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# Chapter 7

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### Chapter 7

1. 1. Warm Up Problem of the Day Lesson Presentation 7-1 Ratios and Proportions Course 3
2. 2. Warm Up Write each fraction in lowest terms. 7 8 3 8 1 8 3 8 Course 3 7-1 Ratios and Proportions 14 16 1. 9 72 3. 24 64 2. 45 120 4.
3. 3. Problem of the Day A magazine has page numbers from 1 to 80. What fraction of those page numbers include the digit 5? 17 80 Course 3 7-1 Ratios and Proportions
4. 4. Learn to find equivalent ratios to create proportions. Course 3 7-1 Ratios and Proportions
5. 5. Vocabulary ratio equivalent ratio proportion Insert Lesson Title Here Course 3 7-1 Ratios and Proportions
6. 6. Relative density is the ratio of the density of a substance to the density of water at 4°C. The relative density of silver is 10.5. This means that silver is 10.5 times as heavy as an equal volume of water. The comparisons of water to silver in the table are ratios that are all equivalent. Course 3 7-1 Ratios and Proportions 42 g 31.5 g 21 g 10.5 g Silver 4 g 3 g 2 g 1 g Water Comparisons of Mass of Equal Volumes of Water and Silver
7. 7. Course 3 7-1 Ratios and Proportions Ratios can be written in several ways. A colon is often used. 90:3 and name the same ratio. Reading Math 90 3
8. 8. A ratio is a comparison of two quantities by division. In one rectangle, the ratio of shaded squares to unshaded squares is 7:5. In the other rectangle, the ratio is 28:20. Both rectangles have equivalent shaded areas. Ratios that make the same comparison are equivalent ratios . Course 3 7-1 Ratios and Proportions
9. 9. Additional Example 1: Finding Equivalent Ratios Find two ratios that are equivalent to each given ratio. B. 18 54 1 3 128 48 8 3 A. Multiply or divide the numerator and denominator by the same nonzero number. Course 3 7-1 Ratios and Proportions = 9 27 = 9 • 2 27 • 2 = 9 ÷ 9 27 ÷ 9 9 27 = Two ratios equivalent to are and . 9 27 18 54 1 3 Two ratios equivalent to are and . 64 24 128 48 8 3 = 64 • 2 24 • 2 = 64 ÷ 8 24 ÷ 8 64 24 = 64 24 =
10. 10. Try This : Example 1 Find two ratios that are equivalent to each given ratio. B. 16 32 2 4 64 32 4 2 A. Multiply or divide the numerator and denominator by the same nonzero number. Course 3 7-1 Ratios and Proportions = 8 16 = 8 • 2 16 • 2 = 8 ÷ 4 16 ÷ 4 8 16 = Two ratios equivalent to are and . 8 16 16 32 2 4 Two ratios equivalent to are and . 32 16 64 32 4 2 = 32 • 2 16 • 2 = 32 ÷ 8 16 ÷ 8 32 16 = 32 16 =
11. 11. Ratios that are equivalent are said to be proportional , or in proportion . Equivalent ratios are identical when they are written in simplest form. Course 3 7-1 Ratios and Proportions
12. 12. Additional Example 2: Determining Whether Two Ratios are in Proportion Simplify to tell whether the ratios form a proportion. 1 9 1 9 4 5 3 4 Course 3 7-1 Ratios and Proportions 12 15 B. and 27 36 3 27 A. and 2 18 Since , the ratios are in proportion. 1 9 = 1 9 = 3 ÷ 3 27 ÷ 3 3 27 = = 2 ÷ 2 18 ÷ 2 2 18 = = 12 ÷ 3 15 ÷ 3 12 15 = = 27 ÷ 9 36 ÷ 9 27 36 = Since , the ratios are not in proportion. 4 5  3 4
13. 13. Try This : Example 2 Simplify to tell whether the ratios form a proportion. 1 5 1 5 2 7 4 9 Course 3 7-1 Ratios and Proportions 14 49 B. and 16 36 Since , the ratios are in proportion. 1 5 = 1 5 = 3 ÷ 3 15 ÷ 3 3 15 = = 9 ÷ 9 45 ÷ 9 9 45 = = 14 ÷ 7 49 ÷ 7 14 49 = = 16 ÷ 4 36 ÷ 4 16 36 = Since , the ratios are not in proportion. 2 7  4 9 3 15 A. and 9 45
14. 14. Additional Example 3: Earth Science Application At 4°C, four cubic feet of silver has the same mass as 42 cubic feet of water. At 4°C, would 210 cubic feet of water have the same mass as 20 cubic feet of silver? Course 3 7-1 Ratios and Proportions 4 ÷ 2 42 ÷ 2 ? = 20 ÷ 10 210 ÷ 10 2 21 = 2 21 4 42 ? = 20 210 Since , 210 cubic feet of water would have the same mass at 4°C as 20 cubic feet of silver. 2 21 = 2 21
15. 15. Try This : Example 3 At 4°C, two cubic feet of silver has the same mass as 21 cubic feet of water. At 4°C, would 105 cubic feet of water have the same mass as 10 cubic feet of silver? Course 3 7-1 Ratios and Proportions ? = 10 ÷ 5 105 ÷ 5 2 21 2 21 = 2 21 2 21 ? = 10 105 Since , 105 cubic feet of water would have the same mass at 4°C as 10 cubic feet of silver. 2 21 = 2 21
16. 16. Lesson Quiz: Part 1 Insert Lesson Title Here Find two ratios that are equivalent to each given ratio. Simplify to tell whether the ratios form a proportion. and and Course 3 7-1 Ratios and Proportions 8 5 8 5 = ; yes 4 15 1. 8 21 2. 16 10 3. 36 24 4. 8 30 12 45 Possible answer: , 16 42 24 63 Possible answer: , 32 20 28 18 3 2 14 9  ; no
17. 17. Lesson Quiz: Part 2 Insert Lesson Title Here 5. Kate poured 8 oz of juice from a 64 oz bottle. Brian poured 16 oz of juice from a 128 oz bottle. What ratio of juice is missing from each bottle? Are the ratios proportional? Course 3 7-1 Ratios and Proportions 8 64 16 128 and ; yes, both equal 1 8
18. 18. Warm Up Problem of the Day Lesson Presentation 7-2 Ratios, Rates, and Unit Rates Course 3
19. 19. Warm Up Divide. Round answers to the nearest tenth. 1. 2. 3. 4. 23.3 3.5 23.8 23.9 Course 3 7-2 Ratios, Rates, and Unit Rates 420 18 73 21 380 16 430 18
20. 20. Problem of the Day There are 3 bags of flour for every 2 bags of sugar in a freight truck. A bag of flour weighs 60 pounds, and a bag of sugar weighs 80 pounds. Which part of the truck’s cargo is heavier, the flour or the sugar? flour Course 3 7-2 Ratios, Rates, and Unit Rates
21. 21. Learn to work with rates and ratios. Course 3 7-2 Ratios, Rates, and Unit Rates
22. 22. Vocabulary rate unit rate unit price Insert Lesson Title Here Course 3 7-2 Ratios, Rates, and Unit Rates
23. 23. An aspect ratio describes a screen by comparing its width to its height. Movie and television screens range in shape from almost perfect squares to wide rectangles. Common aspect ratios are 4:3, 37:20, 16:9, and 47:20. Course 3 7-2 Ratios, Rates, and Unit Rates
24. 24. Additional Example 1A: Ordering Ratios A. Order the ratios 4:3, 23:10, 13:9, and 47:20 from the least to greatest. Insert Lesson Title Here 4:3 = 23:10 = 13:9 = 47:20 = = 2.3 = 2.35 The ratios in order from least to greatest are 4:3, 13:9, 23:10, and 47:20. Course 3 7-2 Ratios, Rates, and Unit Rates 4 3 4 3 Divide. = 1.3 1 = 1.3 23 10 13 9 47 20 = 1.4 The decimals in order are 1.3, 1.4, 2.3, and 2.35.
25. 25. B. A television has screen width 20 in. and height 15 in. What is the aspect ratio of this screen? Insert Lesson Title Here Additional Example 1B: Ordering Ratios The ratio of the width to the height is 20:15. The screen has the aspect ratio 4:3. Course 3 7-2 Ratios, Rates, and Unit Rates The ratio can be simplified: 20 15 5(4) 5(3) 4 3 20 15 = = .
26. 26. A. Order the ratios 2:3, 35:14, 5:3, and 49:20 from the least to greatest. Insert Lesson Title Here Try This : Example 1A 2:3 = 35:14 = 5:3 = 49:20 = = 2.5 = 2.45 The ratios in order from least to greatest are 2:3, 5:3, 49:20, 35:14. Course 3 7-2 Ratios, Rates, and Unit Rates 2 3 2 3 Divide. = 0.6 1 = 0.6 35 14 5 3 49 20 = 1.6 The decimals in order are 0.6, 1.6, 2.45, and 2.5.
27. 27. B. A movie theater has a screen width 36 ft. and height 20ft. What is the aspect ratio of this screen? Insert Lesson Title Here Try This : Example 1B The ratio of the width to the height is 36:20. The screen has the aspect ratio 9:5. Course 3 7-2 Ratios, Rates, and Unit Rates The ratio can be simplified: 36 20 4(9) 4(5) 9 5 36 20 = = .
28. 28. Ratio: 90 3 Rate: 90 miles 3 hours Read as “ 90 miles per 3 hours.” A rate is a comparison of two quantities that have different units. A ratio is a comparison of two quantities. Course 3 7-2 Ratios, Rates, and Unit Rates
29. 29. Unit rates are rates in which the second quantity is 1. unit rate: 30 miles , 1 hour or 30 mi/h 90 3 = 30 1 Course 3 7-2 Ratios, Rates, and Unit Rates The ratio 90 3 can be simplified by dividing:
30. 30. Insert Lesson Title Here Additional Example 2: Using a Bar Graph to Determine Rates Use the bar graph to find the number of acres, to the nearest acre, destroyed in Nevada and Alaska per week. Nevada =  12,308 acres 1 week 640,000 acres 52 weeks Course 3 7-2 Ratios, Rates, and Unit Rates
31. 31. Insert Lesson Title Here Additional Example 2 Continued Use the bar graph to find the number of acres, to the nearest acre, destroyed in Nevada and Alaska per week. Alaska =  14,423 acres 1 week 750,000 acres 52 weeks Course 3 7-2 Ratios, Rates, and Unit Rates
32. 32. Insert Lesson Title Here Montana =  18,269 acres 1 week 950,000 acres 52 weeks Try This : Example 2 Use the bar graph to find the number of acres, to the nearest acre, destroyed in Montana and Idaho per week. Course 3 7-2 Ratios, Rates, and Unit Rates
33. 33. Insert Lesson Title Here Idaho =  26,923 acres 1 week 1,400,000 acres 52 weeks Try This : Example 2 Continued Use the bar graph to find the number of acres, to the nearest acre, destroyed in Montana and Idaho per week. Course 3 7-2 Ratios, Rates, and Unit Rates
34. 34. Unit price is a unit rate used to compare costs per item. Course 3 7-2 Ratios, Rates, and Unit Rates
35. 35. Pens can be purchased in a 5-pack for \$1.95 or a 15-pack for \$6.20. Which is the better buy? Additional Example 3A: Finding Unit Prices to Compare Costs Divide the price by the number of pens. price for package number of pens  \$1.95 5 = \$0.39 price for package number of pens = \$6.20 15  \$0.41 The better buy is the 5-pack for \$1.95. Course 3 7-2 Ratios, Rates, and Unit Rates
36. 36. Jamie can buy a 15-oz jar peanut butter for \$2.19 or a 20-oz jar for \$2.78. Which is the better buy? Additional Example 3B: Finding Unit Prices to Compare Costs  \$2.19 15 = \$0.15 = \$2.78 20  \$0.14 The better buy is the 20-oz jar for \$2.78. Divide the price by the number of ounces. Course 3 7-2 Ratios, Rates, and Unit Rates price for jar number of ounces price for jar number of ounces
37. 37. Golf balls can be purchased in a 3-pack for \$4.95 or a 12-pack for \$18.95. Which is the better buy? Try This : Example 3A Divide the price by the number of balls. price for package number of balls  \$4.95 3 = \$1.65 price for package number of balls = \$18.95 12  \$1.58 The better buy is the 12-pack for \$18.95. Course 3 7-2 Ratios, Rates, and Unit Rates
38. 38. Try This : Example 3B John can buy a 24 oz bottle of ketchup for \$2.19 or a 36 oz bottle for \$3.79. Which is the better buy?  \$2.19 24 = \$0.09 = \$3.79 36  \$0.11 The better buy is the 24-oz jar for \$2.19. Divide the price by the number of ounces. Course 3 7-2 Ratios, Rates, and Unit Rates price for bottle number of ounces price for bottles number of ounces
39. 39. Lesson Quiz 1. At a family golf outing, a father drove the ball 285 ft. His daughter drove the ball 95 ft. Express the ratio of the father’s distance to his daughter’s in simplest terms. 2. Find the unit price of 6 stamps for \$2.22. 3. Find the unit rate of 8 heartbeats in 6 seconds. 4. What is the better buy, a half dozen carnations for \$4.75 or a dozen for \$9.24? 5. Which is the better buy, four pens for \$5.16 or a ten-pack for \$12.90? \$0.37 per stamp 3:1 Insert Lesson Title Here  1.3 beats/s a dozen They cost the same. Course 3 7-2 Ratios, Rates, and Unit Rates
40. 40. Warm Up Problem of the Day Lesson Presentation 7-3 Analyze Units Course 3
41. 41. Warm Up Find each unit rate. 1. Jump rope 192 times in 6 minutes 2. Four pounds of bananas for \$2.36 3. 16 anchor bolts for \$18.56 4. 288 movies on 9 shelves 32 jumps/min \$0.59/lb \$1.16/bolt 32 movies/shelf Course 3 7-3 Analyze Units
42. 42. Problem of the Day Replace each • with a digit from 0 to 6 to make equivalent ratios. Use each digit only once. •• •• • •• = Course 3 7-3 Analyze Units Possible answer: 13 65 4 20 =
43. 43. Learn to use one or more conversion factors to solve rate problems. Course 3 7-3 Analyze Units
44. 44. Vocabulary conversion factor Insert Lesson Title Here Course 3 7-3 Analyze Units
45. 45. You can measure the speed of an object by using a strobe lamp and a camera in a dark room. Each time the lamp flashes, the camera records the object’s position. Problems often require dimensional analysis , also called unit analysis , to convert from one unit to another unit. Course 3 7-3 Analyze Units
46. 46. To convert units, multiply by one or more ratios of equal quantities called conversion factors . For example, to convert inches to feet you would use the ratio below as a conversion factor. Course 3 7-3 Analyze Units 1 ft 12 in.
47. 47. Multiplying by a conversion factor is like multiplying by a fraction that reduces to 1, such as . 5 5 12 in. 12 in. 1 ft 1 ft = , or = 1 Course 3 7-3 Analyze Units 1 ft 12 in.
48. 48. Course 3 7-3 Analyze Units <ul><li>The conversion factor </li></ul><ul><li>must introduce the unit desired in the answer and </li></ul><ul><li>must cancel the original unit so that the unit desired is all that remains. </li></ul>Helpful Hint
49. 49. Find the appropriate factor for each conversion. Additional Example 1: Finding Conversion Factors A. feet to yards B. pounds to ounces Course 3 7-3 Analyze Units 1 yd 3 ft There are 3 feet in 1 yard. To convert feet to yards, multiply the number of feet by . 16 oz 1 lb There are 16 ounces in 1 pound. To convert pounds to ounces, multiply the number of pounds by .
50. 50. Try This : Example 1 Find the appropriate factor for each conversion. Insert Lesson Title Here A. minutes to seconds B. hours to days Course 3 7-3 Analyze Units 60 sec 1 min There are 60 seconds in 1 minute. To convert minutes to seconds, multiply the number of minutes by . 1 day 24 h There are 24 hours in 1 day. To convert hours to days, multiply the number hours by .
51. 51. The average American uses 580 pounds of paper per year. Find the number of pounds of paper the average American uses per month, to the nearest tenth. Additional Example 2: Using Conversion Factors to Solve Problems The problem gives the ratio 580 pounds to 1 year and asks for an answer in pounds per month. Multiply the ratio by the conversion factor Cancel yr units. Divide 580 by 12. Course 3 7-3 Analyze Units 580 lb 1 yr 1 yr 12 mo 580 lb 12 mo = = 48.3 lb per month
52. 52. The average American uses 580 pounds of paper per year. Find the number of pounds of paper the average American uses per month, to the nearest tenth. Additional Example 2 Continued Course 3 7-3 Analyze Units The average American uses 48.3 pounds of paper per month.
53. 53. Try This : Example 2 Sam drives his car 23,000 miles per year. Find the number of miles he drives per month. Insert Lesson Title Here The problem gives the ratio 23,000 miles to 1 year and asks for an answer in miles per month. Multiply the ratio by the conversion factor Cancel yr units. Divide 23,000 by 12. Sam drives his car about 1917 miles per month. Course 3 7-3 Analyze Units 23,000 mi 1 yr 1 yr 12 mo 23,000 mi 12 mo = = 1916.6 per month
54. 54. Additional Example 3: Problem Solving Application A car traveled 60 miles on a road in 2 hours. How many feet per second was the car traveling? The problem is stated in units of miles and hours . The question asks for the answer in units of feet and seconds . You will need to use several conversion factors. List the important information: • Miles to feet 5280 ft 1 mi • Hours to minutes • Minutes to seconds 1 min 60 s 1 Understand the Problem Course 3 7-3 Analyze Units 1 h 60 min
55. 55. Multiply by each conversion factor separately, or simplify the problem and multiply by several conversion factors at once. Additional Example 3 Continued Course 3 7-3 Analyze Units 2 Make a Plan
56. 56. First, convert 60 miles in 2 hours into a unit rate. 60 mi 2 h = (60÷2) mi (2÷2) h = 30 mi 1 h Create a single conversion factor to convert hours directly to seconds: hours to seconds = • 1 min 60 s Set up the conversion factors. Course 3 7-3 Analyze Units Solve 3 minutes to seconds 1 min 60 s hours to minutes 1 h 60 min 1 h 60 min 1 h 3600 s = 30 mi 1 h • 5280 ft 1 mi • 1 h 3600 s
57. 57. Solve Do not include the numbers yet. Notice what happens to the units. 44 ft 1 s The car was traveling 44 feet per second. Multiply. Course 3 7-3 Analyze Units 3 30 • 5280 ft • 1 1 • 1 • 3600 s = 158,400 ft 3600 s = Simplify. Only remains. ft s mi h • ft mi • h s • • 30 mi 1 h 5280 ft 1 mi 1 h 3600 s
58. 58. A rate of 44 ft/s is less than 50 ft/s. A rate of 60 miles in 2 hours is 30 mi/h or 0.5 mi/min. Look Back Since 0.5 mi/min is less than 3000 ft/ 60 s or 50 ft/s and 44 ft/s is less than 50 ft/s, then 44 ft/s is a reasonable answer. Course 3 7-3 Analyze Units 4
59. 59. Try This : Example 3 A train traveled 180 miles on a railroad track in 4 hours. How many feet per second was the train traveling? The problem is stated in units of miles and hours . The question asks for the answer in units of feet and seconds . You will need to use several conversion factors. List the important information: • Miles to feet 5280 ft 1 mi • Hours to minutes • Minutes to seconds 1 min 60 s 1 Understand the Problem Course 3 7-3 Analyze Units 1 h 60 min
60. 60. Multiply by each conversion factor separately, or simplify the problem and multiply by several conversion factors at once. Try This : Example 3 Continued Course 3 7-3 Analyze Units 2 Make a Plan
61. 61. First, convert 180 miles in 4 hours into a unit rate. 180 mi 4 h = (180 ÷ 4) mi (4 ÷ 4) h = 45 mi 1 h Create a single conversion factor to convert hours directly to seconds: hours to seconds = • 1 min 60 s Set up the conversion factors. Course 3 7-3 Analyze Units Solve 3 minutes to seconds 1 min 60 s hours to minutes 1 h 60 min 1 h 60 min 1 h 3600 s = 45 mi 1 h • 5280 ft 1 mi • 1 h 3600 s
62. 62. Solve Do not include the numbers yet. Notice what happens to the units. 66 ft 1 s The train was traveling 66 feet per second. Multiply. Course 3 7-3 Analyze Units 3 45 • 5280 ft • 1 1 • 1 • 3600 s = 237,600 ft 3600 s = Simplify. Only remains. ft s mi h • ft mi • h s • • 45 mi 1 h 5280 ft 1 mi 1 h 3600 s
63. 63. A rate of 66 ft/s is more than 50 ft/s. A rate of 180 miles in 4 hours is 45 mi/h or 0.75 mi/min. Look Back Since 0.75 mi/min is more than 3000 ft/60 s or 50 ft/s and 66 ft/s is more than 50 ft/s, then 66 ft/s is a reasonable answer. Course 3 7-3 Analyze Units 4
64. 64. Additional Example 4: Physical Science Application Course 3 7-3 Analyze Units A strobe lamp can be used to measure the speed of an object. The lamp flashes every of a second. A camera records the object moving 52 cm between flashes. How fast is the object moving in m/s? 1 100 distance . time Use rate = 52 cm 1 100 s
65. 65. It may help to eliminate the fraction first. Additional Example 4 Continued Multiply top and bottom by 100. Course 3 7-3 Analyze Units 1 100 5200 cm 1 s = 52 cm 1 100 s = 100 • 52 cm 1 100 s 100 •
66. 66. Now convert centimeters to meters. Additional Example 4 Continued 5200 cm 1 s Multiply by the conversion factor. The object is traveling 52 m/s. Course 3 7-3 Analyze Units 5200 m 100 s = 52 m 1 s = 5200 cm 1 s = • 1 m 100 cm
67. 67. Try This : Example 4 Course 3 7-3 Analyze Units A strobe lamp can be used to measure the speed of an object. The lamp flashes every of a second. A camera records the object moving 65 cm between flashes. How fast is the object moving in m/s? 1 100 distance . time Use rate = 65 cm 1 100 s
68. 68. It may help to eliminate the fraction first. Try This : Example 4 Continued Multiply top and bottom by 100. Course 3 7-3 Analyze Units 1 100 6500 cm 1 s = 65 cm 1 100 s = 100 • 65 cm 1 100 s 100 •
69. 69. Now convert centimeters to meters. Try This : Example 4 Continued 6500 cm 1 s Multiply by the conversion factor. The object is traveling 65 m/s. Course 3 7-3 Analyze Units 6500 m 100 s = 65 m 1 s = 6500 cm 1 s = • 1 m 100 cm
70. 70. The rate 1 knot equals 1 nautical mile per hour. One nautical mile is 1852 meters. What is the speed in kilometers per hour of a ship traveling at 5 knots? Additional Example 5: Transportation Application 5 knots = 5 nautical mi/h Examine the units. The ship is traveling 9.26 km/h. Course 3 7-3 Analyze Units Set up the units to obtain in your answer. km h 5 • 1852 • 1 km 1 h • 1 • 1000 = = 9260 km 1000 h = 9.26 km 1 h nautical mi h • m nautical mi • km m 5 nautical mi 1 h • • 1852 m 1 nautical mi 1 km 1000 m
71. 71. Try This : Example 5 The rate 1 knot equals 1 nautical mile per hour. One nautical mile is 1852 meters. What is the speed in kilometers per hour of a ship traveling at 9 knots? 9 knots = 9 nautical mi/h Examine the units. The ship is traveling about 16.67 km/h. Course 3 7-3 Analyze Units Set up the units to obtain in your answer. km h 9 • 1852 • 1 km 1 h • 1 • 1000 = = 16,668 km 1000 h  16.67 km 1 h nautical mi h • m nautical mi • km m 9 nautical mi 1 h • • 1852 m 1 nautical mi 1 km 1000 m
72. 72. Lesson Quiz Find the appropriate factor for each conversion. 1. kilograms to grams 2. pints to gallons 3. You drive 136 miles from your house to your aunt’s house at the lake. You use 8 gallons of gas. How many yards does your car get to the gallon? 4. A cheetah was timed running 200 yards in 6 seconds. What was the average speed in miles per hour? Insert Lesson Title Here 1000 g kg 1 gal 8 pt 29,920 yd gal ≈ 68 mi/h Course 3 7-3 Analyze Units
73. 73. Warm Up Problem of the Day Lesson Presentation 7-4 Solving Proportions Course 3
74. 74. Warm Up Find two ratios that are equivalent to each given ratio. 3 5 1. 45 30 3. 90 60 10 12 2. 20 24 8 9 4. 24 27 Possible answers: Course 3 7-4 Solving Proportions 3 2 , 5 6 , 16 18 , 9 15 6 10 ,
75. 75. Problem of the Day Replace each • with a digit from 1 to 7 to write a proportion. Use each digit once. The digits 2 and 3 are already shown. Course 3 7-4 Solving Proportions • • • 2 3 •• = 1 4 7 2 3 56 = Possible answer:
76. 76. Learn to solve proportions. Course 3 7-4 Solving Proportions
77. 77. Vocabulary cross product Insert Lesson Title Here Course 3 7-4 Solving Proportions
78. 78. Unequal masses will not balance on a fulcrum if they are an equal distance from it; one side will go up and the other side will go down. Unequal masses will balance when the following proportions is true: Course 3 7-4 Solving Proportions mass 2 length 1 mass 1 length 2 = Mass 1 Mass 2 Fulcrum Length 1 Length 2
79. 79. One way to find whether ratios, such as those on the previous slide, are equal is to find a common denominator. The ratios are equal if their numerators are equal after the fractions have been rewritten with a common denominator. Course 3 7-4 Solving Proportions 72 96 6 8 = 72 96 9 12 = 9 12 6 8 =
80. 80. Course 3 7-4 Solving Proportions
81. 81. Course 3 7-4 Solving Proportions The cross product represents the numerator of the fraction when a common denominator is found by multiplying the denominators. Helpful Hint
82. 82. Tell whether the ratios are proportional. A. Since the cross products are equal, the ratios are proportional. 60 Additional Example 1A: Using Cross Products to Identify Proportions 60 = 60 Find cross products. 60 Course 3 7-4 Solving Proportions 4 10 6 15 = ? 4 10 6 15
83. 83. A mixture of fuel for a certain small engine should be 4 parts gasoline to 1 part oil. If you combine 5 quarts of oil with 15 quarts of gasoline, will the mixture be correct? 4 • 5 = 20 1 • 15 = 15 20 ≠ 15 The ratios are not equal. The mixture will not be correct. Set up ratios. Find the cross products. Additional Example 1B: Using Cross Products to Identify Proportions Course 3 7-4 Solving Proportions 4 parts gasoline 1 part oil = ? 15 quarts gasoline 5 quarts oil
84. 84. Tell whether the ratios are proportional. Try This : Example 1A Since the cross products are equal, the ratios are proportional. 20 20 = 20 Find cross products. 20 A. Course 3 7-4 Solving Proportions Course 3 7-4 Solving Proportions 2 4 5 10 2 4 5 10 = ?
85. 85. A mixture for a certain brand of tea should be 3 parts tea to 1 part sugar. If you combine 4 tablespoons of sugar with 12 tablespoons of tea, will the mixture be correct? Try This : Example 1B 3 • 4 = 12 1 • 12 = 12 12 = 12 The ratios are equal. The mixture will be correct. Set up ratios. Find the cross products. Course 3 7-4 Solving Proportions 3 parts tea 1 part sugar = ? 12 tablespoons tea 4 tablespoons sugar
86. 86. When you do not know one of the four numbers in a proportion, set the cross products equal to each other and solve. Course 3 7-4 Solving Proportions
87. 87. Solve the proportion. 6 p = 12 • 5 p = 10 6 p = 60 Find the cross products. Solve. Additional Example 2: Solving Proportions Course 3 7-4 Solving Proportions 5 6 p 12 =  ; the proportion checks. 5 6 10 12 =
88. 88. Solve the proportion. 14 • 3 = 2 g 21 = g 42 = 2 g Find the cross products. Solve. Try This : Example 2 Course 3 7-4 Solving Proportions 2 3 14 g =  ; the proportion checks. 2 3 14 21 =
89. 89. Allyson weighs 55 lbs and sits on a seesaw 4 ft away from its center. If Marco sits 5 ft away from the center and the seesaw is balanced, how much does Marco weigh? 44 = x Set up the proportion. Let x represent Marco’s weight. Find the cross products. Multiply. Solve. Divide both sides by 5. Marco weighs 44 lb. Additional Example 3: Physical Science Application 220 = 5 x 55 • 4 = 5 x Course 3 7-4 Solving Proportions 5 x 5 220 5 = x 4 55 5 = mass 1 length 2 = mass 2 length 1
90. 90. Robert weighs 90 lbs and sits on a seesaw 5 ft away from its center. If Sharon sits 6 ft away from the center and the seesaw is balanced, how much does Sharon weigh? Try This : Example 3 75 = x Set up the proportion. Let x represent Sharon’s weight. Find the cross products. Multiply. Solve. Divide both sides by 6. Sharon weighs 75 lb. 450 = 6 x 90 • 5 = 6 x Course 3 7-4 Solving Proportions 6 x 6 450 6 = x 5 90 6 = mass 1 length 2 = mass 2 length 1
91. 91. Lesson Quiz Insert Lesson Title Here Tell whether each pair of ratios is proportional. 1. 2. Solve each proportion. 3. 4. 5. Two weights are balanced on a fulcrum. If a 6lb weight is positioned 1.5 ft from the fulcrum, at what distance from the fulcrum must an 18 lb weight be placed to keep the weights balanced? yes no n = 30 n = 16 0.5 ft Course 3 7-4 Solving Proportions 48 42 = ? 16 14 20 15 = ? 3 4 45 18 n 12 = n 24 6 9 =
92. 92. Warm Up Problem of the Day Lesson Presentation 7-5 Dilations Course 3
93. 93. Warm Up Multiply. 1. 4  2. 12  3. 24  4. –36  3 18 10 9 – 27 30 5. 4  2.5 6. 12  2.5 Course 3 7-5 Dilations 3 4 3 4 3 4 3 4
94. 94. Problem of the Day Every day, a plant grows to three times its size. Every night, it shrinks to half its size. After three days and nights, it is 6.75 in. tall. How tall was the plant at the start? 2 in. Course 3 7-5 Dilations
95. 95. Learn to identify and create dilations of plane figures. Course 3 7-5 Dilations
96. 96. Vocabulary dilation scale factor center of dilation Insert Lesson Title Here Course 3 7-5 Dilations
97. 97. Your pupils are the black areas in the center of your eyes. When you go to the eye doctor, the doctor may dilate your pupils, which makes them larger. Course 3 7-5 Dilations
98. 98. Translations, reflections, and rotations are transformations that do not change the size or shape of a figure. A dilation is a transformation that changes the size, but not the shape, of a figure. A dilation can enlarge or reduce a figure. Course 3 7-5 Dilations
99. 99. A scale factor describes how much a figure is enlarged or reduced. A scale factor can be expressed as a decimal, fraction, or percent. A 10% increase is a scale factor of 1.1, and a 10% decrease is a scale factor of 0.9. Course 3 7-5 Dilations
100. 100. Insert Lesson Title Here Course 3 7-5 Dilations A scale factor between 0 and 1 reduces a figure. A scale factor greater than 1 enlarges it. Helpful Hint
101. 101. Tell whether each transformation is a dilation. Insert Lesson Title Here The transformation is a dilation. The transformation is not a dilation. The figure is distorted. Additional Example 1A & 1B: Identifying Dilations A. B. Course 3 7-5 Dilations
102. 102. Tell whether each transformation is a dilation. Insert Lesson Title Here The transformation is a dilation. The transformation is not a dilation. The figure is distorted. Additional Example 1C & 1D: Identifying Dilations C. D. Course 3 7-5 Dilations
103. 103. Tell whether each transformation is a dilation. A. The transformation is a dilation. The transformation is not a dilation. The figure is distorted. Try This : Example 1A & 1B B. Course 3 7-5 Dilations A ' B ' C ' A B C B A C A ' B ' C '
104. 104. Tell whether each transformation is a dilation. C. The transformation is a dilation. The transformation is not a dilation. The figure is distorted. Try This : Example 1C & 1D D. Course 3 7-5 Dilations A ' B ' C ' A B C A ' B ' C ' A B C
105. 105. Insert Lesson Title Here Every dilation has a fixed point that is the center of dilation . To find the center of dilation, draw a line that connects each pair of corresponding vertices. The lines intersect at one point. This point is the center of dilation . Course 3 7-5 Dilations
106. 106. Dilate the figure by a scale factor of 1.5 with P as the center of dilation. Insert Lesson Title Here Additional Example 2: Dilating a Figure Multiply each side by 1.5. Course 3 7-5 Dilations
107. 107. Insert Lesson Title Here Dilate the figure by a scale factor of 0.5 with G as the center of dilation. G F H 2 cm 2 cm 2 cm Multiply each side by 0.5. Try This : Example 2 Course 3 7-5 Dilations G F H 2 cm 2 cm 2 cm F’ H’ 1 cm 1 cm 1 cm
108. 108. Insert Lesson Title Here Additional Example 3A: Using the Origin as the Center of Dilation Dilate the figure in Example 3A on page 363 by a scale factor of 2. What are the vertices of the image? Multiply the coordinates by 2 to find the vertices of the image. The vertices of the image are A ’(8, 16), B ’(6, 4), and C ’(10, 4). Course 3 7-5 Dilations A (4, 8) A ’(4  2 , 8  2 ) A ’(8, 16) B (3, 2) B ’(3  2 , 2  2 ) B ’(6, 4) C (5, 2) C ’(5  2 , 2  2 ) C ’(10, 4) ABC A’B’C’
109. 109. Insert Lesson Title Here Additional Example 3B: Using the Origin as the Center of Dilation The vertices of the image are A ’(1, 3), B ’(3, 2), and C ’(2, 1). Course 3 7-5 Dilations Dilate the figure in Example 3B by a scale factor of . What are the vertices of the image? 1 3 ABC A’B’C’ A (3, 9) A ’(3  , 9  ) A ’(1, 3) 1 3 1 3 B (9, 6) B ’(9  , 6  ) B ’(3, 2) 1 3 1 3 C (6, 3) C ’(6  , 3  ) C ’(2, 1) 1 3 1 3 Multiply the coordinates by to find the vertices of the image. 1 3
110. 110. Insert Lesson Title Here Try This : Example 3A Dilate the figure by a scale factor of 2. What are the vertices of the image? 2 4 2 4 6 8 10 0 6 8 10 B C A Course 3 7-5 Dilations
111. 111. Insert Lesson Title Here Try This : Example 3A Continued The vertices of the image are A’(4, 4), B’(8, 4), and C’(4, 8). Course 3 7-5 Dilations A (2, 2) A ’(2  2 , 2  2 ) A ’(4, 4) B (4, 2) B ’(4  2 , 2  2 ) B ’(8, 4) C (2, 4) C ’(2  2 , 4  2 ) C ’(4, 8) ABC A’B’C’
112. 112. Insert Lesson Title Here Try This : Example 3A Continued 2 4 2 4 6 8 10 0 6 8 10 B C A Course 3 7-5 Dilations B’ C’ A’
113. 113. Insert Lesson Title Here Try This : Example 3B Dilate the figure by a scale factor of 0.5. What are the vertices of the image? 2 4 2 4 6 8 10 0 6 8 10 B C A Course 3 7-5 Dilations
114. 114. Insert Lesson Title Here Try This : Example 3B Continued The vertices of the image are A’(2, 2.5), B’(4, 2.5), and C’(2, 4.5). Course 3 7-5 Dilations A (4, 5) A ’(4  0.5 , 5  0.5 ) A ’(2, 2.5) B (8, 5) B ’(8  0.5 , 5  0.5 ) B ’(4, 2.5) C (4, 9) C ’(4  0.5 , 9  0.5 ) C ’(2, 4.5) ABC A’B’C’
115. 115. Insert Lesson Title Here Try This : Example 3B Continued 2 4 2 4 6 8 10 0 6 8 10 B C A Course 3 7-5 Dilations B’ C’ A’
116. 116. Lesson Quiz Insert Lesson Title Here 2 4 6 -2 -4 -6 2. Dilate the figure by a scale factor of 1.5 with P as the center of dilation. 3. Dilate the figure by a scale factor of 2 with the origin as the center of dilation. What are the coordinates of the image? A (2,4) B (5,6) C (6,1) A ’(4,8) B ’(10,12) C ’(12,2) yes Course 3 7-5 Dilations 1. Tell whether the transformation is a dilation. A(0, 4) B(5,5) C(3,3) A’(0, 8) B’(10, 10) C’(6, 6) P A B C C ’ B ’ A ’
117. 117. Warm Up Problem of the Day Lesson Presentation 7-6 Similar Figures Course 3
118. 118. Warm Up Solve each proportion. 1. 2. 3. 4. b = 10 y = 8 p = 3 m = 52 Course 3 7-6 Similar Figures b 30 3 9 = 56 35 y 5 = 4 12 p 9 = 56 m 28 26 =
119. 119. Problem of the Day A plane figure is dilated and gets 50% larger. What scale factor should you use to dilate the figure back to its original size? ( Hint: The answer is not ). 1 2 2 3 Course 3 7-6 Similar Figures
120. 120. Learn to determine whether figures are similar, to use scale factors, and to find missing dimensions in similar figures. Course 3 7-6 Similar Figures
121. 121. Vocabulary similar Insert Lesson Title Here Course 3 7-6 Similar Figures
122. 122. The heights of letters in newspapers and on billboards are measured using points and picas . There are 12 points in 1 pica and 6 picas in one inch. A letter 36 inches tall on a billboard would be 216 picas, or 2592 points. The first letter in this paragraph is 12 points. Course 3 7-6 Similar Figures
123. 123. Congruent figures have the same size and shape. Similar figures have the same shape, but not necessarily the same size. The A ’s in the table are similar. The have the same shape, but they are not the same size. The ratio formed by the corresponding sides is the scale factor. Course 3 7-6 Similar Figures
124. 124. Additional Example 1: Using Scale Factors to Find Missing Dimensions A picture 10 in. tall and 14 in. wide is to be scaled to 1.5 in. tall to be displayed on a Web page. How wide should the picture be on the Web page for the two pictures to be similar? To find the scale factor, divide the known measurement of the scaled picture by the corresponding measurement of the original picture. 0.15 Then multiply the width of the original picture by the scale factor. 2.1 14 • 0.15 = 2.1 The picture should be 2.1 in. wide. Course 3 7-6 Similar Figures 0.15 1.5 10 =
125. 125. Try This : Example 1 A painting 40 in. tall and 56 in. wide is to be scaled to 10 in. tall to be displayed on a poster. How wide should the painting be on the poster for the two pictures to be similar? To find the scale factor, divide the known measurement of the scaled painting by the corresponding measurement of the original painting. 0.25 Then multiply the width of the original painting by the scale factor. 14 56 • 0.25 = 14 The painting should be 14 in. wide. Course 3 7-6 Similar Figures 0.25 10 40 =
126. 126. Additional Example 2: Using Equivalent Ratios to Find Missing Dimensions A T-shirt design includes an isosceles triangle with side lengths 4.5 in, 4.5 in., and 6 in. An advertisement shows an enlarged version of the triangle with two sides that are each 3 ft. long. What is the length of the third side of the triangle in the advertisement? Set up a proportion. 4.5 in. • x ft = 3 ft • 6 in. Find the cross products. Course 3 7-6 Similar Figures 6 in. x ft 4.5 in. 3 ft = 4.5 in. • x ft = 3 ft • 6 in. in. • ft is on both sides
127. 127. 4.5 x = 3 • 6 4.5 x = 18 Cancel the units. Multiply Solve for x. Additional Example 2 Continued The third side of the triangle is 4 ft long. Course 3 7-6 Similar Figures x = = 4 18 4.5
128. 128. Try This: Example 2 Set up a proportion. 18 ft • x in. = 24 ft • 4 in. Find the cross products. A flag in the shape of an isosceles triangle with side lengths 18 ft, 18 ft, and 24 ft is hanging on a pole outside a campground. A camp t-shirt shows a smaller version of the triangle with two sides that are each 4 in. long. What is the length of the third side of the triangle on the t-shirt? Course 3 7-6 Similar Figures 24 ft x in. 18 ft 4 in. = 18 ft • x in. = 24 ft • 4 in. in • ft is on both sides
129. 129. 18 x = 24 • 4 18 x = 96 Cancel the units. Multiply Solve for x. Try This : Example 2 Continued The third side of the triangle is about 5.3 in. long. Course 3 7-6 Similar Figures x =  5.3 96 18
130. 130. The following are matching, or corresponding:  A and  X Remember! A C B Z Y X  C and  Z  B and  Y Course 3 7-6 Similar Figures AB and XY BC and YZ AC and XZ
131. 131. Additional Example 3: Identifying Similar Figures Which rectangles are similar? Since the three figures are all rectangles, all the angles are right angles. So the corresponding angles are congruent. Course 3 7-6 Similar Figures
132. 132. Additional Example 3 Continued 50  48 The ratios are equal. Rectangle J is similar to rectangle K . The notation J ~ K shows similarity. The ratios are not equal. Rectangle J is not similar to rectangle L . Therefore, rectangle K is not similar to rectangle L . 20 = 20 Compare the ratios of corresponding sides to see if they are equal. Course 3 7-6 Similar Figures length of rectangle J length of rectangle K width of rectangle J width of rectangle K 10 5 4 2 ? = length of rectangle J length of rectangle L width of rectangle J width of rectangle L 10 12 4 5 ? =
133. 133. Try This : Example 3 Which rectangles are similar? A 8 ft 4 ft B 6 ft 3 ft C 5 ft 2 ft Since the three figures are all rectangles, all the angles are right angles. So the corresponding angles are congruent. Course 3 7-6 Similar Figures
134. 134. Try This : Example 3 16  20 The ratios are equal. Rectangle A is similar to rectangle B . The notation A ~ B shows similarity. The ratios are not equal. Rectangle A is not similar to rectangle C . Therefore, rectangle B is not similar to rectangle C . 24 = 24 Compare the ratios of corresponding sides to see if they are equal. Course 3 7-6 Similar Figures length of rectangle A length of rectangle B width of rectangle A width of rectangle B 8 6 4 3 ? = length of rectangle A length of rectangle C width of rectangle A width of rectangle C 8 5 4 2 ? =
135. 135. Lesson Quiz Use the properties of similar figures to answer each question. 1. A rectangular house is 32 ft wide and 68 ft long. On a blueprint, the width is 8 in. Find the length on the blueprint. 2. Karen enlarged a 3 in. wide by 5 in. tall photo into a poster. If the poster is 2.25 ft wide, how tall is it? 3. Which rectangles are similar? 17 in. 3.75 ft A and B are similar. Course 3 7-6 Similar Figures
136. 136. Warm Up Problem of the Day Lesson Presentation 7-7 Scale Drawings Course 3
137. 137. Warm Up Evaluate the following for x = 16. 1. 3 x 2. x Evaluate the following for x = . 3. 10 x 4. x 48 12 4 Course 3 7-7 Scale Drawings 3 4 2 5 1 4 1 10
138. 138. Problem of the Day An isosceles triangle with a base length of 6 cm and side lengths of 5 cm is dilated by a scale factor of 3. What is the area of the image? 108 cm 2 Course 3 7-7 Scale Drawings
139. 139. Learn to make comparisons between and find dimensions of scale drawings and actual objects. Course 3 7-7 Scale Drawings
140. 140. Vocabulary scale drawing scale reduction enlargement Insert Lesson Title Here Course 3 7-7 Scale Drawings
141. 141. A scale drawing is a two-dimensional drawing that accurately represents an object. The scale drawing is mathematically similar to the object. A scale gives the ratio of the dimensions in the drawing to the dimensions of the object. All dimensions are reduced or enlarged using the same scale. Scales can use the same units or different units. Course 3 7-7 Scale Drawings
142. 142. Insert Lesson Title Here The scale a : b is read “ a to b .” For example, the scale 1 cm:3 ft is read “one centimeter to three feet.” Reading Math Course 3 7-7 Scale Drawings in. on the drawing is 1 ft. in. = 1 ft 1 cm on the drawing is 1 m. 1 cm: 1 m 1 unit on the drawing is 20 units. 1:20 Interpretation Scale 1 4 1 4
143. 143. A. The length of an object on a scale drawing is 2 cm, and its actual length is 8 m. The scale is 1 cm: __ m. What is the scale? Additional Example 1A: Using Proportions to Find Unknown Scales or Lengths = 1  8 = x  2 Find the cross products. 8 = 2 x Solve the proportion. The scale is 1 cm:4 m. 4 = x Course 3 7-7 Scale Drawings 1 cm x m 2 cm 8 m Set up proportion using scale length . actual length
144. 144. B. The length of an object on a scale drawing is 1.5 inches. The scale is 1 in:6 ft. What is the actual length of the object? Additional Example 1B: Using Proportions to Find Unknown Scales or Lengths = 1  x = 6  1.5 Find the cross products. x = 9 Solve the proportion. The actual length is 9 ft. Course 3 7-7 Scale Drawings 1 in. 6 ft 1.5 in. x ft Set up proportion using scale length . actual length
145. 145. A. The length of an object on a scale drawing is 4 cm, and its actual length is 12 m. The scale is 1 cm: __ m. What is the scale? Try This : Example 1A = 1  12 = x  4 Find the cross products. 12 = 4 x Solve the proportion. The scale is 1 cm:3 m. 3 = x Course 3 7-7 Scale Drawings 1 cm x m 4 cm 12 m Set up proportion using scale length . actual length
146. 146. B. The length of an object on a scale drawing is 2 inches. The scale is 1 in:4 ft. What is the actual length of the object? Try This : Example 1B = 1  x = 4  2 Find the cross products. x = 8 Solve the proportion. The actual length is 8 ft. Course 3 7-7 Scale Drawings 1 in. 4 ft 2 in. x ft Set up proportion using scale length . actual length
147. 147. Insert Lesson Title Here A scale drawing that is smaller than the actual object is called a reduction . A scale drawing can also be larger than the object. In this case, the drawing is referred to as an enlargement . Course 3 7-7 Scale Drawings
148. 148. Under a 1000:1 microscope view, an amoeba appears to have a length of 8 mm. What is its actual length? Additional Example 2: Life Sciences Application 1000  x = 1  8 Find the cross products. x = 0.008 The actual length of the amoeba is 0.008 mm. Solve the proportion. Course 3 7-7 Scale Drawings scale length actual length 1000 1 = 8 mm x mm
149. 149. Under a 10,000:1 microscope view, a fiber appears to have length of 1mm. What is its actual length? Try This : Example 2 10,000  x = 1  1 Find the cross products. x = 0.0001 The actual length of the fiber is 0.0001 mm. Solve the proportion. Course 3 7-7 Scale Drawings scale length actual length 10,000 1 = 1 mm x mm
150. 150. Insert Lesson Title Here Course 3 7-7 Scale Drawings A drawing that uses the scale in. = 1 ft is said to be in in. scale. Similarly, a drawing that uses the scale in. = 1 ft is in in. scale. 1 4 1 4 1 2 1 2
151. 151. Additional Example 3A: Using Scales and Scale Drawings to Find Heights 0.25  x = 1  4 Find the cross products. x = 16 The wall is 16 ft tall. Length ratios are equal. Solve the proportion. Course 3 7-7 Scale Drawings scale length actual length 0.25 in. 1 ft = 4 in. x ft. A. If a wall in a in. scale drawing is 4 in. tall, how tall is the actual wall? 1 4
152. 152. B. How tall is the wall if a in. scale is used? Additional Example 3B: Using Scales and Scale Drawings to Find Heights 0.5  x = 1  4 Find the cross products. x = 8 The wall is 8 ft tall. Length ratios are equal. Solve the proportion. Course 3 7-7 Scale Drawings 1 2 scale length actual length 0.5 in. 1 ft = 4 in. x ft.
153. 153. Try This : Example 3A 0.25  x = 1  0.5 Find the cross products. x = 2 The wall is 2 ft thick. Length ratios are equal. Solve the proportion. Course 3 7-7 Scale Drawings scale length actual length 0.25 in. 1 ft = 0.5 in. x ft. A. If a wall in a in. scale drawing is 0.5 in. thick, how thick is the actual wall? 1 4
154. 154. B. How thick is the wall if a in. scale is used? Try This : Example 3A Continued 0.5  x = 1  0.5 Find the cross products. x = 1 The wall is 1 ft thick. Length ratios are equal. Solve the proportion. Course 3 7-7 Scale Drawings 1 2 scale length actual length 0.5 in. 1 ft = 0.5 in. x ft.
155. 155. 1. What is the scale of a drawing in which a 9 ft wall is 6 cm long? 2. Using a in. = 1 ft scale, how long would a drawing of a 22 ft car be? 3. The height of a person on a scale drawing is 4.5 in. The scale is 1:16. What is the actual height of the person? The scale of a map is 1 in. = 21 mi. Find each length on the map. 4. 147 mi 5. 5.25 mi Lesson Quiz 5.5 in. 1 cm = 1.5 ft Insert Lesson Title Here 72 in. 7 in. 0.25 in. Course 3 7-7 Scale Drawings 1 4