{"id":6030,"date":"2022-03-28T19:55:59","date_gmt":"2022-03-28T14:25:59","guid":{"rendered":"https:\/\/ccssmathanswers.com\/?p=6030"},"modified":"2022-04-07T12:55:10","modified_gmt":"2022-04-07T07:25:10","slug":"go-math-grade-7-answer-key-chapter-9-circumference-area-and-volume","status":"publish","type":"post","link":"https:\/\/bigideasmathanswers.com\/go-math-grade-7-answer-key-chapter-9-circumference-area-and-volume\/","title":{"rendered":"Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume"},"content":{"rendered":"

Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume:<\/strong> Get the solutions to all the questions in this article. Download Go Math Grade 7 Answer Key<\/a> 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.<\/p>\n

Download Go Math Grade 7 Answer Key Chapter 9 Circumference, Area, and Volume Pdf<\/h2>\n

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.<\/p>\n

Guided Practice – Page No. 268<\/h3>\n

Find the circumference of each circle.<\/strong><\/p>\n

Question 1.
\n\"Go
\n________ in<\/p>\n

Answer: 56.57 in<\/p>\n

Explanation:
\nCircumference of the circle = 2\u03c0r = 2 x 22\/7 x 9 = 56.57 in<\/p>\n

Question 2.
\n\"Go
\n________ cm<\/p>\n

Answer: 44 cm<\/p>\n

Explanation:
\nCircumference of the circle = 2\u03c0r = 2 x 22\/7 x 7 = 44 cm<\/p>\n

Find the circumference of each circle. Use 3.14 or \\(\\frac{22}{7}\\) for \u03c0. Round to the nearest hundredth, if necessary.<\/strong><\/p>\n

Question 3.
\n______ m<\/p>\n

Question 4.
\n\"Go
\n______ yd<\/p>\n

Answer: 30.15 yd<\/p>\n

Explanation:
\nCircumference of the circle = 2\u03c0r = 2 x 3.14 x 4.8 = 30.144 yd<\/p>\n

Question 5.
\n\"Go
\n______ in<\/p>\n

Answer: 7.5 in<\/p>\n

Explanation:
\nCircumference of the circle = 2\u03c0r = 2 x 3.14 x 7.5 = 47.1 in<\/p>\n

Question 6.
\nA 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 $0.45 per foot, and Carlos needs 4 feet more than the diameter of the pool. How much will Carlos pay for the rope?
\n$ ______<\/p>\n

Answer: $6.525<\/p>\n

Explanation:
\nCircumference of the swimming pool = 66 feet
\n\u03c0d = 66
\n22\/7 x d = 66
\nd = 66 x 7\/ 22 = 10.5
\nThe diameter of the pool = 10.5 feet
\nCarlos needs 4 feet more than the diameter of the pool.
\nTotal rope needed = 10.5 + 4 = 14.5 feet
\nCost of rope per foot = $0.45
\nTotal cost of the rope = 14.5 x $0.45 = $6.525
\nTherefore the total cost of the rope = $6.525<\/p>\n

Find each missing measurement to the nearest hundredth. Use 3.14 for \u03c0.<\/strong><\/p>\n

Question 7.
\nr =
\nd =
\nC = \u03c0 yd
\nr = ________ yd
\nd = ________ yd<\/p>\n

Answer:
\nr = 0.5 yd
\nd = 1 yd<\/p>\n

Explanation:
\nCircumference = \u03c0 yd
\n2\u03c0r = \u03c0 yd
\nr = 1\/2 yd = 0.5 yd
\nd = 2r = 2 [1\/2] = 1 yd<\/p>\n

Question 8.
\nr \u2248
\nd \u2248
\nC = 78.8 ft
\nr \u2248 ________ ft
\nd \u2248 ________ ft<\/p>\n

Answer:
\nr = 495.31 ft
\nd = 990.62 ft<\/p>\n

Explanation:
\nCircumference = 78.8 ft
\n2\u03c0r = 78.8 ft
\nr = 2 x 22\/7 x 78.8 = 495.31 ft
\nd = 2 x 495.31 = 990.62 ft<\/p>\n

Question 9.
\nr \u2248
\nd \u2248 3.4 in
\nC =
\nr \u2248 ________ in
\nC = ________ in<\/p>\n

Answer:
\nr = 1.7 in
\nc = 10.68 in<\/p>\n

Explanation:
\nDiameter = 3.4 in
\nCircumference = \u03c0d = 22\/7 x 3.4 in = 10.68 in
\nr = d\/2 = 1.7 in<\/p>\n

Essential Question Check-In<\/strong><\/p>\n

Question 10.
\nNorah knows that the diameter of a circle is 13 meters. How would you tell her to find the circumference?
\nType below:
\n____________<\/p>\n

Answer: Circumference = 16.82 meters<\/p>\n

Explanation:
\nGiven,
\nDiameter = 13 meters
\nCircumference = \u03c0d = 22\/7 x 13 = 16.82 meters<\/p>\n

Independent Practice – Page No. 269<\/h3>\n

For 11\u201313, find the circumference of each circle. Use 3.14 or \\(\\frac{22}{7}\\) for \u03c0. Round to the nearest hundredth, if necessary.<\/strong><\/p>\n

Question 11.
\n\"Go
\n_______ ft<\/p>\n

Answer:
\nCicumference = 18.526 ft = 19 ft (approx)<\/p>\n

Explanation:
\nGiven:
\nDiameter = 5.9 ft
\nCicumference = \u03c0d = 3.14 x 5.9 = 18.526 ft = 19 ft (approx)<\/p>\n

Question 12.
\n\"Go
\n_______ cm<\/p>\n

Answer:
\nCicumference =176 cm<\/p>\n

Explanation:
\nGiven:
\nRadius = 56 cm
\nCicumference = \u03c0d = 22\/7 x 56 = 176 cm<\/p>\n

Question 13.
\n\"Go
\n_______ in<\/p>\n

Answer:
\nCicumference = 110 in<\/p>\n

Explanation:
\nGiven:
\nDiameter = 35 in
\nCicumference = \u03c0d = 22\/7 x 35 = 110 in<\/p>\n

Question 14.
\nIn Exercises 11\u201313, for which problems did you use \\(\\frac{22}{7}\\) for \u03c0? Explain your choice.
\nType below:
\n_____________<\/p>\n

Answer:
\n11th question as 3.14 and the 12 and 13 questions as \u03c0<\/p>\n

Explanation:
\nWe can take 3.14 as \u03c0 for 11 th question because the diameter is given in decimal points.
\nAnd in questions 12 and 13 we need to take \u03c0 because the radius and diameter are given in whole number form.<\/p>\n

Question 15.
\nA circular fountain has a radius of 9.4 feet. Find its diameter and circumference to the nearest tenth.
\nd = _________ ft
\nC = _________ ft<\/p>\n

Answer:
\nd = 19 ft
\nC = 59 ft<\/p>\n

Explanation:
\nGiven:
\nRadius = 9.4 ft
\nDiameter = 2r = 2 x 9.4\u00a0 = 18.8 ft = 19 ft (approx)
\nCircumference = \u03c0d = 22\/7 x 18.8 = 59.08 = 59 ft (approx)<\/p>\n

Question 16.
\nFind the radius and circumference of a CD with a diameter of 4.75 inches.
\nr = _________ in
\nC = _________ in<\/p>\n

Answer:
\nr = 2.4 in
\nC = 15 in<\/p>\n

Explanation:
\nGiven:
\nDiameter = 4.75 in
\nRadius = r\/2 = 4.75\/2 = 2.37 in = 2.4 in (approx)
\nCircumference = \u03c0d = 22\/7 x 4.75 = 14.92 in =15 in (approx)<\/p>\n

Question 17.
\nA dartboard has a diameter of 18 inches. What are its radius and circumference?
\nr = _________ in
\nC = _________ in<\/p>\n

Answer:
\nr = 9 in
\nC = 56.6 in<\/p>\n

Explanation:
\nGiven:
\nDiameter = 18 in
\nRadius = r\/2 = 18\/2 = 9 in
\nCircumference = \u03c0d = 22\/7 x 18 = 56.57 in = 56.6 in (approx)<\/p>\n

Question 18.
\nMultistep
\nRandy\u2019s circular garden has a radius of 1.5 feet. He wants to enclose the garden with edging that costs $0.75 per foot. About how much will the edging cost? Explain.
\n$ _______<\/p>\n

Answer:<\/p>\n

Explanation:
\nGiven:
\nThe radius of the garden= 1.5 ft
\nCircumference of the garden = 2\u03c0r = 2 x 22\/7 x 1.5 = 9.42 ft
\nCost of enclosing the garden per foot = $0.75
\nTotal cost of edging = 9.42 x $0.75 = $7.06 = $7 (approx)<\/p>\n

Question 19.
\nRepresent Real-World Problems
\nThe Ferris wheel shown makes 12 revolutions per ride. How far would someone travel during one ride?
\n\"Go
\n_______ ft<\/p>\n

Answer: Total distance travelled in one ride is 4,752 ft<\/p>\n

Explanation:
\nGiven:
\nThe diameter of the Ferris wheel= 63 ft
\nCircumference of the Ferris wheel = 2\u03c0r = 2 x 22\/7 x 63 = 396 ft
\nTotal number of revolutions = 12
\nTotal distance travelled = 12 x 396 = 4,752 ft<\/p>\n

Question 20.
\nThe diameter of a bicycle wheel is 2 feet. About how many revolutions does the wheel make to travel 2 kilometres? Explain. Hint: 1 km \u2248 3,280 ft
\n_______ revolutions<\/p>\n

Answer:
\n1044 revolutions<\/p>\n

Explanation:
\nGiven:
\nDiametre of the bicycle wheel = 2 feet
\nTotal distance travelled = 2 kilometres
\nWe know that,
\n1 km \u2248 3,280 ft
\n2 km = 2 x 3,280 = 6,560 ft
\nCircumference of the bicycle = Distance travelled in one revolution = \u03c0d = 22\/7 x 2 = 6.28 ft = 6.3 ft
\nTotal number of revolutions = Total distance travelled \/ distance travelled in one revolution
\n= 6560 \/ 6.28 = 1044\u00a0 revolutions<\/p>\n

Question 21.
\nMultistep
\nA 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?
\n_______ mi<\/p>\n

Answer:<\/p>\n

Explanation:
\nGiven,
\nThe diameter of the pond = 0.25 mi
\nThe length of the bridge = The diameter of the pond = 0.25 mi
\nThen the distance walked by the man = 0.25 mi
\nDistance travelled by the friend = Halfway around the pond to meet you at the other side of the bridge = \u03c0d\/2
\n= 22\/7 x 0.25\/2\u00a0 = 0.39 = 0.4 mi
\nThe friend travelled more distance compared to the man
\nThe more distance travelled by the friend = 0.39 – 0.25 = 0.14 mi<\/p>\n

Page No. 270<\/h3>\n

Question 22.
\nArchitecture
\nThe Capitol Rotunda connects the House and the Senate sides of the U.S. Capitol. Complete the table. Round your answers to the nearest foot.
\n\"Go
\nType below:
\n_____________<\/p>\n

Answer:
\nRadius = 48 ft
\nDiameter = 96 ft<\/p>\n

Explanation:
\nGiven
\nHeight = 180 ft
\nCircumference = 301.5 ft
\n\u03c0d = 301.5
\n22\/7 x d = 301.5
\nd = 95.93 = 96 ft
\nr = d\/2 = 96\/2 = 48 ft<\/p>\n

H.O.T.<\/strong><\/p>\n

Focus on Higher Order Thinking<\/strong><\/p>\n

Question 23.
\nMultistep
\nA 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 $9.25 per linear foot. About how much will the fencing cost altogether?
\n$ _______<\/p>\n

Answer:
\nThe total cost of fencing = $712<\/p>\n

Explanation:
\nGiven,
\nThe diameter = 30 ft
\nCircumference of the garden in the shape of circle = 2\u03c0r
\nCircumference of the semicircle = \u03c0r = \u03c0d\/2 =\u00a0 22\/7 x 30\/2 = 47.14ft
\nCost of fencing for each foot = $9.25
\nThe total cost of fencing the semicircular garden = 47.14 x $9.25 + 30 x\u00a0 $9.25 \u00a0= $712 (approx)<\/p>\n

Question 24.
\nCritical Thinking
\nSam 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?
\n_______ strands<\/p>\n

Answer: 12 and a half strands of light = 13 strands (approx)<\/p>\n

Explanation:
\nGiven,
\nThe diameter of the circular patio = 18 ft = 216 inch
\nCircumference of the circular patio =\u00a0\u03c0d = 22\/7 x 216 = 678.85 inch
\nThe lights will come in a length (in one strand)= 54 inches
\nTotal number of strands of light required for the circular patio
\n= 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<\/p>\n

Question 25.
\nRepresent Real-World Problems
\nA 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?
\n_______ feet<\/p>\n

Answer: about 12.6 ft<\/p>\n

Explanation:
\nGiven,
\nWidth of the circular path = 2 ft
\nThe inner diameter of the circular path = 150 ft
\nThe outer diameter of the circular path = 150 + 2(2) = 154 ft
\nInner circumference = \u03c0d = 150 \u03c0
\nOuter circumference =\u00a0 \u03c0d = 154\u03c0
\nDistance between the outer and inner edge = 154 \u03c0 – 150 \u03c0 = 4 \u03c0 = 12.6 ft<\/p>\n

Question 26.
\nCritique Reasoning
\nGear 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.
\n_______<\/p>\n

Answer:
\nJustin statement is incorrect.<\/p>\n

Explanation:
\nThe circumference of the larger gear = \u03c0d = 4\u03c0
\nThe circumference of the smaller gear = \u03c0d = 2\u03c0
\nSince, 2 x 2\u03c0 = 4\u03c0, the circumference of the larger gear is two times the circumference of the smaller gear.
\nSince = 4\u03c0 – 2\u03c0 = 2\u03c0 = 6.28
\nTherefore, The larger circumference is not 2 inches more than the smaller circumference<\/p>\n

Question 27.
\nPersevere in Problem Solving
\nConsider 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.
\n_______<\/p>\n

Answer:
\nPool B about 0.9 meters<\/p>\n

Explanation:
\nGiven,
\nPool A has a diameter = 24 ft
\nPool B has a diameter = 7.5 m
\nWe know that,
\n1 ft = 0.3 metres
\n24 ft = 7.2 metres
\nThe pool B has a greater diameter so it has a greater circumference.
\nCircumference of the pool A = 7.2\u03c0
\nCircumference of the pool B = 7.5\u03c0
\nDifference between the circumferences = 7.5\u03c0 – 7.2\u03c0 = 0.9 meters.<\/p>\n

Guided Practice – Page No. 274<\/h3>\n

Find the area of each circle. Round to the nearest tenth if necessary. Use 3.14 for \u03c0.<\/strong><\/p>\n

Question 1.
\n\"Go
\n_______ m2<\/sup><\/p>\n

Answer: 153.9 m2<\/sup><\/p>\n

Explanation:
\nGiven:
\nDiameter = 14 m
\nRadius = 14\/2 = 7 m
\nArea of the circle = \u03c0r2<\/sup>
\n= 3.14 x 7 x 7 = 153.86 = 153.9 m2<\/sup><\/p>\n

Question 2.
\n\"Go
\n_______ mm2<\/sup><\/p>\n

Answer: 452.2 mm2<\/sup><\/p>\n

Explanation:
\nGiven:
\nRadius =12mm
\nArea of the circle = \u03c0r2<\/sup>
\n= 3.14 x 12 x 12 = 3.14(144) = 452.2mm2<\/sup><\/p>\n

Question 3.
\n\"Go
\n_______ yd2<\/sup><\/p>\n

Answer: 314 yd2<\/sup><\/p>\n

Explanation:
\nGiven:
\nDiameter = 20yd
\nRadius = 20\/2 = 10yd
\nArea of the circle = \u03c0r2<\/sup>
\n= 3.14 x 10 x 10 = 3.14(100) = 314yd2<\/sup><\/p>\n

Solve. Use 3.14 for \u03c0.<\/strong><\/p>\n

Question 4.
\nA clock face has a radius of 8 inches. What is the area of the clock face? Round your answer to the nearest hundredth.
\n_______ in2<\/sup><\/p>\n

Answer: 200.96 in2<\/sup><\/p>\n

Explanation:
\nGiven:
\nRadius = 8inches
\nArea of the clock face = \u03c0r2<\/sup>
\n= 3.14 x 8 x 8= 3.14(64) = 200.96 in2<\/sup><\/p>\n

Question 5.
\nA DVD has a diameter of 12 centimeters. What is the area of the DVD? Round your answer to the nearest hundredth.
\n_______ cm2<\/sup><\/p>\n

Answer: 113.04 cm2<\/sup><\/p>\n

Explanation:
\nGiven:
\nDiameter = 12 centimeters
\nRadius = 12\/2 = 6 centimeters
\nArea of the DVD= \u03c0r2<\/sup>
\n= 3.14 x 6 x 6 = 3.14(36) = 113.04 cm2<\/sup><\/p>\n

Question 6.
\nA 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.
\n_______ in2<\/sup><\/p>\n

Answer: 132.67 in2<\/sup><\/p>\n

Explanation:
\nGiven:
\nDiameter = 13 inches
\nRadius = 13\/2 = 6.5 inches
\nArea of each lid= \u03c0r2<\/sup>
\n= 3.14 x 6.5 x 6.5 = 3.14(42.25) = 132.67 in2<\/sup><\/p>\n

Find the area of each circle. Give your answers in terms of \u03c0.<\/strong><\/p>\n

Question 7.
\nC = 4\u03c0
\nA =
\nType below:
\n______________<\/p>\n

Answer: 4\u03c0<\/p>\n

Explanation:
\nGiven:
\nCircumcenter = 4\u03c0
\n2\u03c0r = 4\u03c0
\nRadius = 4\/2 = 2 units
\nArea of the circle = \u03c0r2<\/sup>
\n= \u03c0 x 2 x 2 = \u03c0(4) = 4\u03c0 square units<\/p>\n

Question 8.
\nC = 12\u03c0
\nA =
\nType below:
\n______________<\/p>\n

Answer: 36\u03c0<\/p>\n

Explanation:
\nGiven:
\nCircumcenter = 12\u03c0
\n2\u03c0r = 12\u03c0
\nRadius =6 units
\nArea of the circle = \u03c0r2<\/sup>
\n= \u03c0 x 6 x 6 = \u03c0(36) = 36\u03c0 square units<\/p>\n

Question 9.
\nC = \\(\\frac{\u03c0}{2}\\)
\nA =
\nType below:
\n______________<\/p>\n

Answer: \u03c0\/16<\/p>\n

Explanation:
\nGiven:
\nCircumcenter = \\(\\frac{\u03c0}{2}\\)
\n2\u03c0r = \\(\\frac{\u03c0}{2}\\)
\nRadius = 1\/4 units
\nArea of the circle = \u03c0r2<\/sup>
\n= \u03c0 x 1\/4 x 1\/4 = \u03c0(1\/16) = \u03c0\/16 square units<\/p>\n

Question 10.
\nA circular pen has an area of 64\u03c0 square yards. What is the circumference of the pen? Give your answer in terms of \u03c0
\nType below:
\n______________<\/p>\n

Answer: 16\u03c0<\/p>\n

Explanation:
\nGiven:
\nArea of the circular pen = 64\u03c0 square yards
\n\u03c0r2<\/sup> = 64\u03c0
\nr = 8 yards
\nCircumference of the circle = 2\u03c0r = 2 x 8 x \u03c0 = 16\u03c0 yards<\/p>\n

Essential Question Check-In<\/strong><\/p>\n

Question 11.
\nWhat is the formula for the area A of a circle in terms of the radius r?
\nType below:
\n______________<\/p>\n

Answer: \u03c0r2<\/sup><\/p>\n

Explanation:
\nArea of a circle = \u03c0r2<\/sup><\/p>\n

Independent Practice – Page No. 275<\/h3>\n

Question 12.
\nThe most popular pizza at Pavone\u2019s 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.
\n_______ in2<\/sup><\/p>\n

Answer: 78.5 in2<\/sup><\/p>\n

Explanation:
\nGiven:
\nDiameter = 10 inches
\nRadius = 10\/2 = 5 inches
\nArea of a pizza = \u03c0r2<\/sup>
\n= 3.14 x 5 x 5 = 3.14(25) = 78.5 in2<\/sup><\/p>\n

Question 13.
\nA hubcap has a radius of 16 centimeters. What is the area of the hubcap? Round your answer to the nearest hundredth.
\n\"Go
\n_______ cm2<\/sup><\/p>\n

Answer: 803.84 cm2<\/sup><\/p>\n

Explanation:
\nGiven:
\nRadius = 16 cm
\nArea of the circle = \u03c0r2<\/sup>
\n= 3.14 x 16 x 16 = 3.14(256) = 803.84 cm2<\/sup><\/p>\n

Question 14.
\nA 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.
\n_______ in2<\/sup><\/p>\n

Answer: 508.68 in2<\/sup><\/p>\n

Explanation:
\nArea of the semicircle = 1\/2 \u03c0r2<\/sup> = 1\/2(3.14)(18)(18) = 1\/2 (3.14)(324) = 1.57(324) = 508.68 in 2 <\/sup><\/p>\n

Question 15.
\nAnalyze Relationships
\nThe point (3,0) lies on a circle with the centre at the origin. What is the area of the circle to the nearest hundredth?
\n_______ units2<\/sup><\/p>\n

Answer: 28.26 units2<\/sup><\/p>\n

Explanation:
\nRadius = 3
\nArea of the circle = \u03c0r2<\/sup> = \u03c0(3)2<\/sup> = 3.14(9) = 28.26 units2<\/sup><\/p>\n

Question 16.
\nMultistep
\nA 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.
\n_______ mi2<\/sup><\/p>\n

Answer: 9813 mi2<\/sup><\/p>\n

Explanation:
\nGiven:
\nThe radius of a radio station broadcasted the signal (r) = 50 miles
\nThe greatest radius to which the broadcast can be relayed (R) = 75 miles
\nThe greatest area of the broadcast region when the signal is relayed = \u03c0R2<\/sup>-\u03c0r2<\/sup> = \u03c0(75) (75) – \u03c0 (50) (50)
\n= 5625\u03c0 – 2500\u03c0
\n= 3125\u03c0
\n= 3125(3.14) = 9813 mi2<\/sup>(approx)<\/p>\n

Question 17.
\nMultistep
\nThe 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.
\n_______ m2<\/sup><\/p>\n

Answer:30.96 m2<\/sup><\/p>\n

Explanation:
\nGiven:
\nThe side of the square = 12 meters
\nThe diameter circular area of the field in the centre = The side of the square = 12 meters
\nThe radius of the field = 12\/2 = 6 meters
\nArea of the field which is not reached by the sprinkler = Area of the square – Area of the circular area
\n= (side)2<\/sup>-\u03c0r2<\/sup> = (12)(12) – \u03c0 (6) (6)
\n= 144 – 36 (3.14)
\n= 144 – 113.04
\n= 30.96 m2<\/sup><\/p>\n

Question 18.
\nJustify Reasoning
\nA small silver dollar pancake served at a restaurant has a circumference of 2\u03c0 inches. A regular pancake has a circumference of 4\u03c0 inches. Is the area of the regular pancake twice the area of the silver dollar pancake? Explain.
\n_______<\/p>\n

Answer: No, the area of the regular pancake is 4 times the area of the silver dollar pancake<\/p>\n

Explanation:
\nSilver Dollar pancake:
\nCircumference of the silver Dollar pancake = 2\u03c0 inches
\n2\u03c0r = 2\u03c0
\nr = 1 inch
\nArea of the silver dollar pancake = \u03c0r2<\/sup> = \u03c0 (1) (1) = \u03c0 in2<\/sup><\/p>\n

Regular pancake:
\nCircumference of the regular pancake = 4\u03c0 inches
\n2\u03c0r = 4\u03c0
\nr = 2 inch
\nArea of the silver dollar pancake = \u03c0r2<\/sup> = \u03c0 (2) (2) = 4\u03c0 in2<\/sup><\/p>\n

Therefore, the area of the regular pancake is 4 times the area of the silver dollar pancake<\/p>\n

Question 19.
\nAnalyze Relationships
\nA 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.
\n_______<\/p>\n

Answer: No, the area of the large cake is 9 times the area of the small cake<\/p>\n

Explanation:
\nSmall Cake:
\nThe diameter of the small cake= 8 inches
\nThe radius of the small cake = 8\/2 = 4 inches
\nArea of the small cake\u00a0 = \u03c0r2<\/sup> = \u03c0 (4) (4) = 16 \u03c0 in2<\/sup><\/p>\n

Large Cake:
\nThe diameter of the large cake= 24 inches
\nThe radius of the large cake = 24\/2 = 12 inches
\nArea of the large cake\u00a0 = \u03c0r2<\/sup> = \u03c0 (12) (12) = 144 \u03c0 in2
\n<\/sup><\/p>\n

Since 144 \u03c0\/ 16 \u03c0 = 9
\nTherefore the<\/p>\n

area of the large cake is 9 times the area of the small cake.<\/p>\n

Page No. 276<\/h3>\n

Question 20.
\nCommunicate Mathematical Ideas
\nYou can use the formula A = \\(\\frac{C^{2}}{4\u03c0}\\) to find the area of a circle given the circumference. Describe another way to find the area of a circle when given the circumference.
\nType below:
\n____________<\/p>\n

Answer: Area = C2<\/sup>\/4\u03c0<\/p>\n

Explanation:
\nCircumference of the circle = 2\u03c0r
\nC = 2\u03c0r
\nDivide both sides by 2\u03c0
\nthen, r = C\/2\u03c0
\nArea of the circle = \u03c0r2<\/sup>
\nSubstitute C\/2\u03c0 for r:
\nArea = \u03c0(c\/2\u03c0)2<\/sup> = C2<\/sup>\/4\u03c0<\/p>\n

Question 21.
\nDraw Conclusions
\nMark wants to order a pizza. Which is the better deal? Explain.
\n\"Go
\n_____<\/p>\n

Answer: The pizza of 18 inches is a better deal<\/p>\n

Explanation:
\nGiven:
\nThe diameter of the pizza = 12 inches
\nThe radius of the pizza = 12\/2= 6 inches
\nArea of the circle = \u03c0r2<\/sup>
\n= (3.14)(6)(6) = 113 (approx) in2<\/sup>
\nThe total cost of the pizza = $10
\nCost of the pizza per inch = $10\/113 = $0.09 per square inch<\/p>\n

The diameter of the pizza = 18 inches
\nThe radius of the pizza = 18\/2= 9 inches
\nArea of the circle = \u03c0r2<\/sup>
\n= (3.14)(9)(9) = 254 (approx) in2<\/sup>
\nThe total cost of the pizza = $20
\nCost of the pizza per inch = $20\/254 = $0.08 per inch<\/p>\n

Question 22.
\nMultistep
\nA bear was seen near a campground. Searchers were dispatched to the region to find the bear.
\na. 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 \u03c0. Round your answer to the nearest square mile.
\n_____ mi2<\/sup><\/p>\n

Answer: 201mi2<\/sup><\/p>\n

Explanation:
\nThe bear can walk a distance = 2 x 4 = 8 miles
\nSince it is walking 2 miles per hour for 4 hours
\nThe radius of the bear = 8 miles
\nArea of the circle = \u03c0r2<\/sup>
\n= (3.14)(8)(8) = 201 (approx) mi2<\/sup><\/p>\n

Question 22.
\nb. What If? How much additional area would the searchers have to cover if the bear were last seen 5 hours ago?
\n_____ mi2<\/sup><\/p>\n

Answer: 113mi2<\/sup><\/p>\n

Explanation:
\nIf the bear for 5 hours then,
\nThe bear can walk a distance = 2 x 5 = 10 miles
\nSince it is walking 2 miles per hour for 5 hours
\nThe radius of the bear = 10 miles
\nArea of the circle = \u03c0r2<\/sup>
\n= (3.14)(10)(10) = 314 (approx) mi2<\/sup><\/p>\n

The additional area covered by the searches = 314 – 201 = 113 mi2<\/sup><\/p>\n

H.O.T.<\/strong><\/p>\n

Focus on Higher Order Thinking<\/strong><\/p>\n

Question 23.
\nAnalyze Relationships
\nTwo 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.
\n_____<\/p>\n

Answer: No<\/p>\n

Explanation:
\nIf the radius of two circles is the same.
\nthen, Let the radii of the circles be 1.
\nThe area of each circle =\u00a0 \u03c0 square units
\nThe combined area of 2 circles =\u03c0+\u03c0 = 2\u03c0 square units<\/p>\n

If the radius is doubled.
\nthen, Let the radii of the circles be 2
\nThe area of each circle =\u00a0 4\u03c0 square units
\nThe combined area of 2 circles =\u00a0 4\u03c0+4\u03c0 = 8\u03c0 square units<\/p>\n

Therefore the areas of both cases are not the same.<\/p>\n

Question 24.
\nLook for a Pattern
\nHow does the area of a circle change if the radius is multiplied by a factor of n, where n is a whole number?
\nType below:
\n____________<\/p>\n

Answer: The new area is then n2<\/sup> times the area of the original circle.<\/p>\n

Explanation:
\nIf the radius is multiplied by a factor “n”
\nthen, the new radius = rn
\nThe area of the circle (with radius rn) = \u03c0(rn)2\u00a0<\/sup>= n2<\/sup> (\u03c0r2<\/sup>).
\nTherefore the new area is n2<\/sup> times the area of the original circle.<\/p>\n

Question 25.
\nRepresent Real World Problems
\nThe bull\u2019s-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\u2019s-eye? Explain.
\nType below:
\n____________<\/p>\n

Answer: 1\/25 of the target<\/p>\n

Explanation:
\nBull’s eye:
\nDiameter of Bull’s eye = 3 inches
\nRadius of Bull’s eye = 3\/2 = 1.5 inches
\nArea of the Bull’s eye = \u03c0(r)2\u00a0<\/sup>= \u03c0(1.5)2<\/sup> = 2.25\u03c0
\nTarget:
\nDiameter of the target = 15 inches
\nRadius of the target = 15\/2 = 7.5 inches
\nArea of the target = \u03c0(r)2\u00a0<\/sup>= \u03c0(7.5)2<\/sup> = 56.25\u03c0<\/p>\n

The part of Bull’s eye in the whole target = 2.25\u03c0\/ 56.25\u03c0 = 1\/25<\/p>\n

Therefore the 1\/25th part of the whole target is the Bull’s eye.<\/p>\n

Guided Practice – Page No. 280<\/h3>\n

Question 1.
\nA 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?
\n\"Go
\n_____ cm2<\/sup><\/p>\n

Answer: Area of the irregular shape = 34 cm2<\/sup><\/p>\n

Explanation:
\nSTEP1 First divide the irregular shapes into polygons.
\nSTEP2 The irregular shape can be divided into a triangle, rectangle, parallelogram
\nSTEP3 Areas of the polygons
\nArea of triangle = 1\/2 (base x height) = 1\/2 (4 x 2) = 4 cm2<\/sup>
\nArea of the rectangle = length x breadth = 5 x 3 = 15 cm2<\/sup>
\nArea of the parallelogram = base x height = 5 x 3 = 15 cm2<\/sup>
\nArea of the irregular shape = (15+15+5) cm2<\/sup>= 34cm2<\/sup><\/p>\n

Question 2.
\nShow two different ways to divide the composite figure. Find the area both ways. Show your work below.
\n\"Go
\n_____ cm2<\/sup><\/p>\n

Answer: Area of the figure in both ways = 288 cm2<\/sup><\/p>\n

Explanation:
\nThe first way to divide up the composite shape is to divide it into an 8 by 9 rectangle and a 12 by 18 rectangle.
\nThe area of the first rectangle = Length x breadth = 9 x 8 = 72 cm2<\/sup>
\nThe area of the second rectangle =\u00a0 Length x breadth = 18 x 12 = 216 cm2<\/sup>
\nThe total area of the figure = 72 + 216 = 288 cm2<\/sup><\/p>\n

Question 3.
\nSal is tiling his entryway. The floor plan is drawn on a unit grid. Each unit length represents 1 foot. Tile costs $2.25 per square foot. How much will Sal pay to tile his entryway?
\n\"Go
\n$ _____<\/p>\n

Answer: Sal will pay $97.875<\/p>\n

Explanation:
\nSeparate this figure into trapezium and parallelogram.
\nArea of the trapezium = 1\/2 (a+b)h = 1\/2 (7+4) 5 = 1\/2 (11) 5 = 27.5 ft2<\/sup>
\nArea of the parallelogram = base x height = 4 x 4 = 16 ft2<\/sup><\/p>\n

The total area of the figure = 27.5 + 16 = 43.5ft2<\/sup>
\nCost of each square foot = $2.25
\nAmount paid by Sal = 43.5 x 2.25 = $97.875<\/p>\n

Essential Question Check-In<\/strong><\/p>\n

Question 4.
\nWhat is the first step in finding the area of a composite figure?
\nType below:
\n______________<\/p>\n

Answer:
\nThe first step in finding the area of a composite figure is to divide it up into smaller basic shapes.<\/p>\n

Explanation:
\nThe 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.
\nThen calculate the area of each figure and add them to find the area of the figure.<\/p>\n

Independent Practice – Page No. 281<\/h3>\n

Question 5.
\nA 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 \u03c0.
\n\"Go
\n_____ in2<\/sup><\/p>\n

Answer: 941.33 in2<\/sup><\/p>\n

Explanation:
\nArea of the square = side x side = 26 x 26 = 676 in2<\/sup>
\nArea of the semicircle =1\/2 \u03c0r2<\/sup>= 1\/2 (3.14) (13) (13) = 1\/2 (3.14) (169) = 265.33 in2<\/sup>
\nArea of the figure = 676 + 265.33 = 941.33 in2<\/sup><\/p>\n

Question 6.
\nMultistep
\nErin wants to carpet the floor of her closet. A floor plan of the closet is shown.
\n\"Go
\na. How much carpet does Erin need?
\n_____ ft2<\/sup><\/p>\n

Answer: 61 ft2<\/sup><\/p>\n

Explanation:
\nArea of the rectangle = length x breadth = 4 x 10 = 40 ft
\nArea of the triangle = 1\/2 x base x height = 1\/2 x 6 x 7 = 21 ft
\nThe total area of the figure = 40+21 = 61 ft2<\/sup><\/p>\n

Question 6.
\nb. The carpet Erin has chosen costs $2.50 per square foot. How much will it cost her to carpet the floor?
\n$ _____<\/p>\n

Answer: $152.50<\/p>\n

Explanation:
\nCost per square foot of the carpet = $2.50
\nThe total cost of the carpet on the floor = 61 x $2.50 =$152.50<\/p>\n

Question 7.
\nMultiple Representations
\nHexagon 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?
\n\"Go
\n_____ units2<\/sup><\/p>\n

Answer: The area of the figure is 30 square units<\/p>\n

Explanation:
\nSeparate the figure into a trapezium and a rectangle.
\nArea of a trapezium = 1\/2 (a+b) h= 1\/2 (2+4) x 3 = 1\/2 (6) 3 = 9 square units
\nArea of a rectangle = length x breadth = 7 x 3 = 21 square units
\nThe total area of the figure = 9+21 = 30 square units<\/p>\n

Question 8.
\nA field is shaped like the figure shown. What is the area of the field? Use 3.14 for \u03c0.
\n\"Go
\n_____ m2<\/sup><\/p>\n

Answer: 146.24 m2<\/sup><\/p>\n

Explanation:
\nDivide the figure into a square, triangle and a quarter of a circle.<\/p>\n

Area of a square = side x side = 8 x 8 = 64 m2<\/sup>
\nArea of a quarter of a circle = 1\/4 (\u03c0r2<\/sup>) = 1\/4 (3.14 x 82<\/sup>)
\n= 1\/4 (200.96) = 50.24 m2<\/sup>
\nArea of the triangle = 1\/2 x base x height = 1\/2 x 8 x 8 = 32 m2<\/sup>
\nTotal area of the figure = 64+32+50.24 = 146.24 m2<\/sup><\/p>\n

Question 9.
\nA 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 \u03c0.
\n_____ cm2<\/sup><\/p>\n

Answer: 60.56 cm2<\/sup><\/p>\n

Explanation:
\nThe bookmark is divided into a rectangle, semicircle.
\nArea of the rectangle = length x breadth = 12 x 4 = 48 cm2<\/sup>
\nThe diameter of the semicircle = The width of the rectangle = 4 cm
\nThe radius of the semicircle = 4\/2 = 2 cm
\nThe area of the semicircle = \u03c0r2<\/sup> = 3.14 x 2 x 2 = 12.56 cm2<\/sup>
\nThe total area of the bookmark = 12.56 + 48 = 60.56 cm2<\/sup><\/p>\n

Question 10.
\nMultistep
\nAlex 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 $1.25 per square foot. How much will it cost Alex to make 12 pennants?
\n\"Go
\n$ _____<\/p>\n

Answer: $52.50<\/p>\n

Explanation:
\nEach pendant is made up of a rectangle and a triangle.
\nArea of the rectangle = length x breadth = 3 x 1 = 3 ft2<\/sup>
\nArea of the triangle = 1\/2 x base x height = 1\/2 x 1 x 1 = 0.5 ft2<\/sup>
\nThe total area of the pendant = 3+0.5 = 3.5 ft2<\/sup>
\nNumber of pendants = 12
\nArea of the pendants = 12 x 3.5 = 42 ft2<\/sup>
\nCost of each square feet of the pendant = $1.25
\nTotal cost for all the 12 pendants = 12 x $1.25\u00a0 = $52.50<\/p>\n

Question 11.
\nReasoning
\nA composite figure is formed by combining a square and a triangle. Its total area is 32.5 ft2<\/sup>. The area of the triangle is 7.5 ft2<\/sup>. What is the length of each side of the square? Explain.
\n_____ ft<\/p>\n

Answer: 5 ft<\/p>\n

Explanation:
\nGiven:
\nThe area of the composite figure = 32.5 ft2<\/sup>
\nThe area of the triangle = 7.5 ft2<\/sup>
\nThe area of the square = 32.5 – 7.5 = 25
\nside x side = 25
\nside2<\/sup> = 25
\nside = root 25 = 5 ft<\/p>\n

H.O.T. – Page No. 282<\/h3>\n

Focus on Higher Order Thinking<\/strong><\/p>\n

Question 12.
\nRepresent Real-World Problems
\nChristina 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.
\n\"Go
\n______ carrots<\/p>\n

Answer: 300 carrots<\/p>\n

Explanation:
\nThis shape is divided into two triangles and a square.
\nArea of figure = 2(1\/2 x 2 x 2) + 4(4) = 4 + 16 = 20 square units
\nNumber of carrots per square unit = 300
\nTotal number of carrots = 20 x 15 = 300<\/p>\n

Question 13.
\nAnalyze Relationships
\nThe 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.
\n\"Go
\n_____ in2<\/sup><\/p>\n

Answer: 192 in2<\/sup><\/p>\n

Explanation:
\nGiven:
\nThe perimeter of the figure = 56 inches
\nThe figure is divided into a square and a triangle.
\n10 + 10 + 3s = 56
\n3s = 36
\ns = 12
\nThe area of a triangle = 1\/2 x 12 x 8 = 48 in2<\/sup>
\nThe area of a square = 12 x 12 = 144 in2<\/sup>
\nTotal area of the figure = 144 + 48 = 192 in2<\/sup><\/p>\n

Question 14.
\nCritical Thinking
\nThe pattern for a scarf is shown at right. What is the area of the scarf? Use 3.14 for \u03c0.
\n\"Go
\n_____ in2<\/sup><\/p>\n

Answer: 243 in2<\/sup><\/p>\n

Explanation:
\nArea of the rectangle in the given figure = 28 x 15 = 420 in2<\/sup>
\nArea of two semicircles = 2 (1\/2 \u03c0r2<\/sup> ) = 3.14 x 7.5 x 7.5 = 176.625 in2<\/sup>
\nArea of the shaded region = 420 – 176.625 = 243 in2<\/sup>(approx)<\/p>\n

Question 15.
\nPersevere in Problem Solving
\nThe 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<\/sup>.
\n\"Go
\na. What is the area of the window? Use 3.14 for \u03c0.
\n_____ in2<\/sup><\/p>\n

Answer: a) 2228 in2<\/sup><\/p>\n

Explanation:
\nArea of the square = 100 in2<\/sup>
\nside x side = 100
\nSide = 10 in
\nSince the side of each square is 10 in and there are 4 squares.
\nThe side length of the larger square (s) = 40 in
\nArea of the larger square = side x side = 40 x 40 = 1600 in2<\/sup>
\nSince the side of each square is 10 in and there are 2 squares.
\nThe radius of the semi-circle = 20 in
\nArea of the semi-circle = 1\/2(\u03c0r2<\/sup>) = 1\/2(3.14 x 202<\/sup>) = 628 in2<\/sup>
\nThe area of the window = 1600 + 628 = 2228 in2<\/sup><\/p>\n

Question 15.
\nb. What is the area of the shade? Round your answer to the nearest whole number.
\n_____ in2<\/sup><\/p>\n

Answer: b) 3016 in2<\/sup><\/p>\n

Explanation:
\nThe shade extends 4 inches beyond the shapes so the length of the bottom rectangle is 40+4+4 = 48 in
\nThe length extends below the original square.
\nThe height is now = 40+4 = 44 in
\nThe radius of the semi-circle = 20+4 = 24 in
\nThe new area of the figure = 48(44) + 1\/2(3.14 x 242<\/sup>) = 2112 + 904.32 = 3016.32 = 3016 in2<\/sup><\/p>\n

Guided Practice – Page No. 286<\/h3>\n

Find the surface area of each solid figure.<\/strong><\/p>\n

Question 1.
\n\"Go
\nTotal surface area: _____ ft2<\/sup><\/p>\n

Answer: 150 ft2<\/sup><\/p>\n

Explanation:
\nThe 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
\nThe height of the prism = 7 ft
\nThe base is a triangle.
\nArea of the triangle = 1\/2 (8) (3) = 12 ft2<\/sup>
\nThe surface area formula for a prism is S = Ph + 2b
\nP = Perimeter = 18 h = height = 7 b = base = area of the triangle = 12
\nThe surface area of the prism = 18(7) + 2(12) = 126 + 24 = 150 ft2<\/sup><\/p>\n

Question 2.
\n\"Go
\nTotal surface area: _____ m2<\/sup><\/p>\n

Answer: 503 m2<\/sup><\/p>\n

Explanation:
\nGiven:
\nDimensions of the cuboid:
\nLength = 11 m
\nBreadth = 9 m
\nHeight = 7 m
\nThe surface area of the cuboid = 2(lb+bh+hl) = 2(11 x 9 + 9 x 7 + 7 x 11) = 478m2
\n<\/sup>
\nThe dimensions of the cube:
\nLength of the side = 2.5 m
\nThe surface area of the cube = 6a2<\/sup> = 6 x 2.5 x 2.5 = 37.5 m2<\/sup>
\nThe surface area of the rectangular prism = 2.5 x 2.5 = 6.25
\nThe surface area of the figure = The overlapping area is the area of the base of the cube
\n= 37.5 + 478 – 2(6.25) = 503 m2<\/sup><\/p>\n

Essential Question Check-In<\/strong><\/p>\n

Question 3.
\nHow can you find the surface area of a composite solid made up of prisms?
\nType below:
\n_____________<\/p>\n

Answer: The surface area of the prisms, add them up, and then subtract the overlapping areas twice.<\/p>\n

Explanation:
\nThe 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.<\/p>\n

Independent Practice – Page No. 287<\/h3>\n

Question 4.
\nCarla is wrapping a present in the box shown. How much wrapping paper does she need, not including overlap?
\n\"Go
\n_____ in2<\/sup><\/p>\n

Answer: 164 in2<\/sup><\/p>\n

Explanation:
\nThe surface area of the cuboid excluding the top = 2h(l+b) + lb = 2 x 4 ( 13 ) + 10 x 3 =\u00a0 164 in2<\/sup>
\nThe length of the wrapping paper = The surface area of the cuboid excluding the top = 164 in2<\/sup><\/p>\n

Question 5.
\nDmitri 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?
\n\"Go
\n_____ tiles<\/p>\n

Answer: 3720 tiles<\/p>\n

Explanation:
\nThe surface area of the cuboid excluding the bottom = 2h(l+b) + lb = 2 x 9 (35) + 20 x 15 = 930 cm2<\/sup>
\n5mm = 0.5 cm
\nArea of a tile = Area of the square = a2<\/sup> = 0.5cm x 0.5cm = 0.25 cm2<\/sup>
\nTotal number of tiles = 930\/0.25 = 3720 tiles<\/p>\n

Question 6.
\nShera is building a cabinet. She is making wooden braces for the corners of the cabinet. Find the surface area of each brace.
\n\"Go
\n_____ in2<\/sup><\/p>\n

Answer: 45 in2<\/sup><\/p>\n

Explanation:
\nThe perimeter of the figure = P = 3(3) + 2(1) = 11 in
\nBase = B = 3(2) = 6 in
\nHeight = h = 3
\nThe surface area of the figure = Ph + 2B = 11 x 3 +2(6) = 33 + 12 = 45 in2<\/sup><\/p>\n

Question 7.
\nThe doghouse shown has a floor, but no windows. Find the total surface area of the doghouse, including the door.
\n\"Go
\n_____ ft2<\/sup><\/p>\n

Answer:\u00a0 66ft2<\/sup><\/p>\n

Explanation:
\nPerimeter of the pentagon base (P) = 2(2.5) + 2(2) + 3 = 5 + 4 + 3 = 12
\nArea 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
\nHeight (h) = 2 + 2 = 4
\nThe surface area of the figure = Ph + 2B = 12(4) + 2(9) = 48 + 18 = 66ft2<\/sup><\/p>\n

Eddie built the ramp shown to train his puppy to do tricks. Use the figure for 8\u20139.<\/strong>
\n\"Go<\/p>\n

Question 8.
\nAnalyze Relationships
\nDescribe two ways to find the surface area of the ramp.
\nType below:
\n____________<\/p>\n

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.<\/p>\n

Explanation:
\nThe 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.<\/p>\n

Question 9.
\nWhat is the surface area of the ramp?
\n_____ in2<\/sup><\/p>\n

Answer: 3264 in2<\/sup><\/p>\n

Explanation:
\nP = Perimeter of the figure =\u00a0 16(3) + 2 (20) + 16 = 104
\nB = Base of the figure = 1\/2 (12) (16 + 3(16)) = 6 (16 + 48) = 6 (64) = 384
\nh = Height of the figure = 2
\nSurface area of the figure = Ph + 2B = 104(2) + 2(384) = 2496 + 768 = 3264 in2<\/sup><\/p>\n

Marco and Elaine are building a stand like the one shown to display trophies. Use the figure for 10\u201311.<\/strong>
\n\"Go<\/p>\n

Question 10.
\nWhat is the surface area of the stand?
\n_____ ft2<\/sup><\/p>\n

Answer:\u00a0 58 ft2<\/sup><\/p>\n

Explanation:
\nTop:
\nPerimeter = P = 4(1) = 4
\nBase = B = 1(1) = 1
\nHeight = h = 3
\nTop surface area = Ph + 2B = 4(3) + 2(1) = 14 ft2<\/sup>
\nBottom :
\nPerimeter = P = 2(7) + 2(1) = 14 + 2 = 16
\nBase = B = 7(1) = 7
\nHeight = h = 2
\nTop surface area = Ph + 2B = 16(2) + 2(7) = 46 ft2<\/sup>
\nOverlapping area = 1(1) = 1
\nThe 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<\/sup><\/p>\n

Question 11.
\nCritique Reasoning
\nMarco 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.
\n_____<\/p>\n

Answer: No<\/p>\n

Explanation:
\nSince the surface area is 58 ft2<\/sup>, they will need 3 cans of paint. Since each can paints 25 ft2<\/sup> and we cannot buy a fraction of cans.
\n3 cans would then cost 6.79 x 3 = 20.37 so this is not enough.<\/p>\n

Page No. 288<\/h3>\n

Question 12.
\nHenry wants to cover the box shown with paper without any overlap. How many square centimeters will be covered with paper?
\n\"Go
\n_____ cm2<\/sup><\/p>\n

Answer: 2316\u00a0cm2<\/sup><\/p>\n

Explanation:
\nGiven:
\nLength = 24cm\u00a0 Breadth = 27cm Height = 10cm
\nP = Perimeter = 2(24) + 2(27) = 48 + 54 = 102
\nB = Base = 24(27) = 648
\nh = Height = 10
\nSurface area of the figure = Ph + 2B = 102(10) + 2(648) = 1020 + 1296 = 2316\u00a0cm2<\/sup><\/p>\n

Question 13.
\nWhat If?
\nSuppose the length and width of the box in Exercise 12 double. Does the surface area S double? Explain.
\n_____<\/p>\n

Answer: No<\/p>\n

Explanation:
\nGiven :
\nLength = 24cm x 2 = 48 cm\u00a0 Breadth = 27cm x 2 = 54 cm Height = 10cm
\nP = 2(48) + 2(54) = 96 + 108 = 204
\nB = 48(54) = 2592
\nNew Surface area = Ph + 2B = 204(10) + 2(2592) = 2040 + 5184 = 7224 cm2<\/sup>
\nDouble of surface area = 2 (2316) = 4632 cm2<\/sup>
\nSo the new surface area is not double of the initial area.<\/p>\n

H.O.T.<\/strong><\/p>\n

Focus on Higher Order Thinking<\/strong><\/p>\n

Question 14.
\nPersevere in Problem Solving
\nEnya 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.
\nType below:
\n______________<\/p>\n

Answer: The difference would just be the area in the bottom surface. It would be 6.25 ft2<\/sup> less.<\/p>\n

Explanation:
\nThe 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<\/sup> = 6.25.
\nTherefore she would paint 6.25 ft2<\/sup> less if she painted all but the bottom surface compared to painting the entire prism.<\/p>\n

Question 15.
\nInterpret the Answer
\nThe 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
\n\"Go
\n_____ ft2<\/sup><\/p>\n

Answer: 138.28 in2<\/sup> , 1440 ft2<\/sup><\/p>\n

Explanation:
\nGiven:
\nDimensions of the cylinder:
\nRadius: 2 in
\nHeight: 9 in
\nThe total surface area of the cylinder = 2\u03c0r(r+h) = 2 x 22\/7 x 2 (2 + 9) = 138.28 in2<\/sup><\/p>\n

The total number of square inches needed for 1,500 oatmeal boxes = 1,500 x 138.28 = 207,300 in2<\/sup>
\n1 ft = 12 in
\n(1 ft)2<\/sup> = (12 in)2<\/sup>
\n1 ft2<\/sup> = 144 in2<\/sup>
\nThe total number of square feet needed for 1,500 oatmeal boxes (to the nearest whole number)
\n= 207,300\/144 = 1440 ft2<\/sup><\/p>\n

Question 16.
\nAnalyze Relationships
\nA 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.
\n\"Go
\nType below:
\n______________<\/p>\n

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.<\/p>\n

Explanation:
\nThe 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.<\/p>\n

Guided Practice – Solving Volume Problems – Page No. 292<\/h3>\n

Question 1.
\nFind the volume of the triangular prism.
\n\"Go
\n_____ ft3<\/sup><\/p>\n

Answer: 84 ft3<\/sup><\/p>\n

Explanation:
\nBase area of the prism = 1\/2 x 8 x 3 = 12 ft2<\/sup>
\nHeight of the prism = 7 ft
\nVolume of the prism = (12 x 7) ft3<\/sup><\/p>\n

Question 2.
\nFind the volume of the trapezoidal prism.
\n\"Go
\n_____ m3<\/sup><\/p>\n

Answer: 330 m3<\/sup><\/p>\n

Explanation:
\nBase area of the prism = 1\/2 x (15 + 5) x 3 = 30 m2<\/sup>
\nHeight of the prism = 11 m
\nVolume of the prism = (30 x 11) m3<\/sup> = 330 m3<\/sup><\/p>\n

Question 3.
\nFind the volume of the composite figure.
\n\"Go
\n_____ ft2<\/sup><\/p>\n

Answer: Composite figure: 360 ft3<\/sup><\/p>\n

Explanation:
\nThe volume of the triangular prism:
\nThe base area of the prism = 1\/2 x 4 x 6 = 12 ft2<\/sup>
\nHeight = 6 ft
\nThe volume of the triangular prism = 12 x 6 = 72 ft3<\/sup><\/p>\n

The volume of the rectangular prism:
\nThe base area of the prism = 4 x 6 = 24 ft2<\/sup>
\nHeight = 12 ft
\nThe volume of the triangular prism = 12 x 24 = 288 ft3<\/sup><\/p>\n

Volume of the composite figure = (288 + 72)ft3<\/sup> = 360 ft3<\/sup><\/p>\n

Find the volume of each figure.<\/strong><\/p>\n

Question 4.
\nThe figure shows a barn that Mr. Fowler is building for his farm.
\n\"Go
\n_____ ft3<\/sup><\/p>\n

Answer: 40,000 ft3<\/sup><\/p>\n

Explanation:
\nTriangular prism:
\nB = Base area = 1\/2 x 10 (40) = 200 cm2<\/sup>
\nHeight = 50 cm
\nThe volume of the triangular prism = Bh = 200 x 50 = 10,000 cm3<\/sup>
\nRectangular prism:
\nB = Base area =40 x 15 = 600 cm2<\/sup>
\nHeight = 50 cm
\nThe volume of the triangular prism = Bh = 600 x 50 = 30,000 cm3<\/sup>
\nTotal volume of the prism = 10,000 + 30,000 = 40,000 cm3<\/sup><\/p>\n

Question 5.
\nThe figure shows a container, in the shape of a trapezoidal prism, that Pete filled with sand.
\n\"Go
\n_____ cm3<\/sup><\/p>\n

Answer: 385 cm3<\/sup><\/p>\n

Explanation:
\nB = Base area = 1\/2 x 5 (10 + 12) = 55 cm2<\/sup>
\nHeight = 7 cm
\nThe volume of the container = Bh = 55 x 55 = 385 cm3<\/sup><\/p>\n

Essential Question Check-In<\/strong><\/p>\n

Question 6.
\nHow do you find the volume of a composite solid formed by two or more prisms?
\nType below:
\n______________<\/p>\n

Answer: Finding the volume of each figure adding them up to get the volume of the composite solid.<\/p>\n

Explanation:
\nTo 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.<\/p>\n

Independent Practice – Page No. 293<\/h3>\n

Question 7.
\nA trap for insects is in the shape of a triangular prism. The area of the base is 3.5 in2<\/sup> and the height of the prism is 5 in. What is the volume of this trap?
\n_____ in3<\/sup><\/p>\n

Answer: 17.5 in3<\/sup><\/p>\n

Explanation:
\nThe volume of the trap = Base area x height = 3.5 x 5 = 17.5 in3<\/sup><\/p>\n

Question 8.
\nArletta built a cardboard ramp for her little brothers\u2019 toy cars. Identify the shape of the ramp. Then find its volume.
\n\"Go
\nShape: _________
\nArea: _________ in3<\/sup><\/p>\n

Answer: 525 in3<\/sup><\/p>\n

Explanation:
\nBase area = 1\/2 x 6 x 25 = 75 in2<\/sup>
\nHeight\u00a0 = 7 in
\nVolume of the figure = 75 x 7 = 525 in3<\/sup><\/p>\n

Question 9.
\nAlex 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?
\n\"Go
\n_____ ft3<\/sup><\/p>\n

Answer: 384 ft3<\/sup><\/p>\n

Explanation:
\nBase area = 1\/2 x 4 x 8 = 16 ft2<\/sup>
\nHeight\u00a0 = 24 ft
\nVolume of the figure = 16 x 24 = 384 ft3<\/sup><\/p>\n

Question 10.
\nA 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?
\n_____ in3<\/sup><\/p>\n

Answer: 192 in3<\/sup><\/p>\n

Explanation:
\nBase area = 1\/2 x 4 x (7+5) = 24 in2<\/sup>
\nHeight\u00a0 = 8 in
\nVolume of the figure = 24 x 8 = 192 Base area = 1\/2 x 6 x 25 = 75 in2<\/sup>
\nHeight\u00a0 = 7 in
\nVolume of the figure = 75 x 7 = 525 in3<\/sup><\/p>\n

Question 11.
\nExplain the Error
\nA student wrote this statement: \u201cA 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.\u201d Identify and correct the error.
\nType below:
\n____________<\/p>\n

Answer: The error is measurement unit.<\/p>\n

Explanation:
\nThe volume of the prism is:
\nbase area x height = 20 x 15 = 300 in3<\/sup><\/p>\n

Find the volume of each figure. Round to the nearest hundredth if necessary.<\/strong><\/p>\n

Question 12.
\n\"Go
\n_____ in3<\/sup><\/p>\n

Answer: 97.2 in3<\/sup><\/p>\n

Explanation:
\nThe volume of the hexagonal prism = 23.4 x\u00a0 3 = 70.2 in3<\/sup><\/p>\n

Base area of the rectangular prism = 3 x 3 = 9 in2<\/sup>
\nThe volume of the rectangular prism = Bh = 9 x 3 = 27 in3<\/sup><\/p>\n

Total volume of the figure = 70.2 + 27 = 97.2 in3<\/sup><\/p>\n

Question 13.
\n\"Go
\n_____ m3<\/sup><\/p>\n

Answer: 316.41 m3<\/sup><\/p>\n

Explanation:
\nThe volume of the rectangular prism on the left = Bh = [7.5 x 3.75] (3.75) = 105.47 m3<\/sup>
\nThe volume of the rectangular prism on the right = Bh = [7.5 x 3.75](7.5) = 210.94 m3<\/sup>
\nTotal volume of the composite figure = 105.47 + 210.94 = 316.41 m3<\/sup><\/p>\n

Question 14.
\nMulti-Step
\nJosie 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.
\n_____<\/p>\n

Answer: No<\/p>\n

Explanation:
\nThe volume of the hexagonal prism = 21 x 8 = 168
\nThe total volume of wax, 260 is equal to the sum of the volumes of each prism.
\nB is the base area of the triangular prism.
\n168 + 14B = 260 cm3<\/sup>
\n14B = 260 – 168
\nB = 6.6 cm3<\/sup><\/p>\n

Page No. 294<\/h3>\n

Question 15.
\nA 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.
\n\"Go
\nType below:
\n___________<\/p>\n

Answer: The triangular prism is a better deal since it has a larger volume<\/p>\n

Explanation:
\nThe base area of the trapezoidal prism = 36 in2<\/sup>
\nThe volume of the trapezoidal prism = Bh = 36 x 5 = 175 in3<\/sup><\/p>\n

The base area of the triangular prism = 32 in2<\/sup>
\nThe volume of the rectangular prism = Bh = 32 x 6 = 192 in3<\/sup><\/p>\n

The triangular prism is a better deal since it has a larger volume.<\/p>\n

H.O.T.<\/strong><\/p>\n

Focus on Higher Order Thinking<\/strong><\/p>\n

Question 16.
\nCritical Thinking
\nThe wading pool shown is a trapezoidal prism with a total volume of 286 cubic feet. What is the missing dimension?
\n\"Go
\n______ ft.<\/p>\n

Answer: 3.5 ft<\/p>\n

Explanation:
\nArea of the trapezoidal prism = B = 1\/2 x 13 (2+x)
\nVolume of the figure = 286 cubic feet
\nV = Bh
\n286 = 1\/2 x 13 (2+x)(8)
\n5.5 = (2+x)
\nx = 3.5 ft<\/p>\n

Question 17.
\nPersevere in Problem Solving
\nLynette 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.
\n\"Go
\n______ grams<\/p>\n

Answer: 75 cubic centimeter, 645 grams<\/p>\n

Explanation:
\nV = Bh
\nB = Area of the triangle of base = 10 cm , height = 6 cm = 1\/2 x 10 x 6 = 30 square centimeter
\nV = 30 x 2.5 = 75 cubic centimeter<\/p>\n

1 cubic centimeter = 8.6 grams in mass
\nV = 75 cubic centimeter x 8.6 = 645 grams<\/p>\n

Question 18.
\nAnalyze Relationships
\nWhat effect would tripling all the dimensions of a triangular prism have on the volume of the prism? Explain your reasoning.
\nType below:
\n____________<\/p>\n

Answer: The volume is 27 times the original volume.<\/p>\n

Explanation:
\nThe area of the base = B = 1\/2 (3b) (3h) = 9\/2 (bh)
\nH is the height of the prism
\nThe volume would be = 9\/2 (bh) x (3H) = 27 [ 1\/2 (bhH) ]<\/p>\n

Therefore, The volume is 27 times the original volume.<\/p>\n

Question 19.
\nPersevere in Problem Solving
\nEach of two trapezoidal prisms has a volume of 120 cubic centimetres. The prisms have no dimensions in common. Give possible dimensions for each prism.
\nType below:
\n____________<\/p>\n

Answer: A possible combination of dimension could be the height at 8 cm, base at 2 cm and 3 cm<\/p>\n

Explanation:
\nThe 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.
\nIf the base area is 20, the height of the trapezoid and the length of the bases could be 8,2 and 3 respectively.<\/p>\n

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.
\nIf the base area is 30, the height of the trapezoid and the length of the bases could be 10,1 and 5 respectively.<\/p>\n

9.1, 9.2 Circumference and Area of Circles – Page No. 295<\/h3>\n

Find the circumference and area of each circle. Use 3.14 for \u03c0. Round to the nearest hundredth if necessary.<\/strong><\/p>\n

Question 1.
\n\"Go
\nC = _________ m
\nA = _________ m2<\/sup><\/p>\n

Answer:
\nC = 43.96 m
\nA = 153.86 m2<\/sup><\/p>\n

Explanation:
\nC = 2 \u03c0r = 2 \u03c0(7) = 14 (3.14) = 43.96 m
\nA = \u03c0r2<\/sup> = 3.14 (7)2<\/sup> = 153.86 m2<\/sup><\/p>\n

Question 2.
\n\"Go
\nC = _________ ft
\nA = _________ ft2<\/sup><\/p>\n

Answer:
\nC = 37.68 ft
\nA = 113.04 ft2<\/sup><\/p>\n

Explanation:
\nDiameter = 12 ft
\nRadius = d\/2 = 12\/2 = 6 ft
\nC = 2 \u03c0r = 2 \u03c0(6) = 6 (3.14) = 37.68 ft
\nA = \u03c0r2<\/sup> = 3.14 (6)2<\/sup> = 113.04 ft2<\/sup><\/p>\n

9.3 Area of Composite Figures<\/strong><\/p>\n

Find the area of each figure. Use 3.14 for \u03c0.<\/strong><\/p>\n

Question 3.
\n\"Go
\n______ m2<\/sup><\/p>\n

Answer: 180.48 m2<\/sup><\/p>\n

Explanation:
\nArea of the triangle = 1\/2 x 16 x 10 = 80 m2<\/sup>
\nArea of the semicircle = 1\/2\u00a0\u03c0r2<\/sup> = 1\/2 (3.14) (8)2<\/sup> = 100.48 m2<\/sup>
\nThe total area of the figure = 80 + 100.48 = 180.48 m2<\/sup><\/p>\n

Question 4.
\n\"Go
\n______ cm2<\/sup><\/p>\n

Answer: 200 cm2<\/sup><\/p>\n

Explanation:
\nArea of the parallelogram = 4.5(20) = 90 cm2<\/sup>
\nArea of the rectangle = 20(5.5) = 110 cm2<\/sup>
\nThe total area of the figure = 90 + 110 = 200 cm2<\/sup><\/p>\n

9.4, 9.5 Solving Surface Area and Volume Problems<\/strong><\/p>\n

Find the surface area and volume of each figure.<\/strong><\/p>\n

Question 5.
\n\"Go
\nS = _________ cm2<\/sup>
\nV = _________ cm3<\/sup><\/p>\n

Answer:
\nS = 132 cm2<\/sup>
\nV = 60 cm3<\/sup><\/p>\n

Explanation:
\nPerimeter = 3+4+5 = 12 cm
\nBase area = Area of the triangle = 1\/2 x 3 x 4 = 6
\nS = Ph + 2B = 12(10) + 2(6) = 120 +12 = 132 cm2<\/sup><\/p>\n

V = Bh = 6 x 10 = 60 cm3<\/sup><\/p>\n

Question 6.
\n\"Go
\nS = _________ yd2<\/sup>
\nV = _________ yd3<\/sup><\/p>\n

Answer:
\nS = 54.5 yd2<\/sup>
\nV = 27.5 yd3<\/sup><\/p>\n

Explanation:
\nPerimeter = 2(2.5) + 2(2) + 4 = 13 cm
\nBase area = Area of the triangle + Area of the rectangle = 1\/2 x 1.5 x 4 + 4(2)= 11
\nS = Ph + 2B = 13(2.5) + 2(11) = 32.5 +22 = 54.5 yd2<\/sup><\/p>\n

V = Bh = 11 x 2.5 = 27.5 yd3<\/sup><\/p>\n

Essential Question<\/strong><\/p>\n

Question 7.
\nHow can you use geometry figures to solve real-world problems?
\nType below:
\n______________<\/p>\n

Answer: We can solve real-world problems by finding surface area and volume.
\nExample: We can find the amount of liquid in a tank by calculating its volume.<\/p>\n

Explanation:
\nReal-world problems by finding surface area and volume.
\nExample1: We can find the amount of liquid in a tank by calculating its volume.
\nExample2: We can find the surface area of the house and find the amount of paint required to paint the house.<\/p>\n

Page No. 296<\/h3>\n

Question 1.
\nWhat is the circumference of the circle?
\n\"Go
\na. 34.54 m
\nb. 69.08 m
\nc. 379.94 m
\nd. 1519.76 m<\/p>\n

Answer: b. 69.08 m<\/p>\n

Explanation:
\nCircumference = 2 \u03c0r = 2 \u03c0(11) = 22 (3.14) = 69.08 m<\/p>\n

Question 2.
\nWhat is the area of the circle?
\n\"Go
\nOptions:
\na. 23.55 m2<\/sup>
\nb. 47.1 m2<\/sup>
\nc. 176.625 m2<\/sup>
\nd. 706.5 m2<\/sup><\/p>\n

Answer: c. 176.625 m2<\/sup><\/p>\n

Explanation:
\nDiameter = 15 m
\nRadius = 7.5 m
\nArea of the circle = \u03c0r2<\/sup> = 3.14 (7.5)2<\/sup> = 176.625 m2<\/sup><\/p>\n

Question 3.
\nWhat is the area of the figure?
\n\"Go
\nOptions:
\na. 28.26 m2<\/sup>
\nb. 36 m2<\/sup>
\nc. 64.26 m2<\/sup>
\nd. 92.52 m2<\/sup><\/p>\n

Answer: c. 64.26 m2<\/sup><\/p>\n

Explanation:
\nArea of the square = 6 x 6 = 36 m2<\/sup>
\nRadius = 6 m
\nArea of the quarter circle = 1\/4 \u03c0r2<\/sup> = 1\/4 x 3.14 (6)2<\/sup> = 28.26 m2<\/sup>
\nThe total area of the figure = 36 + 28.26 = 64.26 m2<\/sup><\/p>\n

Question 4.
\nA 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?
\nOptions:
\na. 12x + 150 = 480
\nb. \\(\\frac{x}{12}\\) + 150 = 480
\nc. 12x + 480 = 150
\nd. \\(\\frac{x}{12}\\) + 480 = 150<\/p>\n

Answer: a. 12x + 150 = 480<\/p>\n

Explanation:
\nIf x is the monthly fee, then 12x is the total monthly fees.
\nThe joining fee = $150
\nTotal cost = $480
\nthen,
\n12x + 150 = 480<\/p>\n

Question 5.
\nWhat is the volume of the prism?
\n\"Go
\nOptions:
\na. 192 ft3<\/sup>
\nb. 48 ft3<\/sup>
\nc. 69 ft3<\/sup>
\nd. 96 ft3<\/sup><\/p>\n

Answer: d. 96 ft3<\/sup><\/p>\n

Explanation:
\nB = Base area of the triangle = 1\/2 x 8 x 2 = 8 ft2<\/sup>
\nHeight = 12 ft
\nVolume of the triangular orism = Bh = 8(12) = 96 ft3<\/sup><\/p>\n

Question 6.
\nA 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?
\nOptions:
\na. 16 pounds
\nb. 36 pounds
\nc. 48 pounds
\nd. 120 pounds
\ne. 120 pounds<\/p>\n

Answer: a. 16 pounds<\/p>\n

Explanation:
\n2\/3 is equal to x\/24 then 3 times 8 is equal to 24 and if 2 times 8 is equal to 16.<\/p>\n

Question 7.
\nFind the percent change from $20 to $25.
\nOptions:
\na. 25% decrease
\nb. 25% increase
\nc. 20% decrease
\nd. 20% increase<\/p>\n

Answer: b. 25% increase<\/p>\n

Explanation:
\n25 – 20 = 5 divide by 20 = 1\/4
\nWhen we find the percentage we get 25.
\nSo we can say that there is an increase in 25%<\/p>\n

Question 8.
\nEach dimension of the smaller prism is half the corresponding dimension of the larger prism.
\n\"Go
\na. What is the surface area of the figure?
\n_____ in2<\/sup><\/p>\n

Answer: 856 in2<\/sup><\/p>\n

Explanation:
\nHeight of the top prism = 10\/2 = 5
\nLength of the top prism = 16\/2 = 8
\nWidth of the top prism = 8\/2 = 4
\nPerimeter = 2l + 2w = 2(8) + 2(4) = 16 + 8 = 24 in
\nB = lw = 8(4) = 32 in
\nSurface area of top prism= Ph + 2B = 24(5) + 2(32) = 184 in2<\/sup><\/p>\n

Height of the prism = 10
\nLength of the prism = 16
\nWidth of the prism = 8
\nPerimeter = 2l + 2w = 2(16) + 2(8) = 32 + 16 = 48 in
\nB = lw = 16(8) = 128 in
\nSurface area of bottom prism= Ph + 2B = 48(10) + 2(128) = 736 in2<\/sup><\/p>\n

Area of overlapping region = 32 in2<\/sup><\/p>\n

The total surface area of the prism
\n= Surface area of top prism + Surface area of bottom prism – 2[Area of overlapping region ]
\n= 184 + 736 – 2(32) = 856 in2<\/sup><\/p>\n

Question 8.
\nb. What is the volume of the figure?
\n_____ in3<\/sup><\/p>\n

Answer: 1440 in3<\/sup><\/p>\n

Explanation:
\nVolume of top prism = Bh = 32(5) = 160 in3<\/sup>
\nVolume of bottom prism = Bh = 128(10) = 1280 in3<\/sup>
\nThe total volume of the figure = 160 + 1280 = 1440 in3<\/sup><\/p>\n

EXERCISES – Page No. 298<\/h3>\n

Question 1.
\nIn the scale drawing of a park, the scale is 1 cm: 10 m. Find the area of the actual park.
\n\"Go
\n_____ m2<\/sup><\/p>\n

Answer: 450 m2<\/sup><\/p>\n

Explanation:
\nMultiply the dimensions of the scale drawing by 10 since 1 cm = 10 m
\n3cm by 1.5 cm = 30m by 15 m
\nArea = 30(15) = 450 m2<\/sup><\/p>\n

Question 2.
\nFind the value of y and the measure of \u2220YPS.
\n\"Go
\ny = __________ \u00b0
\nmYPS = __________ \u00b0<\/p>\n

Answer: y = 8
\nmYPS = 40 \u00b0<\/p>\n

Explanation:
\n140 + 5y = 180 [sum of angle on a line = 180\u00b0]
\n5y = 40
\ny = 8<\/p>\n

mYPS = mRPZ = 5y [vertically opposite angles]
\nmYPS = 5(8) = 40\u00b0<\/p>\n

Question 3.
\nKanye 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.
\n\"Go
\n_____<\/p>\n

Answer: No<\/p>\n

Explanation:
\nA 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.
\nThe sum of the first 2 sides = 3+4 = 7 < 8
\nTherefore, he cannot form a triangle unless he cuts the logs.<\/p>\n

Question 4.
\nIn 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?
\n\"Go
\n_____ in2<\/sup><\/p>\n

Answer: 12 in2<\/sup><\/p>\n

Explanation:
\nBase = 4 in
\nHeight = 6 in
\nArea of the triangle = 1\/2 x 6 x 4 = 12 in2<\/sup><\/p>\n

Page No. 300<\/h3>\n

Find the circumference and area of each circle. Round to the nearest hundredth.<\/strong><\/p>\n

Question 1.
\n\"Go
\nC = __________ in
\nA = __________ in2<\/sup><\/p>\n

Answer:
\nC = 69.08 in
\nA = 379.94 in2<\/sup><\/p>\n

Explanation:
\nDiameter = 22 in
\nRadius = d\/2 = 22\/2 = 11 in
\nC = 2 \u03c0r = 2 \u03c0(11) = 22 (3.14) = 69.08 in
\nA = \u03c0r2<\/sup> = 3.14 (11)2<\/sup> = 379.94 in2<\/sup><\/p>\n

Question 2.
\n\"Go
\nC = __________ m
\nA = __________ m2<\/sup><\/p>\n

Answer:
\nC = 28.26 m
\nA = 63.59m2<\/sup><\/p>\n

Explanation:
\nRadius = 4.5 m
\nC = 2 \u03c0r = 2 \u03c0(4.5) = 9 (3.14) = 28.26 m
\nA = \u03c0r2<\/sup> = 3.14 (4.5)2<\/sup> = 63.59 m2<\/sup><\/p>\n

Find the area of each composite figure. Round to the nearest hundredth if necessary.<\/strong><\/p>\n

Question 3.
\n\"Go
\n______ in2<\/sup><\/p>\n

Answer: 99 in2<\/sup><\/p>\n

Explanation:
\nArea of the square = 9 x 9 = 81 in2<\/sup>
\nBase of the triangle = 13 – 9 = 4 in
\nArea of the triangle = 1\/2 x 4 x 9 = 18 in2<\/sup>
\nThe total area of the figure = 81 + 18 = 99 in2<\/sup><\/p>\n

Question 4.
\n\"Go
\n______ cm2<\/sup><\/p>\n

Answer: 420.48 cm2<\/sup><\/p>\n

Explanation:
\nArea of the rectangle = 16 x 20 = 320 cm2<\/sup>
\nDiameter = 16 cm
\nRadius = 16\/2 = 8 cm
\nArea of the semi circle = 1\/2 \u03c0r2<\/sup> = 1\/2 x 3.14 (8)2<\/sup> = 100.48 cm2<\/sup>
\nThe total area of the figure = 320 + 100.48 = 420.48 cm2<\/sup><\/p>\n

Find the volume of each figure.<\/strong><\/p>\n

Question 5.
\n\"Go
\n______ in3<\/sup><\/p>\n

Answer: 420 in3<\/sup><\/p>\n

Explanation:
\nB = 7(5) = 35 in2<\/sup>
\nV = Bh = 35 x 12 = 420 in3<\/sup><\/p>\n

Question 6.
\nThe 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.
\n______ in<\/p>\n

Answer: 5.5 ft<\/p>\n

Explanation:
\nV = Bh
\n264 = 48h
\nh = 264\/48 = 5.5ft<\/p>\n

Page No. 301<\/h3>\n

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.<\/strong>
\n\"Go<\/p>\n

Question 7.
\nWhat is the volume of the paperweight?
\n______ cm3<\/sup><\/p>\n

Answer: 576 cm3<\/sup><\/p>\n

Explanation:
\nPyramid:
\nB = 8 x 8 = 64 cm2<\/sup>
\nV = 1\/3 Bh = 1\/3 x 64 x 3 = 64 cm3<\/sup>
\nPrism:
\nB = 8 x 8 = 64 cm2<\/sup>
\nV = Bh = 64 x 8 = 512 cm3<\/sup><\/p>\n

The total volume of the figure = 64 + 512 = 576 cm3<\/sup><\/p>\n

Question 8.
\nWhat is the total surface area of the paperweight?
\n______ cm2<\/sup><\/p>\n

Answer: 400 cm2<\/sup><\/p>\n

Explanation:
\nPyramid:
\nP = 4(8) = 32 cm
\nS = 1\/2 Pl + B = 80 + 64 = 144 cm2<\/sup><\/p>\n

Prism:
\nP = 4(8) = 32 cm
\nS = Ph + 2B = 32(8) + 2(64) = 384 cm2<\/sup>
\nThe total surface area of the prism
\n= Area of the prism + Area of the pyramid – 2[Area of the overlapping region]
\n= 144 + 384 – 2(64) = 400<\/p>\n

Unit 4 Performance Tasks<\/strong><\/p>\n

Question 9.
\nProduct Design Engineer
\nMiranda 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.
\n\"Go
\na. How many square feet of material does Miranda need to make the tent (including the floor)? Show your work.
\n______ ft2<\/sup><\/p>\n

Answer: 261 3\/4\u00a0 ft2<\/sup><\/p>\n

Explanation:
\nP = 2 x 7 1\/2 + 8 = 22 1\/2
\nB = 4\/2 (8) (6) = 24
\nS = Ph + 2B = 22 1\/2 x 9 1\/2 + 2(24) = 213 3\/4 + 48 = 261 3\/4 ft2<\/sup><\/p>\n

Question 9.
\nb. What is the volume of the tent? Show your work.
\n______ ft3<\/sup><\/p>\n

Answer: 228 ft3<\/sup><\/p>\n

Explanation:
\nV = Bh = 24 x 9 1\/2 = 228 ft3<\/sup><\/p>\n

Question 9.
\nc. 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
\nType below:
\n____________<\/p>\n

Answer: Increase the height to 10.45 ft<\/p>\n

Explanation:
\nNew volume = 1.10 x 228 = 250.8
\n250.8 = 24h
\nh = 10.45 ft<\/p>\n

Unit 4 Performance Tasks (cont’d) – Page No. 302<\/h3>\n

Question 10.
\nLi 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.
\na. What is the volume of the stand? Write your answer in cubic centimeters.
\n______ cm3<\/sup><\/p>\n

Answer: 135,000 cm3<\/sup><\/p>\n

Explanation:
\nB = 45 x 25 = 1125 cm2<\/sup>
\nV = Bh = 1125 x 120 = 135,000 cm3<\/sup><\/p>\n

Question 10.
\nb. 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.
\n______ cm3<\/sup><\/p>\n

Answer: 116,702 cm3<\/sup><\/p>\n

Explanation:
\nWidth = 45 – 2(1) = 43 ft
\nLength = 25 – 2(1) =23ft
\nHeight = 120-2(1) = 118ft
\nB = 43 x 23 = 989 ft2<\/sup>
\nV = Bh = 989 x 118 = 116,702 ft3<\/sup><\/p>\n

Selected Response – Page No. 303<\/h3>\n

Question 1.
\nA school flag is in the shape of a rectangle with a triangle removed as shown.
\n\"Go
\nWhat is the measure of angle x?
\nOptions:
\na. 50\u00b0
\nb. 80\u00b0
\nc. 90\u00b0
\nd. 100\u00b0<\/p>\n

Answer: d. 100\u00b0<\/p>\n

Explanation:
\nx = 50 + 50 = 100\u00b0 [ Sum of two angles created by the 2 lines]<\/p>\n

Question 2.
\nOn a map with a scale of 2 cm = 1 km, the distance from Beau\u2019s house to the beach is 4.6 centimetres. What is the actual distance?
\nOptions:
\na. 2.3 km
\nb. 4.6 km
\nc. 6.5 km
\nd. 9.2 km<\/p>\n

Answer: a. 2.3 km<\/p>\n

Explanation:
\n2\/1 = 4.6\/x
\nx = 4.6\/2 = 2.3 km<\/p>\n

Question 3.
\nLalasa 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?
\nOptions:
\na. 17.5 ft2<\/sup>
\nb. 52.5 ft2<\/sup>
\nc. 39.375 ft2<\/sup>
\nd. 78.75 ft2<\/sup><\/p>\n

Answer: c. 39.375 ft2<\/sup><\/p>\n

Explanation:
\n1in = 1.5ft
\nThe base of the triangle = 5 in = 1.5(5) ft = 7.5 ft
\nHeight = 7 in = 7(1.5) ft = 10.5 ft
\nArea of the triangle = 1\/2 x 7.5 x 10.5 = 39.375 ft2<\/sup><\/p>\n

Question 4.
\nSonya has four straws of different lengths: 2 cm, 8 cm, 14 cm, and 16 cm. How many triangles can she make using the straws?
\nOptions:
\na. no triangle
\nb. one triangle
\nc. two triangles
\nd. more than two triangles<\/p>\n

Answer: b. one triangle<\/p>\n

Explanation:
\nThe third side of a triangle must be smaller than the sum of the other two sides to form a triangle.
\n2+8 = 10<14
\n2+8 = 10<16
\n8+14 = 22>14
\n8+14 = 22>16
\n2+14 = 16=16
\n2+16 = 18>16<\/p>\n

Therefore, only one triangle can be formed using the sides 8, 14, 16.<\/p>\n

Question 5.
\nA 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?
\nOptions:
\na. 15x + 1.25 = 20
\nb. 1.25x + 15 = 20
\nc. 15x \u2212 1.25 = 20
\nd. 1.25x \u2212 15 = 20<\/p>\n

Answer: b. 1.25x + 15 = 20<\/p>\n

Explanation:
\nIf x is the number of additional toppings, then 1.25 x is the cost of the additional toppings.
\nThis gives the total cost is 1.25x + 15
\nthen,
\n1.25x + 15 = 20<\/p>\n

Question 6.
\nA 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?
\nOptions:
\na. $15680
\nb. $17360
\nc. $19040
\nd. $20720<\/p>\n

Answer: c. $19040<\/p>\n

Explanation:
\nSimple interest = 14,000 x 0.12 x 2 = $5,040
\nAmount = $14,000 + $5,040 = $19040<\/p>\n

Question 7.
\nWhat is the volume of a triangular prism that is 75 centimeters long and that has a base with an area of 30 square centimeters?
\nOptions:
\na. 2.5 cm3<\/sup>
\nb. 750 cm3<\/sup>
\nc. 1125 cm3<\/sup>
\nd. 2250 cm3<\/sup><\/p>\n

Answer: d. 2250 cm3<\/sup><\/p>\n

Explanation:
\nV = Bh = 30(75) = 2250cm3<\/sup><\/p>\n

Question 8.
\nConsider the right circular cone shown.
\n\"Go
\nIf a vertical plane slices through the cone to create two identical half cones, what is the shape of the cross section?
\nOptions:
\na. a rectangle
\nb. a square
\nc. a triangle
\nd. a circle<\/p>\n

Answer: c. a triangle<\/p>\n

Explanation:
\nSlicing through the vertex to create 2 identical half cones would create a cross-section that\u00a0 is a triangle.<\/p>\n

Page No. 304<\/h3>\n

Question 9.
\nThe radius of the circle is given in meters. What is the circumference of the circle? Use 3.14 for \u03c0.
\n\"Go
\na. 25.12 m
\nb. 50.24 m
\nc. 200.96 m
\nd. 803.84 m<\/p>\n

Answer: b. 50.24 m<\/p>\n

Explanation:
\nCircumference = 2 \u03c0r = 2 \u03c0(8) = 16 (3.14) = 50.24 m<\/p>\n

Question 10.
\nThe dimensions of the figure are given in millimeters. What is the area of the two-dimensional figure?
\n\"Go
\nOptions:
\na. 39 mm2<\/sup>
\nb. 169 mm2<\/sup>
\nc. 208 mm2<\/sup>
\nd. 247 mm2<\/sup><\/p>\n

Answer: c. 208 mm2<\/sup><\/p>\n

Explanation:
\nArea of the square = 13 x 13 = 169 mm2<\/sup>
\nArea of the triangle = 1\/2 x 13 x 6 = 39 mm2<\/sup>
\nThe total area of the figure = 169 + 39 = 208 mm2<\/sup><\/p>\n

Question 11.
\nA 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\u2019s radius to the nearest tenth of a foot?
\nOptions:
\na. 1.4 ft
\nb. 2.7 ft
\nc. 4.3 ft
\nd. 17.2 ft<\/p>\n

Answer: a. 1.4 ft<\/p>\n

Explanation:
\nCircumference = 2 \u03c0r = 8.6 ft
\nr = 8.6\/2 \u03c0 = 1.4 ft<\/p>\n

Question 12.
\nWhat is the measure in degrees of an angle that is supplementary to a 74\u00b0 angle?
\nOptions:
\na. 16\u00b0
\nb. 74\u00b0
\nc. 90\u00b0
\nd. 106\u00b0<\/p>\n

Answer: d. 106\u00b0<\/p>\n

Explanation:
\nSum of supplementary angles = 180\u00b0
\nx + 74\u00b0 = 180\u00b0
\nx = 106\u00b0<\/p>\n

Question 13.
\nWhat 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?
\nOptions:
\na. 19.7 cm3<\/sup>
\nb. 108.5 cm3<\/sup>
\nc. 217 cm3<\/sup>
\nd. 237.4 cm3<\/sup><\/p>\n

Answer: c. 217 cm3<\/sup><\/p>\n

Explanation:
\nV = Bh
\nB = 6.2 x 3.5 = 21.7 cm2<\/sup>
\nh = 10 cm
\nV = 21.7 x 10 = 217 cm3<\/sup><\/p>\n

Question 14.
\nA patio is the shape of a circle with diameter shown.
\n\"Go
\nWhat is the area of the patio? Use 3.14 for \u03c0.
\nOptions:
\na. 9 m2<\/sup>
\nb. 28.26 m2<\/sup>
\nc. 254.34 m2<\/sup>
\nd. 1017.36 m2<\/sup><\/p>\n

Answer: c. 254.34 m2<\/sup><\/p>\n

Explanation:
\nDiameter = 18 m
\nRadius = 18\/2 = 9 m
\nArea of the patio = \u03c0r2<\/sup> = 3.14 (9)2<\/sup> = 254.34 m2<\/sup><\/p>\n

Question 15.
\nPetra fills a small cardboard box with sand. The dimensions of the box are 3 inches by 4 inches by 2 inches.
\na. What is the volume of the box?
\n______ in3<\/sup><\/p>\n

Answer: 24 in3<\/sup><\/p>\n

Explanation:
\nV = Bh
\nB = 3 x 4 = 12 in2<\/sup>
\nV = 12 x 2 = 24 in3<\/sup><\/p>\n

Question 15.
\nb. 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?
\n______ in2<\/sup><\/p>\n

Answer: 76 in2<\/sup><\/p>\n

Explanation:
\nP = 2(3) + 2(4) = 6 + 8 = 14 in
\nS = Ph + 2B = 14 x 2 + 2 x 24 = 76 in2<\/sup><\/p>\n

Question 15.
\nc. 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?
\nVolume is _________ times greater.
\nSurface area is _________ times greater<\/p>\n

Answer: Surface area is about 2.7 times larger<\/p>\n

Explanation:
\nB = 6 x 8 = 48 in2<\/sup>
\nV = Bh = 48 x 4 = 192 in3<\/sup>
\n192\/24 = 8
\nP = 2(6) + 2(8) = 12 + 16 = 28
\nS = Ph + 2B = 28(4) + 2(48) = 112 + 96 = 208 in2<\/sup>
\n208\/76 = 2.7<\/p>\n

Conclusion:<\/em><\/p>\n

We wish the information provided in this article regarding the Go Math Grade 7 Chapter 9 Circumference, Area, and Volume is beneficial for all the students. Make use of the given links and practice well for the exams. If you have any quieries about HMH Go Math 7th Grade Chapter 9 Circumference, Area, and Volume you can post your comments in the below section.<\/p>\n","protected":false},"excerpt":{"rendered":"

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 … Read more<\/a><\/p>\n","protected":false},"author":6,"featured_media":8807,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"https:\/\/bigideasmathanswers.com\/wp-content\/uploads\/2020\/11\/go-math-grade-7-chapter-9-circumference-area-and-volume-answer-key.jpeg","_links":{"self":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/6030"}],"collection":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/users\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/comments?post=6030"}],"version-history":[{"count":1,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/6030\/revisions"}],"predecessor-version":[{"id":92814,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/6030\/revisions\/92814"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/media\/8807"}],"wp:attachment":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/media?parent=6030"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/categories?post=6030"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/tags?post=6030"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}