1. 1.5 Measurement AS 90130 Internal (3 credits)

2. Calculate the area of the following shapes. 6 cm 3 cm 5 cm 4 cm 3 cm 4 cm 2 cm A = 9 cm 2 A = 12.6 cm 2 A = 45 cm 2

4. 2 rugby fields side by side

5. 5 mL 8L 3760m 750m 2 55m 2 1600cm 3 60W 2m 750mL 80kg 40L -22°C 2000cm 2 1m 9.85s 1.3kg 1500m 160kPa 37.4° 50g 180mm

6. Try These – Metric Conversions 5.5 ha = _______m 2 25 m = ________ cm 345 m = _______km 4321 mm = _______m 0.12 m = ________ mm 237 cm = _________ m 998 km = ____________ cm 199500 cm = ________ km 55000 2500.345 4.321 120 2.37 99 800 000 1.995 5.7 L = ______mL 3400 kg = ______T 0.42 T = _______kg 4300 m 2 = ______ha 3 m 2 = ________cm 2 400000mm 2 ____m 2 4500cm 3 _______ L 5700 3.4 420 0.430 30000 0.4 IWB Fundamentals Ex. 8.02 pg 220 4.5

7. Starter 12 ha of land is subdivided in 650 m 2 lots. How many lots of this size can be created? 12 ha = _______ m 2 120000 120000 m 2 = 184.6 650 m 2 184 lots can be created. A teaspoon holds 5 cm 3. How many teaspoons are needed to measure out exactly 1 m 3 of water? 5 cm 3 = ____ mL = ____ L 1 m 3 = ____ L 5.005 1000 1000 L = 200 000 teaspoons.005 L

8. Note 1: Limits of Accuracy Measurements are never exact. There is a limit to the accuracy with which a measurement can be made. The limits of accuracy of measurement refers to the range of values within which the true value of the measurement lies. The range of values is defined by an upper limit and a lower limit.

9. Limits of Accuracy To find the upper limit, add 5 to the nearest significant place. To find the lower limit, minus 5 to the nearest significant place. e.g. The distance to Bluff on a signpost reads 17 km. The upper limit is 17 + 0.5 = The lower limit is 17 – 0.5 = Therefore the limits of accuracy are 16.5 km ≤ Bluff ≤ 17.5 km e.g. The distance to Bluff on a signpost reads 17 km. The upper limit is 17 + 0.5 = The lower limit is 17 – 0.5 = Therefore the limits of accuracy are 16.5 km ≤ Bluff ≤ 17.5 km 16.5 km 17.5 km

10. e.g.From home to school it is 27.5 km. What are the limits of accuracy for my distance to school? The upper limit is 27.5 + 0.05 = 27.55 km The lower limit is 27.5 – 0.05 = 27.45 km Therefore the limits of accuracy are: 27.45 km ≤ Distance ≤ 27.55 km

11. e.g.At the Otago vs Auckland game at Carisbrook it was reported that 28500 people attended. Give the limits of accuracy for the number of people attending the game? The upper limit is 28500 + 50 = 28550 The lower limit is 28500 – 50 = 28450 Therefore the limits of accuracy are: 28450 ≤ People ≤ 28550

12. Give the limits of accuracy for these measurements: 1.) 68 mm 2.) 397 mm 3.) 4 seconds 4.) 50 g 5.) 5890 kg 6.) 820 cm 7.) 92 kg 8.) 89.1° 67.5 mm ≤ x ≤ 68.5 mm 396.5 mm ≤ x ≤ 397.5 mm 3.5 seconds ≤ x ≤ 4.5 seconds 45 g ≤ x ≤ 55 g 5885 kg ≤ x ≤ 5895 kg 815 cm ≤ x ≤ 825 cm 91.5 kg ≤ x ≤ 92.5 kg 89.05° ≤ x ≤ 89.15°

13. Note 2: Areas ShapeArea FormulaExample RectangleArea = b × h 4 cm 2 cm Area = 4 x 2 = 8 cm 2 SquareArea = l × l 3 cm Area = 3 x 3 = 9 cm 2 TriangleArea = ½ b × h = ½ × 6 × 2 = 6 mm 2 6 mm 2 mm

14. Note 2: Areas ShapeArea FormulaExample ParallelogramArea = b × h 7 m 2 m Area = 7 x 2 = 14 m 2 Trapezium Area = 1/2 (a+b)× h 2 cm 6 cm 4 cm Area= ½ (4+2)×6= 18 cm 2 CircleArea = πr 2 Area = πr 2 = π × (5) 2 = 78.5 cm 2 5 cm

15. Note 3: Perimeter & Circumference The perimeter of a figure is the total length of its sides. 11 cm 5 cm The perimeter of this kite is: 11 cm + 5 cm + 11 cm +5 cm = 32 cm

16. The circumference of a circle is the total distance around it. C = 2πr or C = πd e.g. Calculate the circumference of a circle which has a radius of 32 cm 32 cm C = 2πr = 2 × π × 32 = 201.1 cm (4 sf)

17. To calculate the radius, when given the circumference, we need to rearrange the formula to make r the subject. C = 2πr r = C 2π e.g. Calculate the radius of a circle that has a circumference of 11.5 m r = 11.5 2π r = 1.83 m

18. Find the perimeter of this shape that is formed using 3 semicircles (2dp) 12.4 cm (π x 6.2) + (π x 6.2) = 2 x π x 6.2 r = 6.2 (for large circle) = 38.96 cm

19. Ex. 9.03 pg 250-254 Ex. 9.04 pg 258 Ex 9.05 pg 263-265 Find the perimeter if each square is exactly half the dimensions of the preceding square. 24 mm (24 x 3) + (12 x 3) + (6 x 3) + (3 x 3) + (1.5 x 4) = 141 cm

20. Starter A cage is to be constructed entirely of steel bars as shown. Steel bar costs $4.35/m and you have $350 to spend on steel. The cage consists of steel bar uprights and a circular hoop top and bottom. The structure is to be twice as wide as it is high. Calculate the height (x) and diameter (2x) Hint – there are 21 uprights and 2 circles 2x x Length of steel required = 21x + 2(2πx) = 33.57 x Amount of steel to purchase = $350 $4.35 = 80.46 m 33.57 x = 80.46 x = 2.40 m = 4.8 m

21. Note 5: Compound Areas Compound Areas are made up of more than one mathematical shape To find the area of a compound shape, find the areas of each individual shape and either add or subtract as required.

22. e.g. Find the area of this shape 4 cm 2 cm 5 cm Area of compound shape: = area of Rectangle + area of Triangle + area of semi-circle Area = b × h + ½ b × h + ½ π × r 2 = 4 × 5 + ½ 4 × 2 + 0.5 π × (2) 2 = 30.3 cm 2 (1 dp)

23. e.g. Think of a typical running track. What is the perimeter? How long are the straight sections? Calculate the area enclosed by the track. 400 m 100 m d = ? 100 m A circle + A rectangle = A total 3183.1 + 6366.2 = 9549.3 m 2 (1dp)

24. Starter A glass porthole on a ship has a diameter of 28 cm. It is completely surrounded by a wooden ring that is 3 cm wide. a.) Calculate the area of glass in the porthole b.) Calculate the area of the wooden ring A = πr 2 r = 14 cm A = π (14) 2 A = 616 cm 2 Area of porthole = πr 2, r = 17 cm (including frame) = 908 cm 2 Area of frame = 908-616 = 292 cm 2

25. Note 6: Area of a Sector/ Arc Length The sector is part of a circle Area of a sector = x° × πr 2 360° x° r e.g. 6 cm 35° Area of a sector = 35° × π × (6 cm) 2 360° = 11.0 cm 2 Recall: Area of a circle = πr2πr2

26. Note 6: Area of a Sector/ Arc Length Arc length = x° × 2πr 360° x° r e.g. 6 cm 75° Arc length = 75° × 2π × 6 360° = 7.85 cm (3 sf)

27. e.g. The length of the minor arc of a circle is 3π cm and the length of the major arc is 15π cm = length of minor arc × 360° total circumference a.) Find the radius of the circle 3π cm 15π cm Total circumference = 3π + 15π = 18π So 18π = 2 × π × r 18 = 2 r r = 9 cm b.) Find the angle of the minor sector = × 360° π π = 60° 60°

28. Note 7: Calculating the Radius/ Diameter from the Area or Circumference When we know the circumference or area of a circle, we can rearrange the equation to calculate the diameter or radius of the circle. e.g. The circumference of a circle is 25.4 cm. Calculate the radius. C = 2πr 25.4 cm = 2π × r r = 25.4 cm 2π r = 4.0 cm (1 dp)

29. e.g. The area of a circle is 35.6 cm 2 Calculate the diameter. A = πr 2 35.6 = π × r 2 35.6 = r 2 π r = 35.6 π √ r = 3.37 cm Diameter = 6.7 cm (1 dp) IWB Fundamentals Ex 9.04 pg 258-259 IWB Fundamentals Ex 9.04 pg 258-259

30. Starter A concrete courtyard is designed as in the diagram below. There are four circular gardens (all the same size) and one square garden. The rest of the courtyard is to have a pattern moulded into the concrete to look like bricks. The cost of each brick mould is 83¢ and a moulded paver measures 250 mm x 150 mm. Estimate the minimum number of moulds for the entire courtyard (assume no wastage), giving your answer to the appropriate level of precision and then calculate the total price. 18 m 24 m 2 m 2.5 m

31. Merit - Starter The rest of the courtyard is to have a pattern moulded into the concrete to look like bricks. The cost of each brick mould is 83¢ and a moulded paver measures 250 mm x 150 mm. Estimate the minimum number of moulds for the entire courtyard (assume no wastage), giving your answer to the appropriate level of precision and then calculate the total price. 18 m 24 m 2 m 2.5 m Area courtyard = (24 ×18) – 4(π×1.25 2 ) – 2 2 No. of moulds = 408.37 m 2 / (.25 ×.15) Total cost = 10890 × $ 0.83 ≈ $9100 = 408.37 m 2 = 10 890

32. Note 8: Surface Area The surface area of a solid is the sum of the areas of each face or curved surface. It is helpful to picture surface area as the net of the shape SA (cube) = 6a 2 a a

33. Surface Area 4 cm 5 cm 7 cm Calculate the surface area of this triangular prism How many faces are there? 5 2 Triangles: 2 × (1/2 b × h) = 2 × 0.5 × 5cm × 4 cm = 20 cm 2 Left Rectangle: b × h = 7 cm × 4 cm = 28 cm 2 Base Rectangle: b × h = 7 cm × 5 cm = 35 cm 2 Right Rectangle: b × h = 7 cm × 6.4 cm = 44.8 cm 2 6.4 cm Total surface Area = 127.8 cm 2

34. Surface Area SA (cone) = the curved surface + the circular base = π r l + π r 2 SA (sphere) = 4π r 2 l r r

35. Surface Area (cylinder) SA (cylinder) = area of rectangle + 2 ×(area of circle) = 2πr × h + 2 ×(πr 2 ) = 2πrh + 2πr 2 SA (cylinder) = 2πr (h + r) C = 2πr h r Surface area depends on whether the ends are open or not. Open cylinders are called pipes. Textbook Ex 12.04 pg 156-157 Ex 12.05 pg 157-160 Textbook Ex 12.04 pg 156-157 Ex 12.05 pg 157-160 IWB Fundamentals Ex 9.06 pg 271-273 Ex 9.07 pg 275-279 IWB Fundamentals Ex 9.06 pg 271-273 Ex 9.07 pg 275-279

36. Starter Grass seed is sold in two sizes. 500 g bags which cost $5.95/bag and cover an area of 13 square metres and 1 kg bags which cost $8.95 and cover double the area of the 500g bag. What quantity and combination of bags should the gardener buy to ensure he has enough to cover the required land, and at the best possible price? 22.0 m 15 m 4.1 m 9.2 m 7.0 m 7.6 m Total Area = 22×15 – (4.1×9.2) – (7.0×7.6) = 239.08 m 2 239.08 m 2 / 26 = 9.2 1kg bags 9 x $8.95 + 1 x $5.95 = $86.50 A gardener wishes to sow grass seed on his land. = 239 m 2 (3 sf)

37. Note 9: Volume of Prisms Volume = Area of cross section × Length Area L The volume of a solid figure is the amount of space it occupies. It is measured in cubic centimetres, cm 3 or cubic metres, m 3

38. Examples: Volume = (b × h) × L = 4 m × 4 m × 10 m = 160 m 3 4 m 10 m 4 cm 5 cm 7 cm Volume = (1/2 b×h) × L = ½ × 4cm ×5cm × 10 cm = 70 cm 3

39. Cylinder – A circular Prism Volume (cylinder) = πr 2 × h Volume = Area of cross section × Length 8 cm 1.2 cm V =πr 2 × h = π(1.2cm) 2 × 8cm = 36.19 cm 3 (4 sf)

40. Merit This tent-shaped plastic hothouse is to change its air five times every hour. What volume of air per minute is required from the fan to achieve this? Round appropriately. 8.0 m 13 m 12 m Volume = ½ × 13 × 12 × 8 = 624 m 3 624 m 3 × 5 = 3120 m 3 /hr 3 × = 52 m 3 /min

41. Starter 125 cm B 1 cm 8 cm 10cm A 1.) Calculate the volume of these two cubes /cuboids of ice 2.) Which would melt faster if left outside on a hot day? 3.) Calculate the total area of the six faces for both pieces. A = 1000 cm 3 B = 1000 cm 3

42. Merit The walls of a lounge are to be wallpapered. The room’s dimensions are depicted below. Each roll of wallpaper is 50 cm wide and 8 m long. Each roll of wallpaper costs $34.95. Calculate how many rolls of wallpaper are needed to wallpaper the room, and the cost. You must hang complete strips of wallpaper to cover the 2.5 m height, part pieces cannot be pasted together. Ignore door, windows and pattern matching. 2.5 m 9 m 4.2 m Long Wall – 9 m ÷ 0.5 m = 18 strips across Strips per roll – 8 m ÷ 2.5 m = 3 / roll 18 ÷3 = 6 rolls per side= 12 rolls

43. Merit The walls of a lounge are to be wallpapered. The room’s dimensions are depicted below. Each roll of wallpaper is 50 cm wide and 8 m long. Each roll of wallpaper costs $34.95. Calculate how many rolls of wallpaper are needed to wallpaper the room, and the cost. You must hang complete strips of wallpaper to cover the 2.5 m height, part pieces cannot be pasted together. Ignore door, windows and pattern matching. 2.5 m 9 m 4.2 m Short Wall – 4.2 m ÷ 0.5 m = 8.4 (9) strips across 9 ÷ 3 (strips per roll) = 3 rolls per side = 6 rolls+ 12 rolls (long) = 18 18 x $34.95= $629.10

44. Note 10: Volume of Pyramids, cones & Spheres V = 1/3 A × h A = area of base h = perpendicular height V = 1/3 A × h = 1/3 (5m×4m)×8m 5 m 4 m Apex = 53.3 m 3

45. Volume of Cones V = 1/3 × πr 2 × h A = πr 2 (area of base) h = perpendicular height = 1/3 × π×(1.5cm) 2 × 9cm = 21.21 cm 3 (4 sf) Vertex A cone is a pyramid on a circular base 1.5 cm 9 cm V = 1/3 × πr 2 × h

46. Spheres A sphere is a perfectly round ball. It has only one measurement: the radius, r. The volume of a sphere is: V = 4/3πr 3 Ex 13.02 pg 165-167 IWB Fundamentals Ex 10.02 pg 296-299 IWB Fundamentals Ex 10.02 pg 296-299

47. Starter The sonar of a whale can be heard within a sphere of diameter 0.150 km. How many litres of water are contained in this sphere? = 1.77 x 10 9 L V = 4/3 π(75 m) 3 V = 1767146 m 3 = 1.77 x 10 6 kL V =4/3 πr 3

48. Merit An excavator is digging a drainage trench. The shape is twice as wide as it is deep. This particular trench has a width across the top of 3.2 m and a length 245 m. What is the best model to calculate the volume of material removed? What volume of material must be moved to make this trench? 1.6 m 3.2 m * diagram not to scale Possible modelsHalf cylinder Trapezoidal Prism Volume = ½ x π (1.6) 2 x 245 = 985.2 m 3 IWB Fundamentals Ex 10.03 pg 301-304 IWB Fundamentals Ex 10.03 pg 301-304

49. Starter V with peel = 4/3 π (45) 3 = 381 704 mm 3 mm. V no peel = 4/3 π (40) 3 = 268 083 mm 3 V peel = V with peel – V no peel = 113 621 mm 3 = 114 000 mm 3

50. Note 11: Liquid Volume (Capacity) There are 2 ways in which we measure volume: Solid shapes have volume measured in cubic units (cm 3, m 3 …) Liquids have volume measured in litres or millilitres (mL) WeightLiquid VolumeEquivalent Solid Volume 1 gram1 mL1 cm 3 1 kg1 litre1000 cm 3 Metric system – Weight/volume conversions for water.

51. e.g. 600 ml = $ 0.83/0.6 L = $ 1.383 / L 1 L = $ 1.39/ L 2 L = $ 2.76/ 2 L = $ 1.38 / L

52. e.g. S.A. = 2(55 × 42) + 2(18 × 42) + (55 x 18) 55 cm 42 cm a.) Calculate the area of glass required for the fish tank b.) Calculate the volume of water in the tank. Give your answer to the nearest litre = 7122 cm 2 V tank = 55 × 42 × 18 = 41580 cm 3 V water = 4/5 (41580) = 33264 cm 3 = 33 L

53. Example: Estimate the weight of a 333 mL can of soda. How many cans would it take to fill a container which has a volume of 4300 cm 3 ? 333 grams + can Number of cans = 4300 333 = 12.91 ≈ 13 cans Ex 14.01 pg 173-175 IWB Fundamentals Ex 10.05 pg 315-318 IWB Fundamentals Ex 10.05 pg 315-318

54. Starter A time capsule is buried in the foundations of a new classroom block at JMC. It consists of a 20 cm cylinder, fitted at each end with a hemisphere. The total length is 28 cm. What is the capacity (in L) of the time capsule? 28 cm 20 cm Radius = 4 cm Volume (sphere) = 4/3πr 3 Volume (cylinder) = πr 2 x h = 268.08 cm 3 = 1005.31 cm 3 Total volume = 268.08 + 1005.31 = 1273.39 cm 3 Capacity = 1273.39 mL = 1.273 L

55. Which of these buckets has the largest mass? (assume volumes are the same) Water Popcorn Gold Coins

56. Note 12: Density The density of an object describes the ratio of its mass to a standard volume. The density of water is exactly 1. (1 cm 3 = 1 g) DV M M = V × D V = M D D = M V M = mass, V = Volume, D = Density

57. Density Examples: Concrete has a density of 2.3 g/cm 3. What is the mass of a block of concrete that measures 15 cm x 12 cm x 10 cm? Ans: Volume = 15 x 12 x 10 = 1800 cm 3 Mass = V x D = 1800 x 2.3 DV M = 4140 g (4.14 kg) M = V × D

58. Density Examples: A 10 kg block of gold has a volume of 518 cm 3. Calculate the density of gold in g/cm 3. Density = 10000 g 518 g/cm 3 10 kg = 10000 g = 19.3 g/cm 3 DV M D = M V Gamma Textbook Ex 14.02 pg 176-177 Gamma Textbook Ex 14.02 pg 176-177 IWB Fundamentals Ex 11.03 pg 331 - 334 Ex 11.01 pg 323 - 325 IWB Fundamentals Ex 11.03 pg 331 - 334 Ex 11.01 pg 323 - 325

59. Starter A gold wedding band with diameter 16 mm & a cross section as shown below shows the band is semi circular with a radius of 4 mm. Estimate the volume by imagining the ring cut and opened up. 16 mm 4 mm Density of gold is 19.3 g/cm 3 Estimate the value of the ring if it was melted down and recovered. Assume gold is currently traded for $59.70/g Length of ‘opened up’ ring = π × 1.6 cm = 5.0265 cm Volume of ‘opened up’ ring = ½ × π × 0.4 2 × 5.0265 cm = 1.263 cm 3 Mass = 1.263 cm 3 × 19.3g/cm 3 = 24.38 g Value = 24.38 g x $59.70 = $1456

60. Note 13: Time Equivalent times (in seconds) 1 minute = 1 hour = 1 day = 60 seconds 60 minutes = 60 x 60 seconds = 3600 seconds 24 hours = 24 x 60 x 60 seconds 24 hour time is represented using 4 digits 12 hour clock times are followed by am or pm e.g 0630 hours 6:30 am There are 365.242 days in a year. We account for this by adding an extra day every 4 years so that this error does not build up. = 86 400 seconds

61. International Time Zones

63. Measuring and Modelling Practical Situations. Apply the Skills we have learned Represent the situation using a Diagram – Assign appropriate measurements – Simplify and Clarify Use appropriate Units – Be consistent Identify regular shape(s) from your diagram Do Calculations Relate your answer back to the original problem