each equation, which means that satisfies all the equations. Therefore, it is considered to be the solution to that system.

Solving Systems of Linear Equations by Graphing – Page No. 233

Question 5.
Vocabulary
A_________________ is a set of equations that have the same variables.
______________

Answer:
system of equations

Explanation:
A system of equations is a set of equations that have the same variables.

Question 6.
Eight friends started a business. They will wear either a baseball cap or a shirt imprinted with their logo while working. They want to spend exactly $36 on the shirts and caps. Shirts cost $6 each and caps cost $3 each.
a. Write a system of equations to describe the situation. Let x represent the number of shirts and let y represent the number of caps.
______________

Answer:
6x + 3y = 36

Explanation:
The sum of caps and shirts is 8. The total cost of caps and shirts is $36.
x + y = 8
6x + 3y = 36

Question 6.
b. Graph the system. What is the solution and what does it represent?
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson1: Solving Systems of Linear Equations by Graphing img 4
Type below:
______________

Answer:
The solution is (4, 4)
grade 8 chapter 8 image 4

Explanation:
x + y – x = -x + 8
y = -x + 8
6x + 3y – 6x = -6x + 36
3y = -6x + 36
y = -6/2 . x + 36/3
y = -2x + 12
(4, 4). They should order 4 shirts and 4 caps.

Question 7.
Multistep The table shows the cost for bowling at two bowling alleys.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson1: Solving Systems of Linear Equations by Graphing img 5
a. Write a system of equations, with one equation describing the cost to bowl at Bowl-o-Rama and the other describing the cost to bowl at Bowling Pinz. For each equation, let x represent the number of games played and let y represent the total cost.
Type below:
______________

Answer:
y = 2.5x + 2
y = 2x + 4

Explanation:
Cost at Bowl-o-Rama => y = 2.5x + 2
Cost at Bowling Pinz => y = 2x + 4

Question 7.
b. Graph the system. What is the solution and what does it represent?
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson1: Solving Systems of Linear Equations by Graphing img 6
Type below:
______________

Answer:
grade 8 chapter 8 image 5

Explanation:
The solution of the linear system of equations is the intersection of the two equations.
(4, 12)
When 4 games are played, the total cost is $12.

Solving Systems of Linear Equations by Graphing – Page No. 234

Question 8.
Multi-Step Jeremy runs 7 miles per week and increases his distance by 1 mile each week. Tony runs 3 miles per week and increases his distance by 2 miles each week. In how many weeks will Jeremy and Tony be running the same distance? What will that distance be?
Type below:
______________

Answer:
After 4 weeks Jeremy and Tony will be running the same distance and that distance would be 11 miles.

Explanation:
Multi-Step Jeremy runs 7 miles per week and increases his distance by 1 mile each week.
y = x + 7
Tony runs 3 miles per week and increases his distance by 2 miles each week.
y = 2x + 3
grade 8 chapter 8 image 6
The solution of the system of linear equation is (4, 11) which means that after 4 weeks Jeremy and Tony will be running the same distance and that distance would be 11 miles.

Question 9.
Critical Thinking Write a real-world situation that could be represented by the system of equations shown below.
\(\left\{\begin{array}{l}y=4 x+10 \\y=3x+15\end{array}\right.\)
Type below:
______________

Answer:
The entry fee of the first gym is $10 and for every hour that you spend there, you pay an extra $4. If we denote with x the number of hours that somebody spends at the gym and with y the total cost is
y = 4x + 10
The entry fee of the second gym is $15 and for every hour that you spend there, you pay an extra $3. If we denote with x the number of hours that somebody spends at the gym and with y the total cost is
y = 3x + 15
y = 4x + 10
y = 3x + 15

FOCUS ON HIGHER ORDER THINKING

Question 10.
Multistep The table shows two options provided by a high-speed Internet provider.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson1: Solving Systems of Linear Equations by Graphing img 7
a. In how many months will the total cost of both options be the same? What will that cost be?
________ months
$ ________

Answer:
5 months
$ 200

Explanation:
Let y be the total cost after x month
y = 30x + 50
Let y be the total cost after x month
y = 40x
Substitute y = 40x in y = 30x + 50
40x = 30x + 50
40x – 30x = 50
10x = 50
x = 50/10
x = 5
The total cost of both options will be the same after 5 months. Total cost would be y = 40(5) = $200.

Question 10.
b. If you plan to cancel your Internet service after 9 months, which is the cheaper option? Explain.
______________

Answer:
When x = 9 months
y = 30(9) + 50 = $320
y = 40(9) = $360
$320 < $360
Option 1 is cheaper as the total cost is less for option 1

Question 11.
Draw Conclusions How many solutions does the system formed by x − y = 3 and ay − ax + 3a = 0 have for a nonzero number a? Explain.
Type below:
______________

Answer:
x – y = 3
ay – ax + 3a =0
ay – ax + 3a – 3a = 0 – 3a
ay – ax = – 3a
a(y – x) = -3a
y – x = -3
x – y = 3
Both equations are the same. The system of linear equations have infinitely many solutions.

Guided Practice – Solving Systems by Substitution – Page No. 240

Solve each system of linear equations by substitution.

Question 1.
\(\left\{\begin{array}{l}3x-2y=9 \\y=2x-7\end{array}\right.\)
x = ________
y = ________

Answer:
x = 5
y = 3

Explanation:
\(\left\{\begin{array}{l}3x-2y=9 \\y=2x-7\end{array}\right.\)
Substitute 2x – 7 in 3x – 2y = 9
3x – 2(2x – 7) = 9
3x – 4x + 14 = 9
-x + 14 = 9
-x + 14 – 14 = 9 – 14
-x = -5
x = -5/-1 = 5
y = 2(5) – 7 = 3
Solution is (5, 3)

Question 2.
\(\left\{\begin{array}{l}y=x-4 \\2x+y=5\end{array}\right.\)
x = ________
y = ________

Answer:
x = 3
y = -1

Explanation:
\(\left\{\begin{array}{l}y=x-4 \\2x+y=5\end{array}\right.\)
2x + x – 4 = 5
3x – 4 = 5
3x – 4 + 4 = 5 + 4
3x = 9
x = 9/3 = 3
y = 3 – 4 = -1
The solution is (3, -1)

Question 3.
\(\left\{\begin{array}{l}x+4y=6 \\y=-x+3\end{array}\right.\)
x = ________
y = ________

Answer:
x = 2
y = 1

Explanation:
\(\left\{\begin{array}{l}x+4y=6 \\y=-x+3\end{array}\right.\)
Substitute y = -x + 3 in x + 4y = 6
x + 4(-x + 3) = 6
x – 4x + 12 = 6
-3x + 12 = 6
-3x + 12 – 12 = 6 – 12
-3x = -6
x = -6/-3 = 2
y = -2 + 3 = 1
The solution is (2, 1)

Question 4.
\(\left\{\begin{array}{l}x+2y=6 \\x-y=3\end{array}\right.\)
x = ________
y = ________

Answer:
x = 4
y = 1

Explanation:
\(\left\{\begin{array}{l}x+2y=6 \\x-y=3\end{array}\right.\)
y = x – 3
Substitute y = x – 3 in x + 2y = 6
x + 2(x – 3) = 6
x + 2x – 6 = 6
3x = 12
x = 12/3
x = 4
4 – y = 3
-y = 3 – 4
-y = -1
y = 1
The solution is (4, 1)

Solve each system. Estimate the solution first.

Question 5.
\(\left\{\begin{array}{l}6x+y=4 \\x-4y=19\end{array}\right.\)
Estimate ______________
Solution ______________
Type below:
______________

Answer:
Estimate (2, -5)
Solution (1.4, -4.4)

Explanation:
\(\left\{\begin{array}{l}6x+y=4 \\x-4y=19\end{array}\right.\)
Let’s find the estimation by graphing the equations
Estimate: (2, -5)
grade 8 chapter 8 image 7
x = 4y + 19
6(4y + 19) + y = 4
24y + 114 + y = 4
25y + 114 = 4
25y = 4 – 114
25y = -110
y = -110/25
y = -4.4
x + 4(-4.4) = 19
x + 17.6 = 19
x = 19 – 17.6
x = 1.4
The solution is (1.4, -4.4)

Question 6.
\(\left\{\begin{array}{l}x+2y=8 \\3x+2y=6\end{array}\right.\)
Estimate ______________
Solution ______________
Type below:
______________

Answer:
Estimate (-1, 5)
Solution (-1, 4.5)

Explanation:
\(\left\{\begin{array}{l}x+2y=8 \\3x+2y=6\end{array}\right.\)
Let’s find the estimation by graphing the equations
Estimate: (-1, 5)
grade 8 chapter 8 image 8
x = -2y + 8
Substitute the equation x = -2y + 8 in 3x + 2y = 6
3(-2y + 8) + 2y = 6
-6y + 24 + 2y = 6
-4y = 6 – 24
-4y = -18
y = -18/-4
y = 4.5
x + 2(4.5) = 8
x + 9 = 8
x = 8 – 9
x = -1
The solution is (-1, 4.5)

Question 7.
\(\left\{\begin{array}{l}3x+y=4 \\5x-y=22\end{array}\right.\)
Estimate ______________
Solution ______________
Type below:
______________

Answer:
Estimate (3, -6)
Solution (3.25, -5.75)

Explanation:
\(\left\{\begin{array}{l}3x+y=4 \\5x-y=22\end{array}\right.\)
Find the Estimation using graphing the equations.
Estimate: (3, -6)
grade 8 chapter 8 image 9
y = -3x + 4
Substitute y = -3x + 4 in 5x – y = 22
5x – (-3x + 4) = 22
5x + 3x -4 = 22
8x = 26
x = 26/8
x = 3.25
3(3.25) + y = 4
9.75 + y = 4
y = 4 – 9.75
y = -5.75
The solution is (3.25, -5.75)

Question 8.
\(\left\{\begin{array}{l}2x+7y=2 \\x+y=-1\end{array}\right.\)
Estimate ______________
Solution ______________
Type below:
______________

Answer:
Estimate (-2, 1)
Solution (-1.8, 0.8)

Explanation:
\(\left\{\begin{array}{l}2x+7y=2 \\x+y=-1\end{array}\right.\)
Find the Estimation using graphing the equations.
Estimate: (-2, 1)
grade 8 chapter 8 image 10
y = -x -1
Substitute y = -x – 1 in 2x + 7y = 2
2x + 7(-x – 1) = 2
2x – 7x -7 = 2
-5x = 2 + 7
-5x = 9
x = -9/5
x = -1.8
-1.8 + y = -1
y = -1 + 1.8
y = 0.8
The solution is (-1.8, 0.8)

Question 9.
Adult tickets to Space City amusement park cost x dollars. Children’s tickets cost y dollars. The Henson family bought 3 adult and 1 child tickets for $163. The Garcia family bought 2 adult and 3 child tickets for $174.
a. Write equations to represent the Hensons’ cost and the Garcias’ cost.
Hensons’ cost: ________________
Garcias’ cost:__________________
Type below:
______________

Answer:
Hensons’ cost: 3x + y = 163
Garcias’ cost: 2x + 3y = 174

Explanation:
Henson’s cost
3x + y = 163
Garcia’s cost
2x + 3y = 174

Question 9.
b. Solve the system.
adult ticket price: $ _________
Garcias’ cost: $ _________

Answer:
adult ticket price: $ 45
Garcias’ cost: $ 28

Explanation:
y = -3x + 163
Substitute y = -3x + 163 in 2x + 3y = 174
2x + 3(-3x + 163) = 174
2x -9x + 489 = 174
-7x = -315
x = -315/-7 = 45
3(45) + y = 163
135 + y = 163
y = 163 – 135
y = 28
adult ticket price: $ 45
Garcias’ cost: $ 28

ESSENTIAL QUESTION CHECK-IN

Question 10.
How can you decide which variable to solve for first when you are solving a linear system by substitution?
Type below:
______________

Answer:
The variable with the unit coefficient should be solved first when solving a linear system by substitution.

8.2 Independent Practice – Solving Systems by Substitution – Page No. 241

Question 11.
Check for Reasonableness Zach solves the system
\(\left\{\begin{array}{l}x+y=-3 \\x-y=1\end{array}\right.\)
and finds the solution (1, -2). Use a graph to explain whether Zach’s solution is reasonable.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 2: Solving Systems by Substitution img 8
Type below:
______________

Answer:
grade 8 chapter 8 image 11

Explanation:
\(\left\{\begin{array}{l}x+y=-3 \\x-y=1\end{array}\right.\)
The x coordinate of the solution is negative, hence Zach’s solution is not reasonable.

Represent Real-World Problems Angelo bought apples and bananas at the fruit stand. He bought 20 pieces of fruit and spent $11.50. Apples cost $0.50 and bananas cost $0.75 each.
a. Write a system of equations to model the problem. (Hint: One equation will represent the number of pieces of fruit. A second equation will represent the money spent on the fruit.)
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 2: Solving Systems by Substitution img 9
Type below:
______________

Answer:
x + y = 20
0.5x + 0.75y = 11.5

Explanation:
x + y = 20
0.5x + 0.75y = 11.5
where c is the number of Apples and y is the number of Bananas.

Question 12.
b. Solve the system algebraically. Tell how many apples and bananas Angelo bought.
________ apples
________ bananas

Answer:
14 apples
6 bananas

Explanation:
y = -x + 20
Substitute y = -x + 20 in 0.5x + 0.75y = 11.5
0.5x + 0.75(-x + 20) = 11.5
0.5x – 0.75x + 15 = 11.5
-0.25x + 15 = 11.5
-0.25x = 11.5 – 15
-0.25x = -3.5
x = -3.5/-0.25
x = 14
14 + y = 20
y = 6
Angelo bought 14 apples and 6 bananas.

Question 13.
Represent Real-World Problems A jar contains n nickels and d dimes. There is a total of 200 coins in the jar. The value of the coins is $14.00. How many nickels and how many dimes are in the jar?
________ nickels
________ dimes

Answer:
120 nickels
80 dimes

Explanation:
A jar contains n nickels and d dimes. There is a total of 200 coins in the jar. The value of the coins is $14.00.
$14 = 1400 cents
n + d = 200
5n + 10d = 1400
d = -n + 200
5n + 10(-n + 200) = 1400
5n – 10n + 2000 = 1400
-5n = -600
n = -600/-5
n = 120
120 + d = 200
d = 200 – 120
d = 80
There are 120 nickles and 80 dimes in the jar.

Question 14.
Multistep The graph shows a triangle formed by the x-axis, the line 3x−2y=0, and the line x+2y=10. Follow these steps to find the area of the triangle.
a. Find the coordinates of point A by solving the system
\(\left\{\begin{array}{l}3x-2y=0 \\x-2y=10\end{array}\right.\)
Point A: ____________________
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 2: Solving Systems by Substitution img 10
Type below:
______________

Answer:
Point A: (2.5, 3.75)Coordinate of A is (2.5, 3.75)

Explanation:
\(\left\{\begin{array}{l}3x-2y=0 \\x-2y=10\end{array}\right.\)
x = -2y + 10
Substitute x = -2y + 10 in 3x – 2y = 0
3(-2y + 10) -2y = 0
-6y + 30 – 2y = 0
-8y = -30
y = -30/-8 = 3.75
x + 2(3.75) = 10
x + 7.5 = 10
x = 10 – 7.5
x = 2.5
Coordinate of A is (2.5, 3.75)

Question 14.
b. Use the coordinates of point A to find the height of the triangle.
height:__________________
height: \(\frac{□}{□}\) units

Answer:
height: 3.75
height: \(\frac{15}{4}\) units

Explanation:
Height of the triangle is the y coordinate of A
Height = 3.75

Question 14.
c. What is the length of the base of the triangle?
base:________________
base: ______ units

Answer:
base: 10 units

Explanation:
Length of the base = 10

Question 14.
d. What is the area of the triangle?
A = ______ \(\frac{□}{□}\) square units

Answer:
A = 18.75 square units
A = 18 \(\frac{3}{4}\) square units

Explanation:
Area of the triangle = 1/2 . Height . Base
Area = 1/2 . 3.75 . 10 = 18.75

Solving Systems by Substitution – Page No. 242

Question 15.
Jed is graphing the design for a kite on a coordinate grid. The four vertices of the kite are at A(−\(\frac{4}{3}\), \(\frac{2}{3}\)), B(\(\frac{14}{3}\), −\(\frac{4}{3}\)), C(\(\frac{14}{3}\), −\(\frac{16}{3}\)), and D(\(\frac{2}{3}\), −\(\frac{16}{3}\)). One kite strut will connect points A and C. The other will connect points B and D. Find the point where the struts cross.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 2: Solving Systems by Substitution img 11
Type below:
______________

Answer:
The struts cross as (8/3, 10/3)

Explanation:
1. From AC
Slope = (y2 – y1)/(x2 – x1) = [(-16/3)-(2/3)] ÷ [(14/3) – (-4/3)] = (-18/3) ÷ (18/3) = -1
y = mx + b
2/3 = -1(-4/3) + b
2/3 = 4/3 + b
1. From BD
Slope = (y2 – y1)/(x2 – x1) = [(-16/3)-(-4/3)] ÷ [(2/3) – (144/3)] = (-12/3) ÷ (-12/3) = 1
y = mx + b
-4/3 = 1(14/3) + b
-4/3 = 14/3 + b
-18/3 = b
-6 = b
y = mx + b
y = x -6
3. y = -x -2/3
y = x – 6
4. y = -x – 2/3
x – 6 = -x – 2/3
x = -x – 2/3 + 6
x = – x + 16/3
2x = 16/3
x = 16/6
x = 8/3
then y = x – 6
y = 8/3 – 18/3
y = -10/3
The struts cross as (8/3, 10/3)

FOCUS ON HIGHER ORDER THINKING

Question 16.
Analyze Relationships Consider the system
\(\left\{\begin{array}{l}6x-3y=15 \\x+3y=-8\end{array}\right.\)
Describe three different substitution methods that can be used to solve this system. Then solve the system.
Type below:
______________

Answer:
(1, -3) is the answer.

Explanation:
As there are three different substitution methods, we can write
Solve for y in the first equation, then substitute that value into the second equation.
Solve for x in the second equation, then substitute that value into the first equation.
Solve either equation for 3y, then substitute that value into the other equation.
From the Second method,
x + 3y = -8
x = -3y – 8
6x – 3y = 15
6 (-3y – 8) -3y = 15
-18y – 48 -3y = 15
-21y – 48 = 15
-21y = 63
y = -3
x + 3y = -8
x + 3(-3) = -8
x – 9 = -8
x = 1
(1, -3) is the answer.

Question 17.
Communicate Mathematical Ideas Explain the advantages, if any, that solving a system of linear equations by substitution has over solving the same system by graphing.
Type below:
______________

Answer:
The advantage of solving a system of linear equations by graphing is that it is relatively easy to do and requires very little algebra.

Question 18.
Persevere in Problem Solving Create a system of equations of the form
\(\left\{\begin{array}{l}Ax+By=C \\Dx+Ey=F\end{array}\right.\)
that has (7, −2) as its solution. Explain how you found the system.
Type below:
______________

Answer:
x + y = 5
x – y = 9
solves in :
x = (5+9)/2 = 7
y = 5-9)/2 = -2
A=1, B=2, C= 5
D=1, E= -1, F=9
x = 7
y = -2
IS a system (even if it is a trivial one) of equations so this answer would be acceptable.
The target for a system is to find it SOLUTION SET and not to conclude with x=a and y=b

Guided Practice – Solving Systems by Elimination – Page No. 248

Question 1.
Solve the system
\(\left\{\begin{array}{l}4x+3y=1 \\x-3y=-11\end{array}\right.\)
by adding.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 3: Solving Systems by Elimination img 12
Type below:
______________

Answer:
4x + 3y = 1
x – 3y = -11
Add the above two equations
4x + 3y = 1
+(x – 3y = -11)
Add to eliminate the variable y
5x + 0y = -10
Simplify and solve for x
5x = -10
Divide both sided by 5
x = -10/5 = -2
Substitute into one of the original equations and solve for y.
4(-2) + 3y = 1
-8 + 3y = 1
3y = 9
y = 9/3 = 3
So, (-2, 3) is the solution of the system.

Solve each system of equations by adding or subtracting.

Question 2.
\(\left\{\begin{array}{l}x+2y=-2 \\-3x+2y=-10\end{array}\right.\)
x = ________
y = ________

Answer:
x = 2
y = -2

Explanation:
\(\left\{\begin{array}{l}x+2y=-2 \\-3x+2y=-10\end{array}\right.\)
Subtract the equations
x + 2y = -2
-(-3x + 2y = -10)
y is eliminated as it has reversed coefficients. Solve for x
x + 2y + 3x – 2y = -2 + 10
4x = 8
x = 8/4 = 2
Substituting x in either of the equation to find y
2 + 2y = -2
2 + 2y -2 = -2 -2
2y = -4
y = -4/2 = -2
(2, -2) is the answer.

Question 3.
\(\left\{\begin{array}{l}3x+y=23 \\3x-2y=8\end{array}\right.\)
(________ , ________)

Answer:
(6, 5)

Explanation:
\(\left\{\begin{array}{l}3x+y=23 \\3x-2y=8\end{array}\right.\)
Subtract the equations
3x + y = 23
-(3x – 2y = 8)
x is eliminated as it has reversed coefficients. Solve for y
3x + y – 3x + 2y = 23 – 8
3y = 15
y = 15/3 = 5
Substituting y in either of the equation to find x
3x + 5 = 23
3x + 5 – 5 = 23 – 5
3x = 18
x = 18/3 = 6
Solution is (6, 5)

Question 4.
\(\left\{\begin{array}{l}-4x-5y=7 \\3x+5y=-14\end{array}\right.\)
(________ , ________)

Answer:
(7, -7)

Explanation:
\(\left\{\begin{array}{l}-4x-5y=7 \\3x+5y=-14\end{array}\right.\)
Add the equations
-4x – 5y = 7
+(3x + 5y = -14)
y is eliminated as it has reversed coefficients. Solve for x
-4x -5y +3x + 5y = 7 -14
-x = -7
x = -7/-1 = 7
Substituting x in either of the equation to find y
3(7) + 5y = -14
21 + 5y -21 = -14 -21
5y = -35
y = -35/5 = -7
The answer is (7, -7)

Question 5.
\(\left\{\begin{array}{l}x-2y=-19 \\5x+2y=1\end{array}\right.\)
(________ , ________)

Answer:
(-3, 8)

Explanation:
\(\left\{\begin{array}{l}x-2y=-19 \\5x+2y=1\end{array}\right.\)
Add the equations
x – 2y = -19
+(5x + 2y = 1)
y is eliminated as it has reversed coefficients. Solve for x
x – 2y + 5x + 2y = -19 + 1
6x = -18
x = -18/6 = -3
Substituting x in either of the equation to find y
-3 -2y = -19
-3 -2y + 3 = -19 + 3
-2y = -16
y = -16/-2 = 8
The answer is (-3, 8)

Question 6.
\(\left\{\begin{array}{l}3x+4y=18 \\-2x+4y=8\end{array}\right.\)
(________ , ________)

Answer:
(2, 3)

Explanation:
\(\left\{\begin{array}{l}3x+4y=18 \\-2x+4y=8\end{array}\right.\)
Subtract the equations
3x + 4y = 18
-(-2x + 4y = 8)
y is eliminated as it has reversed coefficients. Solve for x
3x + 4y + 2x – 4y = 18 – 8
5x = 10
x = 10/5 = 2
Substituting x in either of the equation to find y
3(2) + 4y = 18
6 + 4y – 6 = 18 – 6
4y = 12
y = 12/4 =3
Solution is (2, 3)

Question 7.
\(\left\{\begin{array}{l}-5x+7y=11 \\-5x+3y=19\end{array}\right.\)
(________ , ________)

Answer:
(-5, -2)

Explanation:
\(\left\{\begin{array}{l}-5x+7y=11 \\-5x+3y=19\end{array}\right.\)
Subtract the equations
-5x + 7y = 11
-(-5x + 3y = 19)
x is eliminated as it has reversed coefficients. Solve for y
-5x + 7y + 5x – 3y = 11 – 19
4y = -8
y = -8/4 = -2
Substituting y in either of the equation to find x
-5x + 7(-2) = 11
-5x -14 + 14 = 11 + 14
-5x = 25
x = 25/-5 = -5
Solution is (-5, -2)

Question 8.
The Green River Freeway has a minimum and a maximum speed limit. Tony drove for 2 hours at the minimum speed limit and 3.5 hours at the maximum limit, a distance of 355 miles. Rae drove 2 hours at the minimum speed limit and 3 hours at the maximum limit, a distance of 320 miles. What are the two speed limits?
a. Write equatios to represent Tony’s distance and Rae’s distance.
Type below:
______________

Answer:
Tony’s distance: 2x + 3.5y = 355
Rae’s distance: 2x + 3y = 320
where x is the minimum speed and y is the maximum speed.

Question 8.
b. Solve the system.
minimum speed limit:______________
maximum speed limit______________
minimum speed limit: ________ mi/h
maximum speed limit: ________ mi/h

Answer:
minimum speed limit:55
maximum speed limit70
minimum speed limit: 55mi/h
maximum speed limit: 70mi/h

Explanation:
Subtract the equations
2x + 3.5y = 355
-(2x + 3y = 320)
x is eliminated as it has reversed coefficients. Solve for y
2x + 3.5y – 2x – 3y = 355 – 320
0.5y = 35
y = 35/0.5 = 70
Substituting y in either of the equation to find x
2x + 3(70) = 320
2x + 210 – 210 = 320 – 210
2x = 110
x = 110/2 = 55
Minimum speed limit: 55 miles per hour
Maximum speed limit: 70 miles per hour

ESSENTIAL QUESTION CHECK-IN

Question 9.
Can you use addition or subtraction to solve any system? Explain.
________

Answer:
No. One of the variables should have the same coefficient in order to add or subtract the system.

8.3 Independent Practice – Solving Systems by Elimination – Page No. 249

Question 10.
Represent Real-World Problems Marta bought new fish for her home aquarium. She bought 3 guppies and 2 platies for a total of $13.95. Hank also bought guppies and platies for his aquarium. He bought 3 guppies and 4 platies for a total of $18.33. Find the price of a guppy and the price of a platy.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 3: Solving Systems by Elimination img 13
Guppy: $ ________
Platy: $ ________

Answer:
Guppy: $ 3.19
Platy: $ 2.19

Explanation:
3x + 2y = 13.95
3x + 4y = 18.33
where x is the unit price of guppy and y is the unit price of platy
Subtract the equations
3x + 2y = 13.95
-(3x + 4y = 18.33)
x is eliminated as it has reversed coefficients. Solve for y
3x + 2y – 3x – 4y = 13.95 – 18.33
-2y = -4.38
y = -4.38/-2 = 2.19
Substituting y in either of the equation to find x
3x + 2(2.19) = 13.95
3x + 4.38 – 4.38 = 13.95 – 4.38
3x = 9.57
x = 9.57/3 = 3.19
The price of a guppy is $3.19 and price of a platy is $2.19

Question 11.
Represent Real-World Problems The rule for the number of fish in a home aquarium is 1 gallon of water for each inch of fish length. Marta’s aquarium holds 13 gallons and Hank’s aquarium holds 17 gallons. Based on the number of fish they bought in Exercise 10, how long is a guppy and how long is a platy?
Length of a guppy = ________ inches
Length of a platy = ________ inches

Answer:
Length of a guppy = 3 inches
Length of a platy = 2 inches

Explanation:
3x + 2y = 13
3x + 4y = 17
where x is the length of guppy and y is the length of a platy
Subtract the equations
3x + 2y = 13
-(3x + 4y = 17)
x is eliminated as it has reversed coefficients. Solve for y
3x + 2y – 3x – 4y = 13 – 17
-2y = -4
y = -4/-2 = 2
Substituting y in either of the equation to find x
3x + 2(2) = 13
3x + 4 – 4 = 13 – 4
3x = 9
x = 9/3 = 3
The length of a guppy is 3 inches and price of a platy is 2 inches

Question 12.
Line m passes through the points (6, 1) and (2, -3). Line n passes through the points (2, 3) and (5, -6). Find the point of intersection of these lines.
Type below:
________________

Answer:
The intersection of these lines is (3.5, -1.5)

Explanation:
Find the slope of line m = (y2 – y1)/(x2 – x1) where (x2, y2) = (2, -3) and (x1, y1) = (6, 1)
Slope = (-3 -1)/(2 – 6) = -4/-4 = 1
Substitute the value of m and any of the given ordered pair (x, y) in point-slope form of equation: y – y1 = m(x – x1)
y – 1 = 1(x – 6)
y – 1 = x – 6
y = x – 6 + 1
x – y = 5
Find the slope of line n = (y2 – y1)/(x2 – x1) where (x2, y2) = (5, -6) and (x1, y1) = (2, 3)
Slope = (-6 -3)/(5 – 2) = -9/3 = -3
Substitute the value of m and any of the given ordered pair (x, y) in point-slope form of equation: y – y1 = m(x – x1)
y – 3 = -3(x – 2)
y – 3 = -3x + 6
y = -3x + 6 + 3
3x + y = 9
Add the equations
x – y = 5
+(3x + y = 9)
y is eliminated as it has reversed coefficients. Solve for x
x – y + 3x + y = 5 + 9
4x = 14
x = 14/4 = 3.5
Substituting x in either of the equation to find y
3.5 – y = 5
3.5 – y – 3.5 = 5 – 3.5
-y = 1.5
y = -1.5
The intersection of these lines is (3.5, -1.5)

Question 13.
Represent Real-World Problems Two cars got an oil change at the same auto shop. The shop charges customers for each quart of oil plus a flat fee for labor. The oil change for one car required 5 quarts of oil and cost $22.45. The oil change for the other car required 7 quarts of oil and cost $25.45. How much is the labor fee and how much is each quart of oil?
Labor fee: $ ________
Quart of oil: $ ________

Answer:
Labor fee: $ 14.95
Quart of oil: $ 1.5

Explanation:
5x + y = 22.45
7x + y = 25.45
where x is the unit cost of quarts of oil and y is the flat fee for labor
Subtract the equations
5x + y = 22.45
-(7x + y = 25.45)
y is eliminated as it has reversed coefficients. Solve for x
5x + y – 7x – y = 22.45 – 25.45
-2x = -3
x = -3/-2 = 1.5
Substituting x in either of the equation to find y
5(1.5) + y = 22.45
7.5 + y – 7.5 = 22.45 – 7.5
y = 14.95
Labor fee is $14.95 and unit cost of quart of oil is $1.5

Question 14.
Represent Real-World Problems A sales manager noticed that the number of units sold for two T-shirt styles, style A and style B, was the same during June and July. In June, total sales were $2779 for the two styles, with A selling for $15.95 per shirt and B selling for $22.95 per shirt. In July, total sales for the two styles were $2385.10, with A selling at the same price and B selling at a discount of 22% off the June price. How many T-shirts of each style were sold in June and July combined?
________ T-shirts of style A and style B were sold in June and July.

Answer:
15.95x + 22.95y = 2779
15.95x + 17.9y = 2385.10
where x is number of style A shirt and y is the number of style B shirt
In July, the price of style B shirt is 22% of the price of style B shirt in June, hence 0.78(22.95) = 17.90
Subtract the equations
15.95x + 22.95y = 2779
-(15.95x + 17.9y = 2385.10)
x is eliminated as it has reversed coefficients. Solve for y
15.95x + 22.95 – 15.95x – 17.9y = 2779 – 2385.10
5.05y = 393.9
y = 393.9/5.05 = 78
Substituting y in either of the equation to find x
15.95x +22.95(78) = 2779
15.95x + 1790.1 – 1790.1 = 2779 – 1790.1
15.95x = 988.9
x = 988.9/15.95 = 62
The number of style A T shirt sold in June is 62.
Since the number of T-shirts sold in both numbers is the same, the total number = 2. 62 = 124.
The number of style B T-shirts sold in June is 78.
Since the number of T-shirts sold in both numbers is the same, the total number = 2. 78 = 156.

Question 15.
Represent Real-World Problems Adult tickets to a basketball game cost $5. Student tickets cost $1. A total of $2,874 was collected on the sale of 1,246 tickets. How many of each type of ticket were sold?
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 3: Solving Systems by Elimination img 14img 14
________ adult tickets
________ student tickets

Answer:
407 adult tickets
839 student tickets

Explanation:
x + y = 1246
5x + y = 2874
where x is the number of adult tickets sold and y is the number of student tickets sold.
Subtract the equations
x + y = 1246
-(5x + y = 2874)
y is eliminated as it has reversed coefficients. Solve for x
x + y – 5x – y = 1246 – 2874
-4x = -1628
x = -1628/-4 = 407
Substituting x in either of the equation to find y
407 + y = 1246
407 + y – 407 = 1246 – 407
y = 839
The number of adult tickets sold is 407 and student tickets sold is 839.

FOCUS ON HIGHER ORDER THINKING – Solving Systems by Elimination – Page No. 250

Question 16.
Communicate Mathematical Ideas Is it possible to solve the system
\(\left\{\begin{array}{l}3x-2y=10 \\x+2y=6\end{array}\right.\)
by using substitution? If so, explain how. Which method, substitution or elimination, is more efficient? Why?
________

Answer:
The system can be solved by substitution as x in equation 2 can be isolated.
3x – 2y = 10
x + 2y = 6
Solve the equation for x in the equation.
x = -2y + 6
Substitute the expression for x in the other equation and solve.
3(-2y + 6) -2y = 10
-6y + 18 – 2y = 10
-8y + 18 = 10
-8y = -8
y = -8/-8 = 1
Substitute the values of y into one of the equations and solve for the other variable x.
x + 2(1) = 6
x = 4
The solution is (4, 1)
As the cofficient if variable y is opposite, it will be eliminated and solved for x in less number of steps.
Elimination would be more efficient.

Question 17.
Jenny used substitution to solve the system
\(\left\{\begin{array}{l}2x+y=8 \\x-y=1\end{array}\right.\). Her solution is shown below.
Step 1: y = -2x + 8               Solve the first equation for y.
Step 2: 2x + (-2x + 8) = 8     Substitute the value of y in an original equation.
Step 3: 2x – 2x + 8 = 8          Use the Distributive Property.
Step 4: 8 = 8                         Simplify.
a. Explain the Error Explain the error Jenny made. Describe how to correct it.
Type below:
______________

Answer:
2x + y = 8
x – y = 1
Rewritten equation should be substituted in the other original equation
Error is that Jenny solved for y in the first equation and substitute it in the original equation.
x – (-2x + 8) = 1
3x – 8 = 1
3x = 9
x = 9/3 = 3
x = 3

Question 17.
b. Communicate Mathematical Ideas Would adding the equations have been a better method for solving the system? If so, explain why.
________

Answer:
Yes

Explanation:
As the coefficient, if variable y is the opposite, it will be eliminated and solved for x in less number of steps.

Guided Practice – Solving Systems by Elimination with Multiplication – Page No. 256

Question 1.
Solve the system
\(\left\{\begin{array}{l}3x-y=8 \\-2x+4y=-12\end{array}\right.\)
by multiplying and adding.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 4: Solving Systems by Elimination with Multiplication img 15
Type below:
______________

Answer:
\(\left\{\begin{array}{l}3x-y=8 \\-2x+4y=-12\end{array}\right.\)
Multiply each term in the first equation by 4 to get opposite coefficients for the y-terms.
4(3x – y = 8)
12x – 4y = 32
Add the second equation to the new equation
12x – 4y = 32
+(-2x + 4y = -12)
Add to eliminate the variable y
10x = 20
Divide both sides by 10
x = 20/10 = 2
Substitue into one of the original equations and solve for y
y = 3(2) – 8 = -1
S0, (2, -2)is the solution of the system.

Solve each system of equations by multiplying first.

Question 2.
\(\left\{\begin{array}{l}x+4y=2 \\2x+5y=7\end{array}\right.\)
(________ , ________ )

Answer:
(6, -1)

Explanation:
x + 4y = 2
2x + 5y = 7
To eliminate x terms, multiply the 2nd equation by 2
2(x + 4y = 2)
2x + 8y = 4
Subtract the equations
2x + 8y = 4
-(2x + 5y = 7)
x is eliminated as it has reversed coefficients. Solve for y
2x + 8y – 2x – 5y = 4 – 7
3y = -3
y = -3/3 = -1
Substituting y in either of the equation to find x
x + 4(-1) = 2
x – 4 + 4 = 2 + 4
x = 6
Solution: (6, -1)

Question 3.
\(\left\{\begin{array}{l}3x+y=-1 \\2x+3y=18\end{array}\right.\)
(________ , ________ )

Answer:
(-3, 8)

Explanation:
\(\left\{\begin{array}{l}3x+y=-1 \\2x+3y=18\end{array}\right.\)
To eliminate y terms, multiply the 1st equation by 3
3(3x + y = -1)
9x + 3y = -3
Subtract the equations
9x + 3y = -3
-(2x + 3y = 18)
y is eliminated as it has reversed coefficients. Solve for x
9x + 3y – 2x – 3y = -3 -18
7x = -21
x = -21/7
x = -3
Substituting x in either of the equation to find y
3(-3) + y = -1
-9 + y + 9 = -1 + 9
y = 8
Solution: (-3, 8)

Question 4.
\(\left\{\begin{array}{l}2x+8y=21 \\6x-4y=14\end{array}\right.\)
Type below:
______________

Answer:
The soultion is (3.5, 1.75)

Explanation:
\(\left\{\begin{array}{l}2x+8y=21 \\6x-4y=14\end{array}\right.\)
To eliminate y terms, multiply the 2nd equation by 2
2(6x – 4y = 14)
2x + 8y = 21
Add the equations
2x + 8y = 21
+(12x – 8y = 28)
y is eliminated it has reversed coefficients. Solve for x
2x + 8y + 12x – 8y = 21 + 28
14x = 49
x = 49/14 = 3.5
Substituting x in either of the equation to find y
6(3.5) – 4y = 14
21 – 4y – 21 = 14 – 21
-4y = -7
y = -7/-4 = 1.75
The soultion is (3.5, 1.75)

Question 5.
\(\left\{\begin{array}{l}2x+y=3 \\-x+3y=-12\end{array}\right.\)
(________ , ________ )

Answer:

Explanation:
\(\left\{\begin{array}{l}2x+y=3 \\-x+3y=-12\end{array}\right.\)
To eliminate x terms, multiply the 2nd equation by 2
2(-x + 3y = -12)
-2x + 6y = -24
Add the equations
2x + y = 3
+(-2x + 6y = -24)
x is eliminated it has reversed coefficients. Solve for y
2x + y – 2x + 6y = 3 – 24
7y = -21
y = -21/7 = -3
Substituting y in either of the equation to find x
-x + 3(-3) = -12
-x -9 + 9 = -12 + 9
-x = -3
x = 3
The soultion is (3, -3)

Question 6.
\(\left\{\begin{array}{l}6x+5y=19 \\2x+3y=5\end{array}\right.\)
(________ , ________ )

Answer:
The soultion is (4, -1)

Explanation:
\(\left\{\begin{array}{l}6x+5y=19 \\2x+3y=5\end{array}\right.\)
To eliminate x terms, multiply the 2nd equation by 3
3(2x + 3y = 5)
6x + 9y = 15
Subtract the equations
6x + 5y = 19
-(6x + 9y = 15)
x is eliminated it has reversed coefficients. Solve for y
6x + 5y – 6x – 9y = 19 – 15
-4y = 4
y = 4/-4 = -1
Substituting y in either of the equation to find x
2x + 3(-1) = 5
2x – 3 + 3 = 5 + 3
2x = 8
x = 8/2 = 4
The soultion is (4, -1)

Question 7.
\(\left\{\begin{array}{l}2x+5y=16 \\-4x+3y=20\end{array}\right.\)
(________ , ________ )

Answer:
The soultion is (-2, 4)

Explanation:
\(\left\{\begin{array}{l}2x+5y=16 \\-4x+3y=20\end{array}\right.\)
To eliminate x terms, multiply the 1st equation by 2
2(2x + 5y = 16)
4x + 10y = 32
Add the equations
4x + 10y = 32
+(-4x + 3y = 20)
x is eliminated it has reversed coefficients. Solve for y
10y + 3y = 32 + 20
13y = 52
y = 52/13 = 4
Substituting y in either of the equation to find x
2x + 5(4) = 16
2x + 20 – 20 = 16 – 20
2x = -4
x = -4/2 = -2
The soultion is (-2, 4)

Question 8.
Bryce spent $5.26 on some apples priced at $0.64 each and some pears priced at $0.45 each. At another store he could have bought the same number of apples at $0.32 each and the same number of pears at $0.39 each, for a total cost of $3.62. How many apples and how many pears did Bryce buy?
a. Write equations to represent Bryce’s expenditures at each store
First store: _____________
Second store: _____________
Type below:
_____________

Answer:
First store: 0.64x + 0.45y = 5.26
Second store: 0.32x + 0.39y = 3.62

Explanation:
First store = 0.64x + 0.45y = 5.26
Second store = 0.32x + 0.39y = 3.62
where x is the number of apples and y is the number of pears.

Question 8.
b. Solve the system.
Number of apples: _______
Number of pears: _______

Answer:
Number of apples: 4
Number of pears: 6

Explanation:
First store = 0.64x + 0.45y = 5.26
Second store = 0.32x + 0.39y = 3.62
Multiply by 100
64x + 45y = 526
32x + 39y = 362
To eliminate x terms, multiply the 2nd equation by 2
2(32x + 39y = 362)
64x + 45y = 526
Subtract the equations
64x + 45y = 526
-(64x + 78y = 724)
x is eliminated it has reversed coefficients. Solve for y
64x + 45y – 64x – 78y = 526 – 724
-33y = -198
y = -198/-33 = 6
Substituting y in either of the equation to find x
32x + 39(6) = 362
32x + 234 – 234 = 362 – 234
32x = 128
x = 128/32 = 4
He bought 4 apples and 6 pears.

ESSENTIAL QUESTION CHECK-IN

Question 9.
When solving a system by multiplying and then adding or subtracting, how do you decide whether to add or subtract?
Type below:
_____________

Answer:
If the variable with the same coefficient but reversed sign, we add and if they have the same sign, we subtract.

Solving Systems by Elimination with Multiplication – Page No. 257

Question 10.
Explain the Error Gwen used elimination with multiplication to solve the system
\(\left\{\begin{array}{l}2x+6y=3 \\x-3y=-1\end{array}\right.\)
Her work to find x is shown. Explain her error. Then solve the system.
2(x − 3y) = -1
2x − 6y = -1
+2x + 6y = 3
_____________
4x + 0y = 2
x = \(\frac{1}{2}\)
Type below:
____________

Answer:
2x + 6y = 3
x – 3y = -1
To eliminate x terms, multiply the 2nd equation by 2
2(x – 3y = -1)
2x – 6y = -2
Error is the Gnew did not multiply the entire expression with 2.
Add the equations
2x + 6y = 3
+(2x – 6y = -2)
y is eliminated it has reversed coefficients. Solve for x
2x + 6y + 2x – 6y = 3 – 2
4x = 1
x = 1/4
Substituting x in either of the equation to find y
x – 3y = -1
1/4 – 3y – 1/4 = -1 -1/4
-3y = -5/4
y = -5/4(-3) = 5/12

Question 11.
Represent Real-World Problems At Raging River Sports, polyester-fill sleeping bags sell for $79. Down-fill sleeping bags sell for $149. In one week the store sold 14 sleeping bags for $1,456.
a. Let x represent the number of polyester-fill bags sold and let y represent the number of down-fill bags sold. Write a system of equations you can solve to find the number of each type sold.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 4: Solving Systems by Elimination with Multiplication img 16
Type below:
____________

Answer:
x + y = 14
79x + 149y = 1456
where x is the polyster-fill bags and y is the number of down-fill bags

Question 11.
b. Explain how you can solve the system for y by multiplying and subtracting.
Type below:
____________

Answer:
x + y = 14
79x + 149y = 1456
Multiply the second equation by 79. Subtract the new equation from the first equation and solve the resulting equation for y.

Question 11.
c. Explain how you can solve the system for y using substitution.
Type below:
____________

Answer:
Solve the second equation for x. Substitute the expression for x , in the first equation and solve the resulting equation for y.

Question 11.
d. How many of each type of bag were sold?
_______ polyester-fill
_______ down-fill

Answer:
9 polyester-fill
5 down-fill

Explanation:
x + y = 14
79x + 149y = 1456
To eliminate x terms, multiply the 2nd equation by 2
79(x + y = 14)
79x + 149y = 1456
Subtract the equations
79x + 79y = 1106
-(79x + 149y = 1456)
x is eliminated it has reversed coefficients. Solve for y
79x + 79y – 79x – 149y = 1106 – 1456
-70y = -350
y = -350/-70 = 5
Substituting y in either of the equation to find x
x + 5 = 14
x = 14 – 5
x = 9
There were 9 polyster-fill bags and 5 down-fill bags sold.

Question 12.
Twice a number plus twice a second number is 310. The difference between the numbers is 55. Find the numbers by writing and solving a system of equations. Explain how you solved the system.
x = _______
y = _______

Answer:
x = 105
y = 50

Explanation:
2x + 2y = 310
x – y = 55
To eliminate y terms, multiply the 2nd equation by 2
2(x – y = 55)
2x – 2y = 110
Add the equations
2x + 2y = 310
+ (2x – 2y = 110)
y is eliminated it has reversed coefficients. Solve for x
2x + 2y + 2x – 2y = 310 + 110
4x = 420
x = 420/4 = 105
Substituting x in either of the equation to find y
105 – y = 55
y = 105 – 55
y = 50
The solution is (105, 50)

Solving Systems by Elimination with Multiplication – Page No. 258

Question 13.
Represent Real-World Problems A farm stand sells apple pies and jars of applesauce. The table shows the number of apples needed to make a pie and a jar of applesauce. Yesterday, the farm picked 169 Granny Smith apples and 95 Red Delicious apples. How many pies and jars of applesauce can the farm make if every apple is used?
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 4: Solving Systems by Elimination with Multiplication img 17
_______ pies
_______ jars of applesauce

Answer:
21 pies
16 jars of applesauce

Explanation:
5x + 4y = 169
3x + 2y = 95
where x is the number of apples needed for pie and y is the number of apples for jar of applesauce
To eliminate y terms, multiply the 2nd equation by 2
2(3x + 2y = 95)
6x + 4y = 190
Subtract the equations
5x + 4y = 169
– (6x + 4y = 190)
y is eliminated it has reversed coefficients. Solve for x
5x + 4y – 6x – 4y = 169 – 190
-x = -21
x = -21/-1 = 21
Substituting x in either of the equation to find y
5(21) + 4y = 169
105 + 4y – 105 = 169 – 105
4y = 64
y = 64/4 = 16
The number of apples needed for pie is 21 and the number of apples for jar of applesauce is 16.

FOCUS ON HIGHER ORDER THINKING

Question 14.
Make a Conjecture Lena tried to solve a system of linear equations algebraically and in the process found the equation 5 = 9. Lena thought something was wrong, so she graphed the equations and found that they were parallel lines. Explain what Lena’s graph and equation could mean.
Type below:
____________

Answer:
Lena’s graph is a parallel line which means the graph does not intersect each other, hence they have no solutions. Equation 5 = 9 means variables are eliminated and this statement is not true. This linear system has no solution.

Question 15.
Consider the system
\(\left\{\begin{array}{l}2x+3y=6 \\3x+7y=-1\end{array}\right.\)
a. Communicate Mathematical Ideas Describe how to solve the system by multiplying the first equation by a constant and subtracting. Why would this method be less than ideal?
Type below:
____________

Answer:
Multiplying the first equation by a constant and subtracting
2x + 3y = 6
3x + 7y = -1
Multiply the first equation by 1.5 and subtract. This would be less than ideal because you would introduce decimals into the solution process.

Question 15.
b. Draw Conclusions Is it possible to solve the system by multiplying both equations by integer constants? If so, explain how.
Type below:
____________

Answer:
Yes

Explanation:
Multiply the first equation by 3 and the second equation by 2. Both x-term coefficients would be 6. Solve by eliminating the x-terms using subtraction.

Question 15.
c. Use your answer from part b to solve the system.
(_______ , _______)

Answer:
(9, -4)

Explanation:
2x + 3y = 6
3x + 7y = -1
Multiply the first equation by 3 and the second equation by 2.
3(2x + 3y = 6)
2(3x + 7y = -1)
Subtract the equations
6x + 9y = 18
-(6x + 14y = -2)
x is eliminated it has reversed coefficients. Solve for y
6x + 9y – 6x – 14y = 18 + 2
-5y = 20
y = 20/-5 = -4
Substituting y in either of the equation to find x
2x + 3(-4) = 6
2x = 18
x = 18/2 = 9
The solution is (9, -4)

Guided Practice – Solving Solving Special Systems – Page No. 262

Use the graph to solve each system of linear equations

Question 1.
A. \(\left\{\begin{array}{l}4x-2y=-6 \\2x-y=4\end{array}\right.\)
B. \(\left\{\begin{array}{l}4x-2y=-6 \\x+y=6\end{array}\right.\)
C. \(\left\{\begin{array}{l}2x-y=4 \\6x-3y=-12\end{array}\right.\)
STEP 1 Decide if the graphs of the equations in each system intersect, are parallel, or are the same line.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 5: Solving Solving Special Systems img 18
System A: The graphs __________
System B: The graphs __________
System C: The graphs __________

Answer:
System A: The graphs are parallel
System B: The graphs are intersecting
System C: The graphs are  the same line

Explanation:
System A: 4x – 2y = -6
2x – y = 4
System B: 4x – 2y = -6
x + y = 6
System C: 2x – y = 4
6x – 3y = 12

Question 1.
STEP 2 Decide how many points the graphs have in common.
a. Intersecting lines have _______________ point(s) in common.
b. Parallel lines have _______________ point(s) in common.
c. The same lines have ___________ point(s) in common.
a. __________
b. __________
c. __________

Answer:
a. Intersecting lines have one point(s) in common.
b. Parallel lines have no point(s) in common.
c. The same lines have infinitely many points (s) in common.

Explanation:
From the graphs,
Intersecting lines have one point(s) in common
Parallel lines have no point(s) in common
The same lines have infinitely many points (s) in common

Question 1.
STEP 3 Solve each system.
System A has __________ points in common, so it has __________ solution.
System B has __________ point in common. That point is the solution, __________.
System C has __________ points in common. ________ ordered pairs on the line will make both equations true.
Type below:
___________

Answer:
System A has no points in common, so it has no solution. System B has one point in common. That point is the solution, (1,5). System C has an infinite number of points in common. All ordered pairs on the line will make both equations true.

Explanation:
Number of solutions for each system
System A has no points in common, so it has no solution. System B has one point in common. That point is the solution, (1,5). System C has an infinite number of points in common. All ordered pairs on the line will make both equations true.

Solve each system. Tell how many solutions each system has.

Question 2.
\(\left\{\begin{array}{l}x-3y=4 \\-5x+15y=-20\end{array}\right.\)
___________

Answer:
infinitely many solutions

Explanation:
x – 3y = 4
-5x + 15y = -20
To eliminate y terms, multiply the 1st equation by 5
5(x – 3y = 4)
5x – 15y = 20
Add the equations
5x – 15y = 20
+(-5x + 15y = -20)
x and y is eliminated as it has reversed coefficients.
5x – 15y – 5x + 15y = 20 – 20
0 = 0
The statement is true, hence the solution has infinitely many solutions.

Question 3.
\(\left\{\begin{array}{l}6x+2y=-4 \\3x+y=4\end{array}\right.\)
___________

Answer:
no solution

Explanation:
6x + 2y = -4
3x + y = 4
To eliminate y terms, multiply the 2nd equation by 5
2(3x + y = 4)
6x + 2y = 8
Subtract the equations
6x + 2y = -4
-(6x + 2y = 8)
x and y is eliminated as it has reversed coefficients.
6x + 2y – 6x – 2y = -4 -8
0 = -12
The statement is false, hence the solution has no solution.

Question 4.
\(\left\{\begin{array}{l}6x-2y=-10 \\3x+4y=-25\end{array}\right.\)
___________

Answer:
one solution

Explanation:
6x – 2y = -10
3x + 4y = -25
To eliminate y terms, multiply the 1st equation by 2
2(6x – 2y = -10)
12x – 4y = -20
Add the equations
12x – 4y = -20
+(3x + 4y = -25)
y is eliminated as it has reversed coefficients. Solve for x.
12x – 4y + 3x + 4y = -20 – 25
15x = -45
x = -45/15 = -3
Substitute x in any one of the original equations and solve for y
3(-3) + 4y = -25
-9 + 4y + 9 = -25 + 9
4y = -16
y = -16/4
y = -4
There is one solution, (-3, -4)

ESSENTIAL QUESTION CHECK-IN

Question 5.
When you solve a system of equations algebraically, how can you tell whether the system has zero, one, or an infinite number of solutions?
Type below:
___________

Answer:
When x and y are eliminated and the statement is true, the system has infinitely many solutions.
When x and y are eliminated and the statement is false, the system has no solutions.
When the system has one solution by solving, the system has one solution.

8.5 Independent Practice – Solving Solving Special Systems – Page No. 263

Solve each system by graphing. Check your answer algebraically.

Question 6.
\(\left\{\begin{array}{l}-2x+6y=12 \\x-3y=3\end{array}\right.\)
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 5: Solving Solving Special Systems img 19
Solution: ______________
___________

Answer:
\(\left\{\begin{array}{l}-2x+6y=12 \\x-3y=3\end{array}\right.\)
Graph the equations on same coordinate plane
No solution as equations are parallel
grade 8 chapter 8 image 1
To eliminate y terms, multiply the 2nd equation by 2
2(x – 3y = 3)
2x – 6y = 6
Add the equations
-2x + 6y = 12
2x – 6y = 6
x and y is eliminated as it has reversed coefficients.
-2x + 6y + 2x – 6y = 12 + 6
0 = 18
The statement is false, hence the system has no solution.

Question 7.
\(\left\{\begin{array}{l}15x+5y=5 \\3x+y=1\end{array}\right.\)
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 5: Solving Solving Special Systems img 20
Solution: ______________
___________

Answer:
\(\left\{\begin{array}{l}15x+5y=5 \\3x+y=1\end{array}\right.\)
Graph the equations on same coordinate plane
grade 8 chapter 8 image 2
Infinitely many solutions as equations are overlapping
To eliminate y terms, multiply the 2nd equation by 5
5(3x + y = 1)
15x + 5y = 5
Subtarct the equations
15x + 5y = 5
-(15x + 5y = 5)
x and y is eliminated as it has reversed coefficients.
15x + 5y -15x – 5y = 5 – 5
0 = 0
The statement is true, hence the system has infinitely many solutions.

For Exs. 8–

14, state the number of solutions for each system of linear equations

Question 8.
a system whose graphs have the same slope but different y-intercepts
___________

Answer:
No solutions

Explanation:
Equations are parallel
No solutions

Question 9.
a system whose graphs have the same y-intercepts but different slopes
___________

Answer:
One solution

Explanation:
Equations are intersecting
One solution

Question 10.
a system whose graphs have the same y-intercepts and the same slopes
___________

Answer:
Infinitely many solutions

Explanation:
Equations are overlapping
Infinitely many solutions

Question 11.
a system whose graphs have different y-intercepts and different slopes
___________

Answer:
One solution

Explanation:
Equations are intersecting
One solution

Question 12.
the system
\(\left\{\begin{array}{l}y=2 \\y=-3\end{array}\right.\)
___________

Answer:
No solutions

Explanation:
Equations are parallel
No solutions

Question 13.
the system
\(\left\{\begin{array}{l}y=2 \\y=-3\end{array}\right.\)
___________

Answer:
One solution

Explanation:
Equations are intersecting
One solution

Question 14.
the system whose graphs were drawn using these tables of values:
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 5: Solving Solving Special Systems img 21
___________

Answer:
No solutions

Explanation:
Equations are parallel The slope is the same for both equations but the y-intercept is different.
No solutions

Question 15.
Draw Conclusions The graph of a linear system appears in a textbook. You can see that the lines do not intersect on the graph, but also they do not appear to be parallel. Can you conclude that the system has no solution? Explain.
___________

Answer:

No; although the lines do not intersect on the graph, they intersect at a point that is not on the graph. To prove that a system has no solution, you must do so algebraically

Solving Solving Special Systems – Page No. 264

Question 16.
Represent Real-World Problems Two school groups go to a roller skating rink. One group pays $243 for 36 admissions and 21 skate rentals. The other group pays $81 for 12 admissions and 7 skate rentals. Let x represent the cost of admission and let y represent the cost of a skate rental. Is there enough information to find values for x and y? Explain.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 5: Solving Solving Special Systems img 22
___________

Answer:
36x + 21y = 243
12x + 7y = 81
where x is the cost of admission and y is the cost of stake rentals.
Although the information can be used to develop a system of linear equation, where each equation has two variables when the system is solved, the number of solutions is infinite, Mee the values of x and y cannot be determined.
No

Question 17.
Represent Real-World Problems Juan and Tory are practicing for a track meet. They start their practice runs at the same point, but Tory starts 1 minute after Juan. Both run at a speed of 704 feet per minute. Does Tory catch up to Juan? Explain.
___________

Answer:
No; Both Juan and Tory-run at the same rate, so the lines representing the distances each has run are parallel. There is no solution to the system

FOCUS ON HIGHER ORDER THINKING

Question 18.
Justify Reasoning A linear system with no solution consists of the equation y = 4x − 3 and a second equation of the form y = mx + b. What can you say about the values of m and b? Explain your reasoning.
Type below:
___________

Answer:
y = 4x – 3
y = mx + b
Since the system has no solutions, the two equations are parallel. The value of the slope, m would be the same i.e. 4. The value of y-intercept, b can be any number except -3 as b is different for parallel lines.

Question 19.
Justify Reasoning A linear system with infinitely many solutions consists of the equation 3x + 5 = 8 and a second equation of the form Ax + By = C. What can you say about the values of A, B, and C? Explain your reasoning.
Type below:
___________

Answer:
3x + 5 = 8
Ax + By = C
Since the system has infinitely many solutions, the values of A, B, and C must all be the same multiple of 3, 5, and 8, respectively. The two equations represent a single line, so the coefficients and constants of one equation must be a multiple of the other.

Question 20.
Draw Conclusions Both the points (2, -2) and (4, -4) are solutions of a system of linear equations. What conclusions can you make about the equations and their graphs?
Type below:
___________

Answer:
If a system has more than one solution, the equations represent the same line and have infinitely many solutions.

Ready to Go On? – Model Quiz – Page No. 265

8.1 Solving Systems of Linear Equations by Graphing

Solve each system by graphing.

Question 1.
\(\left\{\begin{array}{l}y=x-1 \\y=2x-3\end{array}\right.\)
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Model Quiz img 23
(________ , ________)

Answer:
(2, 1)

Explanation:
y = x – 1
y = 2x – 3
Graph the equations on the same coordinate plane
grade 8 chapter 8 image 3
The solution of the system is the point of intersection
The solution is (2, 1)

Question 2.
\(\left\{\begin{array}{l}x+2y=1 \\-x+y=2\end{array}\right.\)
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Model Quiz img 24
(________ , ________)

Answer:
(-1, 1)

Explanation:
x + 2y = 1
-x + y = 2
Graph the equations on same coordinate plane
grade 8 chapter 8 image 4
The solution of the system is the point of intersection
The solution is (-1, 1)

8.2 Solving Systems by Substitution

Solve each system of equations by substitution.

Question 3.
\(\left\{\begin{array}{l}y=2x \\x+y=-9\end{array}\right.\)
(________ , ________)

Answer:
(-3, -6)

Explanation:
y = 2x
x + y = -9
Substitute y from equation 1 in the other equation.
x + 2x = -9
3x = -9
x = -9/3
x = -3
Then, y = 2(-3) = -6
The Solution is (-3, -6)

Question 4.
\(\left\{\begin{array}{l}3x-2y=11 \\x+2y=9\end{array}\right.\)
(________ , ________)

Answer:
(5, 2)

Explanation:
3x – 2y = 11
x + 2y = 9
Solve for x in equation 2
x = – 2y + 9
Substitute x from equation 2 in the other equation
3(-2y + 9) – 2y = 11
-6y + 27 -2y = 11
-8y = -16
y = -16/-8 = 2
Substitute y in any of the equations to find x
x + 2(2) = 9
x + 4 – 4 = 9 – 4
x = 5
The solution is (5, 2)

8.3 Solving Systems by Elimination

Solve each system of equations by adding or subtracting.

Question 5.
\(\left\{\begin{array}{l}3x+y=9 \\2x+y=5\end{array}\right.\)
(________ , ________)

Answer:
(4, -3)

Explanation:
\(\left\{\begin{array}{l}3x+y=9 \\2x+y=5\end{array}\right.\)
Subtract the equations
3x + y = 9
-(2x + y = 5)
y is eliminated as it has reversed coefficients. Solve for x
3x + y – 2x – y = 9 – 5
x = 4
Substituting x in either of the equation to find y
2(4) + y = 5
8 + y – 8 = 5 – 8
y = -3
The solution is (4, -3)

Question 6.
\(\left\{\begin{array}{l}-x-2y=4 \\3x+2y=4\end{array}\right.\)
(________ , ________)

Answer:
(4, -4)

Explanation:
\(\left\{\begin{array}{l}-x-2y=4 \\3x+2y=4\end{array}\right.\)
Add the equations
-x – 2y = 4
+(3x + 2y = 4)
y is eliminated as it has reversed coefficients. Solve for x
-x – 2y + 3x + 2y = 4 + 4
2x = 8
x = 8/2 = 4
Substituting x in either of the equation to find y
3(4) + 2y = 4
12 + 2y – 12 = 4 – 12
2y = -8
y = -8/2 = -4
The solution is (4, -4)

8.4 Solving Systems by Elimination with Multiplication

Solve each system of equations by multiplying first.

Question 7.
\(\left\{\begin{array}{l}x+3y=-2 \\3x+4y=-1\end{array}\right.\)
(________ , ________)

Answer:
(1, -1)

Explanation:
\(\left\{\begin{array}{l}x+3y=-2 \\3x+4y=-1\end{array}\right.\)
Subtract the equations
3x + 9y = -6
-(3x + 4y = -1)
x is eliminated as it has reversed coefficients. Solve for y
3x + 9y – 3x – 4y = -6 + 1
5y = -5
y = -5/5
y = -1
Substituting y in either of the equation to find x
x + 3(-1) = -2
x – 3 = -2
x = -2 + 3
x = 1
The solution is (1, -1)

Question 8.
\(\left\{\begin{array}{l}2x+8y=22 \\3x-2y=5\end{array}\right.\)
(________ , ________)

Answer:
(3, 2)

Explanation:
\(\left\{\begin{array}{l}2x+8y=22 \\3x-2y=5\end{array}\right.\)
Multiply equation 2 by 4 so that y can be eliminated
4(3x – 2y = 5)
12x – 8y = 20
Add the equations
2x + 8y = 22
+(12x – 8y = 20)
y is eliminated as it has reversed coefficients. Solve for x
2x + 8y + 12x – 8y = 22 + 20
14x = 42
x = 42/14
x = 3
Substituting y in either of the equation to find x
2(3) + 8y = 22
6 + 8y = 22
8y = 22 – 6
8y = 16
y = 16/8
y = 2
The solution is (3, 2)

8.5 Solving Special Systems

Solve each system. Tell how many solutions each system has.

Question 9.
\(\left\{\begin{array}{l}-2x+8y=5 \\x-4y=-3\end{array}\right.\)
_____________

Answer:
no solution

Explanation:
\(\left\{\begin{array}{l}-2x+8y=5 \\x-4y=-3\end{array}\right.\)
Multiply equation 2 by 2 so that y can be eliminated
2(x – 4y = -3)
2x – 8y = -6
Add the equations
-2x + 8y = 5
+(2x – 8y = -6)
x and y is eliminated
-2x + 8y + 2x – 8y = 5 – 6
0 = -1
The statement is false. Hence, the system has no solution.

Question 10.
\(\left\{\begin{array}{l}6x+18y=-12 \\x+3y=-2\end{array}\right.\)
_____________

Answer:
infinitely many solutions

Explanation:
\(\left\{\begin{array}{l}6x+18y=-12 \\x+3y=-2\end{array}\right.\)
Multiply equation 2 by 6 so that x can be eliminated
6(x + 3y = -2)
6x + 18y = -12
Subtract the equations
6x + 18y = -12
-(6x + 18y = -12)
x and y is eliminated
6x + 18y -6x -18y = -12 + 12
0 = 0
The statement is true. Hence, the system has infinitely many solutions.

ESSENTIAL QUESTION

Question 11.
What are the possible solutions to a system of linear equations, and what do they represent graphically?
Type below:
___________

Answer:
System of linear equations can have no solution, which is represented by parallel lines; one solution, which is represented by intersecting lines; and infinitely many solutions, which is represented by overlapping lines.

Selected Response – Mixed Review – Page No. 266

Question 1.
The graph of which equation is shown?
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Mixed Review img 25
Options:
A. y = −2x + 2
B. y = −x + 2
C. y = 2x + 2
D. y = 2x + 1

Answer:
C. y = 2x + 2

Explanation:
Option A and B are eliminated as the slope of the graph is 2.
Option D is eliminated as the y-intercept from the graph should be 2.
Option C is the equation of the graph

Question 2.
Which best describes the solutions to the system
\(\left\{\begin{array}{l}x+y=-4 \\-2x-2y=0\end{array}\right.\)
Options:
A. one solution
B. no solution
C. infinitely many
D. (0, 0)

Answer:
B. no solution

Explanation:
\(\left\{\begin{array}{l}x+y=-4 \\-2x-2y=0\end{array}\right.\)
Multply equation 1 by 2 so that x can be eliminated
2(x + y = -4)
2x + 2y = -8
Add the equations
2x + 2y = -8
-2x – 2y = 0
x and y is eliminated
2x + 2y – 2x -2y = -8 + 0
0 = -8
The statement is false. Hence, the system has no solution.

Question 3.
Which of the following represents 0.000056023 written in scientific notation?
Options:
A. 5.6023 × 105
B. 5.6023 × 104
C. 5.6023 × 10-4
D. 5.6023 × 10-5

Answer:
D. 5.6023 × 10-5

Explanation:
Move the decimal 5 points right to get the equation.
D. 5.6023 × 10-5

Question 4.
Which is the solution to
\(\left\{\begin{array}{l}2x-y=1 \\4x+y=11\end{array}\right.\)
Options:
A. (2, 3)
B. (3, 2)
C. (-2, 3)
D. (3, -2)

Answer:
A. (2, 3)

Explanation:
\(\left\{\begin{array}{l}2x-y=1 \\4x+y=11\end{array}\right.\)
Add the equations
2x – y = 1
4x + y = 11
y is eliminated as it has reversed coefficients. Solve for x.
2x – y + 4x + y = 1 + 11
6x = 12
x = 12/6 = 2
Substituting x in either of the equation to find y
4(2) + y = 11
8 + y = 11
y = 11 – 8
y = 3
The solution is (2, 3)

Question 5.
Which expression can you substitute in the indicated equation to solve
\(\left\{\begin{array}{l}3x-y=5 \\x+2y=4\end{array}\right.\)
Options:
A. 2y – 4 for x in 3x – y = 5
B. 4 – x for y in 3x – y = 5
C. 3x – 5 for y in 3x – y = 5
D. 3x – 5 for y in x + 2y = 4

Answer:
D. 3x – 5 for y in x + 2y = 4

Explanation:
\(\left\{\begin{array}{l}3x-y=5 \\x+2y=4\end{array}\right.\)
Solve for y in equation 1
y = 3x – 5
Substitute in other equation x + 2y = 4

Question 6.
What is the solution to the system of linear equations shown on the graph?
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Mixed Review img 26
Options:
A. -1
B. -2
C. (-1, -2)
D. (-2, -1)

Answer:
C. (-1, -2)

Explanation:
The point of intersection is (-1, -2), which is the solution of the system

Question 7.
Which step could you use to start solving
\(\left\{\begin{array}{l}x-6y=8 \\2x-5y=3\end{array}\right.\)
Options:
A. Add 2x – 5y = 3 to x – 6y = 8.
B. Multiply x – 6y = 8 by 2 and add it to 2x – 5y = 3.
C. Multiply x – 6y = 8 by 2 and subtract it from 2x – 5y = 3.
D. Substitute x = 6y – 8 for x in 2x – 5y = 3.

Answer:
C. Multiply x – 6y = 8 by 2 and subtract it from 2x – 5y = 3.

Explanation:
x – 6y = 8
2x – 5y = 3
Multiply the 1st equation by 2 so that the coefficient of variable x is the same in both equations
Subtract the equations as x has the same sign.

Mini-Task

Question 8.
A hot-air balloon begins rising from the ground at 4 meters per second at the same time a parachutist’s chute opens at a height of 200 meters. The parachutist descends at 6 meters per second.
a. Define the variables and write a system that represents the situation.
Type below:
_____________

Answer:
y represents the distance from the ground and x represents the time in seconds
y = 4x
y = -6x + 200

Question 8.
b. Find the solution. What does it mean?
Type below:
_____________

Answer:
Substitute y from the equation 1 in the equation 2
4x = -6x + 200
4x + 6x = -6x + 200 + 6x
10x = 200
x = 200/10 = 20
Substitute x in any one of the equations and solve for x
y = 4(20) = 80
The solution is (20, 80)
The ballon and parachute meets after 20sec at 80m from the ground.

Final Words:

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Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume

go-math-grade-7-chapter-9-circumference-area-and-volume-answer-key

Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume: Get the solutions to all the questions in this article. Download Go Math Grade 7 Answer Key for Chapter 9 Circumference, Area and Volume pdf for free. Know how and where to use the formulas with the help of the HMH Go Math Grade 7 Solution Key Chapter 9 Circumference, Area, and Volume.

Download Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume Pdf

The pupils who are in search of solutions of grade 7 chapter 9 circumference, area, and volume can get them on Go Math Answer Key. The students of 7th Grade can know how to find the area, circumference, and volume of various shapes here. Learn the different methods to solve the problems in Chapter 9 Circumference, Area and Volume using the formulas. We have provided the solutions as per the topics in the below sections.

Guided Practice – Page No. 268

Find the circumference of each circle.

Question 1.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 1
________ in

Answer: 56.57 in

Explanation:
Circumference of the circle = 2πr = 2 x 22/7 x 9 = 56.57 in

Question 2.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 2
________ cm

Answer: 44 cm

Explanation:
Circumference of the circle = 2πr = 2 x 22/7 x 7 = 44 cm

Find the circumference of each circle. Use 3.14 or \(\frac{22}{7}\) for π. Round to the nearest hundredth, if necessary.

Question 3.
______ m

Question 4.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 3
______ yd

Answer: 30.15 yd

Explanation:
Circumference of the circle = 2πr = 2 x 3.14 x 4.8 = 30.144 yd

Question 5.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 4
______ in

Answer: 7.5 in

Explanation:
Circumference of the circle = 2πr = 2 x 3.14 x 7.5 = 47.1 in

Question 6.
A round swimming pool has a circumference of 66 feet. Carlos wants to buy a rope to put across the diameter of the pool. The rope costs $0.45 per foot, and Carlos needs 4 feet more than the diameter of the pool. How much will Carlos pay for the rope?
$ ______

Answer: $6.525

Explanation:
Circumference of the swimming pool = 66 feet
πd = 66
22/7 x d = 66
d = 66 x 7/ 22 = 10.5
The diameter of the pool = 10.5 feet
Carlos needs 4 feet more than the diameter of the pool.
Total rope needed = 10.5 + 4 = 14.5 feet
Cost of rope per foot = $0.45
Total cost of the rope = 14.5 x $0.45 = $6.525
Therefore the total cost of the rope = $6.525

Find each missing measurement to the nearest hundredth. Use 3.14 for π.

Question 7.
r =
d =
C = π yd
r = ________ yd
d = ________ yd

Answer:
r = 0.5 yd
d = 1 yd

Explanation:
Circumference = π yd
2πr = π yd
r = 1/2 yd = 0.5 yd
d = 2r = 2 [1/2] = 1 yd

Question 8.
r ≈
d ≈
C = 78.8 ft
r ≈ ________ ft
d ≈ ________ ft

Answer:
r = 495.31 ft
d = 990.62 ft

Explanation:
Circumference = 78.8 ft
2πr = 78.8 ft
r = 2 x 22/7 x 78.8 = 495.31 ft
d = 2 x 495.31 = 990.62 ft

Question 9.
r ≈
d ≈ 3.4 in
C =
r ≈ ________ in
C = ________ in

Answer:
r = 1.7 in
c = 10.68 in

Explanation:
Diameter = 3.4 in
Circumference = πd = 22/7 x 3.4 in = 10.68 in
r = d/2 = 1.7 in

Essential Question Check-In

Question 10.
Norah knows that the diameter of a circle is 13 meters. How would you tell her to find the circumference?
Type below:
____________

Answer: Circumference = 16.82 meters

Explanation:
Given,
Diameter = 13 meters
Circumference = πd = 22/7 x 13 = 16.82 meters

Independent Practice – Page No. 269

For 11–13, find the circumference of each circle. Use 3.14 or \(\frac{22}{7}\) for π. Round to the nearest hundredth, if necessary.

Question 11.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 5
_______ ft

Answer:
Cicumference = 18.526 ft = 19 ft (approx)

Explanation:
Given:
Diameter = 5.9 ft
Cicumference = πd = 3.14 x 5.9 = 18.526 ft = 19 ft (approx)

Question 12.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 6
_______ cm

Answer:
Cicumference =176 cm

Explanation:
Given:
Radius = 56 cm
Cicumference = πd = 22/7 x 56 = 176 cm

Question 13.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 7
_______ in

Answer:
Cicumference = 110 in

Explanation:
Given:
Diameter = 35 in
Cicumference = πd = 22/7 x 35 = 110 in

Question 14.
In Exercises 11–13, for which problems did you use \(\frac{22}{7}\) for π? Explain your choice.
Type below:
_____________

Answer:
11th question as 3.14 and the 12 and 13 questions as π

Explanation:
We can take 3.14 as π for 11 th question because the diameter is given in decimal points.
And in questions 12 and 13 we need to take π because the radius and diameter are given in whole number form.

Question 15.
A circular fountain has a radius of 9.4 feet. Find its diameter and circumference to the nearest tenth.
d = _________ ft
C = _________ ft

Answer:
d = 19 ft
C = 59 ft

Explanation:
Given:
Radius = 9.4 ft
Diameter = 2r = 2 x 9.4  = 18.8 ft = 19 ft (approx)
Circumference = πd = 22/7 x 18.8 = 59.08 = 59 ft (approx)

Question 16.
Find the radius and circumference of a CD with a diameter of 4.75 inches.
r = _________ in
C = _________ in

Answer:
r = 2.4 in
C = 15 in

Explanation:
Given:
Diameter = 4.75 in
Radius = r/2 = 4.75/2 = 2.37 in = 2.4 in (approx)
Circumference = πd = 22/7 x 4.75 = 14.92 in =15 in (approx)

Question 17.
A dartboard has a diameter of 18 inches. What are its radius and circumference?
r = _________ in
C = _________ in

Answer:
r = 9 in
C = 56.6 in

Explanation:
Given:
Diameter = 18 in
Radius = r/2 = 18/2 = 9 in
Circumference = πd = 22/7 x 18 = 56.57 in = 56.6 in (approx)

Question 18.
Multistep
Randy’s circular garden has a radius of 1.5 feet. He wants to enclose the garden with edging that costs $0.75 per foot. About how much will the edging cost? Explain.
$ _______

Answer:

Explanation:
Given:
The radius of the garden= 1.5 ft
Circumference of the garden = 2πr = 2 x 22/7 x 1.5 = 9.42 ft
Cost of enclosing the garden per foot = $0.75
Total cost of edging = 9.42 x $0.75 = $7.06 = $7 (approx)

Question 19.
Represent Real-World Problems
The Ferris wheel shown makes 12 revolutions per ride. How far would someone travel during one ride?
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 8
_______ ft

Answer: Total distance travelled in one ride is 4,752 ft

Explanation:
Given:
The diameter of the Ferris wheel= 63 ft
Circumference of the Ferris wheel = 2πr = 2 x 22/7 x 63 = 396 ft
Total number of revolutions = 12
Total distance travelled = 12 x 396 = 4,752 ft

Question 20.
The diameter of a bicycle wheel is 2 feet. About how many revolutions does the wheel make to travel 2 kilometres? Explain. Hint: 1 km ≈ 3,280 ft
_______ revolutions

Answer:
1044 revolutions

Explanation:
Given:
Diametre of the bicycle wheel = 2 feet
Total distance travelled = 2 kilometres
We know that,
1 km ≈ 3,280 ft
2 km = 2 x 3,280 = 6,560 ft
Circumference of the bicycle = Distance travelled in one revolution = πd = 22/7 x 2 = 6.28 ft = 6.3 ft
Total number of revolutions = Total distance travelled / distance travelled in one revolution
= 6560 / 6.28 = 1044  revolutions

Question 21.
Multistep
A map of a public park shows a circular pond. There is a bridge along a diameter of the pond that is 0.25 mi long. You walk across the bridge, while your friend walks halfway around the pond to meet you at the other side of the bridge. How much farther does your friend walk?
_______ mi

Answer:

Explanation:
Given,
The diameter of the pond = 0.25 mi
The length of the bridge = The diameter of the pond = 0.25 mi
Then the distance walked by the man = 0.25 mi
Distance travelled by the friend = Halfway around the pond to meet you at the other side of the bridge = πd/2
= 22/7 x 0.25/2  = 0.39 = 0.4 mi
The friend travelled more distance compared to the man
The more distance travelled by the friend = 0.39 – 0.25 = 0.14 mi

Page No. 270

Question 22.
Architecture
The Capitol Rotunda connects the House and the Senate sides of the U.S. Capitol. Complete the table. Round your answers to the nearest foot.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 9
Type below:
_____________

Answer:
Radius = 48 ft
Diameter = 96 ft

Explanation:
Given
Height = 180 ft
Circumference = 301.5 ft
πd = 301.5
22/7 x d = 301.5
d = 95.93 = 96 ft
r = d/2 = 96/2 = 48 ft

H.O.T.

Focus on Higher Order Thinking

Question 23.
Multistep
A museum groundskeeper is creating a semicircular statuary garden with a diameter of 30 feet. There will be a fence around the garden. The fencing costs $9.25 per linear foot. About how much will the fencing cost altogether?
$ _______

Answer:
The total cost of fencing = $712

Explanation:
Given,
The diameter = 30 ft
Circumference of the garden in the shape of circle = 2πr
Circumference of the semicircle = πr = πd/2 =  22/7 x 30/2 = 47.14ft
Cost of fencing for each foot = $9.25
The total cost of fencing the semicircular garden = 47.14 x $9.25 + 30 x  $9.25  = $712 (approx)

Question 24.
Critical Thinking
Sam is placing rope lights around the edge of a circular patio with a diameter of 18 feet. The lights come in lengths of 54 inches. How many strands of lights does he need to surround the patio edge?
_______ strands

Answer: 12 and a half strands of light = 13 strands (approx)

Explanation:
Given,
The diameter of the circular patio = 18 ft = 216 inch
Circumference of the circular patio = πd = 22/7 x 216 = 678.85 inch
The lights will come in a length (in one strand)= 54 inches
Total number of strands of light required for the circular patio
= Circumference of the circular patio/ The lights will come in a length (in one strand) = 678.85/54 = 12.57 = 12 and a half strands of light

Question 25.
Represent Real-World Problems
A circular path 2 feet wide has an inner diameter of 150 feet. How much farther is it around the outer edge of the path than around the inner edge?
_______ feet

Answer: about 12.6 ft

Explanation:
Given,
Width of the circular path = 2 ft
The inner diameter of the circular path = 150 ft
The outer diameter of the circular path = 150 + 2(2) = 154 ft
Inner circumference = πd = 150 π
Outer circumference =  πd = 154π
Distance between the outer and inner edge = 154 π – 150 π = 4 π = 12.6 ft

Question 26.
Critique Reasoning
Gear on a bicycle has the shape of a circle. One gear has a diameter of 4 inches, and a smaller one has a diameter of 2 inches. Justin says that the circumference of the larger gear is 2 inches more than the circumference of the smaller gear. Do you agree? Explain your answer.
_______

Answer:
Justin statement is incorrect.

Explanation:
The circumference of the larger gear = πd = 4π
The circumference of the smaller gear = πd = 2π
Since, 2 x 2π = 4π, the circumference of the larger gear is two times the circumference of the smaller gear.
Since = 4π – 2π = 2π = 6.28
Therefore, The larger circumference is not 2 inches more than the smaller circumference

Question 27.
Persevere in Problem Solving
Consider two circular swimming pools. Pool A has a radius of 12 feet, and Pool B has a diameter of 7.5 meters. Which pool has a greater circumference? How much greater? Justify your answers.
_______

Answer:
Pool B about 0.9 meters

Explanation:
Given,
Pool A has a diameter = 24 ft
Pool B has a diameter = 7.5 m
We know that,
1 ft = 0.3 metres
24 ft = 7.2 metres
The pool B has a greater diameter so it has a greater circumference.
Circumference of the pool A = 7.2π
Circumference of the pool B = 7.5π
Difference between the circumferences = 7.5π – 7.2π = 0.9 meters.

Guided Practice – Page No. 274

Find the area of each circle. Round to the nearest tenth if necessary. Use 3.14 for π.

Question 1.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 10
_______ m2

Answer: 153.9 m2

Explanation:
Given:
Diameter = 14 m
Radius = 14/2 = 7 m
Area of the circle = πr2
= 3.14 x 7 x 7 = 153.86 = 153.9 m2

Question 2.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 11
_______ mm2

Answer: 452.2 mm2

Explanation:
Given:
Radius =12mm
Area of the circle = πr2
= 3.14 x 12 x 12 = 3.14(144) = 452.2mm2

Question 3.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 12
_______ yd2

Answer: 314 yd2

Explanation:
Given:
Diameter = 20yd
Radius = 20/2 = 10yd
Area of the circle = πr2
= 3.14 x 10 x 10 = 3.14(100) = 314yd2

Solve. Use 3.14 for π.

Question 4.
A clock face has a radius of 8 inches. What is the area of the clock face? Round your answer to the nearest hundredth.
_______ in2

Answer: 200.96 in2

Explanation:
Given:
Radius = 8inches
Area of the clock face = πr2
= 3.14 x 8 x 8= 3.14(64) = 200.96 in2

Question 5.
A DVD has a diameter of 12 centimeters. What is the area of the DVD? Round your answer to the nearest hundredth.
_______ cm2

Answer: 113.04 cm2

Explanation:
Given:
Diameter = 12 centimeters
Radius = 12/2 = 6 centimeters
Area of the DVD= πr2
= 3.14 x 6 x 6 = 3.14(36) = 113.04 cm2

Question 6.
A company makes steel lids that have a diameter of 13 inches. What is the area of each lid? Round your answer to the nearest hundredth.
_______ in2

Answer: 132.67 in2

Explanation:
Given:
Diameter = 13 inches
Radius = 13/2 = 6.5 inches
Area of each lid= πr2
= 3.14 x 6.5 x 6.5 = 3.14(42.25) = 132.67 in2

Find the area of each circle. Give your answers in terms of π.

Question 7.
C = 4π
A =
Type below:
______________

Answer: 4π

Explanation:
Given:
Circumcenter = 4π
2πr = 4π
Radius = 4/2 = 2 units
Area of the circle = πr2
= π x 2 x 2 = π(4) = 4π square units

Question 8.
C = 12π
A =
Type below:
______________

Answer: 36π

Explanation:
Given:
Circumcenter = 12π
2πr = 12π
Radius =6 units
Area of the circle = πr2
= π x 6 x 6 = π(36) = 36π square units

Question 9.
C = \(\frac{π}{2}\)
A =
Type below:
______________

Answer: π/16

Explanation:
Given:
Circumcenter = \(\frac{π}{2}\)
2πr = \(\frac{π}{2}\)
Radius = 1/4 units
Area of the circle = πr2
= π x 1/4 x 1/4 = π(1/16) = π/16 square units

Question 10.
A circular pen has an area of 64π square yards. What is the circumference of the pen? Give your answer in terms of π
Type below:
______________

Answer: 16π

Explanation:
Given:
Area of the circular pen = 64π square yards
πr2 = 64π
r = 8 yards
Circumference of the circle = 2πr = 2 x 8 x π = 16π yards

Essential Question Check-In

Question 11.
What is the formula for the area A of a circle in terms of the radius r?
Type below:
______________

Answer: πr2

Explanation:
Area of a circle = πr2

Independent Practice – Page No. 275

Question 12.
The most popular pizza at Pavone’s Pizza is the 10-inch personal pizza with one topping. What is the area of a pizza with a diameter of 10 inches? Round your answer to the nearest hundredth.
_______ in2

Answer: 78.5 in2

Explanation:
Given:
Diameter = 10 inches
Radius = 10/2 = 5 inches
Area of a pizza = πr2
= 3.14 x 5 x 5 = 3.14(25) = 78.5 in2

Question 13.
A hubcap has a radius of 16 centimeters. What is the area of the hubcap? Round your answer to the nearest hundredth.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 13
_______ cm2

Answer: 803.84 cm2

Explanation:
Given:
Radius = 16 cm
Area of the circle = πr2
= 3.14 x 16 x 16 = 3.14(256) = 803.84 cm2

Question 14.
A stained glass window is shaped like a semicircle. The bottom edge of the window is 36 inches long. What is the area of the stained glass window? Round your answer to the nearest hundredth.
_______ in2

Answer: 508.68 in2

Explanation:
Area of the semicircle = 1/2 πr2 = 1/2(3.14)(18)(18) = 1/2 (3.14)(324) = 1.57(324) = 508.68 in 2

Question 15.
Analyze Relationships
The point (3,0) lies on a circle with the centre at the origin. What is the area of the circle to the nearest hundredth?
_______ units2

Answer: 28.26 units2

Explanation:
Radius = 3
Area of the circle = πr2 = π(3)2 = 3.14(9) = 28.26 units2

Question 16.
Multistep
A radio station broadcasts a signal over an area with a radius of 50 miles. The station can relay the signal and broadcast over an area with a radius of 75 miles. How much greater is the area of the broadcast region when the signal is relayed? Round your answer to the nearest square mile.
_______ mi2

Answer: 9813 mi2

Explanation:
Given:
The radius of a radio station broadcasted the signal (r) = 50 miles
The greatest radius to which the broadcast can be relayed (R) = 75 miles
The greatest area of the broadcast region when the signal is relayed = πR2-πr2 = π(75) (75) – π (50) (50)
= 5625π – 2500π
= 3125π
= 3125(3.14) = 9813 mi2(approx)

Question 17.
Multistep
The sides of a square field are 12 meters. A sprinkler in the center of the field sprays a circular area with a diameter that corresponds to a side of the field. How much of the field is not reached by the sprinkler? Round your answer to the nearest hundredth.
_______ m2

Answer:30.96 m2

Explanation:
Given:
The side of the square = 12 meters
The diameter circular area of the field in the centre = The side of the square = 12 meters
The radius of the field = 12/2 = 6 meters
Area of the field which is not reached by the sprinkler = Area of the square – Area of the circular area
= (side)2-πr2 = (12)(12) – π (6) (6)
= 144 – 36 (3.14)
= 144 – 113.04
= 30.96 m2

Question 18.
Justify Reasoning
A small silver dollar pancake served at a restaurant has a circumference of 2π inches. A regular pancake has a circumference of 4π inches. Is the area of the regular pancake twice the area of the silver dollar pancake? Explain.
_______

Answer: No, the area of the regular pancake is 4 times the area of the silver dollar pancake

Explanation:
Silver Dollar pancake:
Circumference of the silver Dollar pancake = 2π inches
2πr = 2π
r = 1 inch
Area of the silver dollar pancake = πr2 = π (1) (1) = π in2

Regular pancake:
Circumference of the regular pancake = 4π inches
2πr = 4π
r = 2 inch
Area of the silver dollar pancake = πr2 = π (2) (2) = 4π in2

Therefore, the area of the regular pancake is 4 times the area of the silver dollar pancake

Question 19.
Analyze Relationships
A bakery offers a small circular cake with a diameter of 8 inches. It also offers a large circular cake with a diameter of 24 inches. Does the top of the large cake have three times the area of that of the small cake? If not, how much greater is its area? Explain.
_______

Answer: No, the area of the large cake is 9 times the area of the small cake

Explanation:
Small Cake:
The diameter of the small cake= 8 inches
The radius of the small cake = 8/2 = 4 inches
Area of the small cake  = πr2 = π (4) (4) = 16 π in2

Large Cake:
The diameter of the large cake= 24 inches
The radius of the large cake = 24/2 = 12 inches
Area of the large cake  = πr2 = π (12) (12) = 144 π in2

Since 144 π/ 16 π = 9
Therefore the

area of the large cake is 9 times the area of the small cake.

Page No. 276

Question 20.
Communicate Mathematical Ideas
You can use the formula A = \(\frac{C^{2}}{4π}\) to find the area of a circle given the circumference. Describe another way to find the area of a circle when given the circumference.
Type below:
____________

Answer: Area = C2/4π

Explanation:
Circumference of the circle = 2πr
C = 2πr
Divide both sides by 2π
then, r = C/2π
Area of the circle = πr2
Substitute C/2π for r:
Area = π(c/2π)2 = C2/4π

Question 21.
Draw Conclusions
Mark wants to order a pizza. Which is the better deal? Explain.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 14
_____

Answer: The pizza of 18 inches is a better deal

Explanation:
Given:
The diameter of the pizza = 12 inches
The radius of the pizza = 12/2= 6 inches
Area of the circle = πr2
= (3.14)(6)(6) = 113 (approx) in2
The total cost of the pizza = $10
Cost of the pizza per inch = $10/113 = $0.09 per square inch

The diameter of the pizza = 18 inches
The radius of the pizza = 18/2= 9 inches
Area of the circle = πr2
= (3.14)(9)(9) = 254 (approx) in2
The total cost of the pizza = $20
Cost of the pizza per inch = $20/254 = $0.08 per inch

Question 22.
Multistep
A bear was seen near a campground. Searchers were dispatched to the region to find the bear.
a. Assume the bear can walk in any direction at a rate of 2 miles per hour. Suppose the bear was last seen 4 hours ago. How large an area must the searchers cover? Use 3.14 for π. Round your answer to the nearest square mile.
_____ mi2

Answer: 201mi2

Explanation:
The bear can walk a distance = 2 x 4 = 8 miles
Since it is walking 2 miles per hour for 4 hours
The radius of the bear = 8 miles
Area of the circle = πr2
= (3.14)(8)(8) = 201 (approx) mi2

Question 22.
b. What If? How much additional area would the searchers have to cover if the bear were last seen 5 hours ago?
_____ mi2

Answer: 113mi2

Explanation:
If the bear for 5 hours then,
The bear can walk a distance = 2 x 5 = 10 miles
Since it is walking 2 miles per hour for 5 hours
The radius of the bear = 10 miles
Area of the circle = πr2
= (3.14)(10)(10) = 314 (approx) mi2

The additional area covered by the searches = 314 – 201 = 113 mi2

H.O.T.

Focus on Higher Order Thinking

Question 23.
Analyze Relationships
Two circles have the same radius. Is the combined area of the two circles the same as the area of a circle with twice the radius? Explain.
_____

Answer: No

Explanation:
If the radius of two circles is the same.
then, Let the radii of the circles be 1.
The area of each circle =  π square units
The combined area of 2 circles =π+π = 2π square units

If the radius is doubled.
then, Let the radii of the circles be 2
The area of each circle =  4π square units
The combined area of 2 circles =  4π+4π = 8π square units

Therefore the areas of both cases are not the same.

Question 24.
Look for a Pattern
How does the area of a circle change if the radius is multiplied by a factor of n, where n is a whole number?
Type below:
____________

Answer: The new area is then n2 times the area of the original circle.

Explanation:
If the radius is multiplied by a factor “n”
then, the new radius = rn
The area of the circle (with radius rn) = π(rn)= n2 (πr2).
Therefore the new area is n2 times the area of the original circle.

Question 25.
Represent Real World Problems
The bull’s-eye on a target has a diameter of 3 inches. The whole target has a diameter of 15 inches. What part of the whole target is the bull’s-eye? Explain.
Type below:
____________

Answer: 1/25 of the target

Explanation:
Bull’s eye:
Diameter of Bull’s eye = 3 inches
Radius of Bull’s eye = 3/2 = 1.5 inches
Area of the Bull’s eye = π(r)= π(1.5)2 = 2.25π
Target:
Diameter of the target = 15 inches
Radius of the target = 15/2 = 7.5 inches
Area of the target = π(r)= π(7.5)2 = 56.25π

The part of Bull’s eye in the whole target = 2.25π/ 56.25π = 1/25

Therefore the 1/25th part of the whole target is the Bull’s eye.

Guided Practice – Page No. 280

Question 1.
A tile installer plots an irregular shape on grid paper. Each square on the grid represents 1 square centimeter. What is the area of the irregular shape?
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 15
_____ cm2

Answer: Area of the irregular shape = 34 cm2

Explanation:
STEP1 First divide the irregular shapes into polygons.
STEP2 The irregular shape can be divided into a triangle, rectangle, parallelogram
STEP3 Areas of the polygons
Area of triangle = 1/2 (base x height) = 1/2 (4 x 2) = 4 cm2
Area of the rectangle = length x breadth = 5 x 3 = 15 cm2
Area of the parallelogram = base x height = 5 x 3 = 15 cm2
Area of the irregular shape = (15+15+5) cm2= 34cm2

Question 2.
Show two different ways to divide the composite figure. Find the area both ways. Show your work below.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 16
_____ cm2

Answer: Area of the figure in both ways = 288 cm2

Explanation:
The first way to divide up the composite shape is to divide it into an 8 by 9 rectangle and a 12 by 18 rectangle.
The area of the first rectangle = Length x breadth = 9 x 8 = 72 cm2
The area of the second rectangle =  Length x breadth = 18 x 12 = 216 cm2
The total area of the figure = 72 + 216 = 288 cm2

Question 3.
Sal is tiling his entryway. The floor plan is drawn on a unit grid. Each unit length represents 1 foot. Tile costs $2.25 per square foot. How much will Sal pay to tile his entryway?
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 17
$ _____

Answer: Sal will pay $97.875

Explanation:
Separate this figure into trapezium and parallelogram.
Area of the trapezium = 1/2 (a+b)h = 1/2 (7+4) 5 = 1/2 (11) 5 = 27.5 ft2
Area of the parallelogram = base x height = 4 x 4 = 16 ft2

The total area of the figure = 27.5 + 16 = 43.5ft2
Cost of each square foot = $2.25
Amount paid by Sal = 43.5 x 2.25 = $97.875

Essential Question Check-In

Question 4.
What is the first step in finding the area of a composite figure?
Type below:
______________

Answer:
The first step in finding the area of a composite figure is to divide it up into smaller basic shapes.

Explanation:
The first step in finding the area of a composite figure is to divide it up into smaller basic shapes such as triangles, squares, rectangles, parallelograms, circles and trapezium.
Then calculate the area of each figure and add them to find the area of the figure.

Independent Practice – Page No. 281

Question 5.
A banner is made of a square and a semicircle. The square has side lengths of 26 inches. One side of the square is also the diameter of the semicircle. What is the total area of the banner? Use 3.14 for π.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 18
_____ in2

Answer: 941.33 in2

Explanation:
Area of the square = side x side = 26 x 26 = 676 in2
Area of the semicircle =1/2 πr2= 1/2 (3.14) (13) (13) = 1/2 (3.14) (169) = 265.33 in2
Area of the figure = 676 + 265.33 = 941.33 in2

Question 6.
Multistep
Erin wants to carpet the floor of her closet. A floor plan of the closet is shown.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 19
a. How much carpet does Erin need?
_____ ft2

Answer: 61 ft2

Explanation:
Area of the rectangle = length x breadth = 4 x 10 = 40 ft
Area of the triangle = 1/2 x base x height = 1/2 x 6 x 7 = 21 ft
The total area of the figure = 40+21 = 61 ft2

Question 6.
b. The carpet Erin has chosen costs $2.50 per square foot. How much will it cost her to carpet the floor?
$ _____

Answer: $152.50

Explanation:
Cost per square foot of the carpet = $2.50
The total cost of the carpet on the floor = 61 x $2.50 =$152.50

Question 7.
Multiple Representations
Hexagon ABCDEF has vertices A(-2, 4), B(0, 4), C(2, 1), D(5, 1), E(5, -2), and F(-2, -2). Sketch the figure on a coordinate plane. What is the area of the hexagon?
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 20
_____ units2

Answer: The area of the figure is 30 square units

Explanation:
Separate the figure into a trapezium and a rectangle.
Area of a trapezium = 1/2 (a+b) h= 1/2 (2+4) x 3 = 1/2 (6) 3 = 9 square units
Area of a rectangle = length x breadth = 7 x 3 = 21 square units
The total area of the figure = 9+21 = 30 square units

Question 8.
A field is shaped like the figure shown. What is the area of the field? Use 3.14 for π.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 21
_____ m2

Answer: 146.24 m2

Explanation:
Divide the figure into a square, triangle and a quarter of a circle.

Area of a square = side x side = 8 x 8 = 64 m2
Area of a quarter of a circle = 1/4 (πr2) = 1/4 (3.14 x 82)
= 1/4 (200.96) = 50.24 m2
Area of the triangle = 1/2 x base x height = 1/2 x 8 x 8 = 32 m2
Total area of the figure = 64+32+50.24 = 146.24 m2

Question 9.
A bookmark is shaped like a rectangle with a semicircle attached at both ends. The rectangle is 12 cm long and 4 cm wide. The diameter of each semicircle is the width of the rectangle. What is the area of the bookmark? Use 3.14 for π.
_____ cm2

Answer: 60.56 cm2

Explanation:
The bookmark is divided into a rectangle, semicircle.
Area of the rectangle = length x breadth = 12 x 4 = 48 cm2
The diameter of the semicircle = The width of the rectangle = 4 cm
The radius of the semicircle = 4/2 = 2 cm
The area of the semicircle = πr2 = 3.14 x 2 x 2 = 12.56 cm2
The total area of the bookmark = 12.56 + 48 = 60.56 cm2

Question 10.
Multistep
Alex is making 12 pendants for the school fair. The pattern he is using to make the pennants is shown in the figure. The fabric for the pennants costs $1.25 per square foot. How much will it cost Alex to make 12 pennants?
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 22
$ _____

Answer: $52.50

Explanation:
Each pendant is made up of a rectangle and a triangle.
Area of the rectangle = length x breadth = 3 x 1 = 3 ft2
Area of the triangle = 1/2 x base x height = 1/2 x 1 x 1 = 0.5 ft2
The total area of the pendant = 3+0.5 = 3.5 ft2
Number of pendants = 12
Area of the pendants = 12 x 3.5 = 42 ft2
Cost of each square feet of the pendant = $1.25
Total cost for all the 12 pendants = 12 x $1.25  = $52.50

Question 11.
Reasoning
A composite figure is formed by combining a square and a triangle. Its total area is 32.5 ft2. The area of the triangle is 7.5 ft2. What is the length of each side of the square? Explain.
_____ ft

Answer: 5 ft

Explanation:
Given:
The area of the composite figure = 32.5 ft2
The area of the triangle = 7.5 ft2
The area of the square = 32.5 – 7.5 = 25
side x side = 25
side2 = 25
side = root 25 = 5 ft

H.O.T. – Page No. 282

Focus on Higher Order Thinking

Question 12.
Represent Real-World Problems
Christina plotted the shape of her garden on graph paper. She estimates that she will get about 15 carrots from each square unit. She plans to use the entire garden for carrots. About how many carrots can she expect to grow? Explain.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 23
______ carrots

Answer: 300 carrots

Explanation:
This shape is divided into two triangles and a square.
Area of figure = 2(1/2 x 2 x 2) + 4(4) = 4 + 16 = 20 square units
Number of carrots per square unit = 300
Total number of carrots = 20 x 15 = 300

Question 13.
Analyze Relationships
The figure shown is made up of a triangle and a square. The perimeter of the figure is 56 inches. What is the area of the figure? Explain.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 24
_____ in2

Answer: 192 in2

Explanation:
Given:
The perimeter of the figure = 56 inches
The figure is divided into a square and a triangle.
10 + 10 + 3s = 56
3s = 36
s = 12
The area of a triangle = 1/2 x 12 x 8 = 48 in2
The area of a square = 12 x 12 = 144 in2
Total area of the figure = 144 + 48 = 192 in2

Question 14.
Critical Thinking
The pattern for a scarf is shown at right. What is the area of the scarf? Use 3.14 for π.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 25
_____ in2

Answer: 243 in2

Explanation:
Area of the rectangle in the given figure = 28 x 15 = 420 in2
Area of two semicircles = 2 (1/2 πr2 ) = 3.14 x 7.5 x 7.5 = 176.625 in2
Area of the shaded region = 420 – 176.625 = 243 in2(approx)

Question 15.
Persevere in Problem Solving
The design for the palladium window shown includes a semicircular shape at the top. The bottom is formed by squares of equal size. A shade for the window will extend 4 inches beyond the perimeter of the window, shown by the dashed line around the window. Each square in the window has an area of 100 in2.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 26
a. What is the area of the window? Use 3.14 for π.
_____ in2

Answer: a) 2228 in2

Explanation:
Area of the square = 100 in2
side x side = 100
Side = 10 in
Since the side of each square is 10 in and there are 4 squares.
The side length of the larger square (s) = 40 in
Area of the larger square = side x side = 40 x 40 = 1600 in2
Since the side of each square is 10 in and there are 2 squares.
The radius of the semi-circle = 20 in
Area of the semi-circle = 1/2(πr2) = 1/2(3.14 x 202) = 628 in2
The area of the window = 1600 + 628 = 2228 in2

Question 15.
b. What is the area of the shade? Round your answer to the nearest whole number.
_____ in2

Answer: b) 3016 in2

Explanation:
The shade extends 4 inches beyond the shapes so the length of the bottom rectangle is 40+4+4 = 48 in
The length extends below the original square.
The height is now = 40+4 = 44 in
The radius of the semi-circle = 20+4 = 24 in
The new area of the figure = 48(44) + 1/2(3.14 x 242) = 2112 + 904.32 = 3016.32 = 3016 in2

Guided Practice – Page No. 286

Find the surface area of each solid figure.

Question 1.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 27
Total surface area: _____ ft2

Answer: 150 ft2

Explanation:
The base is a triangle with side lengths of 8 ft, 5 ft, 5 ft so the perimeter of the base = P = 8+5+5 = 18 ft
The height of the prism = 7 ft
The base is a triangle.
Area of the triangle = 1/2 (8) (3) = 12 ft2
The surface area formula for a prism is S = Ph + 2b
P = Perimeter = 18 h = height = 7 b = base = area of the triangle = 12
The surface area of the prism = 18(7) + 2(12) = 126 + 24 = 150 ft2

Question 2.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 28
Total surface area: _____ m2

Answer: 503 m2

Explanation:
Given:
Dimensions of the cuboid:
Length = 11 m
Breadth = 9 m
Height = 7 m
The surface area of the cuboid = 2(lb+bh+hl) = 2(11 x 9 + 9 x 7 + 7 x 11) = 478m2

The dimensions of the cube:
Length of the side = 2.5 m
The surface area of the cube = 6a2 = 6 x 2.5 x 2.5 = 37.5 m2
The surface area of the rectangular prism = 2.5 x 2.5 = 6.25
The surface area of the figure = The overlapping area is the area of the base of the cube
= 37.5 + 478 – 2(6.25) = 503 m2

Essential Question Check-In

Question 3.
How can you find the surface area of a composite solid made up of prisms?
Type below:
_____________

Answer: The surface area of the prisms, add them up, and then subtract the overlapping areas twice.

Explanation:
The surface area of a composite solid is made up of prisms by finding the surface areas of the prisms, adding them up, and then up, and then subtracting the overlapping areas.

Independent Practice – Page No. 287

Question 4.
Carla is wrapping a present in the box shown. How much wrapping paper does she need, not including overlap?
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 29
_____ in2

Answer: 164 in2

Explanation:
The surface area of the cuboid excluding the top = 2h(l+b) + lb = 2 x 4 ( 13 ) + 10 x 3 =  164 in2
The length of the wrapping paper = The surface area of the cuboid excluding the top = 164 in2

Question 5.
Dmitri wants to cover the top and sides of the box shown with glass tiles that are 5 mm square. How many tiles does he need?
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 30
_____ tiles

Answer: 3720 tiles

Explanation:
The surface area of the cuboid excluding the bottom = 2h(l+b) + lb = 2 x 9 (35) + 20 x 15 = 930 cm2
5mm = 0.5 cm
Area of a tile = Area of the square = a2 = 0.5cm x 0.5cm = 0.25 cm2
Total number of tiles = 930/0.25 = 3720 tiles

Question 6.
Shera is building a cabinet. She is making wooden braces for the corners of the cabinet. Find the surface area of each brace.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 31
_____ in2

Answer: 45 in2

Explanation:
The perimeter of the figure = P = 3(3) + 2(1) = 11 in
Base = B = 3(2) = 6 in
Height = h = 3
The surface area of the figure = Ph + 2B = 11 x 3 +2(6) = 33 + 12 = 45 in2

Question 7.
The doghouse shown has a floor, but no windows. Find the total surface area of the doghouse, including the door.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 32
_____ ft2

Answer:  66ft2

Explanation:
Perimeter of the pentagon base (P) = 2(2.5) + 2(2) + 3 = 5 + 4 + 3 = 12
Area of the pentagon base by adding the area of the triangle and the area of the rectangle (B) = 1/2(3)(2) + 2(3) = 9
Height (h) = 2 + 2 = 4
The surface area of the figure = Ph + 2B = 12(4) + 2(9) = 48 + 18 = 66ft2

Eddie built the ramp shown to train his puppy to do tricks. Use the figure for 8–9.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 33

Question 8.
Analyze Relationships
Describe two ways to find the surface area of the ramp.
Type below:
____________

Answer: One way is to use the formula S = Ph + 2B. Another way is to find the area of each face of the prism and add them up to get the total surface area.

Explanation:
The very first way to use the formula S = Ph + 2B where the trapeziums are the base. The second way is to find the area of each face of the prism and then add them up to get the total surface area.

Question 9.
What is the surface area of the ramp?
_____ in2

Answer: 3264 in2

Explanation:
P = Perimeter of the figure =  16(3) + 2 (20) + 16 = 104
B = Base of the figure = 1/2 (12) (16 + 3(16)) = 6 (16 + 48) = 6 (64) = 384
h = Height of the figure = 2
Surface area of the figure = Ph + 2B = 104(2) + 2(384) = 2496 + 768 = 3264 in2

Marco and Elaine are building a stand like the one shown to display trophies. Use the figure for 10–11.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 34

Question 10.
What is the surface area of the stand?
_____ ft2

Answer:  58 ft2

Explanation:
Top:
Perimeter = P = 4(1) = 4
Base = B = 1(1) = 1
Height = h = 3
Top surface area = Ph + 2B = 4(3) + 2(1) = 14 ft2
Bottom :
Perimeter = P = 2(7) + 2(1) = 14 + 2 = 16
Base = B = 7(1) = 7
Height = h = 2
Top surface area = Ph + 2B = 16(2) + 2(7) = 46 ft2
Overlapping area = 1(1) = 1
The surface area of the figure = The surface area of the top + The surface area of the bottom – the overlapping area = 14 + 46 – 2 = 60 – 2 = 58 ft2

Question 11.
Critique Reasoning
Marco and Elaine want to paint the entire stand silver. A can of paint covers 25 square feet and costs $6.79. They set aside $15 for paint. Is that enough? Explain.
_____

Answer: No

Explanation:
Since the surface area is 58 ft2, they will need 3 cans of paint. Since each can paints 25 ft2 and we cannot buy a fraction of cans.
3 cans would then cost 6.79 x 3 = 20.37 so this is not enough.

Page No. 288

Question 12.
Henry wants to cover the box shown with paper without any overlap. How many square centimeters will be covered with paper?
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 35
_____ cm2

Answer: 2316 cm2

Explanation:
Given:
Length = 24cm  Breadth = 27cm Height = 10cm
P = Perimeter = 2(24) + 2(27) = 48 + 54 = 102
B = Base = 24(27) = 648
h = Height = 10
Surface area of the figure = Ph + 2B = 102(10) + 2(648) = 1020 + 1296 = 2316 cm2

Question 13.
What If?
Suppose the length and width of the box in Exercise 12 double. Does the surface area S double? Explain.
_____

Answer: No

Explanation:
Given :
Length = 24cm x 2 = 48 cm  Breadth = 27cm x 2 = 54 cm Height = 10cm
P = 2(48) + 2(54) = 96 + 108 = 204
B = 48(54) = 2592
New Surface area = Ph + 2B = 204(10) + 2(2592) = 2040 + 5184 = 7224 cm2
Double of surface area = 2 (2316) = 4632 cm2
So the new surface area is not double of the initial area.

H.O.T.

Focus on Higher Order Thinking

Question 14.
Persevere in Problem Solving
Enya is building a storage cupboard in the shape of a rectangular prism. The rectangular prism has a square base with side lengths of 2.5 feet and a height of 3.5 feet. Compare the amount of paint she would use to paint all but the bottom surface of the prism to the amount she would use to paint the entire prism.
Type below:
______________

Answer: The difference would just be the area in the bottom surface. It would be 6.25 ft2 less.

Explanation:
The difference in the amount of paint would just be the area of the bottom surface. The area of the bottom surface is (2.5)2 = 6.25.
Therefore she would paint 6.25 ft2 less if she painted all but the bottom surface compared to painting the entire prism.

Question 15.
Interpret the Answer
The oatmeal box shown is shaped like a cylinder. Use a net to find the surface area S of the oatmeal box to the nearest tenth. Then find the number of square feet of cardboard needed for 1,500 oatmeal boxes. Round your answer to the nearest whole number
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 36
_____ ft2

Answer: 138.28 in2 , 1440 ft2

Explanation:
Given:
Dimensions of the cylinder:
Radius: 2 in
Height: 9 in
The total surface area of the cylinder = 2πr(r+h) = 2 x 22/7 x 2 (2 + 9) = 138.28 in2

The total number of square inches needed for 1,500 oatmeal boxes = 1,500 x 138.28 = 207,300 in2
1 ft = 12 in
(1 ft)2 = (12 in)2
1 ft2 = 144 in2
The total number of square feet needed for 1,500 oatmeal boxes (to the nearest whole number)
= 207,300/144 = 1440 ft2

Question 16.
Analyze Relationships
A prism is made of centimeter cubes. How can you find the surface area of the prism in Figure 1 without using a net or a formula? How does the surface area change in Figures 2, 3, and 4? Explain.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 37
Type below:
______________

Answer: The surface area for the first 3 figures are the same. The surface area for figure 4 is greater than the surface area of the figures 1 – 3.

Explanation:
The surface area of the first 3 figures is the same. The 3 new faces on figure 2 have the same areas as the 3 visible faces that were removed when the top corner cube was removed. The surface area is then the same as it is for figure 1. Similarly, the areas of the new visible faces in figure 3 are equal to the areas of the visible faces removed from removing the corner cubes so the surface areas are the same as in figure 1. The surface area for figure 4 is greater than the surface areas of the figures 1 – 3. Removing the cube removed 2 of the visible faces (one from the top and one from the front side) but added 4 visible faces so the surface area increases.

Guided Practice – Solving Volume Problems – Page No. 292

Question 1.
Find the volume of the triangular prism.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume Lesson 5: Solving Volume Problems img 38
_____ ft3

Answer: 84 ft3

Explanation:
Base area of the prism = 1/2 x 8 x 3 = 12 ft2
Height of the prism = 7 ft
Volume of the prism = (12 x 7) ft3

Question 2.
Find the volume of the trapezoidal prism.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume Lesson 5: Solving Volume Problems img 39
_____ m3

Answer: 330 m3

Explanation:
Base area of the prism = 1/2 x (15 + 5) x 3 = 30 m2
Height of the prism = 11 m
Volume of the prism = (30 x 11) m3 = 330 m3

Question 3.
Find the volume of the composite figure.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume Lesson 5: Solving Volume Problems img 40
_____ ft2

Answer: Composite figure: 360 ft3

Explanation:
The volume of the triangular prism:
The base area of the prism = 1/2 x 4 x 6 = 12 ft2
Height = 6 ft
The volume of the triangular prism = 12 x 6 = 72 ft3

The volume of the rectangular prism:
The base area of the prism = 4 x 6 = 24 ft2
Height = 12 ft
The volume of the triangular prism = 12 x 24 = 288 ft3

Volume of the composite figure = (288 + 72)ft3 = 360 ft3

Find the volume of each figure.

Question 4.
The figure shows a barn that Mr. Fowler is building for his farm.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume Lesson 5: Solving Volume Problems img 41
_____ ft3

Answer: 40,000 ft3

Explanation:
Triangular prism:
B = Base area = 1/2 x 10 (40) = 200 cm2
Height = 50 cm
The volume of the triangular prism = Bh = 200 x 50 = 10,000 cm3
Rectangular prism:
B = Base area =40 x 15 = 600 cm2
Height = 50 cm
The volume of the triangular prism = Bh = 600 x 50 = 30,000 cm3
Total volume of the prism = 10,000 + 30,000 = 40,000 cm3

Question 5.
The figure shows a container, in the shape of a trapezoidal prism, that Pete filled with sand.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume Lesson 5: Solving Volume Problems img 42
_____ cm3

Answer: 385 cm3

Explanation:
B = Base area = 1/2 x 5 (10 + 12) = 55 cm2
Height = 7 cm
The volume of the container = Bh = 55 x 55 = 385 cm3

Essential Question Check-In

Question 6.
How do you find the volume of a composite solid formed by two or more prisms?
Type below:
______________

Answer: Finding the volume of each figure adding them up to get the volume of the composite solid.

Explanation:
To find the volume of the composite figure that can be divided into 2 or more prisms, find the volume of each prism and add them up to get the volume of the composite solid.

Independent Practice – Page No. 293

Question 7.
A trap for insects is in the shape of a triangular prism. The area of the base is 3.5 in2 and the height of the prism is 5 in. What is the volume of this trap?
_____ in3

Answer: 17.5 in3

Explanation:
The volume of the trap = Base area x height = 3.5 x 5 = 17.5 in3

Question 8.
Arletta built a cardboard ramp for her little brothers’ toy cars. Identify the shape of the ramp. Then find its volume.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 43
Shape: _________
Area: _________ in3

Answer: 525 in3

Explanation:
Base area = 1/2 x 6 x 25 = 75 in2
Height  = 7 in
Volume of the figure = 75 x 7 = 525 in3

Question 9.
Alex made a sketch for a homemade soccer goal he plans to build. The goal will be in the shape of a triangular prism. The legs of the right triangles at the sides of his goal measure 4 ft and 8 ft, and the opening along the front is 24 ft. How much space is contained within this goal?
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 44
_____ ft3

Answer: 384 ft3

Explanation:
Base area = 1/2 x 4 x 8 = 16 ft2
Height  = 24 ft
Volume of the figure = 16 x 24 = 384 ft3

Question 10.
A gift box is in the shape of a trapezoidal prism with base lengths of 7 inches and 5 inches and a height of 4 inches. The height of the gift box is 8 inches. What is the volume of the gift box?
_____ in3

Answer: 192 in3

Explanation:
Base area = 1/2 x 4 x (7+5) = 24 in2
Height  = 8 in
Volume of the figure = 24 x 8 = 192 Base area = 1/2 x 6 x 25 = 75 in2
Height  = 7 in
Volume of the figure = 75 x 7 = 525 in3

Question 11.
Explain the Error
A student wrote this statement: “A triangular prism has a height of 15 inches and a base area of 20 square inches. The volume of the prism is 300 square inches.” Identify and correct the error.
Type below:
____________

Answer: The error is measurement unit.

Explanation:
The volume of the prism is:
base area x height = 20 x 15 = 300 in3

Find the volume of each figure. Round to the nearest hundredth if necessary.

Question 12.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 45
_____ in3

Answer: 97.2 in3

Explanation:
The volume of the hexagonal prism = 23.4 x  3 = 70.2 in3

Base area of the rectangular prism = 3 x 3 = 9 in2
The volume of the rectangular prism = Bh = 9 x 3 = 27 in3

Total volume of the figure = 70.2 + 27 = 97.2 in3

Question 13.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 46
_____ m3

Answer: 316.41 m3

Explanation:
The volume of the rectangular prism on the left = Bh = [7.5 x 3.75] (3.75) = 105.47 m3
The volume of the rectangular prism on the right = Bh = [7.5 x 3.75](7.5) = 210.94 m3
Total volume of the composite figure = 105.47 + 210.94 = 316.41 m3

Question 14.
Multi-Step
Josie has 260 cubic centimeters of candle wax. She wants to make a hexagonal prism candle with a base area of 21 square centimeters and a height of 8 centimeters. She also wants to make a triangular prism candle with a height of 14 centimeters. Can the base area of the triangular prism candle be 7 square centimeters? Explain.
_____

Answer: No

Explanation:
The volume of the hexagonal prism = 21 x 8 = 168
The total volume of wax, 260 is equal to the sum of the volumes of each prism.
B is the base area of the triangular prism.
168 + 14B = 260 cm3
14B = 260 – 168
B = 6.6 cm3

Page No. 294

Question 15.
A movie theater offers popcorn in two different containers for the same price. One container is a trapezoidal prism with a base area of 36 square inches and a height of 5 inches. The other container is a triangular prism with a base area of 32 square inches and a height of 6 inches. Which container is the better deal? Explain.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 47
Type below:
___________

Answer: The triangular prism is a better deal since it has a larger volume

Explanation:
The base area of the trapezoidal prism = 36 in2
The volume of the trapezoidal prism = Bh = 36 x 5 = 175 in3

The base area of the triangular prism = 32 in2
The volume of the rectangular prism = Bh = 32 x 6 = 192 in3

The triangular prism is a better deal since it has a larger volume.

H.O.T.

Focus on Higher Order Thinking

Question 16.
Critical Thinking
The wading pool shown is a trapezoidal prism with a total volume of 286 cubic feet. What is the missing dimension?
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 48
______ ft.

Answer: 3.5 ft

Explanation:
Area of the trapezoidal prism = B = 1/2 x 13 (2+x)
Volume of the figure = 286 cubic feet
V = Bh
286 = 1/2 x 13 (2+x)(8)
5.5 = (2+x)
x = 3.5 ft

Question 17.
Persevere in Problem Solving
Lynette has a metal doorstop with the dimensions shown. Each cubic centimeter of the metal in the doorstop has a mass of about 8.6 grams. Find the volume of the metal in the doorstop. Then find the mass of the doorstop.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 49
______ grams

Answer: 75 cubic centimeter, 645 grams

Explanation:
V = Bh
B = Area of the triangle of base = 10 cm , height = 6 cm = 1/2 x 10 x 6 = 30 square centimeter
V = 30 x 2.5 = 75 cubic centimeter

1 cubic centimeter = 8.6 grams in mass
V = 75 cubic centimeter x 8.6 = 645 grams

Question 18.
Analyze Relationships
What effect would tripling all the dimensions of a triangular prism have on the volume of the prism? Explain your reasoning.
Type below:
____________

Answer: The volume is 27 times the original volume.

Explanation:
The area of the base = B = 1/2 (3b) (3h) = 9/2 (bh)
H is the height of the prism
The volume would be = 9/2 (bh) x (3H) = 27 [ 1/2 (bhH) ]

Therefore, The volume is 27 times the original volume.

Question 19.
Persevere in Problem Solving
Each of two trapezoidal prisms has a volume of 120 cubic centimetres. The prisms have no dimensions in common. Give possible dimensions for each prism.
Type below:
____________

Answer: A possible combination of dimension could be the height at 8 cm, base at 2 cm and 3 cm

Explanation:
The numbers that multiply to get 120 are 20 and 6 so let the first prism have a base area of 20 square centimetres and the height of 6 cm.
If the base area is 20, the height of the trapezoid and the length of the bases could be 8,2 and 3 respectively.

The other numbers that multiply to get 120 are 4 and 30 so let the second prism have a base area of 30 square centimetres and the height of 4 cm.
If the base area is 30, the height of the trapezoid and the length of the bases could be 10,1 and 5 respectively.

9.1, 9.2 Circumference and Area of Circles – Page No. 295

Find the circumference and area of each circle. Use 3.14 for π. Round to the nearest hundredth if necessary.

Question 1.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 50
C = _________ m
A = _________ m2

Answer:
C = 43.96 m
A = 153.86 m2

Explanation:
C = 2 πr = 2 π(7) = 14 (3.14) = 43.96 m
A = πr2 = 3.14 (7)2 = 153.86 m2

Question 2.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 51
C = _________ ft
A = _________ ft2

Answer:
C = 37.68 ft
A = 113.04 ft2

Explanation:
Diameter = 12 ft
Radius = d/2 = 12/2 = 6 ft
C = 2 πr = 2 π(6) = 6 (3.14) = 37.68 ft
A = πr2 = 3.14 (6)2 = 113.04 ft2

9.3 Area of Composite Figures

Find the area of each figure. Use 3.14 for π.

Question 3.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 52
______ m2

Answer: 180.48 m2

Explanation:
Area of the triangle = 1/2 x 16 x 10 = 80 m2
Area of the semicircle = 1/2 πr2 = 1/2 (3.14) (8)2 = 100.48 m2
The total area of the figure = 80 + 100.48 = 180.48 m2

Question 4.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 53
______ cm2

Answer: 200 cm2

Explanation:
Area of the parallelogram = 4.5(20) = 90 cm2
Area of the rectangle = 20(5.5) = 110 cm2
The total area of the figure = 90 + 110 = 200 cm2

9.4, 9.5 Solving Surface Area and Volume Problems

Find the surface area and volume of each figure.

Question 5.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 54
S = _________ cm2
V = _________ cm3

Answer:
S = 132 cm2
V = 60 cm3

Explanation:
Perimeter = 3+4+5 = 12 cm
Base area = Area of the triangle = 1/2 x 3 x 4 = 6
S = Ph + 2B = 12(10) + 2(6) = 120 +12 = 132 cm2

V = Bh = 6 x 10 = 60 cm3

Question 6.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 55
S = _________ yd2
V = _________ yd3

Answer:
S = 54.5 yd2
V = 27.5 yd3

Explanation:
Perimeter = 2(2.5) + 2(2) + 4 = 13 cm
Base area = Area of the triangle + Area of the rectangle = 1/2 x 1.5 x 4 + 4(2)= 11
S = Ph + 2B = 13(2.5) + 2(11) = 32.5 +22 = 54.5 yd2

V = Bh = 11 x 2.5 = 27.5 yd3

Essential Question

Question 7.
How can you use geometry figures to solve real-world problems?
Type below:
______________

Answer: We can solve real-world problems by finding surface area and volume.
Example: We can find the amount of liquid in a tank by calculating its volume.

Explanation:
Real-world problems by finding surface area and volume.
Example1: We can find the amount of liquid in a tank by calculating its volume.
Example2: We can find the surface area of the house and find the amount of paint required to paint the house.

Page No. 296

Question 1.
What is the circumference of the circle?
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 56
a. 34.54 m
b. 69.08 m
c. 379.94 m
d. 1519.76 m

Answer: b. 69.08 m

Explanation:
Circumference = 2 πr = 2 π(11) = 22 (3.14) = 69.08 m

Question 2.
What is the area of the circle?
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 57
Options:
a. 23.55 m2
b. 47.1 m2
c. 176.625 m2
d. 706.5 m2

Answer: c. 176.625 m2

Explanation:
Diameter = 15 m
Radius = 7.5 m
Area of the circle = πr2 = 3.14 (7.5)2 = 176.625 m2

Question 3.
What is the area of the figure?
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 58
Options:
a. 28.26 m2
b. 36 m2
c. 64.26 m2
d. 92.52 m2

Answer: c. 64.26 m2

Explanation:
Area of the square = 6 x 6 = 36 m2
Radius = 6 m
Area of the quarter circle = 1/4 πr2 = 1/4 x 3.14 (6)2 = 28.26 m2
The total area of the figure = 36 + 28.26 = 64.26 m2

Question 4.
A one-year membership to a health club costs $480. This includes a $150 fee for new members that is paid when joining. Which equation represents the monthly cost x in dollars for a new member?
Options:
a. 12x + 150 = 480
b. \(\frac{x}{12}\) + 150 = 480
c. 12x + 480 = 150
d. \(\frac{x}{12}\) + 480 = 150

Answer: a. 12x + 150 = 480

Explanation:
If x is the monthly fee, then 12x is the total monthly fees.
The joining fee = $150
Total cost = $480
then,
12x + 150 = 480

Question 5.
What is the volume of the prism?
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 59
Options:
a. 192 ft3
b. 48 ft3
c. 69 ft3
d. 96 ft3

Answer: d. 96 ft3

Explanation:
B = Base area of the triangle = 1/2 x 8 x 2 = 8 ft2
Height = 12 ft
Volume of the triangular orism = Bh = 8(12) = 96 ft3

Question 6.
A school snack bar sells a mix of granola and raisins. The mix includes 2 pounds of granola for every 3 pounds of raisins. How many pounds of granola are needed for a mix that includes 24 pounds of raisins?
Options:
a. 16 pounds
b. 36 pounds
c. 48 pounds
d. 120 pounds
e. 120 pounds

Answer: a. 16 pounds

Explanation:
2/3 is equal to x/24 then 3 times 8 is equal to 24 and if 2 times 8 is equal to 16.

Question 7.
Find the percent change from $20 to $25.
Options:
a. 25% decrease
b. 25% increase
c. 20% decrease
d. 20% increase

Answer: b. 25% increase

Explanation:
25 – 20 = 5 divide by 20 = 1/4
When we find the percentage we get 25.
So we can say that there is an increase in 25%

Question 8.
Each dimension of the smaller prism is half the corresponding dimension of the larger prism.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 60
a. What is the surface area of the figure?
_____ in2

Answer: 856 in2

Explanation:
Height of the top prism = 10/2 = 5
Length of the top prism = 16/2 = 8
Width of the top prism = 8/2 = 4
Perimeter = 2l + 2w = 2(8) + 2(4) = 16 + 8 = 24 in
B = lw = 8(4) = 32 in
Surface area of top prism= Ph + 2B = 24(5) + 2(32) = 184 in2

Height of the prism = 10
Length of the prism = 16
Width of the prism = 8
Perimeter = 2l + 2w = 2(16) + 2(8) = 32 + 16 = 48 in
B = lw = 16(8) = 128 in
Surface area of bottom prism= Ph + 2B = 48(10) + 2(128) = 736 in2

Area of overlapping region = 32 in2

The total surface area of the prism
= Surface area of top prism + Surface area of bottom prism – 2[Area of overlapping region ]
= 184 + 736 – 2(32) = 856 in2

Question 8.
b. What is the volume of the figure?
_____ in3

Answer: 1440 in3

Explanation:
Volume of top prism = Bh = 32(5) = 160 in3
Volume of bottom prism = Bh = 128(10) = 1280 in3
The total volume of the figure = 160 + 1280 = 1440 in3

EXERCISES – Page No. 298

Question 1.
In the scale drawing of a park, the scale is 1 cm: 10 m. Find the area of the actual park.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 61
_____ m2

Answer: 450 m2

Explanation:
Multiply the dimensions of the scale drawing by 10 since 1 cm = 10 m
3cm by 1.5 cm = 30m by 15 m
Area = 30(15) = 450 m2

Question 2.
Find the value of y and the measure of ∠YPS.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 62
y = __________ °
mYPS = __________ °

Answer: y = 8
mYPS = 40 °

Explanation:
140 + 5y = 180 [sum of angle on a line = 180°]
5y = 40
y = 8

mYPS = mRPZ = 5y [vertically opposite angles]
mYPS = 5(8) = 40°

Question 3.
Kanye wants to make a triangular flower bed using logs with the lengths shown below to form the border. Can Kanye form a triangle with the logs without cutting any of them? Explain.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 63
_____

Answer: No

Explanation:
A side of a triangle must be greater than the difference of the other two sides and smaller than the sum of the other 2 sides.
The sum of the first 2 sides = 3+4 = 7 < 8
Therefore, he cannot form a triangle unless he cuts the logs.

Question 4.
In shop class, Adriana makes a pyramid with a 4-inch square base and a height of 6 inches. She then cuts the pyramid vertically in half as shown. What is the area of each cut surface?
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 64
_____ in2

Answer: 12 in2

Explanation:
Base = 4 in
Height = 6 in
Area of the triangle = 1/2 x 6 x 4 = 12 in2

Page No. 300

Find the circumference and area of each circle. Round to the nearest hundredth.

Question 1.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 65
C = __________ in
A = __________ in2

Answer:
C = 69.08 in
A = 379.94 in2

Explanation:
Diameter = 22 in
Radius = d/2 = 22/2 = 11 in
C = 2 πr = 2 π(11) = 22 (3.14) = 69.08 in
A = πr2 = 3.14 (11)2 = 379.94 in2

Question 2.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 66
C = __________ m
A = __________ m2

Answer:
C = 28.26 m
A = 63.59m2

Explanation:
Radius = 4.5 m
C = 2 πr = 2 π(4.5) = 9 (3.14) = 28.26 m
A = πr2 = 3.14 (4.5)2 = 63.59 m2

Find the area of each composite figure. Round to the nearest hundredth if necessary.

Question 3.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 67
______ in2

Answer: 99 in2

Explanation:
Area of the square = 9 x 9 = 81 in2
Base of the triangle = 13 – 9 = 4 in
Area of the triangle = 1/2 x 4 x 9 = 18 in2
The total area of the figure = 81 + 18 = 99 in2

Question 4.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 68
______ cm2

Answer: 420.48 cm2

Explanation:
Area of the rectangle = 16 x 20 = 320 cm2
Diameter = 16 cm
Radius = 16/2 = 8 cm
Area of the semi circle = 1/2 πr2 = 1/2 x 3.14 (8)2 = 100.48 cm2
The total area of the figure = 320 + 100.48 = 420.48 cm2

Find the volume of each figure.

Question 5.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 69
______ in3

Answer: 420 in3

Explanation:
B = 7(5) = 35 in2
V = Bh = 35 x 12 = 420 in3

Question 6.
The volume of a triangular prism is 264 cubic feet. The area of a base of the prism is 48 square feet. Find the height of the prism.
______ in

Answer: 5.5 ft

Explanation:
V = Bh
264 = 48h
h = 264/48 = 5.5ft

Page No. 301

A glass paperweight has a composite shape: a square pyramid fitting exactly on top of an 8 centimeter cube. The pyramid has a height of 3 cm. Each triangular face has a height of 5 centimeters.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 70

Question 7.
What is the volume of the paperweight?
______ cm3

Answer: 576 cm3

Explanation:
Pyramid:
B = 8 x 8 = 64 cm2
V = 1/3 Bh = 1/3 x 64 x 3 = 64 cm3
Prism:
B = 8 x 8 = 64 cm2
V = Bh = 64 x 8 = 512 cm3

The total volume of the figure = 64 + 512 = 576 cm3

Question 8.
What is the total surface area of the paperweight?
______ cm2

Answer: 400 cm2

Explanation:
Pyramid:
P = 4(8) = 32 cm
S = 1/2 Pl + B = 80 + 64 = 144 cm2

Prism:
P = 4(8) = 32 cm
S = Ph + 2B = 32(8) + 2(64) = 384 cm2
The total surface area of the prism
= Area of the prism + Area of the pyramid – 2[Area of the overlapping region]
= 144 + 384 – 2(64) = 400

Unit 4 Performance Tasks

Question 9.
Product Design Engineer
Miranda is a product design engineer working for a sporting goods company. She designs a tent in the shape of a triangular prism. The dimensions of the tent are shown in the diagram.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 71
a. How many square feet of material does Miranda need to make the tent (including the floor)? Show your work.
______ ft2

Answer: 261 3/4  ft2

Explanation:
P = 2 x 7 1/2 + 8 = 22 1/2
B = 4/2 (8) (6) = 24
S = Ph + 2B = 22 1/2 x 9 1/2 + 2(24) = 213 3/4 + 48 = 261 3/4 ft2

Question 9.
b. What is the volume of the tent? Show your work.
______ ft3

Answer: 228 ft3

Explanation:
V = Bh = 24 x 9 1/2 = 228 ft3

Question 9.
c. Suppose Miranda wants to increase the volume of the tent by 10%. The specifications for the height (6 feet) and the width (8 feet) must stay the same. How can Miranda meet this new requirement? Explain
Type below:
____________

Answer: Increase the height to 10.45 ft

Explanation:
New volume = 1.10 x 228 = 250.8
250.8 = 24h
h = 10.45 ft

Unit 4 Performance Tasks (cont’d) – Page No. 302

Question 10.
Li is making a stand to display a sculpture made in art class. The stand will be 45 centimeters wide, 25 centimeters long, and 1.2 meters high.
a. What is the volume of the stand? Write your answer in cubic centimeters.
______ cm3

Answer: 135,000 cm3

Explanation:
B = 45 x 25 = 1125 cm2
V = Bh = 1125 x 120 = 135,000 cm3

Question 10.
b. Li needs to fill the stand with sand so that it is heavy and stable. Each piece of wood is 1 centimeter thick. The boards are put together as shown in the figure, which is not drawn to scale. How many cubic centimeters of sand does she need to fill the stand? Explain how you found your answer.
______ cm3

Answer: 116,702 cm3

Explanation:
Width = 45 – 2(1) = 43 ft
Length = 25 – 2(1) =23ft
Height = 120-2(1) = 118ft
B = 43 x 23 = 989 ft2
V = Bh = 989 x 118 = 116,702 ft3

Selected Response – Page No. 303

Question 1.
A school flag is in the shape of a rectangle with a triangle removed as shown.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 72
What is the measure of angle x?
Options:
a. 50°
b. 80°
c. 90°
d. 100°

Answer: d. 100°

Explanation:
x = 50 + 50 = 100° [ Sum of two angles created by the 2 lines]

Question 2.
On a map with a scale of 2 cm = 1 km, the distance from Beau’s house to the beach is 4.6 centimetres. What is the actual distance?
Options:
a. 2.3 km
b. 4.6 km
c. 6.5 km
d. 9.2 km

Answer: a. 2.3 km

Explanation:
2/1 = 4.6/x
x = 4.6/2 = 2.3 km

Question 3.
Lalasa and Yasmin are designing a triangular banner to hang in the school gymnasium. They first draw the design on paper. The triangle has a base of 5 inches and a height of 7 inches. If 1 inch on the drawing is equivalent to 1.5 feet on the actual banner, what will the area of the actual banner be?
Options:
a. 17.5 ft2
b. 52.5 ft2
c. 39.375 ft2
d. 78.75 ft2

Answer: c. 39.375 ft2

Explanation:
1in = 1.5ft
The base of the triangle = 5 in = 1.5(5) ft = 7.5 ft
Height = 7 in = 7(1.5) ft = 10.5 ft
Area of the triangle = 1/2 x 7.5 x 10.5 = 39.375 ft2

Question 4.
Sonya has four straws of different lengths: 2 cm, 8 cm, 14 cm, and 16 cm. How many triangles can she make using the straws?
Options:
a. no triangle
b. one triangle
c. two triangles
d. more than two triangles

Answer: b. one triangle

Explanation:
The third side of a triangle must be smaller than the sum of the other two sides to form a triangle.
2+8 = 10<14
2+8 = 10<16
8+14 = 22>14
8+14 = 22>16
2+14 = 16=16
2+16 = 18>16

Therefore, only one triangle can be formed using the sides 8, 14, 16.

Question 5.
A one-topping pizza costs $15.00. Each additional topping costs $1.25. Let x be the number of additional toppings. You have $20 to spend. Which equation can you solve to find the number of additional toppings you can get on your pizza?
Options:
a. 15x + 1.25 = 20
b. 1.25x + 15 = 20
c. 15x − 1.25 = 20
d. 1.25x − 15 = 20

Answer: b. 1.25x + 15 = 20

Explanation:
If x is the number of additional toppings, then 1.25 x is the cost of the additional toppings.
This gives the total cost is 1.25x + 15
then,
1.25x + 15 = 20

Question 6.
A bank offers a home improvement loan with simple interest at an annual rate of 12%. J.T. borrows $14,000 over a period of 3 years. How much will he pay back altogether?
Options:
a. $15680
b. $17360
c. $19040
d. $20720

Answer: c. $19040

Explanation:
Simple interest = 14,000 x 0.12 x 2 = $5,040
Amount = $14,000 + $5,040 = $19040

Question 7.
What is the volume of a triangular prism that is 75 centimeters long and that has a base with an area of 30 square centimeters?
Options:
a. 2.5 cm3
b. 750 cm3
c. 1125 cm3
d. 2250 cm3

Answer: d. 2250 cm3

Explanation:
V = Bh = 30(75) = 2250cm3

Question 8.
Consider the right circular cone shown.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 73
If a vertical plane slices through the cone to create two identical half cones, what is the shape of the cross section?
Options:
a. a rectangle
b. a square
c. a triangle
d. a circle

Answer: c. a triangle

Explanation:
Slicing through the vertex to create 2 identical half cones would create a cross-section that  is a triangle.

Page No. 304

Question 9.
The radius of the circle is given in meters. What is the circumference of the circle? Use 3.14 for π.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 74
a. 25.12 m
b. 50.24 m
c. 200.96 m
d. 803.84 m

Answer: b. 50.24 m

Explanation:
Circumference = 2 πr = 2 π(8) = 16 (3.14) = 50.24 m

Question 10.
The dimensions of the figure are given in millimeters. What is the area of the two-dimensional figure?
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 75
Options:
a. 39 mm2
b. 169 mm2
c. 208 mm2
d. 247 mm2

Answer: c. 208 mm2

Explanation:
Area of the square = 13 x 13 = 169 mm2
Area of the triangle = 1/2 x 13 x 6 = 39 mm2
The total area of the figure = 169 + 39 = 208 mm2

Question 11.
A forest ranger wants to determine the radius of the trunk of a tree. She measures the circumference to be 8.6 feet. What is the trunk’s radius to the nearest tenth of a foot?
Options:
a. 1.4 ft
b. 2.7 ft
c. 4.3 ft
d. 17.2 ft

Answer: a. 1.4 ft

Explanation:
Circumference = 2 πr = 8.6 ft
r = 8.6/2 π = 1.4 ft

Question 12.
What is the measure in degrees of an angle that is supplementary to a 74° angle?
Options:
a. 16°
b. 74°
c. 90°
d. 106°

Answer: d. 106°

Explanation:
Sum of supplementary angles = 180°
x + 74° = 180°
x = 106°

Question 13.
What is the volume in cubic centimeters of a rectangular prism that has a length of 6.2 centimeters, a width of 3.5 centimeters, and a height of 10 centimeters?
Options:
a. 19.7 cm3
b. 108.5 cm3
c. 217 cm3
d. 237.4 cm3

Answer: c. 217 cm3

Explanation:
V = Bh
B = 6.2 x 3.5 = 21.7 cm2
h = 10 cm
V = 21.7 x 10 = 217 cm3

Question 14.
A patio is the shape of a circle with diameter shown.
Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume img 76
What is the area of the patio? Use 3.14 for π.
Options:
a. 9 m2
b. 28.26 m2
c. 254.34 m2
d. 1017.36 m2

Answer: c. 254.34 m2

Explanation:
Diameter = 18 m
Radius = 18/2 = 9 m
Area of the patio = πr2 = 3.14 (9)2 = 254.34 m2

Question 15.
Petra fills a small cardboard box with sand. The dimensions of the box are 3 inches by 4 inches by 2 inches.
a. What is the volume of the box?
______ in3

Answer: 24 in3

Explanation:
V = Bh
B = 3 x 4 = 12 in2
V = 12 x 2 = 24 in3

Question 15.
b. Petra decides to cover the box by gluing on wrapping paper. How much wrapping paper does she need to cover all six sides of the box?
______ in2

Answer: 76 in2

Explanation:
P = 2(3) + 2(4) = 6 + 8 = 14 in
S = Ph + 2B = 14 x 2 + 2 x 24 = 76 in2

Question 15.
c. Petra has a second, larger box that is 6 inches by 8 inches by 4 inches. How many times larger is the volume of this second box? The surface area?
Volume is _________ times greater.
Surface area is _________ times greater

Answer: Surface area is about 2.7 times larger

Explanation:
B = 6 x 8 = 48 in2
V = Bh = 48 x 4 = 192 in3
192/24 = 8
P = 2(6) + 2(8) = 12 + 16 = 28
S = Ph + 2B = 28(4) + 2(48) = 112 + 96 = 208 in2
208/76 = 2.7

Conclusion:

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