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Nc Math 3.Indd CHAPTER 9 Geometric Modeling 9.1 Modeling with Area ..........................................................................................341 9.2 Modeling with Volume ......................................................................................347 9.3 Cross Sections of Solids ..................................................................................355 9.4 Solids of Revolution .........................................................................................361 Copyright © Big Ideas Learning, LLC 339 All rights reserved. Name _________________________________________________________ Date __________ Chapter Maintaining Mathematical Proficiency 9 Find the area of the circle or regular polygon. 1. 2. 3.7 ft 4.7 in. 5 ft 3. a circle with a diameter of 74.6 centimeters 4. a regular hexagon with a perimeter of 42 yards and an apothem of 4.25 yards 5. a circle with a circumference of 24π meters Find the surface area and volume of the solid. 6. 7. 35 cm 8 in. 6 in. 28 cm 12 in. 8. 9. 37 m 58 yd 23 m Copyright © Big Ideas Learning, LLC All rights reserved. 340 Name _________________________________________________________ Date _________ Modeling with Area 9.1 For use with Exploration 9.1 Essential Question How can you use the population and area of a region to describe how densely the region is populated? 1 EXPLORATION: Exploring Population and Area Work with a partner. Use the Internet to find the population and land area of each county in California. Then find the number of people per square mile for each county. a. Mendocino County b. Lake County c. Yolo County d. Napa County e. Sonoma County f. Marin County Copyright © Big Ideas Learning, LLC 341 All rights reserved. Name _________________________________________________________ Date __________ 9.1 Modeling with Area (continued) 2 EXPLORATION: Analyzing Population and Area Work with a partner. The six counties in Exploration 1 appear on a map as shown. a. Without calculating, how would you expect the number of people per square mile in the entire 6-county region to compare to the values for each individual county in Exploration 1? b. Use the populations and land areas in Exploration 1 to justify your answer in part (a). Communicate Your Answer 3. How can you use the population and area of a region to describe how densely the region is populated? 4. Find the population and land area of the county in which you live. How densely populated is your county compared to the counties in Exploration 1? 5. In Exploration 1, the two northern counties are less densely populated than the other four. What factors do you think might influence how densely a region is populated? Copyright © Big Ideas Learning, LLC All rights reserved. 342 Name _________________________________________________________ Date _________ Notetaking with Vocabulary 9.1 For use after Lesson 9.1 In your own words, write the meaning of each vocabulary term. Name _________________________________________________________ Date _________ population density Practice 9.1 For use after Lesson 9.1 In your own words, write the meaning of each vocabulary term. Notes: population density Worked-Out Examples number of people Chapter 1 Population density = —— area of land Notes:Example #1 210, 000 = — 452.39 Step 3 Sketch a diagram of the corral that includes the gate Find the indicated measure. opening, as shown. 4. The area of the region is About 650, 000 people≈ 464.20 live in a circular region with a 6-mile radius. Find the 2 barn Thepopulation population density density in people is about per 464 square people mile. per square mile. 5. The area of the region is π62 = 36π ≈ 113.1 square miles. 20 m corral The population density is: number of people 650, 000 Population density = —— = — area of land 113.1 37 m 3 m ≈ 5747.1 The population density is about 5747 people per square mile. So, you need 2w + ℓ − 3 = 2(20) + 40 − 3 = 77 6. 12. The orange island has numbera greater of population animals density. If the meters of fencing. = —— area of land Example #2 4. islandsame number has a smaller of people area, live 445then on there both areislands, more and people the orangeper = — ≈ 0.0002 square mile. 2, 220, 000 Chapter 1 Smaller bearing Larger bearing MODELING WITH MATHEMATICS A soccer field of length ℓ and width w has a perimeter of 320 yards. d = 12 mm d = 18 mm Dimensions 12 18 a. Write an expression that represents the area of the scoccer field in terms of ℓ. 8. Find the area of Central r = — Park. = 6 mmBecause it ris = approximately — = 9 mm 2 2 shaped like a rectangle, use the formula for the area of a b. Use your expression from part (a) to determine the dimensions of the field that 2 2 rectangle. S = 4πr S = 4πr maximize the area. What do you notice? 4 (6)2 4 (9)2 Surface area = π 2 = π A = ℓw = 2.5(0.5) = 1.25 mi 2 2 = 144π mm 1 = 324π mm So, the area of Central Park is 1.252 ÷ — = 1.25(640) 2= 13. Thea. Use population what you density know aboutis about the 0.0002 area and grizzly perimeter bear perof the acre. ≈ 452.39 mm ( 640 ≈) 1017.88 mm 800 acres. Elk:soccer fi eld to fi nd an expression that represents the area of the fi eld. number of animals Use the formula for population density. Let324 pπ represent9 3the2 Population density = —— The surface area of the larger bearing is — = — = — The perimeter P of a fi eldarea of of length land ℓ and width w is number of people in the park that afternoon.144π 4 ( 2) P = 2ℓ + 2w. So, 32020, = 000 2ℓ + 2w. times the surface area numberof the smaller of people bearing. = — ≈ 0.009 Population density = —— 2, 220, 000 area of3 land2 9 320 = 2ℓ + 2w no; The larger bearing needsp — = — times the amount of The population density is about 0.009 elk per acre. 15 = — (2) 4 320 − 2ℓ = 2w grease to coat it. 800 Mule deer: 320 − 2ℓ 2w — = — number of animals 1.1 Exercises (pp.15(800) 7–8) = p Population density2 = —— 2 area of land 12, 000 = p 160 − ℓ =w Copyright © Big Ideas Learning, LLC Vocabulary and Core Concept Check 2400 337 All rights reserved. So, there are about 12,000 people in Central Park that Solving for w gives= —w = 160 − ≈ℓ .0.001 1. Sample answer: The population of a state is the total number 2, 220, 000 afternoon. of people who live in that state. The population density of a The So, population an exp ression density that is represents about 0.001 the mule area deerof the per soccer acre. fi eld is A = ℓw = ℓ(160 −ℓ). 9. Usestate the is theformula number for of population people per density. square Let mile r representin that state. the Bighorn sheep: Use the table feature of a graphing calculator to create radius of the region. b. number of animals 2. The expression that doesnumber not belong of people with the other three is Populationa table of density values =to ——fi nd the length ℓ Copyright © Big Ideas Learning, LLC Population density = —— area of land “video gamers per city.” Aarea city of is land not a measurement of area. value of (160 ). 343 All rights reserved. ℓ − ℓ 260 79, 000 = — ≈ 0.0001 XY1 2, 220, 000 513 = —2 Monitoring Progress andπ Modelingr with Mathematics 65 6175 70 6300 The population density is about 0.0001 bighorn sheep per 3. Step 1 Find the area2 of Kansas. It is approximately shaped 75 6375 513πr = 79, 000 80 6400 acre.85 6375 like a rectangle. Use the formula for the area of a 90 6300 2 79, 000 95 6175 rectangler to = estimate — the area of Kansas. X=80 513π 7. Use the formula for population density. Let p represent the 2 A = ℓw = 400(210)— = 84, 000 mi population The length of thethat region. maximizes the value of ℓ(160 − ℓ) 79, 000 Step 2 Find the populationr = — density. is 80 yards. So, the width of the fi eld is √ 513 number of people π Populationw 160 density =160 —— 80 80 yards. Because the width number of people = −ℓ = −area of= land Populationr density≈ 7 = —— and the length are both 80 yards, the area of the fi eld is area of land p maximized6366 when = it —is a square.2 The radius of the circular region2, 850, is about 000 7 miles. π(4) = — ≈ 34 84, 000 6366(16π) = p 10. Use the formula for population density. Let r represent the 14. a. Find the lateral area of the cylindrical roller, including the radius of The the population region. density is about 34 people per square 101,856π =1 p 5 1 13 13 nap. The radius is — 1 — = — — = — . mile. number of people 319,990.1 2p⋅ 8 2 ⋅ 8 16 Population density = —— ≈ area of land L = 2πrh There are about 319,990 people who live in the region. 1,150,000 13 1 18, 075 = — L = 2π — + — 9 πr2 ⋅ (16 4 ) ⋅ 18,075 r2 1,150,000 17 π = L = 2π — 9 ⋅ ( 16) ⋅ Copyright © Big Ideas Learning,1,150,000 LLC Integrated Mathematics III 3 r2 = — 153 All rights reserved. 18,075π L = — π ≈ 60.1Worked-Out Solutions — 8 ⋅ 1,150,000 So, the roller covers about 60.1 square inches in one r = — √ 18,075π revolution. r ≈ 4.5 b. Find the circumference of the cylindrical roller, including the nap. The radius of the circular region is about 4.5 kilometers. C = 2πr 11. The length of the radius should be used, not the length of 13 1 C = 2π — + — the diameter. The radius of the region is ⋅ ( 16 4) 7.5 ÷ 2 = 3.75 miles. 17 x C = 2π — 1550 = — ⋅ (16) π 3.752 ⋅ 17 x C = — π ≈ 6.7 1550 = — 14.0625 8 ⋅ π So, in one revolution, you are about 6.7 inches up the 68, 477 ≈ x wall, and in 8 full revolutions, you are about So, about 68,477 people live in the region.
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