go math grade 6 answer key homework book

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Go Math Grade 6 Answer Key of All Chapters | Middle School Grade 6 Solutions Key

Go Math Grade 6 Answer Key: In Today’s World learning has become demanding than ever before. Finding a reliable source has become a tedious task for anyone out there who wants to upgrade their skills. Sharpen your Math Skills taking the help of 6th Grade Go Math Middle School Solutions Key en route to your math Journey. Middle School Go Math Grade 6 Solutions Key includes Worked Out Solutions for all the Problems in Go Math Textbooks.

Access the Chapterwise Solutions for all the Questions in your Grade 6 Go Math Textbooks and make your learning effective. Grade 6 HMH Go Math Answer Keys develops mathematical understanding among students and also creates interest in the subject. With consistent practice, you can learn and clear the standard tests with flying colors.

Grade 6th Go Math Answer Key

All the solutions of Middle School Go Math Books for Grade 6 are prepared by subject experts. Go Math Books prevailing for 6th Standard are prepared to meet both the content and intent of Middle School. You can witness the Mathematical Concepts explained in a concise manner making it easier for you to have a good grip on the subject. HMH Go Math Grade 6 Answer Key includes the solved examples and practice questions to strengthen your Mathematical Concepts.

Grade 6 HMH Go Math – Answer Keys

• Chapter 1: Divide Multi-Digit Numbers
• Chapter 2: Fractions and Decimals
• Chapter 3: Understand Positive and Negative Numbers
• Chapter 4: Model Ratios
• Chapter 5: Model Percents
• Chapter 6: Convert Units of Length
• Chapter 7: Exponents
• Chapter 8: Solutions of Equations
• Chapter 9: Independent and Dependent Variables
• Chapter 10: Area of Parallelograms
• Chapter 11: Surface Area and Volume
• Chapter 12: Data Displays and Measures of Center
• Chapter 13: Variability and Data Distributions

Grade 6 McGraw Hill Glencoe – Answer Keys

• Chapter 1: Ratios and Rates
• Chapter 2: Fractions, Decimals, and Percents
• Chapter 3: Compute with Multi-Digit Numbers
• Chapter 4: Multiply and Divide Fractions
• Chapter 5: Integers and Coordinate Plane
• Chapter 6: Expressions
• Chapter 7: Equations
• Chapter 8: Functions and Inequalities
• Chapter 9: Area
• Chapter 10: Volume and Surface Area
• Chapter 11: Statistical Measures
• Chapter 12: Statistical Display

Go Math Middle School Grade 6 Answer Key of all Chapters

Avail Grade 6 Solutions provided over here and understand the concepts in a better way.  Identify the Knowledge Gap and allot time to the areas you feel difficult. Detailed description provided in the Go Math Grade 6th Solutions Key reflects more of the topics in your Middle School Textbooks. You can use them during your Homework or while preparing for Tests. Tap on the respective chapter you wish to practice and clarify all your concerns at one go.

Why to read Go Math 6th Std. Solutions Key?

There are plenty of benefits that come with solving the Go Math 6th Standard Answer Key. Refer to them and know the need of practicing through Grade 6 HMH Go Math Answer Key. They are as follows

• Go Math Answer Key for Grade 6 ensures success for every learner.
• Middle School Go Math Solutions Key makes learning easier for both Students and Teachers.
• Elaborate Explanation provided to all the Math Practice Problems helps you enhance your subject knowledge.
• Go Math Grade 6 Answer Key lays a stronger foundation of Fundamentals of your Mathematical Concepts.

FAQs on Grade 6 Go Math Answer Key

1.  Where Can I get Go Math Grade 6 Answer Key PDF?

You can get Go Math Grade 6th Answer Key PDF for all the chapters on our page.

2. Which Website offers the best resources on Go Math Grade 6 Answer Key?

ccssmathanswers.com is a trustworthy site for all your needs and provides reliable information on Go Math Answer Key for Class 6th. Take your preparation to the next level and score better grades in your exams.

3. Where can I find Chapterwise Solutions of Go Math 6th Grade Problems?

You can find the Go Math 6th Std Solutions Key of all Chapters on our page. Simply prepare the Chapter you want to access through the direct links and learn accordingly.

4. How to Learn 6th Grade Math Concepts easily?

Learning concepts from the Go Math Middle School Answer Key for Grade 6 makes you familiar with a variety of questions. Thereby, you can prepare effectively and clear your tests with higher grades.

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Go Math! 6 Common Core Edition with Online Resources

Go math 6 common core edition, by houghton mifflin harcourt, published date: 04-11-2024.

Go Math! 6 Common Core Edition is a educational Book By Houghton Mifflin Harcourt.

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Go Math! Student Edition Set Grade 6

• Material Type: Teacher Materials
• Format: Kit
• ISBN-13/EAN: 9780544433403
• ISBN-10: 0544433408
• Product Code: 1596246
• National/State: National
• Copyright Year: 2015
• Language: English

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Go Math! Getting Ready for Grade 6 Math Homework/ Extra Practice

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• Grade 6 HMH Go Math - Answer Keys

Explanation:

Order the numbers from least to greatest.

4, \(^{-}3\), \(^{-}7\)

• Type below:

0, \(^{-}1\), 3

\(^{-}5\), \(^{-}3\), \(^{-}9\)

Order the numbers from greatest to least.

\(^{-}1\), \(^{-}4\), 2

\(^{-}5\), \(^{-}4\), \(^{-}3\)

  On Your Own  

2, 1, \(^{-}1\)

\(^{-}6\), \(^{-}12\), 30

15, \(^{-}9\), \(^{-}20\)

Order the number from greatest to least.

\(^{-}13\), 14, \(^{-}14\)

\(^{-}20\), \(^{-}30\), \(^{-}40\)

9, \(^{-}37\), 0

Saturday’s low temperature was −6°F. Sunday’s low temperature was 3°F. Monday’s low temperature was −2°F. Tuesday’s low temperature was 5°F. Which day’s low temperature was closest to 0°F?

Use Symbols Write a comparison using < or > to show that South America’s Valdes Peninsula (elevation −131 ft) is lower than Europe’s Caspian Sea (elevation −92 ft).

Yes, email page to my online tutor. ( if you didn't add a tutor yet, you can add one here )

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Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms

Students of 6th Grade can get the detailed explanations for Chapter 10 Area of Parallelograms from here. So, Download Go Math Grade 6 Answer Key Chapter 10 Area of Parallelograms pdf for free. You can understand the concepts of areas from Go Math Grade 6 Answer Key. Click on the link in which you are lagging and allot some time to learn the concepts.

Get the answers for Go Math Grade 6 Chapter 10 Area of Parallelograms Practice Test here. So, the who are preparing for the exams can practice the questions and check the answer from the below links. Check out the topics of Chapter 10 Area of Parallelograms before you start your preparation.

The Chapter Area of Parallelograms includes topics like the area of triangles, Area of Trapezoids, Area of Regular Polygons, Composite Figures. To make you understand in a simple manner we have provided the images and graphs in the explanation.

Lesson 1: Algebra • Area of Parallelograms

Share and Show – Page No. 535

Problem solving + applications – page no. 536, area of parallelograms – page no. 537, lesson check – page no. 538.

Lesson 2: Investigate • Explore Area of Triangles

Share and Show – Page No. 541

Sense or nonsense – page no. 542, explore area of triangles – page no. 543, lesson check – page no. 544.

Lesson 3: Algebra • Area of Triangles

Share and Show – Page No. 547

Unlock the problem – page no. 548, area of triangles – page no. 549, lesson check – page no. 550.

Lesson 4: Investigate • Explore Area of Trapezoids

Share and Show – Page No. 553

What’s the error – page no. 554, explore area of trapezoids – page no. 555, lesson check – page no. 556.

Lesson 5: Algebra • Area of Trapezoids

Share and Show – Page No. 559

Problem solving + applications – page no. 560, area of trapezoids – page no. 561, lesson check – page no. 562.

Mid-Chapter Checkpoint

Mid-Chapter Checkpoint – Vocabulary – Page No. 563

• Mid-Chapter Checkpoint – Page No. 564

Lesson 6: Area of Regular Polygons

Share and Show – Page No. 567

Page no. 568, area of regular polygons – page no. 569, lesson check – page no. 570.

Lesson 7: Composite Figures

Share and Show – Page No. 573

Unlock the problem – page no. 574, composite figures – page no. 575, lesson check – page no. 576.

Lesson 8: Problem Solving • Changing Dimensions

Share and Show – Page No. 579

On your own – page no. 580, problem solving changing dimensions – page no. 581, lesson check – page no. 582.

Lesson 9: Figures on the Coordinate Plane

Share and Show – Page No. 585

• Problem Solving + Applications – Page No. 586

Figures on the Coordinate Plane – Page No. 587

Lesson check – page no. 588.

Chapter 10 Review/Test

Chapter 10 Review/Test – Page No. 589

• Chapter 10 Review/Test – Page No. 590
• Chapter 10 Review/Test – Page No. 591
• Chapter 10 Review/Test – Page No. 592
• Chapter 10 Review/Test – Page No. 593
• Chapter 10 Review/Test – Page No. 594

Find the area of the parallelogram or square.

Answer: 9.96

Explanation: Given that Base = 8.3 m Height = 1.2 m We know that the area of the parallelogram is base × height A = bh A = 8.3 m × 1.2 m A = 9.96 square meters Thus the area of the parallelogram for the above figure is 9.96 m²

Explanation: Given, Base = 15 ft Height = 6 ft Area = ? We know that, Area of the parallelogram = bh A = 15 ft × 6 ft A = 90 square feet Thus the area of the parallelogram for the above figure is 90 ft²

Answer: 6.25

Explanation: The above figure is a square The side of the square is a × a A = 2.5 mm × 2.5 mm A = 6.25 square mm Thus the area of the square is 6.25 mm²

Answer: 1/2

Explanation: Given Base = 3/4 ft Height = 2/3 ft Area of the parallelogram is base × height A = bh A = 3/4 × 2/3 A = 1/2 Thus the area of the above parallelogram is 1/2 ft²

Find the unknown measurement for the parallelogram.

Explanation: Given, A = 11 yd² B = 5 1/2 yd We know that A = bh 11 = 5 1/2 × h 11 = 11/2 × h 22 = 11 × h H = 2 yd Thus the height of the above figure is 2 yards.

Answer: 8 yd

Explanation: Given Area = 32 yd 2 Base = 4 yd Height = ? We know that A = b × h 32 = 4 yd × h H = 32/4 H = 8 yd Therefore the height of the above figure is 8 yards.

On Your Own

Find the area of the parallelogram.

Answer: 58.24

Explanation: Given Base = 9.1 m Height = 6.4 m A = b × h A = 9.1 m × 6.4 m A = 58.24 square meters Thus the area of the parallelogram for the above figure is 58.24 m²

Answer: 168

Explanation: Given Base = 21 ft Height = 8ft We know that the area of the parallelogram is  base × height A = 21 ft × 8ft A = 168 square feet Therefore the area of the above figure is 168 ft²

Find the unknown measurement for the figure.

Question 9. square A = ? s = 15 ft A = _______ ft

Answer: 225

Explanation: Given, S = 15 ft The area of the square is s × s A = 15 ft × 15 ft A = 225 ft² Thus the area of the square is 225 square feet.

Question 10. parallelogram A = 32 m 2 b = ? h = 8 m b = _______ m

Explanation: Given A = 32 m² H = 8m B = ? To find the base we have to use the area of the parallelogram formula A = bh 32 m² = b × 8 m B = 32/8 B = 4 m Thus the base is 4 meters

Lesson 1 Extra Practice Area of Parallelograms Answers Question 11. parallelogram A = 51 \(\frac{1}{4}\) in. 2 b = 8 \(\frac{1}{5}\) in. h = ? ________ \(\frac{□}{□}\) in.

Answer: 6 \(\frac{1}{4}\) in.

Explanation: Given, A = 51 \(\frac{1}{4}\) in. 2 b = 8 \(\frac{1}{5}\) in. H = ? We know that the area of the parallelogram is  base × height A = bh 51 \(\frac{1}{4}\) = h × 8 \(\frac{1}{5}\) in. h = 51 \(\frac{1}{4}\) ÷ 8 \(\frac{1}{5}\) in. h = 205/4 ÷ 41/5 h = 1025/164 h = 6 \(\frac{1}{4}\) in. Thus the height of the parallelogram is 6 \(\frac{1}{4}\) in.

Question 12. parallelogram A = 121 mm 2 b = 11 mm h = ? ________ mm

Answer: 11 mm

Explanation: Given A = 121 mm² B = 11 mm H = ? We know that A = b × h 121 mm² = 11 mm × h H = 121/11 H = 11 mm Thus the height is 11 mm.

Question 13. The height of a parallelogram is four times the base. The base measures 3 \(\frac{1}{2}\) ft. Find the area of the parallelogram. ________ ft 2

Explanation: Given B= 3 \(\frac{1}{2}\) H = 4b H = 4 × 3 \(\frac{1}{2}\) H = 4 × 7/2 H = 14 A = bh A = 7/2 × 14 A = 7 × 7 = 49 Thus the area of the parallelogram is 49 ft²

Answer: 2250

Explanation: Jane’s backyard is shaped like a parallelogram. The base of the parallelogram is 90 feet, and the height is 25 feet. A = bh A = 90 ft × 25 ft A = 2250 square feet Therefore the area of the parallelogram for the above figure is 2250 ft 2

Answer: 104

Explanation: Jack made a parallelogram by putting together two congruent triangles and a square, like the figures shown at the right. The triangles have the same height as the square. Base = 8 cm + 5 cm = 13 cm Height = 8 cm Area = bh A = 13 cm × 5 cm A = 104 square cm Thus the area of the parallelogram is 104 cm 2

Question 16. The base of a parallelogram is 2 times the parallelogram’s height. If the base is 12 inches, what is the area? ________ ft 2

Explanation: The base of a parallelogram is 2 times the parallelogram’s height. Base = 12 ft Height = 12/2 = 6 ft The area of parallelogram is base × height A = bh A = 12 ft × 6 ft A = 72 ft 2 Thus the area of the parallelogram is 72 ft 2

Explanation: Base = 3 in Height = 3 in A = bh A = 3 in × 3 in A = 9 square inches Therefore the area of the above figure is 9 in²

Explanation: Base = 12 in H = 5 in A = bh A = 12 in × 5 in A = 60 square inches A = 60 in²

Find the area of the figure.

Answer: 126

Explanation: The base of the figure is 18 ft Height = 7 ft The area of the parallelogram is bh A = 18 ft × 7 ft A = 126 square feet Thus the area of the parallelogram is 126 ft 2

Explanation: Base = 7 cm Height = 5 cm A = bh A = 7 cm × 5 cm A = 35 square cm A = 35 cm 2

Question 3. parallelogram A = 9.18 m 2 b = 2.7 m h = ? h = ________ m

Answer: 3.4

Explanation: A = 9.18 m 2 b = 2.7 m h = ? A = bh 9.18 m 2 = 2.7 m × h h = 9.18/2.7 A = 3.4 m

Question 4. parallelogram A = ? b = 4 \(\frac{3}{10}\) m h = 2 \(\frac{1}{10}\) m A = ________ \(\frac{□}{□}\) m 2

Explanation: b = 4 \(\frac{3}{10}\) m h = 2 \(\frac{1}{10}\) m A = ? A = bh A = 4 \(\frac{3}{10}\) m × 2 \(\frac{1}{10}\) m A = \(\frac{43}{10}\) m × \(\frac{21}{10}\) m A = \(\frac{903}{100}\) m² A = 9 \(\frac{3}{100}\) m²

Question 5. square A = ? s = 35 cm A = ________ cm 2

Answer: 1225

Explanation: s = 35 cm A = s × s A = 35 cm × 35 cm A = 1225 cm 2 Area of the parallelogram is 1225 cm 2

Question 6. parallelogram A = 6.3 mm 2 b = ? h = 0.9 mm b = ________ mm

Explanation: A = 6.3 mm 2 b = ? h = 0.9 mm A = bh 6.3 mm 2 = b × 0.9 mm b = 6.3/0.9 b = 7 mm Thus the base of the parallelogram is 7 mm.

Problem Solving

Question 7. Ronna has a sticker in the shape of a parallelogram. The sticker has a base of 6.5 cm and a height of 10.1 cm. What is the area of the sticker? ________ cm 2

Answer: 65.65

Explanation: Ronna has a sticker in the shape of a parallelogram. The sticker has a base of 6.5 cm and a height of 10.1 cm. A = bh A = 6.5 cm × 10.1 cm A = 65.65 cm 2

Question 8. A parallelogram-shaped tile has an area of 48 in. 2 . The base of the tile measures 12 in. What is the measure of its height? ________ in.

Explanation: A parallelogram-shaped tile has an area of 48 in. 2 The base of the tile measures 12 in. A = bh 48 = 12 × h h = 48/12 = 4 in Therefore the height of the parallelogram is 4 inches

Question 9. Copy the two triangles and the square in Exercise 15 on page 536. Show how you found the area of each piece. Draw the parallelogram formed when the three figures are put together. Calculate its area using the formula for the area of a parallelogram. Type below: _______________

Question 1. Cougar Park is shaped like a parallelogram and has an area of \(\frac{1}{16}\) square mile. Its base is \(\frac{3}{8}\) mile. What is its height? \(\frac{□}{□}\) mile

Answer: \(\frac{1}{6}\) mile

Explanation: Cougar Park is shaped like a parallelogram and has an area of \(\frac{1}{16}\) square mile. Its base is \(\frac{3}{8}\) mile. A = bh \(\frac{1}{16}\) = \(\frac{3}{8}\) × h \(\frac{1}{16}\) × \(\frac{8}{3}\) = h h =  \(\frac{1}{6}\) mile

Question 2. Square County is a square-shaped county divided into 16 equal-sized square districts. If the side length of each district is 4 miles, what is the area of Square County? ________ square miles

Answer: 256 square miles

Explanation: Square County is a square-shaped county divided into 16 equal-sized square districts. If the side length of each district is 4 miles 4 × 4 = 16 A = 16 × 16 = 256 square miles

Spiral Review

Question 3. Which of the following values of y make the inequality y < – 4 true? y = – 4     y = – 6      y = 0    y = – 8    y = 2 Type below: _______________

Answer: y = -6

Practice and Homework Lesson 10.2 Answer Key Question 4. On a winter’s day, 9°F is the highest temperature recorded. Write an inequality that represents the temperature t in degrees Fahrenheit at any time on this day. Type below: _______________

Answer: t ≤ 9

Explanation: On a winter’s day, 9°F is the highest temperature recorded. t will be less than or equal to 9. The inequality is t ≤ 9

Question 5. In 2 seconds, an elevator travels 40 feet. In 3 seconds, the elevator travels 60 feet. In 4 seconds, the elevator travels 80 feet. Write an equation that gives the relationship between the number of seconds x and the distance y the elevator travels. Type below: _______________

Answer: y = 20x

Explanation: x represents the number of seconds y represents the distance the elevator travels. The elevator travels 20 feet per second. Thus the equation is y = 20x

Question 6. The linear equation y = 4x represents the number of bracelets y that Jolene can make in x hours. Which ordered pair lies on the graph of the equation? Type below: _______________

Answer: (4, 16)

Explanation: y = 4x If x = 4 Then y = 4(4) y = 16 Thus the ordered pairs are (4, 16)

Answer: Base = 9 units Height = 4 units Area of the parallelogram = base × height A = 9 × 4 A = 36 sq. units Area of the triangle = ab/2 A = (9 × 4)/2 A = 18 sq. units Area of another triangle = ab/2 A = (9 × 4)/2 A = 18 sq. units

Find the area of each triangle.

Explanation: The area of the right triangle is bh/2 A = (8 × 10)/2 A = 80/2 A = 40 in. 2 Thus the area of the triangle for the above figure is 40 in. 2

Answer: 180

Explanation: The area of the right triangle is bh/2 A = (18 × 20)/2 A = 360/2 A = 180 ft 2

Explanation: The area of the right triangle is bh/2 A = (4 × 11)/2 A = 44/2 A = 22 A = 22 yd 2 Thus the area of the triangle is 22 yd 2

Answer: 495

Explanation: The area of the right triangle is bh/2 A = (30 × 33)/2 A = 990/2 A = 495 mm 2 Thus the area of the triangle is 495 mm 2

Answer: 190

Explanation: The area of the right triangle is bh/2 A = (19 × 20)/2 A = 380/2 A = 190 in. 2 Thus the area of the triangle is 190 in. 2

Explanation: The area of the right triangle is bh/2 A = (16 × 12)/2 A = 192/2 A = 96 Sq. cm Thus the area of the triangle is 96 Sq. cm

Problem Solving + Applications

Question 8. Communicate Describe how you can use two triangles of the same shape and size to form a parallelogram. Type below: _______________

Answer: Put them together like a puzzle. if the sides are parallel then it would be a parallelogram.

Question 9. A school flag is in the shape of a right triangle. The height of the flag is 36 inches and the base is \(\frac{3}{4}\) of the height. What is the area of the flag? _______ in. 2

Answer: 486 in. 2

Explanation: A school flag is in the shape of a right triangle. The height of the flag is 36 inches and the base is \(\frac{3}{4}\) of the height. B = 36 × \(\frac{3}{4}\) B = 27 Area of the triangle = bh/2 A = (36 × 27)/2 A = 486 sq. in Thus the area of the triangle is 486 in. 2

Answer: Tyson’s Model makes sense. The base of the figure is 30 in. The height of the figure is 40 in Area of the triangle = bh/2 A = (30 × 40)/2 A = 1200/2 = 600 sq. in Cyndi’s Model doesn’t make sense because there is no base for the triangle.

Answer: 7.5 ft. 2

Explanation: A flag is separated into two different colors. B = 5 ft H = 3 ft Area of the triangle = bh/2 A = (3 × 5)/2 A = 15/2 A = 7.5 sq. ft

Explanation: Given, Base = 6 ft Height = 10 ft Area of the triangle = bh/2 A = (6 ft × 10 ft)/2 A = 60 sq. ft/2 A = 30 ft 2 Thus the area of the triangle for the above figure is 0 ft 2

Answer: 925

Explanation: Given, Base = 50 cm Height = 37 cm Area of the triangle = bh/2 A = (50 × 37)/2 A = 1850/2 A = 925 sq. cm Therefore the area of the above figure is 925 cm 2

Answer: 400

Explanation: Given, Base = 40 mm Height = 20 mm Area of the triangle = bh/2 A = (40 × 20)/2 A = 800/2 A = 400 mm 2 Therefore the area of the above figure is 400 mm 2

Explanation: Given, Base = 12 in. Height = 30 in. Area of the triangle = bh/2 A = (12 × 30)/2 A = 360/2 A = 180 in. 2 Therefore the area of the above figure is 180 in. 2

Explanation: Given, Base = 15 cm Height = 30 cm Area of the triangle = bh/2 A = (15 × 30)/2 A = 450/2 A = 225 cm 2 Therefore the area of the above figure is 225 cm 2

Answer: 450

Explanation: Given, Base = 20 cm Height = 45 cm Area of the triangle = bh/2 A = (20 × 45)/2 A = 900/2 A = 450 cm 2 Therefore the area of the above figure is 450 cm 2

Question 7. Fabian is decorating a triangular pennant for a football game. The pennant has a base of 10 inches and a height of 24 inches. What is the total area of the pennant? _______ in. 2

Answer: 120

Explanation: Fabian is decorating a triangular pennant for a football game. The pennant has a base of 10 inches and a height of 24 inches. Area of the triangle = bh/2 A = (10 × 24)/2 A = 240/2 A = 120 in. 2 Therefore the area of the above figure is 120 in. 2

Question 8. Ryan is buying a triangular tract of land. The triangle has a base of 100 yards and a height of 300 yards. What is the area of the tract of land? _______ yd 2

Answer: 15000

Explanation: Given, Base = 100 yards Height = 300 yards Area of the triangle = bh/2 A = (100 × 300)/2 A = 30000/2 A = 15000 yd 2 Therefore the area of the above figure is 15000 yd 2

Question 9. Draw 3 triangles on grid paper. Draw appropriate parallelograms to support the formula for the area of the triangle. Tape your drawings to this page. Type below: _______________

Question 1. What is the area of a triangle with a height of 14 feet and a base of 10 feet? _______ ft 2

Explanation: Given, Base = 10 feet Height = 14 feet Area of the triangle = bh/2 A = (14 × 10)/2 A = 140/2 A = 70 ft 2 Therefore the area of the triangle is 70 ft 2

10.2 Area of Parallelograms and Triangles Question 2. What is the area of a triangle with a height of 40 millimeters and a base of 380 millimeters? _______ mm 2

Answer: 7600

Explanation: Given, Base = 380 millimeters Height = 40 millimeters Area of the triangle = bh/2 A = (380 × 40)/2 A = 15200/2 A = 7600 mm 2

Question 3. Jack bought 3 protein bars for a total of $4.26. Which equation could be used to find the cost c in dollars of each protein bar? Type below: _______________

Answer: 3c = 4.26

Explanation: Jack bought 3 protein bars for a total of $4.26. c represents the cost of each protein bar 3c = 4.26

Question 4. Coach Herrera is buying tennis balls for his team. He can solve the equation 4c = 92 to find how many cans c of balls he needs. How many cans does he need? _______ cans

Explanation: Coach Herrera is buying tennis balls for his team. 4c = 92 c = 92/4 c = 23 Therefore he need 23 cans.

Question 5. Sketch the graph of y ≤ – 7 on a number line. Type below: _______________

Question 6. A square photograph has a perimeter of 20 inches. What is the area of the photograph? _______ in. 2

Explanation: A square photograph has a perimeter of 20 inches. p = 4s 20 = 4s s = 20/4 s = 5 in. Area of the square is s × s A = 5 × 5 = 25 Thus the area of square photograph = 25 in. 2

Explanation: B = 14 cm H = 8 cm Area of the triangle = bh/2 A = (14 × 8)/2 A = 14 × 4 A = 56 sq. cm Thus the area of the above figure is 56 cm 2

Explanation: B = 22 in. H = ? A = 132 in. 2 Area of the triangle = bh/2 132 sq. in  = 22 in × h h = 132 sq. in/22 in h = 12 in Thus the height of the above figure is 12 in.

Find the area of the triangle.

Answer: 540

Explanation: B = 27 mm H = 40 mm Area of the triangle = bh/2 A = (27 × 40)/2 A = 27 × 20 = 540 A = 540 mm 2 Therefore the area of the above figure is 540 mm 2

Explanation: B = 5.5 mm H = 4 mm Area of the triangle = bh/2 A = (5.5 mm × 4 mm)/2 A = 5.5 mm × 2 mm A = 11 mm 2 Therefore the area of the above figure is 11 mm 2

Explanation: B = 5 in H =? A = 52.5 sq. in Area of the triangle = bh/2 52.5 sq. in = (5 × h)/2 52.5 sq. in × 2 = 5h h = 21 in Thus the height of the above figure is 21 in

Answer: 4.3

Explanation: B = 80 mm = 8 cm H = ? A = 17.2 sq. cm Area of the triangle = bh/2 17.2 sq. cm = (8 cm × h)/2 17.2 × 2 = 8 × h h = 4.3 cm Thus the height of the above figure is 4.3 cm

Lesson 3 Area of Trapezoids Answer Key Question 7. Verify the Reasoning of Others The height of a triangle is twice the base. The area of the triangle is 625 in. 2 . Carson says the base of the triangle is at least 50 in. Is Carson’s estimate reasonable? Explain. Type below: _______________

Answer: A = 625 in. 2 B = 50 in H = 2b H = 2 × 50 in H = 100 in Area of the triangle = bh/2 625 in. 2 = (50 × 100)/2 625 in. 2 = 2500 No Carson’s estimation is not reasonable.

Answer: Base: 14 in Height: 14 in

Explanation: Given that, Each shelf is 14 inches deep. Height = 14 inches By seeing the above figure we can say that the base of the shelves is 14 inches Base = 14 inches

Question 8. b. What formula can you use to find the area of a triangle? Type below: _______________

Answer: The formula to find the Area of the triangle = bh/2

Question 8. c. Explain how you can find the area of one triangular support. Type below: _______________

Answer: We can find the area of one triangle support by substituting the base and height in the formula. A = (14 × 14)/2 A = 98 sq. in

Question 8. d. How many triangular supports are needed to build 4 shelves? _______ supports

Answer: 8 By seeing the above figure we can say that 8 triangular supports are needed to build 4 shelves.

Question 8. e. How many square inches of wood will Alani need to make all the supports? _______ in. 2

Answer: 784

Explanation: The depth of each shelf made by the Alamo is 14 inches. So the base of the right isosceles triangular supporter is 14 inches. So one equal side is 14 cm. Now by using the Pythagoras theorem, we can calculate the other side of the supporter = = 19.8 inches. The area of the right isosceles triangle is given by × base × height. Here the base and height are equal to 14 inches. Therefore the area of each right isosceles triangular supporter is A = (14 × 14)/2 A = 98 sq. in Each shelf would require two such supporters and there are 4 such shelves. Thus the total number of supporters required is 8. Square inches of wood necessary for 8 right isosceles triangular supporters = 98 × 8 = 784 square inches.

Question 9. The area of a triangle is 97.5 cm 2 . The height of the triangle is 13 cm. Find the base of the triangle. Explain your work. b = _______ cm

Answer: 15 cm

Explanation: Given, The area of a triangle is 97.5 cm 2 . The height of the triangle is 13 cm. Area of the triangle = bh/2 97.5 cm 2 = (b × 13 cm)/2 b = 2 × 97.5cm 2 /13 cm b = 15 cm Therefore the base of the triangle is 15 cm

Question 10. The area of a triangle is 30 ft 2 . For numbers 10a–10d, select Yes or No to tell if the dimensions given could be the height and base of the triangle. 10a. h = 3, b = 10 10b. h = 3, b = 20 10c. h = 5, b = 12 10d. h = 5, b = 24 10a. ___________ 10b. ___________ 10c. ___________ 10d. ___________

Answer: 10a. No 10b. yes 10c. Yes 10d. No

Explanation: The area of a triangle is 30 ft 2 . 10a. h = 3, b = 10 Area of the triangle = bh/2 A = (3 × 10)/2 A = 15 ft 2 . Thus the answer is no. 10b. h = 3, b = 20 Area of the triangle = bh/2 A = (3 × 20)/2 A = 30 ft 2 . Thus the answer is yes. 10c. h = 5, b = 12 Area of the triangle = bh/2 A = (5 × 12)/2 A = 30 ft 2 . Thus the answer is yes. 10d. h = 5, b = 24 Area of the triangle = bh/2 A = (5 × 24)/2 A = 60 ft 2 . Thus the answer is no.

Find the area.

Explanation: Given, Base = 15 in. Height = 6 in. Area of the triangle = bh/2 A = (15 × 6)/2 A = 90/2 A = 45 in. 2

Answer: 0.36

Explanation: Given, Base = 1.2 m Height = 0.6 m Area of the triangle = bh/2 A = (1.2 × 0.6)/2 A = 0.72/2 A = 0.36 m 2

Explanation: Given, Base = 4 1/2 ft Height = 2 2/3 ft Area of the triangle = bh/2 A = (4 1/2 × 2 2/3)/2 A = 12/2 A = 6 ft 2

Find the unknown measurement for the triangle.

Question 4. A = 0.225 mi 2 b = 0.6 mi h = ? h = _______ mi

Answer: 0.75

Explanation: Given, A = 0.225 mi 2 b = 0.6 mi h = ? Area of the triangle = bh/2 0.225 = (0.6 × h)/2 0.450 = 0.6 × h h = 0.450/0.6 h = 0.75 mi

Answer Key 10.3 Practice A Geometry Answers Question 5. A = 4.86 yd 2 b = ? h = 1.8 yd b = _______ yd

Answer: 5.4 yd

Explanation: Given, A = 4.86 yd 2 b = ? h = 1.8 yd Area of the triangle = bh/2 4.86 yd 2 = (b × 1.8 yd)/2 4.86 × 2 = b × 1.8 9.72 = b × 1.8 b = 9.72/1.8 b = 5.4 yd

Question 6. A = 63 m 2 b = ? h = 12 m b = _______ m

Answer: 10.5

Explanation: Given, A = 63 m 2 b = ? h = 12 m Area of the triangle = bh/2 63 = (b × 12)/2 63 = b × 6 b = 63/6 b = 10.5 m

Lesson 3 Skills Practice Area of Trapezoids Answer Key Question 7. A = 2.5 km 2 b = 5 km h = ? h = _______ km

Explanation: Given, A = 2.5 km 2 b = 5 km h = ? Area of the triangle = bh/2 2.5 = (5 km × h)/2 2.5 km 2 = 2.5 km × h h = 2.5/2.5 h = 1 km

Question 8. Bayla draws a triangle with a base of 15 cm and a height of 8.5 cm. If she colors the space inside the triangle, what area does she color? _______ cm 2

Answer: 63.75 cm 2

Explanation: Bayla draws a triangle with a base of 15 cm and a height of 8.5 cm. B = 15 cm h = 8.5 cm Area of the triangle = bh/2 A = (15 cm × 8.5 cm)/2 A = 7.5 cm × 8.5 cm A = 63.75 cm 2

Question 9. Alicia is making a triangular sign for the school play. The area of the sign is 558 in. 2 . The base of the triangle is 36 in. What is the height of the triangle? _______ in.

Explanation: Given, Alicia is making a triangular sign for the school play. The area of the sign is 558 in. 2 The base of the triangle is 36 in. Area of the triangle = bh/2 558 = (36 × h)/2 558 = 18 × h h = 558/18 h = 31 inches

Question 10. Describe how you would find how much grass seed is needed to cover a triangular plot of land. Type below: _______________

You will need to find the area A=height multiplied by the base divided by 2 Area of the triangle = bh/2

Question 1. A triangular flag has an area of 187.5 square inches. The base of the flag measures 25 inches. How tall is the triangular flag? _______ in.

Answer: 15 in.

Explanation: A triangular flag has an area of 187.5 square inches. The base of the flag measures 25 inches. Area of the triangle = bh/2 187.5 square inches = (25 inches × h)/2 187.5 sq. in × 2 = 25h 375 sq. in = 25h h = 375 sq. in/25 h = 15 inches

Lesson 3 Area of Triangles Answer Key Question 2. A piece of stained glass in the shape of a right triangle has sides measuring 8 centimeters, 15 centimeters, and 17 centimeters. What is the area of the piece? _______ cm 2

Explanation: A piece of stained glass in the shape of a right triangle has sides measuring 8 centimeters, 15 centimeters, and 17 centimeters. b = 8 cm h = 15 cm Area of the triangle = bh/2 A = (8 × 15)/2 A = 4 cm × 15 cm A = 60 sq. cm

Question 3. Tina bought a T-shirt and sandals. The total cost was $41.50. The T-shirt cost $8.95. The equation 8.95 + c = 41.50 can be used to find the cost of c in dollars of the sandals. How much did the sandals cost? $ _______

Answer: $32.55

Explanation: Tina bought a T-shirt and sandals. The total cost was $41.50. The T-shirt cost $8.95. 8.95 + c = 41.50 c = 41.50 – 8.95 c = $32.55

Question 4. There are 37 paper clips in a box. Carmen places more paper clips in the box. Write an equation to show the total number of paper clips p in the box after Carmen places n more paper clips in the box. Type below: _______________

Answer: 37 + n = p

Explanation: There are 37 paper clips in a box. Carmen places more paper clips in the box. n represents the number of paper clips in the box The equation is 37 + n = p

Answer: The ordered pairs are (1, 6), (2, 12), (3, 18), (4, 16)

Answer: 58.5

Explanation: Given, b = 13 cm h = 9 cm Area of the triangle = bh/2 A = (13 × 9)/2 A = 117/2 A = 58.5 cm 2

Answer: Figure 1: Base 1 = 3 units Base 2= 7 units Height = 4 units Area of the trapezium = (b1 + b2)h/2 A = (3 + 7)4/2 A = 10 × 2 A = 20 sq. units Figure 2: Base 1 = 7 units Base 2= 3 units Height = 4 units Area of the trapezium = (b1 + b2)h/2 A = (7 + 3)4/2 A = 10 × 2 A = 20 sq. units

Find the area of the trapezoid.

Explanation: Base 1 = 6 cm Base 2 = 10 cm Height = 5 cm We know that the Area of the trapezium is the sum of bases into height divided by 2. Area of the trapezium = (b1 + b2)h/2 A = (6 cm + 10 cm)5 /2 A = (16 × 5)/2 A = 40 sq. cm

Explanation: b1 = 3 in b2 = 9 in. h = 8 in. We know that the Area of the trapezium is the sum of bases into height divided by 2. Area of the trapezium = (b1 + b2)h/2 A = (3 + 9)8/2 A = 12 × 4 A = 48 sq. in

Explanation: b1 = 11 ft b2 = 5 ft h = 8 ft We know that the Area of the trapezium is the sum of bases into height divided by 2. Area of the trapezium = (b1 + b2)h/2 A = (11 + 5)8/2 A = 16 × 4 A = 64 sq. ft

Answer: 266

Explanation: b1 = 16 cm b2 = 22 cm h = 14 cm We know that the Area of the trapezium is the sum of bases into height divided by 2. Area of the trapezium = (b1 + b2)h/2 A = (16 + 22)14/2 A = 38 × 7 A = 266 sq. cm

Answer: 71.5

Explanation: b1 = 8 mm b2 = 14 mm h = 6.5 mm We know that the Area of the trapezium is the sum of bases into height divided by 2. Area of the trapezium = (b1 + b2)h/2 A = (8 + 14)6.5/2 A = 11 × 6.5 A = 71.5 sq. mm

Answer: 31.5

Explanation: b1 = 3 1/2 in. b2 = 8 1/2 in. h = 5 1/4 in. We know that the Area of the trapezium is the sum of bases into height divided by 2. Area of the trapezium = (b1 + b2)h/2 b = 3 1/2 + 8 1/2 b = 12 A = 5 1/4 × 12/2 A = 5 1/4 × 6 A = 31.5 sq. in

Question 8. Describe a Method Explain one way to find the height of a trapezoid if you know the area of the trapezoid and the length of both bases. Type below: _______________

Answer: 1) Add the length of both bases: [Total Length = Length 1 + Length 2] 2) Divide the length that you found by 2. [Average Length = Total Length ÷ 2] 3) Divide the Area with the length found [Height = Area ÷ average length]

Lesson 3 Area of Composite Figures Answer Key Question 9. A patio is in the shape of a trapezoid. The length of the longer base is 18 feet. The length of the shorter base is two feet less than half the longer base. The height is 8 feet. What is the area of the patio? _______ ft 2

Answer: 100

Explanation: trapezoid area = ((sum of the bases) ÷ 2) × height long base = 18 short base = 7 height = 8 trapezoid area = [(18 + 7) / 2] × 8 trapezoid area = [(12.5)] × 8 trapezoid area = 100 square feet

Question 10. Except for a small region near its southeast corner, the state of Nevada is shaped like a trapezoid. The map at the right shows the approximate dimensions of the trapezoid. Sabrina used the map to estimate the area of Nevada. Look at how Sabrina solved the problem. Find her error. Two copies of the trapezoid can be put together to form a rectangle. length of rectangle: 200 + 480 = 680 mi width of rectangle: 300 mi A = lw A = 680 × 300 A = 204,000 The area of Nevada is about 204,000 square miles. Describe the error. Find the area of the trapezoid to estimate the area of Nevada. Type below: _______________

Answer: The area of Nevada is she didn’t divide by 2. Area of the trapezium = (b1 + b2)h/2 A = (200 + 480)300/2 A = 680 × 150 A = 102000 sq. miles

Explanation: b1= 3 in b2 = 7 in h = 6 in. Area of the trapezium = (b1 + b2)h/2 A = (3 + 7)6/2 A = 10 × 3 A = 30 sq. in Thus the area of the trapezium is 30 in. 2

Answer: Figure 1: b1 = 2 units b2 = 6 units h = 3 units Area of the trapezium = (b1 + b2)h/2 A = (2 + 6)3/2 A = (8)(3)/2 A = 24/2 = 12 A = 12 sq. units Figure 2: b1 = 6 units b2 = 2 units h = 3 units Area of the trapezium = (b1 + b2)h/2 A = (6 + 2)3/2 A = (8)(3)/2 A = 24/2 = 12 The area of figure 2 is 12 sq. units

Answer: 38.5

Explanation: Given, b1 = 9 in b2 = 2 in h = 7 in Area of the trapezium = (b1 + b2)h/2 A = (9 + 2)7/2 A = (11 × 7)/2 A = 77/2 = 38.5 in. 2

Answer: 3600

Explanation: Given, b1 = 24 yd b2 = 48 yd h = 100 yd Area of the trapezium = (b1 + b2)h/2 A = (24 + 48)100/2 A = 72 × 50 A = 3600 yd 2

Explanation: Given, b1 = 4.5 ft b2 = 11.5 ft h = 8 ft Area of the trapezium = (b1 + b2)h/2 A = (4.5 + 11.5)8/2 A = 16 × 4 A = 64 sq. ft

Question 5. A cake is made out of two identical trapezoids. Each trapezoid has a height of 11 inches and bases of 9 inches and 14 inches. What is the area of one of the trapezoid pieces? _______ in. 2

Answer: 126.5

Explanation: Given, A cake is made out of two identical trapezoids. Each trapezoid has a height of 11 inches and bases of 9 inches and 14 inches. Area of the trapezium = (b1 + b2)h/2 A = (9 + 14)11/2 A = 23 × 11/2 A = 126.5 in. 2

Go Math Grade 6 Chapter 10 Review Test Answer Key Question 6. A sticker is in the shape of a trapezoid. The height is 3 centimeters, and the bases are 2.5 centimeters and 5.5 centimeters. What is the area of the sticker? _______ cm 2

Explanation: Given, A sticker is in the shape of a trapezoid. The height is 3 centimeters, and the bases are 2.5 centimeters and 5.5 centimeters. Area of the trapezium = (b1 + b2)h/2 A = (2.5 + 5.5)3/2 A = 8 × 3/2 A = 4 × 3 A = 12 sq. cm

Question 7. Find the area of a trapezoid that has bases that are 15 inches and 20 inches and a height of 9 inches. _______ in. 2

Answer: 157.5

Explanation: b1 = 15 inches b2 = 20 inches h = 9 inches Area of the trapezium = (b1 + b2)h/2 A = (15 + 20)9/2 A = (35 × 9)/2 A = 157.5 sq. in

Explanation: b1 = 9 yd b2 = 15 yd h = 7 yd Area of the trapezium = (b1 + b2)h/2 A = (9 + 15)7/2 A = 24 × 3.5 A = 84 sq. yd

Question 2. Maggie colors a figure in the shape of a trapezoid. The trapezoid is 6 inches tall. The bases are 4.5 inches and 8 inches. What is the area of the figure that Maggie colored? _______ in. 2

Answer: 37.5

Explanation: Maggie colors a figure in the shape of a trapezoid. The trapezoid is 6 inches tall. The bases are 4.5 inches and 8 inches. b1 = 4.5 in b2 = 8 in h = 6 in Area of the trapezium = (b1 + b2)h/2 A = (4.5 in + 8 in)6/2 A = 12.5 in × 3 A = 37.5 sq. in

Question 3. Cassandra wants to solve the equation 30 = \(\frac{2}{5}\)p. What operation should she perform to isolate the variable? Type below: _______________

Answer: Divide two sides by \(\frac{2}{5}\)

Explanation: In order to make p independent We have to divide \(\frac{2}{5}\) on both sides. 30 = \(\frac{2}{5}\)p 30 ÷ \(\frac{2}{5}\) p ÷ \(\frac{2}{5}\) p = 75

Question 4. Ginger makes pies and sells them for $14 each. Write an equation that represents the situation, if y represents the money that Ginger earns and x represents the number of pies sold. Type below: _______________

Answer: y = 14x

Explanation: Ginger makes pies and sells them for $14 each. y represents the money that Ginger earns x represents the number of pies sold The equation is y = 14x

Answer: y = 2x By seeing the graph we can say that y = 2x

Question 6. Cesar made a rectangular banner that is 4 feet by 3 feet. He wants to make a triangular banner that has the same area as the other banner. The triangular banner will have a base of 4 feet. What should its height be? _______ feet

Explanation: 6 Because 4×3=12 and (4× 6)/2=12

Explanation: Given, b1 = 6 cm b2 = 3 cm h = 4 cm We know that, Area of the trapezium = (b1 + b2)h/2 A = (6 cm + 3 cm)4 cm/2 A = 9 cm × 2 cm A = 18 sq. cm Therefore the area of the trapezoid is 18 cm 2

Explanation: b1 = 10 ft b2 = 8 ft The area of the trapezoid is 45 ft 2 We know that, Area of the trapezium = (b1 + b2)h/2 45 ft 2 = (10 ft + 8 ft)h/2 90 = 18 × h h = 90/18 h = 5 ft Thus the height of the above figure is 5 ft.

Explanation: b1 = 17 mm b2 = 43 mm h = 18 mm We know that, Area of the trapezium = (b1 + b2)h/2 A = (17 + 43)18/2 A = 60 mm × 9 mm A = 540 sq. mm Thus the area of the trapezoid is 540 mm 2

Explanation: Given, b1 = 17 in b2 = 21 in h = 14 in We know that, Area of the trapezium = (b1 + b2)h/2 A = (17 in + 21 in)14/2 A = 38 in × 7 in A = 266 sq. in Therefore Area of the trapezium is 266 in. 2

Answer: 25.2 m 2

Explanation: Given, b1 = 9.2 m b2 = 2.8 m h = 4.2 m We know that, Area of the trapezium = (b1 + b2)h/2 A = (9.2 + 2.8)4.2/2 A = 12 × 2.1 A = 25.2 sq. m Therefore the area of the trapezium is 25.2 m 2

Find the height of the trapezoid.

Explanation: Given, b1 = 27.5 in b2 = 12.5 in h = ? A = 500 sq. in We know that, Area of the trapezium = (b1 + b2)h/2 500 sq. in = (27.5 in + 12.5 in)h/2 500 sq. in = 40 × h/2 500 sq. in = 20h h = 500/20 h = 25 inches Thus the height of the above figure is 25 inches.

Explanation: A = 99 sq. cm b1 = 3.2 cm b2 = 10 cm h = ? We know that, Area of the trapezium = (b1 + b2)h/2 99 sq. cm = (3.2 cm+ 10 cm)h/2 99 sq. cm = (13.2 cm)h/2 99 sq. cm = 6.6 × h h = 99 sq. cm/6.6 cm h = 15 cm

Question 8. A baseball home plate can be divided into two trapezoids with the dimensions shown in the drawing. Find the area of the home plate. _______ in. 2

Answer: 21.75

Explanation: The bases of the trapezoid area are 8.5 in and 17 in and the height is 8.5 in. We know that, Area of the trapezium = (b1 + b2)h/2 A = 1/2 (8.5 + 17)8.5 A = (25.5)(8.5)/2 A = 1/2 × 216.75 The area of the home plate is double the area of a trapezoid. So, the area of the home plate is 216.75 sq. in.

Question 9. Suppose you cut the home plate along the dotted line and rearranged the pieces to form a rectangle. What would the dimensions and the area of the rectangle be? Type below: _______________

Answer: The dimensions of the rectangle would be 25.5 in by 8.5 in. The area would be 216.75 sq. in.

Answer: 200 sq. cm

Explanation: A side of the inner square measures 10 cm, and a side of the outer square measures 30 cm. The bases of the trapezoid are 10 cm and 30 cm and the height of the trapezoid is 10 cm. We know that, Area of the trapezium = (b1 + b2)h/2 A = (10 + 30)10/2 A = 40 cm × 5 cm A = 200 sq. cm So, the area of one of the yellow trapezoid tiles is 200 sq. cm

Question 11. Verify the Reasoning of Others A trapezoid has a height of 12 cm and bases with lengths of 14 cm and 10 cm. Tina says the area of the trapezoid is 288 cm 2 . Find her error, and correct the error. Type below: _______________

Answer: A trapezoid has a height of 12 cm and bases with lengths of 14 cm and 10 cm. Tina says the area of the trapezoid is 288 cm 2 We know that, Area of the trapezium = (b1 + b2)h/2 A = (14 + 10)12/2 A = 24 cm × 6 cm A = 144 sq. cm The error of Tina is she didn’t divide by 2.

Answer: \(\frac{1}{2}\) × (1.5 + 3.5) × 4

Explanation: b1 = 3.5 ft b2 = 1.5 ft h = 4 ft We know that, Area of the trapezium = (b1 + b2)h/2 A = (3.5 ft + 1.5 ft)4ft/2 A = \(\frac{1}{2}\) × (1.5 + 3.5) × 4 Thus the correct answer is option B.

Answer: 252 cm 2

Explanation: Given that, long base b1 = 17 cm short base b2 = 11 cm h = 18 cm We know that, The Area of the trapezium = (b1 + b2)h/2 A = (17 cm + 11 cm)18 cm/2 A = 28 cm × 9 cm A = 252 cm 2 Thus the area of the trapezium for the above figure is 252 cm 2

Answer: 30 ft 2

Explanation: Given, b1 = 6.5 ft b2 = 5.5 ft h = 5 ft We know that, The Area of the trapezium = (b1 + b2)h/2 A = (6.5 + 5.5)5/2 A = 12 ft × 2.5 ft A = 30 sq. ft Therefore the area of the trapezium is 30 ft 2

Answer: 0.08 cm 2

Explanation: Given, b1 = 0.6 cm b2 = 0.2 cm h = 0.2 cm We know that, The Area of the trapezium = (b1 + b2)h/2 A = (0.6 cm + 0.2 cm)0.2 cm/2 A = 0.8 cm × 0.1 cm A = 0.08 sq. cm Thus the area of the trapezium is 0.08 sq. cm

Answer: 37.5 in. 2

Explanation: Given, b1 = 5 in b2 = 2 1/2 h = 10 in We know that, The Area of the trapezium = (b1 + b2)h/2 A = (5 in + 2 1/2 in)10/2 A = 7 1/2 × 5 A = 37.5 sq. in Thus the area of the trapezium is 37.5 in. 2

Question 5. Sonia makes a wooden frame around a square picture. The frame is made of 4 congruent trapezoids. The shorter base is 9 in., the longer base is 12 in., and the height is 1.5 in. What is the area of the picture frame? _______ in. 2

Explanation: Given, Sonia makes a wooden frame around a square picture. The frame is made of 4 congruent trapezoids. The shorter base is 9 in., the longer base is 12 in., and the height is 1.5 in. We know that, The Area of the trapezium = (b1 + b2)h/2 A = (9 in + 12 in)1.5/2 A = 21 in × 1.5 in/2 A = 63 sq. in Thus the area of the trapezium is 63 in. 2

Question 6. Bryan cuts a piece of cardboard in the shape of a trapezoid. The area of the cutout is 43.5 square centimeters. If the bases are 6 centimeters and 8.5 centimeters long, what is the height of the trapezoid? _______ cm

Answer: 6 cm

Explanation: Given, Bryan cuts a piece of cardboard in the shape of a trapezoid. The area of the cutout is 43.5 square centimeters. If the bases are 6 centimeters and 8.5 centimeters long. We know that, The Area of the trapezium = (b1 + b2)h/2 43.5 sq. cm = (6 + 8.5)h/2 43.5 × 2 = 14.5 × h h = 6 cm Therefore the height of the trapezoid is 6 cm.

Question 7. Use the formula for the area of a trapezoid to find the height of a trapezoid with bases 8 inches and 6 inches and an area of 112 square inches. _______ in.

Answer: 16 in.

Explanation: Given, b1 = 8 inches b2 = 6 in A = 112 sq. in We know that, The Area of the trapezium = (b1 + b2)h/2 112 sq. in = (8 in + 6 in)h/2 112 sq. in = 7 × h h = 112/7 h = 16 in. Thus the height of the trapezoid is 16 in.

Question 1. Dominic is building a bench with a seat in the shape of a trapezoid. One base is 5 feet. The other base is 4 feet. The perpendicular distance between the bases is 2.5 feet. What is the area of the seat? _______ ft 2

Answer: 11.25 sq. ft

Explanation: Given, Dominic is building a bench with a seat in the shape of a trapezoid. One base is 5 feet. The other base is 4 feet. The perpendicular distance between the bases is 2.5 feet. We know that, The Area of the trapezium = (b1 + b2)h/2 A = (5 ft + 4 ft)2.5/2 A = 4.5 ft × 2.5 ft A = 11.25 sq. ft Thus the area of the seat is 11.25 sq. ft

Question 2. Molly is making a sign in the shape of a trapezoid. One base is 18 inches and the other is 30 inches. How high must she make the sign so its area is 504 square inches? _______ in.

Answer: 21 in.

Explanation: Given, Molly is making a sign in the shape of a trapezoid. One base is 18 inches and the other is 30 inches. A = 504 sq. in We know that, The Area of the trapezium = (b1 + b2)h/2 504 sq. in = (18 + 30)h/2 504 sq. in = 24 × h h = 504 sq. in÷ 24 in h = 21 inches Thus the height of the trapezoid is 21 inches.

Question 3. Write these numbers in order from least to greatest. 3 \(\frac{3}{10}\)     3.1       3 \(\frac{1}{4}\) Type below: _______________

Explanation: First, convert the fraction into the decimal. 3 \(\frac{3}{10}\) = 3.3 3 \(\frac{1}{4}\) = 3.25 Now write the numbers from least to greatest. 3.1 3.25 3.3

Question 4. Write these lengths in order from least to greatest. 2 yards       5.5 feet        70 inches Type below: _______________

Answer: 5.5 feet , 70 inches, 2 yards

Explanation: First, convert from inches to feet. 1 feet = 12 inches 70 inches = 5.8 ft 1 yard = 3 feet 2 yards = 2 × 3 ft 2 yards = 6 feet Now write the numbers from least to greatest. 5.5 ft 5.8 ft 6 ft

Question 5. To find the cost for a group to enter the museum, the ticket seller uses the expression 8a + 3c in which a represents the number of adults and c represents the number of children in the group. How much should she charge a group of 3 adults and 5 children? $ _______

Explanation: The expression is 8a + 3c where, a represents the number of adults. c represents the number of children in the group. a = 3 c = 5 8a + 3c = 8(3) + 3(5) = 24 + 15 = $39

Question 6. Brian frosted a cake top shaped like a parallelogram with a base of 13 inches and a height of 9 inches. Nancy frosted a triangular cake top with a base of 15 inches and a height of 12 inches. Which cake’s top had the greater area? How much greater was it? Type below: _______________

Explanation: Parallelogram Formula = Base × Height A=bh A=13 × 9=117 in Triangle Formula= A=1/2bh A=1/2 × 15 × 12 = 90 in Brian’s cake top has a greater area, and by 27 inches.

Question 1. A _____ is a quadrilateral that always has two pairs of parallel sides. Type below: _______________

Answer: A parallelogram is a quadrilateral that always has two pairs of parallel sides.

Question 2. The measure of the number of unit squares needed to cover a surface without any gaps or overlaps is called the _____. Type below: _______________

Answer: The measure of the number of unit squares needed to cover a surface without any gaps or overlaps is called the Area .

Question 3. Figures with the same size and shape are _____. Type below: _______________

Answer: Figures with the same size and shape are Congruent .

Concepts and Skills

Answer: 19.38

Explanation: b = 5.7 cm h = 3.4 cm Area of parallelogram = bh A = 5.7 cm × 3.4 cm A = 19.38 cm 2 Thus the area of the parallelogram is 19.38 cm 2

Answer: 42 \(\frac{1}{4}\) in. 2

Explanation: b = 6 \(\frac{1}{2}\) h = 6 \(\frac{1}{2}\) Area of parallelogram = bh A = 6 \(\frac{1}{2}\) × 6 \(\frac{1}{2}\) A = 42 \(\frac{1}{4}\) in. 2 Thus the area of the parallelogram is 42 \(\frac{1}{4}\) in. 2

Answer: 57.4

Explanation: b = 14 mm h = 8.2 mm A = bh/2 A = (14 mm × 8.2 mm)/2 A = 57.4 mm 2

Answer: 139.5

Explanation: b1 = 13 cm b2= 18 cm h = 9 cm Area of the trapezium = (b1 + b2)h/2 A = (13 + 18)9/2 A = 31 × 4.5 A = 139.5 sq. cm

Question 8. A parallelogram has an area of 276 square meters and a base measuring 12 meters. What is the height of the parallelogram? _______ m

Explanation: A parallelogram has an area of 276 square meters and a base measuring 12 meters. A = bh 276 = 12 × h h = 276/12 h = 23 m

Question 9. The base of a triangle measures 8 inches and the area is 136 square inches. What is the height of the triangle? _______ in.

Explanation: The base of a triangle measures 8 inches and the area is 136 square inches. A = 136 sq. in b = 8 in. h = ? A = bh/2 136 = 8h/2 136 = 4h h = 136/4 h = 34 in

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Question 10. The height of a parallelogram is 3 times the base. The base measures 4.5 cm. What is the area of the parallelogram? _______ cm 2

Answer: 60.75

Explanation: The height of a parallelogram is 3 times the base. The base measures 4.5 cm. A = bh h = 3 × 4.5 h = 13.5 cm b = 4.5 cm A = 13.5 cm × 4.5 cm A = 60.75 cm 2

6th Grade Math Area of Parallelogram Question 11. A triangular window pane has a base of 30 inches and a height of 24 inches. What is the area of the window pane? _______ in. 2

Answer: 360

Explanation: A triangular window pane has a base of 30 inches and a height of 24 inches. b = 30 in h = 24 in A = bh/2 A = (30 × 24)/2 A = 30 × 12 A = 360 in. 2

Question 12. The courtyard behind Jennie’s house is shaped like a trapezoid. The bases measure 8 meters and 11 meters. The height of the trapezoid is 12 meters. What is the area of the courtyard? _______ m 2

Answer: 114

Explanation: Given, The courtyard behind Jennie’s house is shaped like a trapezoid. The bases measure 8 meters and 11 meters. The height of the trapezoid is 12 meters. Area of the trapezium = (b1 + b2)h/2 A = (8 + 11)12/2 A = 19 × 6 A = 114 m 2

Question 13. Rugs sell for $8 per square foot. Beth bought a 9-foot-long rectangular rug for $432. How wide was the rug? _______ feet

Answer: 6 feet

Explanation: If you know the rugs sell for 8$ per square foot and the total spend was $432. You divide 432 by 8 to find the total number of square feet of the rug. To find the total square foot you find the area. So the area of a rectangle is L × W. So 54 = 9 × width. So just divide 54 by 9 and you get the width of the rug. The width is 6 feet. Now you check. A nine by 6 rugs square foot is 54. and then times by 8 and you get 432 total.

Question 14. A square painting has a side length of 18 inches. What is the area of the painting? _______ in. 2

Answer: 324

Explanation: A square painting has a side length of 18 inches. A = s × s A = 18 × 18 A = 324 in. 2

Find the area of the regular polygon.

Explanation: b = 5 cm h = 6 cm Number of congruent figures inside the figure: 8 Area of each triangle = bh/2 A = (5 cm)(6 cm)/2 A = 15 sq. cm Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures. Area of regular octagon = 8 × 15 sq. cm A = 120 sq. cm Therefore the area of the regular octagon for the above figure = 120 sq. cm

Explanation: Given, b = 6 m h = 4 m Number of congruent figures inside the figure: 5 Area of each triangle = bh/2 A = (6 m)(4 m)/2 A = 12 sq. m Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures. Area of regular pentagon = 5 × 12 sq. m A = 60 sq. m Therefore the area of the above figure is 60 sq. m.

Answer: 480

Explanation: Given, b = 8 mm h = 12 mm Number of congruent figures inside the figure: 10 Area of each triangle = bh/2 A = (12 mm)(8 mm)/2 A = 48 sq. mm Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures. Area of regular polygon = 10 × 48 sq. mm A = 480 sq. mm Therefore, the area of the regular polygon is 480 sq. mm

Explanation: Given, b = 8 cm h = 7 cm Number of congruent figures inside the figure: 6 Area of each triangle = bh/2 A = (8 cm)(7 cm)/2 A = 28 sq. cm Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures. Area of regular hexagon = 6 × 28 sq. cm A = 168 sq. cm Thus the area of the above figure is 168 sq. cm

Answer: 6020

Explanation: Given, b = 28 in h = 43 in Number of congruent figures inside the figure: 10 Area of each triangle = bh/2 A = (28 in)(43 in)/2 A = 602 sq. in Now to find the area of the regular polygon we have to multiply the area of each triangle and a number of congruent figures. Area of regular polygon = 10 × Area of each triangle A = 10 × 602 sq. in A = 6020 sq. in Therefore the area of the regular polygon is 6020 sq. in

Area of Parallelogram Answers Key Question 6. Explain A regular pentagon is divided into congruent triangles by drawing a line segment from each vertex to the center. Each triangle has an area of 24 cm 2 . Explain how to find the area of the pentagon Type below: _______________

Explanation: Given, Each triangle has an area of 24 cm 2 . Pentagon has 5 sides. The number of congruent figures is 5. Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures. Area of regular pentagon = 5 × 24 sq. cm A = 120 sq. cm Therefore the area of the pentagon is 120 sq. cm

Answer: 76.8 sq. in

Explanation: b = 4 in h = 4.8 in Number of configured figures of the regular polygon: 8 Area of the triangle = bh/2 A = (4)(4.8)/2 A = 9.6 sq. in. Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures. Area of regular polygon = 8 × area of the triangle A = 8 × 9.6 sq. in. A = 76.8 sq. in Thus the area of the regular polygon is 76.8 sq. in.

Regular polygons are common in nature

One of the best known examples of regular polygons in nature is the small hexagonal cells in honeycombs constructed by honeybees. The cells are where bee larvae grow. Honeybees store honey and pollen in the hexagonal cells. Scientists can measure the health of a bee population by the size of the cells.

Answer: 0.52 cm

Explanation: Since the combined width of 10 cells is 5.2 cm, the width of each cell is 5.2 ÷ 10 = 0.52 cm.

Answer: 0.234 sq. cm

Explanation: The length of the h, the height of the triangle, is half of the width of each cell. Since the width of each cell is 0.52 cm h = 0.52 ÷ 2 = 0.26 cm Area of the triangle = bh/2 A = (0.3)(0.26)/2 A = 0.078/2 A = 0.039 The area of the hexagon is: 6 × 0.039 = 0.234 sq. cm.

Question 10. A rectangular honeycomb measures 35.1 cm by 32.4 cm. Approximately how many cells does it contain? _______ cells

Answer: 4860 cells

Explanation: A = lw A = 35.1 cm × 32.4 cm A = 1137.24 The area of the rectangular honeycomb is 1137.24 sq. cm The honeycomb contains 1137.24 ÷ 0.234 = 4860 cells

Explanation: Given, b = 8 mm h = 7 mm Number of congruent figures inside the figure: 6 Area of each triangle = bh/2 A = (8)(7)/2 A = 28 sq. mm Now to find the area of regular polygon we have to multiply the area of each triangle and number of congruent figures. Area of regular polygon = 6 × 28 sq. mm A = 168 sq. mm Therefore the area of the regular polygon for the above figure is 168 sq. mm

Explanation: Given, b = 9 yd h = 6.2 yd Number of congruent figures inside the figure: 5 Area of each triangle = bh/2 A = (9 yd) (6.2 yd)/2 A = 9 yd × 3.1 yd A = 27.9 sq. yd Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures. Area of regular polygon = 5 × 27.9 sq. yd A = 139.5 sq. yd Thus the area of the regular polygon for the above figure is 139.5 sq. yd.

Answer: 52.8

Explanation: Given, b = 3.3 in h = 4 in Number of congruent figures inside the figure: 8 Area of each triangle = bh/2 A = (3.3 in)(4 in)/2 A = 6.6 sq. in Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures. Area of regular polygon = 8 × 6.6 sq. in A = 52.8 sq. in The area of the regular polygon is 52.8 sq. in

Question 4. Stu is making a stained glass window in the shape of a regular pentagon. The pentagon can be divided into congruent triangles, each with a base of 8.7 inches and a height of 6 inches. What is the area of the window? _______ in. 2

Answer: 130.5

Explanation: Stu is making a stained glass window in the shape of a regular pentagon. The pentagon can be divided into congruent triangles, each with a base of 8.7 inches and a height of 6 inches. Number of congruent figures inside the figure: 5 Area of each triangle = bh/2 A = (8.7 in)(6 in)/2 A = 8.7 in × 3 in A = 26.1 sq. in. Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures. Area of regular polygon = 5 × 26.1 sq. in A = 130.5 sq. in Thus the area of the window is 130.5 sq. in

Area of Parallelogram 6th Grade Question 5. A dinner platter is in the shape of a regular decagon. The platter has an area of 161 square inches and a side length of 4.6 inches. What is the area of each triangle? What is the height of each triangle?

Answer: 7 in

Explanation: A dinner platter is in the shape of a regular decagon. The platter has an area of 161 square inches and a side length of 4.6 inches. Area of each triangle = bh/2 161 sq. in = 4.6 × h/2 161 sq. in = 2.3 × h h = 161 sq. in/2.3 h = 70 sq. in Therefore the height of each triangle is 70 sq. in

Question 6. A square has sides that measure 6 inches. Explain how to use the method in this lesson to find the area of the square. Type below: _______________

Answer: 36 sq. in

Explanation: A square has sides that measure 6 inches. s = 6 in We know that, Area of the square = s × s A = 6 in × 6 in A = 36 sq. in Thus the area of the square is 36 sq. in

Answer: 30 \(\frac{3}{5}\) m 2

Explanation: Given, b = 3 \(\frac{2}{5}\) m h = 3 m Area of each triangle = bh/2 A = 3 \(\frac{2}{5}\) m × 3/2 m A = 5.1 sq. m Now to find the area of the regular polygon we have to multiply the area of each triangle and number of congruent figures. Area of the regular hexagon = 6 × 5.1 = 30.6 = 30 \(\frac{6}{10}\) m 2 = 30 \(\frac{3}{5}\) m 2 Therefore the area of the regular hexagon is 30 \(\frac{3}{5}\) m 2

Question 2. A regular 7-sided figure is divided into 7 congruent triangles, each with a base of 12 inches and a height of 12.5 inches. What is the area of the 7-sided figure? ________ in. 2

Answer: 525 sq. in

Explanation: A regular 7-sided figure is divided into 7 congruent triangles, each with a base of 12 inches and a height of 12.5 inches. Area of each triangle = bh/2 A = (12 in)(12.5 in)/2 A = 12.5 in × 6 in A = 75 sq. inches Thus the area of each triangle = 75 sq. in Now to find the area of the regular polygon we have to multiply the area of each triangle and a number of congruent figures. Area of regular polygon = 7 × 75 sq. in A = 525 sq. in Thus the area of the 7-sided figure is 525 sq. in

Question 3. Which inequalities have b = 4 as one of its solutions? 2 + b ≥ 2      3b ≤ 14 8 − b ≤ 15     b − 3 ≥ 5 Type below: _______________

Answer: b − 3 ≥ 5

Explanation: Substitute b = 4 in the inequality i. 2 + b ≥ 2 2 + 4 ≥ 2 6 ≥ 2 ii. 3b ≤ 14 3(4) ≤ 14 12 ≤ 14 iii. 8 − b ≤ 15 8 – 4 ≤ 15 4 ≤ 15 iv. b − 3 ≥ 5 4 – 3 ≥ 5 1 ≥ 5 1 is not greater than or equal to 5.

Question 4. Each song that Tara downloads costs $1.25. She graphs the relationship that gives the cost y in dollars of downloading x songs. Name one ordered pair that is a point on the graph of the relationship. Type below: _______________

Answer: (2, 2.5)

Explanation: The equation is y = 2x y = 1.25 y = 2 (1.25) y = 2.5 The coordinates of (x,y) is (2, 2.5)

Explanation: b = 6 ft h = 10 ft We know that, Area of each triangle = bh/2 A = (6 ft)(10 ft)/2 A = 60 sq. ft/2 A = 30 sq. ft Therefore the area of triangle ABC is 30 sq. ft

Question 6. Marcia cut a trapezoid out of a large piece of felt. The trapezoid has a height of 9 cm and bases of 6 cm and 11 cm. What is the area of Marcia’s felt trapezoid? ________ cm 2

Answer: 76.5 cm 2

Explanation: Marcia cut a trapezoid out of a large piece of felt. The trapezoid has a height of 9 cm and bases of 6 cm and 11 cm. Area of the trapezium = (b1 + b2)h/2 A = (6 + 11)9/2 A = 17 cm × 4.5 cm A = 76.5 sq. cm Therefore the area of Marcia’s felt trapezoid is 76.5 cm 2

Answer: 126 sq. ft

Explanation: Figure 1: l = 10 ft w = 5 ft A = lw A = 10 ft × 5 ft A = 50 sq. ft Figure 2: l = 10 ft w = 5 ft A = lw A = 10 ft × 5 ft A = 50 sq. ft Figure 3: b = 5 ft + 5 ft + 3 ft b = 13 ft h = 4 ft Area of triangle = bh/2 A = 13 ft × 4 ft/2 A = 13 ft × 2 ft A = 26 sq. ft Add the areas of all the figures = 50 sq. ft + 50 sq. ft + 26 sq. ft Thus the Area of the composite figure is 126 sq. ft.

Answer: 128.2 sq. mm

Explanation: Figure 1: b1 = 11 mm b2 = 11 mm h = 8.2 mm Area of the trapezoid = (b1 + b2)h/2 A = (11 mm + 11 mm)8.2 mm/2 A = 22 mm × 4.1 mm A = 90.2 sq. mm Figure 2: b1 = 11mm b2 = 8mm h = 4mm Area of the trapezoid = (b1 + b2)h/2 A = (11mm + 8mm)4mm/2 A = 19mm × 2mm A = 38 sq. mm Add the areas of both figures = 90.2 sq. mm + 38 sq. mm Thus the area of the figure is 128.2 sq. mm

Answer: 144 sq. m

Explanation: Figure 1: l = 12 m w = 7 m Area of Rectangle = lw A = 12m × 7m A = 84 sq. m Figure 2: Area of right triangle = ab/2 a = 5m b = 12m A = (5m)(12m)/2 A = 30 sq. m Figure 3: Area of right triangle = ab/2 a = 5m b = 12m A = (5m)(12m)/2 A = 30 sq. m Area of all figures = 84 sq. m + 30 sq. m + 30 sq. m = 144 sq. m. Therefore the area of the figure is 144 sq. m

Answer: 184 sq. in

Explanation: Figure 1: b = 8 in h = 6 in Area of right triangle = ab/2 A = 8 in × 6 in/2 A = 24 sq. in Figure 2: Area of Rectangle = lw A = 16 in × 6 in A = 96 sq. in Figure 3: Area of right triangle = ab/2 b = 8 in h = 8 in A = 8 in × 8 in/2 A = 32 sq. in Figure 4: Area of right triangle = ab/2 b = 8 in h = 8 in A = 8 in × 8 in/2 A = 32 sq. in Area of all figures = 24 sq. in + 96 sq. in + 32 sq. in + 32 sq. in = 184 sq. in Thus the area of the figure = 184 sq. in.

Answer: 96.05 sq. m

Explanation: Figure 1: Area of Rectangle = lw A = 12.75 m × 8.8 m A = 112.2 sq. m Figure 2: Area of Rectangle = lw l = 4.25 m w = 3.3 m A = 4.25 m × 3.3 m A = 16.15 sq. m Area of all the figures = 112.2 sq. m + 16.15 sq. m = 90.05 sq. m Therefore the area of the figure = 90.05 sq. m

Answer: I can find the area of the yellow shape by subtracting the areas of the green and red shapes from the area of the entire banner.

Question 6. b. What is the area of the entire banner? Explain how you found it. The area of the banner is ________ in. 2

Answer: 1440 sq. in

Explanation: The banner is a rectangle with a width of 48 inches and a length of 30 inches. A = lw A = 48 in × 30 in A = 1440 sq. in Therefore, the area of the banner is 1440 sq. in.

Question 6. c. What is the area of the red shape? What is the area of each green shape? The area of the red shape is ________ in. 2 The area of each green shape is ________ in. 2

Answer: The area of the red shape is 360 in. 2 The area of each green shape is 360 in. 2

Explanation: The red shape is a triangle with a base of 30 inches and a height of 24 inches. A = bh/2 A = (30)(24)/2 A = 360 sq. in. The area of the red triangle is 360 sq. in. Each green shape is a triangle with a base of 15 inches and a height of 48 inches. A = bh/2 A = 1/2 × 15 × 48 A = 720/2 A = 360 sq. in Therefore the area of each green triangle is 360 sq. in.

Question 6. d. What equation can you write to find A, the area of the yellow shape? Type below: _______________

Answer: A = 1440 – (360 + 360 + 360)

Question 6. e. What is the area of the yellow shape? The area of the yellow shape is ________ in. 2

Answer: 360 sq. in

Explanation: A = bh/2 A = 1/2 × 15 × 48 A = 720/2 A = 360 sq. in Therefore the area of the yellow shape is 360 sq. in

Question 7. There are 6 rectangular flower gardens each measuring 18 feet by 15 feet in a rectangular city park measuring 80 feet by 150 feet. How many square feet of the park are not used for flower gardens? ________ ft 2

Answer: 10380 ft 2

Explanation: 18 × 15=270 270 × 6 flower gardens = 1620 80 × 150=12000 this is the total area of the park 12000 – 1620=10380 ft 2

Answer: 8 × 22 + 130 + \(\frac{1}{2}\) × 10 × 9

Explanation: Figure 1: l = 13 ft w = 10 ft Area of the rectangle = lw A = 13 ft × 10 ft = 130 Figure 2: b = 9 ft h = 10 ft Area of the triangle = bh/2 A = (9)(10)/2 A = 45 sq. ft Figure 3: Area of the rectangle = lw l = 22 ft w = 8 ft The area of the composite figure is 8 × 22 + 130 + \(\frac{1}{2}\) × 10 × 9 Thus the correct answer is option A.

Find the area of the figure

Answer: 37 cm 2

Explanation: Area of square = s × s A = 3 × 3 = 9 sq. cm Area of Triangle = bh/2 A = 2 × 8/2 = 8 sq. cm Area of the trapezoid = (b1 + b2)h/2 A = (5 + 3)5/2 A = 4 × 5 = 20 sq. in Area of composite figure = 9 sq. cm + 8 sq. cm + 20 sq. in A = 37 cm 2

Explanation: Figure 1: b = 9 ft h = 6 ft Area of Triangle = bh/2 A = (9ft)(6ft)/2 A = 27 sq. ft Figure 2: l = 12 ft w = 9 ft Area of the rectangle = lw A = (12ft)(9ft)/2 A = 12 ft × 9 ft A = 108 sq. ft Figure 3: Area of Triangle = bh/2 b = 9 ft h = 10 ft A = (10ft)(9ft)/2 A = 45 sq. ft Area of the composite figure = 27 sq. ft + 108 sq. ft + 45 sq. ft = 180 sq. ft

Answer: 128 yd 2

Explanation: Figure 1: b1 = 7 yd b2 = 14 yd h = 8 yd Area of the trapezoid = (b1 + b2)h/2 A = (7yd + 14yd)8yd/2 A = 21 yd × 4 yd A = 84 sq. yd Figure 2: b = 11 yd h = 4 yd Area of the parallelogram = bh A = 11yd × 4yd = 44 sq. yd Area of the composite figure = 84 sq. yd + 44 sq. yd = 128 sq. yd

Answer: 155 sq. in

Explanation: The poster is a parallelogram, and it’s area is: A = bh A = 20 x 10 A = 200 sq. in The area of the triangle that Janelle cut out of the poster board is: A = 1/2bh A = 1/2 x 10 x 9 A = 90/2 A = 45 sq. in The area of the poster board that she has left is 200 sq. in – 45 sq. in = 155 sq. in

Answer: 204 yd 2

Explanation: The area of the shaded region can be found by finding the total area and subtracting the area of the lap pool. Total area = Area of the trapezium = 1/2 × (Sum of parallel sides) × distance between them Sum of parallel sides = 25 yd + (3 + 12) = 40 yd Distance between them = 12 yd Total area = 1/2 × 40 × 12 = 240 yd² Find the area of the lap pool. Area = length × width = 12 × 3 = 36 yd² Find the area of the shaded region Area to be covered with grass = 240 – 36 = 204 yd²

Question 6. Describe one or more situations in which you need to subtract to find the area of a composite figure. Type below: _______________

Answer: 227 m 2

Explanation: Figure 1: b = 7 m h = 7 m Area of the triangle = bh/2 A = (7m)(7m)/2 A = 24.5 sq. m Figure 2: b1 = 7m b2 = 10m h = 9m Area of the trapezoid = (b1 + b2)h/2 A = (7m + 10m)9m/2 A = 17m × 4.5 m A = 76.5 sq. m Area of the rectangle = lw A = 18m × 7m A = 126 sq. m Area of the figures = 24.5 sq. m + 76.5 sq. m + 126 sq. m = 227 sq. m Thus the area of the figure is 227 sq. m

Answer: 251.5 in. 2

Explanation: Figure 1: l = 21 in w = 15 in Area of triangle = bh/2 A = 21 in × 15 in/2 A = 157.5 sq. in Figure 2: b1 = 12 in b2 = 15 in h = 11 in Area of the trapezoid = (b1 + b2)h/2 A = (12 in + 15 in)11 in/2 A = 27 in × 5.5 in A = 148.5 sq. in Figure 3: b = 13 in h = 14.4 in Area of trinagle = bh/2 A = 13 × 14.4in/2 A = 13in × 7.2 in A = 94 sq. in The area of the shaded region is 94 sq. in + 157.5 sq. in = 251.5 in. 2

Question 3. In Maritza’s family, everyone’s height is greater than 60 inches. Write an inequality that represents the height h, in inches, of any member of Maritza’s family. Type below: _______________

Answer: h > 60

Explanation: Given, Maritza’s family, everyone’s height is greater than 60 inches. The inequality is h > 60

Question 4. The linear equation y = 2x represents the cost y for x pounds of apples. Which ordered pair lies on the graph of the equation? Type below: _______________

Answer: (2, 4)

Explanation: y = 2x put x = 2 y = 2(2) y = 4 The ordered pair is (2,4)

Question 5. Two congruent triangles fit together to form a parallelogram with a base of 14 inches and a height of 10 inches. What is the area of each triangle? ________ in. 2

Answer: 70 in. 2

Explanation: b = 14 in h = 10 in Area of trinagle = bh/2 A = (14 in)(10 in)/2 A = 140/2 A = 70 sq. in Thus the area of the triangle is 70 sq. in.

Question 6. A regular hexagon has sides measuring 7 inches. If the hexagon is divided into 6 congruent triangles, each has a height of about 6 inches. What is the approximate area of the hexagon? ________ in. 2

Answer: 126 in. 2

Explanation: b = 7 in h = 6 in Number of congruent figures: 6 Area of the triangle = bh/2 A = (7in)(6in)/2 A = 21 sq. in Area of regular hexagon = 6 × area of each triangle A = 6 × 21 sq. in A = 126 sq. in Thus the approximate area of the hexagon is 126 sq. in.

Question 1. The dimensions of a 2-cm by 6-cm rectangle are multiplied by 5. How is the area of the rectangle affected? Type below: _______________

Explanation: The dimensions of a 2-cm by 6-cm rectangle are multiplied by 5. Original Area: Area of rectangle = lw A = 2cm × 6cm = 12 sq. cm New dimensions: l = 6 × 5 = 30 cm w = 2 × 5 = 10 cm The new area is: A = 10 cm × 30 cm = 300 sq. cm New Area/ Original Area = 300/12 = 25 So, the new area is 25 times the original area.

Question 2. What if the dimensions of the original rectangle in Exercise 1 had been multiplied by \(\frac{1}{2}\)? How would the area have been affected? Type below: _______________

Answer: The new dimensions are: l = 1/2 × 6 =3cm w = 1/2 × 2 = 1cm The original area is: A = 2 × 6 = 12 sq. cm The new area is: A = 1 × 3 = 3 sq. cm New Area/Original Area = 3/12 = 1/4 So, the new area is 1/4 times the original area.

Question 3. Evan bought two square rugs. The larger one measured 12 ft square. The smaller one had an area equal to \(\frac{1}{4}\) the area of the larger one. What fraction of the side lengths of the larger rug were the side lengths of the smaller one? Type below: _______________

Answer: Since the area of the smaller rug is \(\frac{1}{4}\) times the area of the larger rug, the side lengths of the smaller rug are \(\frac{1}{2}\) of the side lengths of the larger one.

Answer: 45,000

Explanation: Area of triangle= (1/2) (base x height) 1 yard = 3 foot Base of the actual triangle= 100 yards= 300ft Height of the actual triangle= 50 yards= 150ft. Area of the actual triangle= (1/2) (300 x 150) = 45000 square ft The base of the triangle on the map = 2ft Height of the triangle on the map= 1ft Area of the triangle on the map= (1/2) (2 x 1) = 1 square ft. The actual area is 45000 time the area of the map

Question 5. A square game board is divided into smaller squares, each with sides one-ninth the length of the sides of the board. Into how many squares is the game board divided? ________ small squares

Answer: 81 small squares

Explanation: Each side of the game board is divided into 9 lengths. The game board is divided into 9 × 9 = 81 small squares. Thus, the board is divided into 81 small squares.

Question 6. Flynn County is a rectangle measuring 9 mi by 12 mi. Gibson County is a rectangle with an area 6 times the area of Flynn County and a width of 16 mi. What is the length of Gibson County? ________ mi

Answer: 40.5 mi.

Explanation: Flynn County is a rectangle measuring 9 mi by 12 mi. Gibson County is a rectangle with an area 6 times the area of Flynn County and a width of 16 mi. The area of Flynn Country is A = 9 × 12 = 108 sq. mi The area of Gibson Country is A = 6 × 108 = 648 sq. mi A = lw 648 = 16 × l l = 648/16 l = 40.5 mi Therefore the length of Gibson Country is 40.5 miles.

Question 7. Use Diagrams Carmen left her house and drove 10 mi north, 15 mi east, 13 mi south, 11 mi west, and 3 mi north. How far was she from home? ________ miles

Answer: 15 mi – 11 mi = 4 miles Thus Carmen is 4 miles from home.

Question 8. Bernie drove from his house to his cousin’s house in 6 hours at an average rate of 52 mi per hr. He drove home at an average rate of 60 mi per hr. How long did it take him to drive home? ________ hours

Answer: 5.2 hours

Explanation: Given, Bernie drove from his house to his cousin’s house in 6 hours at an average rate of 52 mi per hr. He drove home at an average rate of 60 mi per hr. The distance from Bernie’s house to his cousin’s house is 52 mi/hr × 6hr = 52 × 6mi = 312 miles On the way back, he drove for 312mi ÷ 60mi/hr = 5.2 hours Therefore it takes 5.2 hours for Bernie to drive home.

Question 9. Sophia wants to enlarge a 5-inch by 7-inch rectangular photo by multiplying the dimensions by 3. Find the area of the original photo and the enlarged photo. Then explain how the area of the original photo is affected. Type below: _______________

Answer: Original Area: l = 5 in w = 7 in Area of rectangle = lw A = 5 in × 7 in A = 35 sq. in New dimensions: l = 5 in × 3 = 15 in w = 7 in × 3 = 21 in Area of rectangle = lw A = 15 in × 21 in = 315 sq. in New Area/Original Area = 315 sq. in/35 sq. in = 9 Thus the new area is 9 times the original photo.

Read each problem and solve.

Question 1. The dimensions of a 5-in. by 3-in. rectangle are multiplied by 6. How is the area affected? Type below: _______________

Explanation: Original area: A = 5 × 3 = 15 sq. in new dimensions: l = 6 × 5 = 30 in w = 6 × 3 = 18 in New Area = l × w A = 30 in × 18 in A = 540 sq. in Thus new area = 540 sq. in new area/original area = 540/15 = 36 Thus the area was multiplied by 36.

Question 2. The dimensions of a 7-cm by 2-cm rectangle are multiplied by 3. How is the area affected? Type below: _______________

Explanation: Original area: A = 7 × 2 = 14 sq. cm new dimensions: l = 3 × 7 = 21 cm w = 3 × 2 cm = 6 cm new area: A = 21 cm × 6 cm = 126 sq. cm new area/original area = 126 sq. cm/14 sq. cm The area was multiplied by 9. Thus the answer is 9.

Question 3. The dimensions of a 3-ft by 6-ft rectangle are multiplied by \(\frac{1}{3}\). How is the area affected? Type below: _______________

Answer: 1/9

Explanation: Original area: A = 3 ft × 6 ft = 18 sq. ft new dimensions: l = 3 ft × \(\frac{1}{3}\) = 1 ft w = 6 ft × \(\frac{1}{3}\) = 2 ft New area: A = 1 ft × 2 ft = 2 sq. ft new area/original area = 2/18 = 1/9 The area was multiplied by 1/9.

Question 4. The dimensions of a triangle with base 10 in. and height 4.8 in. are multiplied by 4. How is the area affected? Type below: _______________

Explanation: original area: A = 10 in × 4.8 in = 48 sq. in new dimensions: l = 10 in × 4 = 40 in w = 4.8 in × 4 = 19.2 in new area = l × w A = 40 in × 19.2 in A = 768 sq. in new area/original area = 768/48 Thus the area was multiplied by 16.

Question 5. The dimensions of a 1-yd by 9-yd rectangle are multiplied by 5. How is the area affected? Type below: _______________

Explanation: original area: A = 1 yd × 9 yd = 9 sq. yd new dimensions: l = 1 yd × 5 = 5 yd w = 9 yd × 5 = 45 yd new area = 5 yd × 45 yd = 225 sq. yd new area/original area = 225 sq. yd/9 sq. yd Thus the area was multiplied by 25.

Question 6. The dimensions of a 4-in. square are multiplied by 3. How is the area affected? Type below: _______________

Explanation: original area = 4 in × 4 in = 16 sq. in new dimensions: s = 4 in × 3 = 12 in new area = s × s = 12 in × 12 in = 144 sq. in new area/original area = 144 sq. in/16 sq. in = 9 Thus the area was multiplied by 9.

Question 7. The dimensions of a triangle are multiplied by \(\frac{1}{4}\). The area of the smaller triangle can be found by multiplying the area of the original triangle by what number? Type below: _______________

Answer: 1/16

Explanation: We can find the area of the original triangle by multiplying with \(\frac{1}{4}\) \(\frac{1}{4}\) × \(\frac{1}{4}\) = \(\frac{1}{16}\) Thus the area was multiplied by \(\frac{1}{16}\)

Question 8. Write and solve a word problem that involves changing the dimensions of a figure and finding its area. Type below: _______________

Answer: The dimensions of a triangle with a base 1.5 m and height 6 m are multiplied by 2. How is the area affected? Original area: Area of triangle = bh/2 A = (1.5m)(6m)/2 A = 4.5 sq. m new dimensions: b = 1.5m × 2 = 3 m h = 6 m × 2 = 12 m Area of triangle = bh/2 A = (12m × 3m)/2 A = 6m × 3m A = 18 sq. m new area/original area = 18 sq. m/4.5 sq. m The area was multiplied by 4.

Question 1. The dimensions of Rectangle A are 6 times the dimensions of Rectangle B. How do the areas of the rectangles compare? Type below: _______________

Answer: Area of Rectangle A = 36 × Area of Rectangle B

Explanation: The area of Rectangle A will always be 36 times the area of Rectangle B. If Rectangle B has length 1 and width 2, Rectangle A will have length 6 and width 12. By multiplying, Rectangle A will have an area of 72 and B 2. Divide the two numbers and you will have 36.

Question 2. A model of a triangular piece of jewelry has an area that is \(\frac{1}{4}\) the area of the jewelry. How do the dimensions of the triangles compare? Type below: _______________

Answer: Model dimensions = 1/2 jewelry dimensions

Explanation: The dimensions of the model area 1/4 ÷ 2 = 1/2 times the dimensions of the piece of jewelry.

Question 3. Gina made a rectangular quilt that was 5 feet wide and 6 feet long. She used yellow fabric for 30% of the quilt. What was the area of the yellow fabric? ________ square feet

Answer: 9 square feet

Explanation: Gina made a rectangular quilt that was 5 feet wide and 6 feet long. She used yellow fabric for 30% of the quilt. Area of rectangle = lw A = 5 ft × 6 ft = 30 square ft she used 30% of yellow fabric so 30% of 30 30/x = 100/30 x = 900/100 x = 9 The area of the yellow fabric is 9 square feet.

Question 4. Graph y > 3 on a number line. Type below: _______________

Answer: 1312.5 sq. mm

Explanation: Given, b1 = 25mm b2 = 50mm h = 35mm Area of the trapezoid = (b1 + b2)h/2 A = (25mm + 50mm)35mm/2 A = 75mm × 35mm/2 A = 1312.5 sq. mm Thus the area of the shaded region is 1312.5 sq. mm

Question 6. A rectangle has a length of 24 inches and a width of 36 inches. A square with side length 5 inches is cut from the middle and removed. What is the area of the figure that remains? ________ in. 2

Answer: 839 sq. in

Explanation: Area of rectangle = lw A = 24 in × 36 in A = 864 sq. in Area of square = s × s s = 5 in A = 5 in × 5 in A = 25 sq. in Area of the figure that remains = 864 sq. in – 25 sq. in A = 839 sq. in

Question 1. The vertices of triangle ABC are A(−1, 3), B(−4, −2), and C(2, −2). Graph the triangle and find the length of side \(\overline { BC } \). ________ units

Give the coordinates of the unknown vertex of rectangle JKLM, and graph.

Question 5. The vertices of pentagon PQRST are P(9, 7), Q(9, 3), R(3, 3), S(3, 7), and T(6, 9). Graph the pentagon and find the length of side \(\overline { PQ } \). ________ units

Problem Solving + Applcations – Page No. 586

Question 6. A city planner wants to locate a park where two new roads meet. One of the new roads will go to the mall and be parallel to Lincoln Street which is shown in red. The other new road will go to City Hall and be parallel to Elm Street which is also shown in red. Give the coordinates for the location of the park. Type below: _______________

Question 7. Each unit of the coordinate plane represents 2 miles. How far will the park be from City Hall? ________ miles

Answer: 8 units

Explanation: The distance from City Hall to Park is 4 units. Each unit = 2 miles So, 2 miles × 4 = 8 miles The distance from City Hall to Park is 8 miles.

Question 9. Use Math Vocabulary Quadrilateral WXYZ has vertices with coordinates W(−4, 0), X(−2, 3), Y(2, 3), and Z(2, 0). Classify the quadrilateral using the most exact name possible and explain your answer. Type below: _______________

Question 1. The vertices of triangle DEF are D(−2, 3), E(3, −2), and F(−2, −2). Graph the triangle, and find the length of side \(\overline { DF } \). ________ units

Answer: 5 units

Explanation: Vertical distance of D from 0: |3| = 3 units Vertical Distance of F from 0: |-2| = 2 units The points are in different quadrants, so add to find the distance from D to F: 3 + 2 = 5

Graph the figure and find the length of side \(\overline { BC } \).

Question 2. A(1, 4), B(1, −2), C(−3, −2), D(−3, 3) ________ units

Question 3. A(−1, 4), B(5, 4), C(5, 1), D(−1, 1) ________ units

Question 4. On a map, a city block is a square with three of its vertices at (−4, 1), (1, 1), and (1, −4). What are the coordinates of the remaining vertex? Type below: _______________

Question 5. A carpenter is making a shelf in the shape of a parallelogram. She begins by drawing parallelogram RSTU on a coordinate plane with vertices R(1, 0), S(−3, 0), and T(−2, 3). What are the coordinates of vertex U? Type below: _______________

Question 6. Explain how you would find the fourth vertex of a rectangle with vertices at (2, 6), (−1, 4), and (−1, 6). Type below: _______________

Explanation: Midpoint of AC = (2 + (-1))/2 = 1/2; (6 + 6)/2 = 6 Midpoint of AC = (1/2, 6) Midpoint of BD = (-1 + a)/2 = (-1 + a)/2; (b + 4)/2 (-1 + a)/2 = 1/2 -1 + a = 1 a = 2 (b + 4)/2 = 6 b + 4 = 12 b = 12 – 4 b = 8 So, the fouth vertex D is (2, 8)

Question 1. The coordinates of points M, N, and P are M(–2, 3), N(4, 3), and P(5, –1). What coordinates for point Q make MNPQ a parallelogram? Type below: _______________

Answer: Q (-1, -1)

Question 2. Dirk draws quadrilateral RSTU with vertices R(–1, 2), S(4, 2), T(5, –1), and U( 2, –1). Which is the best way to classify the quadrilateral? Type below: _______________

Answer: The bases and height are not equal. So, the best way to classify the quadrilateral is Trapezoid.

Question 3. Marcus needs to cut a 5-yard length of yarn into equal pieces for his art project. Write an equation that models the length l in yards of each piece of yarn if Marcus cuts it into p pieces. Type below: _______________

Answer: Given, Marcus needs to cut a 5-yard length of yarn into equal pieces for his art project. To find the length we have to divide 5 by p. Thus the equation is l = 5 ÷ p

Question 4. The area of a triangular flag is 330 square centimeters. If the base of the triangle is 30 centimeters long, what is the height of the triangle? ________ cm

Answer: 22 cm

Explanation: Given, A = 330 sq. cm b = 30 h = ? Area of the triangle = bh/2 330 sq. cm = (30 × h)/2 330 sq. cm = 15 × h h = 330 sq. cm/15 cm h = 22 cm

Question 5. A trapezoid is 6 \(\frac{1}{2}\) feet tall. Its bases are 9.2 feet and 8 feet long. What is the area of the trapezoid? ________ ft 2

Answer: 55.9

Explanation: Given that, A trapezoid is 6 \(\frac{1}{2}\) feet tall. Its bases are 9.2 feet and 8 feet long. We know that Area of trapezoid = (b1 + b2)h/2 A = (9.2 + 8)6.5/2 A = (17.2 × 6.5)/2 A = 55.9 ft 2

Answer: 3 × 3 = 9 the area will be multiplied by 9.

Answer: 67.5

Explanation: b = 9 in h = 7.5 in Area of the parallelogram is bh A = 9 in × 7.5 in A = 67.5 sq. in Thus the area of the parallelogram is 67.5 in. 2

Answer: 11 in. 2

Explanation: b = 5.5 in h = 4 in We know that The area of the triangle is bh/2 A = (5.5 in × 4 in)/2 A = 22/2 sq. in A = 11 sq. in Thus the area of one triangle is 11 in. 2

Question 3. The area of a triangle is 36 ft 2 . For numbers 3a–3d, select Yes or No to tell if the dimensions could be the height and base of the triangle. 3a. h = 3 ft, b = 12 ft 3b. h = 3 ft, b = 24 ft 3c. h = 4 ft, b = 18 ft 3d. h = 4 ft, b = 9 ft 3a. ____________ 3b. ____________ 3c. ____________ 3d. ____________

Answer: 3a. No 3b. Yes 3c. Yes 3d. No

Explanation: The area of a triangle is 36 ft 2 . 3a. h = 3 ft, b = 12 ft The area of the triangle is bh/2 A = (12 × 3)/2 A = 6 × 3 = 18 A = 18 sq. ft Thus the answer is no. 3b. h = 3 ft, b = 24 ft The area of the triangle is bh/2 A = (3 × 24)/2 A = 3 × 12 A = 36 sq. ft Thus the answer is yes. 3c. h = 4 ft, b = 18 ft The area of the triangle is bh/2 A = (4 × 18)/2 A = 4 × 9 A = 36 sq. ft Thus the answer is yes. 3d. h = 4 ft, b = 9 ft The area of the triangle is bh/2 A = (4 × 9)/2 A = 2 ft × 9 ft A = 18 sq. ft Thus the answer is no.

Answer: 112

Explanation: b1 = 10 in b2 = 4 in h = 8 in We know that Area of trapezoid = (b1 + b2)h/2 A = (10 in + 4 in)8 in/2 A = 14 in × 4 in A = 56 sq. in Thus the area of the trapezoid for the above figure is 56 sq. in

Chapter 10 Review/Test Page No. 590

Answer: 6, 8

Question 6. A rectangle has an area of 50 cm 2 . The dimensions of the rectangle are multiplied to form a new rectangle with an area of 200 cm 2 . By what number were the dimensions multiplied? Type below: _______________

Explanation: Let A₁ = the original area a and A₂ = the new area and n = the number by which the dimensions were multiplied A₁ = lw A₂ = nl × nw = n²lw A₂/A₁ = (n²lw)/(lw) = 200/50 n² = 4 n = 2

Answer: A trapezoid is a 4-sided figure with one pair of parallel sides. To find the area of a trapezoid, take the sum of its bases, multiply the sum by the height sum by the height of the trapezoid, and then divide the result by 2.

Chapter 10 Review/Test Page No. 591

Question 9. A trapezoid has an area of 32 in. 2 . If the lengths of the bases are 6 in. and 6.8 in., what is the height? ________ in.

Answer: 5 in

Explanation: A trapezoid has an area of 32 in. 2 . If the lengths of the bases are 6 in. and 6.8 in Area of trapezoid = (b1 + b2)h/2 32 sq. in = (6 in + 6.8 in)h/2 32 sq. in = 12.8 in × h/2 32 sq. in =6.4 in × h h = 32 sq. in/6.4 in h = 5 in Thus the height of the trapezium is 5 inches.

Question 10. A pillow is in the shape of a regular pentagon. The front of the pillow is made from 5 pieces of fabric that are congruent triangles. Each triangle has an area of 22 in. 2 . What is the area of the front of the pillow? ________ in. 2

Answer: 110 in. 2

Explanation: Given, Each triangle has an area of 22 in. 2 The front of the pillow is made from 5 pieces of fabric that are congruent triangles. Area of front pillow = 5 × 22 in. 2 = 110 in. 2

Answer: \(\frac{1}{2}\) × (2 + 4.5) × 5

Explanation: b1 = 4.5 in b2 = 2 h = 5 in We know that, Area of trapezoid = (b1 + b2)h/2 A = \(\frac{1}{2}\) × (2 + 4.5) × 5 Thus the correct answer is option B.

Answer: 31 sq. in.

Explanation: b = 5 in h = 6.2 in The area of the triangle is bh/2 A = (5 × 6.2)/2 A = 31/2 A = 15.5 sq. in There are 2 triangles. To find the area of the regular polygon we have to multiply the area of the triangle and number of triangles. A = 15.5 × 2 = 31

Chapter 10 Review/Test Page No. 592

Answer: B, C, D

Here we have to use the Area of the parallelogram, Area of the rectangle, and area of triangle formulas. Thus the suitable answers are 168 + 12 × 14 + 60, 19 × 24 − \(\frac{1}{2}\) × 10 × 12 and 7 × 24 + 12 × 14 + \(\frac{1}{2}\) × 10 × 12.

Question 14. Part B What is the length of side JK? How do you know? Type below: _______________

Answer: By using the above graph we can find the length of JK. The length of the JK is 2 units.

Chapter 10 Review/Test Page No. 593

Question 16. Alex wants to enlarge a 4-ft by 6-ft vegetable garden by multiplying the dimensions of the garden by 2. Part A Find each area. Area of original garden: ________ ft 2 Area of enlarged garden: ________ ft 2

Answer: B = 4 ft w = 6 ft Area of original garden = 4 ft × 6 ft A = 24 sq. ft Now multiply 2 to base and width b = 4 × 2 = 8 ft w = 6 × 2 = 12 ft Area of original garden = bw A = 8 ft × 12 ft A = 96 sq. ft

Answer: Point: (-2, 2) would change to (-2, 1) Rectangle: B = 5 units W = 4 units Area of the rectangle = b × w A = 5 × 4 = 20 A = 20 sq. units

Chapter 10 Review/Test Page No. 594

Answer: Each of the diagonals of a parallelogram divides it into two congruent triangles, as we saw when we proved properties like that the opposite sides are equal to each other or that the two pairs of opposite angles are congruent. Since those two triangles are congruent, their areas are equal. We also saw that the diagonals of the parallelogram bisect each other, and so create two additional pairs of congruent triangles. When comparing the ratio of areas of triangles, we often look for an equal base or an equal height.

Question 19. The roof of Kamden’s house is shaped like a parallelogram. The base of the roof is 13 m and the area is 110.5 m². Choose a number and unit to make a true statement. The height of the roof is _____ __ . Type below: _______________

Answer: 8.5 m

Explanation: A = 110.5 m² b = 13 m Area of the parallelogram is bh 110.5 m² = 13 × h h = 8.5 m

Question 20. Eliana is drawing a figure on the coordinate grid. For numbers 20a–20d, select True or False for each statement. 20a. The point (−1, 1) would be the fourth vertex of a square. 20b. The point (1, 1) would be the fourth vertex of a trapezoid. 20c. The point (2, -1) would be the fourth vertex of a trapezoid. 20d. The point (−1, -1) would be the fourth vertex of a square. 20a. ____________ 20b. ____________ 20c. ____________ 20d. ____________

Answer: 20a. False 20b. False 20c. True 20d. True

Conclusion:

Just tap on the clicks available to access the Go Math 4th Class Answer Key. Refer to the answer provided here while doing your homework. Solve numerous questions to enhance your skills and score maximum marks in the exams. This is the best platform for the students to learn the concepts quickly and easily. Get step-by-step explanations for all the problems from our Go Math Grade 6 Chapter 10 Answer Key.

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