Chapter 5: Measurement of Circles

Chapter 5: Measurement of Circles Getting Started, p. 151 1. a) Perimeter, since the word around is used. b) Area, since since the word wrap is used. c) Perimeter, since the word wrap is used. 2. a) 5 cm < 5 m c) 1.000 m = 1000 mm b) 1 m = 100 cm d) 13 km > 130 m 3. a) perimeter = 4 18 mm = 72 mm area = 18 mm 18 mm = 324 mm 2 b) perimeter = 2 8 m + 2 2 m = 20 m area = 8 m 2 m = 16 m 2 c) perimeter = 5.0 cm + 6.0 cm + 7.8 cm = 18.8 cm area = (5.0 cm 6.0 cm) 2 = 15.0 cm 2 d) perimeter = 12.3 cm + 8.1 cm + 9.1 cm = 29.5 cm area = (6.0 cm 12.3 cm) 2 = 36.9 cm 2 4. perimeter = 2 4.5 cm + 2 7.2 cm = 23.4 cm area = 4.1 cm 7.2 cm = 29.5 cm 2 5. a) For example, the larger rectangle is 3.0 m by 14.1 m, so its area is 3.0 m 14.1 m = 42.3 m 2. The smaller rectangle is 2.0 m by 5.1 m, so its area is 2.0 m 5.1 m = 10.2 m 2. 10.2 m 2 42.3 m 2 The total area is 42.3 m 2 + 10.2 m 2 = 52.5 m 2. b) For example, the larger rectangle is 8.5 m by 3.5 m. Its area is 8.5 m 3.5 m = 29.8 m 2. The smaller rectangle is 4.5 m by 2.0 m. Its area is 4.5 m 2.0 m = 9.0 m 2. 2.0 m 4.5 m 9.0 m2 29.8 m2 8.5 m 3.5 m The total area is 9.0 m 2 + 29.8 m 2 = 38.8 m 2. c) The large rectangle is 7.0 m 14.0 m. Its area is 7.0 m 14.0 m = 98.0 m 2. The right triangle has a height of 4.0 m and a base of 5.4 m. Its area is (7.0 m 14.0 m) 2 = 10.8 m 2 10.8 m 2 98.0 mm 2 2 The total area is 98.0 cm 2 + 10.8 cm 2 = 108.8 m 2. 6. area = (base height) 2 6 cm 2 = (4 cm h) 2 6 cm 2 = 2 cm h 3 cm = h The height of the triangle is 3 cm. 5.3 Calculating Circumference, pp. 158 159 4. a) C = d 3.14 5 cm 16 cm b) C = d 3.14 4.7 cm 14.8 cm 5. a) C = 2 r 2 3.14 10 cm 62.8 cm 63 cm Nelson Mathematics 8 Solutions Manual 5-1

b) C = 2 r = 2 3.14 8.2 m 51.5 m 6. Multiply each diameter by 3.14. a) C = d 3.14 4.5 cm 14.1 cm b) C = d 3.14 1.7 cm 5.3 cm c) C = d 3.14 6.4 cm 20.1 cm d) C = d 3.14 36.0 m 113.1 m e) C = d 3.14 7 mm 22 mm f) C = d 3.14 4.0 cm 12.6 cm 7. Double each radius to find the diameter, and then multiply the diameter by 3.14. a) C = 2 r 2 3.14 7 mm 44 mm b) C = 2 r 2 3.14 19.5 cm 122.5 cm c) C = 2 r 2 3.14 6.3 cm 39.6 cm d) C = 2 r 2 3.14 9.0 cm 56.5 cm e) C = 2 r 2 3.14 23.1 m 145.1 m f) C = 2 r 2 3.14 0.05 m 0.31 m 8. C = d 80 cm 3.14 251 cm The circumference of the wheel is about 251 cm. 9. a) For example, to draw a circle using a compass, you need know the radius of the circle. The radius is half the diameter, so since the diameter is 11 cm, the radius must be 11 cm 2 = 5.5 cm. Since the radius is 5.5 cm, you can set your compass to 5.5 cm and draw the circle. 11 cm 5.5 cm b) The circumference is 3.14 11 cm 35 cm. 10. Since the line forms the diameter, multiply it by to find the length of the chalk line 11.0 m 3.14 34.5 m The chalk line is about 34.5 m long. 11. clock r = 9.0 cm d = 2 r = 2 9.0 cm = 18.0 cm C = 2 r 2 3.14 9.0 cm 56.5 cm round tea bag r = 1.9 cm d = 2 r = 2 1.9 cm = 3.8 cm C = 2 d = 2 3.14 1.9 cm 11.9 cm circle protractor r = 5.9 cm d = 2 r = 2 5.9 cm = 11.8 cm = 3.14 5.9 cm 37.1 cm watch r = d 2 = 36 mm 2 = 18 mm d = 36 mm 3.14 36 mm 113 mm sewer cover r = d 2 = 62 cm 2 = 31 cm d = 62 cm 3.14 62 cm 195 cm electric fan r = d 2 = 201 mm 2 101 mm d = 201 mm 3.14 201 mm 631 mm 12. The 700 m forms the radius of the circle. So, circumference = 2 r = (2 700 m) 3.14 4396 m About 4396 m of fencing is needed to enclose the field. 5-2 Chapter 5: Measurement of Circles

13. The 39.25 m is the circumference, which means that the diameter is 39.25 m 3.14 12.50 m and the radius is half of this, or 6.25 m. The lifeguard must walk 6.25 m from the edge to the centre of the pool. 14. The wheel with a diameter of 120.0 cm has a circumference of 3.14 120.0 cm 376.8 cm. The wheel with a diameter of 150.0 cm has a circumference of 3.14 150.0 cm 471.0 cm. The difference in circumference is 471.0 cm 376.8 cm = 94.2 cm. 15. Since the diameter of each circle is about 9.0 m, the circumference is 3.14 9.0 m. There are 5 circles, so the total length is 5 3.14 9.0 m = 141.3 m. 16. diameter of large circle = circumference 37.7 cm 3.14 12.0 cm diameter of large circle diameter of small circle = 12.0 cm 1.5 cm = 10.5 cm distance from outside edge to inside edge = 10.5 cm 2 5.3 cm 17. C = d 3.14 9.2 m 28.9 m 18. a) For example, about how far does Earth travel in space every year? Since the radius of orbit is about 150 10 6 km, the diameter of orbit must be twice as large, or 300 10 6 km. The circumference of the orbit is about 3.14 (300 10 6 km) 942 10 6 km. b) For example, how far has the tire rolled? For each complete rotation, the tire will move a distance equal to its circumference, or 3.14 50.0 cm 157.0 cm. Since every turn it rolls 157.0 cm, in 15 turns it will roll 15 157.0 cm = 2355.0 cm. 19. Notice that the length of the label is the same as the circumference of the lid. Use 4.0 cm as the diameter and calculate the circumference of the lid. 4.0 cm 3.14 12.6 cm The soup label is about 12.6 cm by 10.0 cm. 20. Separate the perimeter into two parts: two semicircles (or a full circle) of radius 21.7 m, and two straight sections of lengths 56.8 m. The circumference of the full circle is C = d (2 21.7 m) 3.14 136.3 m The combined length of the straight sections is l = 2 56.8 m = 113.6 m total distance around the track = circumference of circle + length of straight section 136.3 m + 113.6 m 249.9 m 21. The red arc is half of the circumference of a circle with the larger diameter. Since the circumference of a circle doubles when the diameter doubles, half of the circumference of the red semi-circle will be equal to the whole circumference of the blue circle. This can be verified algebraically: red arc = larger diameter 2 The larger diameter is equal to twice the smaller diameter, so, red arc = 2 smaller diameter 2 = 2 smaller diameter 2 = smaller diameter = blue circumference The red arc and the blue circle are the same length. Mid-Chapter Review, p. 161 1. radius chord circumference arc diameter 2. From shortest to longest: radius, diameter, circumference. The circumference is about 3.14 times the diameter, which in turn is twice the radius. circumference diameter radius radius diameter circumference 3. An estimate for the circumference of the circle is C 3 3 m 9 m 4. For example, a) about 3 25 cm 75 cm b) about 3 11 m 33 m c) about 3 20 cm 60 cm d) about 3 3 km 9 km Nelson Mathematics 8 Solutions Manual 5-3

5. CD case d = 2 r = 2 6.0 cm = 12.0 cm r = 6.0 cm C = 2 r = 2 3.14 6.0 cm 37.7 cm coaster d = 9.0 cm r = d 2 = 9.0 cm 2 4.5 cm 3.14 9 cm 28.3 cm lock d = 26 mm r = d 2 = 26 mm 2 13 mm 3.14 26 mm 82 mm coin d = 2 r = 2 1.9 cm = 3.8 cm r = 1.9 cm C = 2 d 2 3.14 1.9 cm 11.9 cm 6. a) C = d 15 cm 3.14 47 cm b) C = d 60 cm 3.14 188 cm c) C = d 12.5 cm 3.14 39.3 cm d) C = d 17 cm 3.14 53 cm e) C = d 44 cm 3.14 138 cm f) C = d 20 cm 3.14 63 cm 7. If the diameter of a circle is doubled, the circumference also doubles. For example, the circumference is d, so if the diameter is 2.0 cm, then the circumference is about 2.0 cm 3.14 6.3 cm. If the diameter is 4.0 cm, then the circumference is about 4.0 cm 3.14 12.6 cm, or twice 6.3 cm. d = 2.0 cm C = 6.3 cm d = 4.0 cm C = 12.6 cm 5.5 Calculating Area, pp. 166 167 4. a) area = r 2 3.14 (10.5 cm) 2 346.2 cm 2 b) area = r 2 3.14 (14 cm 2) 2 3.14 (7 cm) 2 153.86 cm 2 154 cm 2 c) area = r 2 3.14 (13 cm) 2 530.66 cm 2 531 cm 2 d) area = r 2 3.14 (2.8 cm 2) 2 3.14 (1.4 cm) 2 6.2 cm 2 5. a) area = r 2 3.14 (7.3 cm 2) 2 3.14 (3.65 cm) 2 41.8 cm 2 b) area = r 2 3.14 (2 cm) 2 13 cm 2 5-4 Chapter 5: Measurement of Circles

c) area = r 2 3.14 (2.7 cm) 2 22.9 cm 2 d) area = r 2 3.14 (1.7 cm 2) 2 3.14 (0.85 cm) 2 2.3 cm 2 6. For example, to estimate: the area of the square outside the circle is 7.0 cm 7.0 cm = 49 cm 2. The area of the square inside the circle is 5.0 cm 5.0 cm = 25 cm 2. The area of the circle is about halfway between these, or about 37 cm 2. To calculate: area = (3.5 cm) 2 38.5 cm 2 7. a) area of circle = r 2 3.14 (2.3 cm 2) 2 4.2 cm 2 b) area of section = area of circle 2 4.2 cm 2 2 2.1 cm 2 8. a) 3.14 (4.1 cm) 2 52.8 cm 2 b) The area of each section is one third of the total area. area of one section 52.8 cm 2 3 17.6 cm 2 9. a) 3.14 (22.0 cm) 2 1519.8 cm 2 b) The area of each section is one quarter of the total area. area of one section 1519.8 cm 2 4 380.0 cm 2 10. a) The area of the square is: A = lw 2 = 10.0 m 10.0 m = 100.0 m 2 b) The area of the circle is: 3.14 (5.0 m) 2 78.5 m 2 c) The area of the red area is the difference between the area of the square and the circle. red area = area of square area of circle 100.0 m 2 78.5 m 2 21.5 m 2 11. Use the circumference to determine the radius. Since the circumference is 10.0 cm, the diameter is 10.0 cm 3.14 3.2 cm. The diameter is about 3.2 cm, so the radius is 3.2 cm 2. Use the radius to calculate the area. 3.14 (1.6 cm 2) 2 8.0 cm 2 12. a) The area of the rectangle is 4.0 cm 2.0 cm = 8.0 cm 2. b) The area of the two circles is 2 r 2. A = 2 r 2 2 3.14 (1.0 cm) 2 6.3 cm 2 The yellow area is the difference between the area of the rectangle and the circles. So, the yellow area has an area of about 8.0 cm 2 6.3 cm 2 1.7 cm 2. 13. For example, what area of plastic is needed to make a DVD? Since the large and small diameters are 12.0 cm and 1.5 cm, their radii are 6.0 cm and 0.8 cm. The area of the DVD is the difference in areas between these two circles. Area of outer circle 3.14 (12.0 cm 2) 2 113.0 cm 2. Area of hole 3.14 (1.5 cm 2) 2 1.8 cm 2 area of DVD = area of outer circle area of hole 113.0 cm 2 1.8 cm 2 111.2 cm 2 The area of the DVD is about 111.2 cm 2. 14. The figure can be separated into two figures, a rectangle and a semicircle. The total area of the figure is the sum of the areas of the rectangle and the semicircle. 5.0 cm 5.0 cm 10.0 cm area of rectangle = length width = 10.0 cm 5.0 cm = 50.0 cm 2 Nelson Mathematics 8 Solutions Manual 5-5

area of semicircle = r 2 2 3.14 (5.0 cm) 2 2 39.3 cm 2 total area of figure = area of rectangle + area of semicircle 50.0 cm 2 + 39.3 cm 2 89.3 cm 2 15. The total area of the target is 3.14 (30.0 cm) 2 2826.0 cm 2 The total target has a radius of 30.0 cm and there are 4 bands of equal width. So the width of each band must be 30.0 cm 4 = 7.5 cm. The area of each band is the difference in areas between the larger and smaller bands. area of small white band = r 2 3.14 (7.5 cm ) 2 176.6 cm 2 area of small red band = r 2 area of small white band 3.14 (15.0 cm) 2 176.6 cm 2 706.5 cm 2 176.6 cm 2 529.9 cm 2 area of large white band = r 2 area of both smaller bands 3.14 (22.5 cm) 2 706.5 cm 2 1589.6 cm 2 706.5 cm 2 883.1 cm 2 area of large red band = r 2 area of three smaller bands 3.14 (30.0 cm) 2 1589.6 cm 2 2826.0 cm 2 1589.6 cm 2 1236.4 cm 2 The total area of the red region is the sum of the areas of the two red bands. red area = 529.9 cm 2 + 1236.4 cm 2 = 1766.3 cm 2 16. a) The total area will be the sum of the area of the square in the middle, plus the area of a full circle with diameter 375.0 m. The radius of the circle is half of the diameter or 375.0 m 2 = 187.5 m. total area of the field = (area of square) + (area of circle) (375.0 m 375.0 m) + 3.14 (187.5 m) 2 251 015.6 m 2 b) 251 015.6 m 2 $1.25/m = $313 769.50 c) The total perimeter is the sum of the two sides measuring 375.0 m and the circumference of the circle with diameter 375.0 m. 2 375.0 m + 3.14 375.0 m 1927.5 m d) 1927.5 m $2.75/m = $5300.63 e) Adding both costs for sod and paving stones together, $313 769.50 + $5300.63 = $319 070.13 5.6 Solve Problems by Working Backward, pp. 170 171 3. a) C = 2 r 62.8 cm = 2 r 31.4 cm = r 31.4 cm 3.14 r 10.0 cm r 20.0 cm d 20 cm b) 314 cm 2 3.14r 2 100 cm 2 r 2 10 cm r 20 cm d 20 cm 5-6 Chapter 5: Measurement of Circles

4. Working backward, use the opposite operations. Start with 32. Subtract 12: 32 12 = 44 Find the opposite: +44 Add 10: 44 + ( 10) = 34 Subtract 12: 34 ( 12) = 34 + 12 = 46 The original number is 46. 5. Use the area of the circle to calculate the radius. area of circle = r 2 78.5 m 2 3.14 r 2 78.5 m 2 3.14 r 2 25.0 m 2 r 2 5.0 m r The height of the triangle is equivalent to the diameter of the circle, or 2 5.0 m = 10.0 m. 6. a) Use the area to determine the radius of the figure. area of figure = r 2 706 cm 2 3.14 r 2 706 cm 2 3.14 r 2 225 cm 2 r 2 15 cm r The large circle has a radius of about 15 cm and a diameter of about 30 cm. Since it is divided into three equal sections, each section has a length of about 10 cm. The radius of the small circle is about 10 cm 2 = 5 cm and the radius of the middle circle is about 20 cm 2 = 10 cm. b) circumference of large circle = 2 r 2 3.14 15 cm 94 cm circumference of middle circle = 2 r 2 3.14 10 cm 63 cm circumference of small circle = 2 r 2 3.14 5 cm 31 cm c) Green area = area of small circle = r 2 3.14 (5 cm) 2 79 cm 2 Purple area = area of middle circle area of small circle r 2 79 cm 2 (10 cm) 2 3.14 79 cm 2 235 cm 2 Blue area = area of large circle area of middle circle 706 314 cm 2 392 cm 2 7. If she increases the amount of weight that she lifts by 2 kg each week, then by the 10th week, that is, after 9 weeks, 2 kg 9 = 18 kg more than she started out with. Since she is now lifting 120 kg, she must have started out by lifting 120 kg 18 kg = 102 kg. 8. For example, to find the original shape, double the figure 5 times: 9. a) The circle is drawn around the square, so the diagonal of the square is equal to the diameter of the circle. 5.0 m 7.1 m 25 m 2 The square has an area of 25.0 m 2, which means the square has side lengths that measure 25.0 m 2 = 5.0 m. The square is 5.0 m by 5.0 m. By Pythagoras, c 2 = a 2 + b 2 c 2 = (5.0 m) 2 + (5.0 m) 2 c 2 = 50.0 m 2 c 7.1 m The diagonal of this square will be about 7.1 m. b) From part a), you know the diameter of the circle is about 7.1 m. The circumference will be 3.14 7.1 m 22.3 m. 10. i) For example, if the wheel rolls 20.5 m, how many times has it rotated? Since the circumference is 2.563 m, you need to find out how many times 2.563 divides 20.5. Since 2.563 m 8 20.5 m, the wheel has rotated 8 times. Nelson Mathematics 8 Solutions Manual 5-7

ii) For example, about how long are the spokes on this bicycle wheel? If the circumference is 256.3 cm, then the diameter is 256.3 3.14 81.6 cm. So, the radius is 40.8 cm, which is also the length of the spokes. iii) For example, if the wheel lies flat on a driveway, how much area does it cover? Since the radius is 40.8 cm, the wheel covers an area of 3.14 (40.8 cm) 2 5227.0 cm 2. 11. Use the area of the circle covered by the sprinkler to determine the radius. 283 m 2 = r 2 283 m 2 3.14 r 2 90 r 2 9 m r The sprinkler can spray about 9 m. 12. The diameter of the coin is 2.8 cm, so its radius is 2.8 cm 2 = 1.4 cm. The total area of the coin is: 3.14 (1.4 cm) 2 6.2 cm 2 area of nickel rim = area of coin area of copper centre 6.2 cm 2 2.0 cm 2 4.2 cm 2 Chapter Self-Test, p. 173 1. diameter arc chord 6.5 cm 2. a) The amount to cover is required. This is the area. b) This is a space that needs to be covered, so it is area. c) Since it is the border, it is a circumference. d) The frame around is needed, so it is circumference. 3. Divide the diameter by 2. 4. a) C = 2 r 2 3.14 2.5 km 15.7 km b) C= 2 r 2 3.14 26 cm 163 cm c) C = d 3.14 3.0 cm 9.4 cm d) C = d 3.14 21 cm 66 cm 5. a) 3.14 (2 cm) 2 13 cm 2 b) 3.14 (11 cm 2) 2 95 cm 2 c) 3.14 (5.7 cm 2) 2 25.5 cm 2 d) 3.14 (6.2 cm) 2 120.7 cm 2 6. a) total area = 15.0 m 15.0 m = 225.0 m 2 b) area of circle = r 2 3.14 (7.5 m) 2 176.6 m 2 c) area of blue region = total area area of circle 225.0 m 2 176.6 m 2 48.4 m 2 7. To determine the area of the orange region, subtract the area of the small circle from the area of the large circle. area of large circle = r 2 3.14 (4.5 cm 2) 2 15.9 cm 2 radius of small circle = 4.5 cm 2 1.5 cm 0.8 cm area of small circle = r 2 3.14 (0.8 cm) 2 2.0 cm 2 5-8 Chapter 5: Measurement of Circles

area of orange region = area of large circle area of small circle 15.9 cm 2 2.0 cm 2 13.9 cm 2 8. Separate the shaded area into two parts. 4.0 cm 8.0 cm The rectangle has an area of 4.0 cm 8.0 cm = 32.0 cm 2. The semicircle has a radius of 4.0 cm. A = r 2 2 3.14 (4.0 cm) 2 2 25.1 cm 2 The total area is about 25.1 cm 2 + 32.0 cm 2 = 57.1 cm 2. 9. a) The area required is the area of the rectangle minus the area of the circle. area of park = area of rectangle area of pool = (l w) ( r 2 ) (6.0 m 14.0 m) 3.14 (2.5 m) 2 64.4 m 2 b) The fence will need to be as long as the circumference of the circle with radius 2.5 m, or 3.14 5.0 m 15.7 m. 10. a) area of circle = r 2 113 cm 2 3.14 r 2 113 cm 2 3.14 r 2 36 cm 2 r 2 6 cm r The radius of the circle is 6 cm. Once you know the radius (or diameter), you can draw the circle. 6 cm b) To calculate the diameter of a circle, divide the circumference by. d = C 17.0 cm 3.14 5.4 cm 5.4 cm 11. Use the area to find the radius. 24.6 m 2 = r 2 24.6 m 2 3.14 r 2 7.8 m 2 r 2 2.8 m r Now calculate the circumference. C = 2 r 2 3.14 2.8 m 17.6 m So, 17.6 m of fencing are needed. Chapter Review, p. 175 1. The circle with the radius of 10.0 cm will have a diameter of 20.0 cm, so you can identify both circles by their size. The larger circle (2) has a radius of 10.0 cm and the smaller circle (1) has the diameter of 10.0 cm. 2. The circumference is just over 3 times as long as the diameter. The exact ratio is a special number called (pi). When doing mental math, it is often useful to use 3 as an approximate value of and then round your answer up. When doing calculations, it is useful to approximate the value of to be 3.14. 3. a) C = 2 r 2 3.14 2.6 cm 16.3 cm b) C = d 3.14 30 cm 94 cm c) C = d 3.14 1.2 km 3.8 km d) C = d 3.14 8.3 cm 26.1 cm 4. Think of the boardwalk as the diameter of the circular wetlands area and the hiking path as its circumference. The length of the hiking path is 3.14 50 m 157 m. 5. For example, you can sketch the pool. Draw a square on the outside of the pool that has side lengths the same as the diameter. Estimate the area of this square. Draw a square inside the pool and estimate the area of this square. The area of the pool is about halfway between the areas of the outside square and the inside square. Nelson Mathematics 8 Solutions Manual 5-9

6. a) 3.14 (50 km 2) 2 1963 km 2 b) A= r 2 3.14 (2 mm) 2 13 mm 2 c) 3.14 (6.5 cm) 2 132.7 cm 2 d) A= r 2 3.14 (11.0 m 2) 2 95.0 m 2 7. The pool s circumference is 25.12 m, so its diameter is about 25.12 m 3.14 8.00 m. The radius is half of that, or about 4.00 m. area = r 2 3.14 (4.00 m) 2 50.24 m 2 8. Every turn of the wheel will move the bicycle forward by a distance equal to its circumference, in this case 197 cm. Calculate the radius. C = 2 r C 2 = r 197 cm 2 = r 31 cm r Use the radius to determine the area covered by the wheel. area covered = r 2 3.14 (31 cm) 2 3018 cm 2 9. a) area of square = 12.0 m 12.0 m = 144.0 m 2 b) area of a quarter circle = r 2 4 3.14 (6.0 m) 2 4 28.3 m 2 c) area of 4 white sections = 4 28.3 m 2 = 113.2 m 2 d) The area of the red region is the area of the square minus the four areas of the white sections. A = 144.0 m 2 113.2 m 2 = 30.8 m 2 10. a) 5.7 cm b) If the circle has an area of 452 cm 2, then the radius can be determined. area of circle = r 2 452 cm 2 3.14 r 2 452 cm 2 3.14 r 2 144 cm 2 r 2 12 cm r Once you know the radius (or diameter), you can draw the circle. 12 cm c) d = C 34.5 cm 3.14 11.0 cm 11.0 cm 5-10 Chapter 5: Measurement of Circles