1. Measurement

2. Note 1 : Measurement Systems
In NZ the measurement system used is the metric system. The units relate directly to each other.

3. Capacity Basic Unit Symbol Distance Mass (weight) Temperature Time
Area
Land Area
Volume
metre m
gram g
litre L °C
degrees celsius
seconds/minutes
s/min
square metres m²
hectares ha
cubic metres m³

4. To change within a unit from one prefix to another prefix, we either multiply or divide by a power of 10.
smaller to larger unit divide by a power of 10
larger to smaller unit multiply by a power of 10
Examples: convert the following
5.76m to cm
489mL to L
cm to km

5. Homework Book
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6. STARTERS Convert the following: 59mL to L 4200kg to tonne
11m465mm to cm
A dairy stores milk in 5 litre containers. How many 350mL milkshakes can be made from one of these containers?

7. Note 2: Derived Units
Derived units show comparisons between two related measures. For example, speed is a measure of how much distance changes over time. The units for speed are m/s or km/h.
Distance
Speed
Time

8. Examples:
A cyclist travels at a steady speed of 24km/h for 40 minutes. How far did the cyclist travel?
40 minutes = 2/3hour
Distance = speed x time
= 24 x 2/3
= 16 km

9. Changing from one speed unit to another
Note: 1km = 1000m
1 hour = 3600sec
Examples:
Change 45km/h into m/s
45km/h = 45 x 1000m/h
= 45000m/h
= 45000m/3600s
= 12.5 m/s

10. Examples:
Change 74m/s into km/h
74m/s = 3600x74m/h
= m/h
= 266.4km/h

11. Homework Book
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12. STARTERS Convert the following: 19m/s to km/h
A truck travels at an average speed of 75km/h for a distance of 300km. What time does the journey take?
A Boeing 747 has a cruising speed of 910km/h. Change this into m/s?

13. Note 3: Perimeter
The perimeter is the distance around the outside of a shape. Start at one corner and work around the shape calculating any missing sides.
5 cm
6 cm
5 cm
2 cm
Perimeter = 5cm + 3cm + 6cm + 2cm + 11cm + 5cm
= 32cm

14. Homework Book
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15. STARTERS Calculate the perimeter of
The plan shows an L-shaped paddock. Calculate the total cost of fencing it at $24/m

16. Note 4: Circumference
The perimeter of a circle is called the circumference.
The formula for the circumference is:
C = πd or C = 2πr
where d = diameter r = radius.
Example: Find the circumference of
C = 2πr
= 2 x π x 8cm
= 50.3cm (1dp)

17. Example: Find the perimeter of the sector
If a sector has an angle at the centre equal to x, then the arc length is x/360 of the circumference.
Example: Find the perimeter of the sector
Angle of sector = 360° ° = 240 °
Arc Length = x/360 x 2πr
= 240/360 x 2 x π x 6m
= 25.1m (1dp)
Perimeter = 2 x 6m m
= 37.1m

18. Homework Book
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19. STARTERS Calculate the perimeter of
Paul goes for a short cycle ride. Each wheel on his bike has a radius of 27cm. His distance counter tells him the wheel has rotated 650 times. Find how far he has travelled in metres.

20. Note 5: Area
Area is measured in square units.

21. Examples converting units:

22. Examples of converting units
5.6cm2 to mm2
Big Small x
5.6cm2 = 5.6 x 100
= 560mm2
396000cm2 to m2
Small Big ÷
396000cm2 = ÷10000
= 39.6m2

23. Examples: Calculate the area of these shapes
Area = ½  base  height
= ½  17  10
= 85 m2
½  12  7 = 42 m2

24. Area = ½ (sum of bases) x height
Radius = 7 ÷ 2 = 3.5 cm
Area = x/360 x π x r²
= 180/360 x π x 3.5²
= 19.2 m² (1dp)
Area = ½ (sum of bases) x height
= ½(9 + 12) x 7
= 73.5 m² (1dp)

25. Homework Book
Page 170 – 171

26. STARTERS Find the area of
A chocolate bar is wrapped in a rectangular piece of foil measuring 10cm by 15cm.
Calculate the area of the piece of foil.
How many pieces could be cut out from a larger sheet of foil measuring 120cm by 75cm?

27. Note 6: Compound Area
Compound shapes are made up of more than one mathematical shape.
To find the area of a compound shape, find the areas of each individual shapes and either add or subtract as you need to.

28. Examples: find the area of
Area splits into a rectangle and a triangle
Area = Area rectangle + area triangle
= b  h + ½ b  h
= 4  5 + ½  4  2
= 24cm2

29. Area splits into a rectangle with another rectangle taken away
Area = area big rectangle – area small rectangle
= b  h - b  h
= 6  4 – 3  2
= 18m2

30. Homework Book
Page 172 – 174

31. STARTERS Find the area of Trapezium = 750 Rectangle = 1000
Half Circle = 628.3
Area =
= cm2

32. Note 7: Finding missing parts of shapes
To find missing sides of shapes, rearrange the formulas .
Example 1: The area of the triangle is 135m2.
Calculate the height of the triangle.
Area = ½  base  height
135 = ½  18  x
135 = 9x
x = 15m

33. Example 2: Calculate the radius of a circle with an area of 65cm2.
Area = π  r2
65 = π  r2
r2 = 65/π
r = √65/π
= 4.5 cm

34. EXERCISES:
Each of these shapes has an area of 60cm2. Calculate the lengths marked x.
10cm
15cm
2.5cm
√60 =7.7cm

35. Calculate the radii of these circles with the given areas.
EXERCISES:
Calculate the radii of these circles with the given areas.
18.7 cm
3.87 m
1.38 cm
0.798 km

36. Radius Diameter circumference EXERCISES:
A circle has an area of 39.47m2. Calculate:
Radius
Diameter
circumference
3.55 m
7.09 m
22.27 m

37. STARTERS Calculate the length if the area is 60cm2
A rotating irrigation jet waters an area of 2600m2. If you did not want to get wet, how far would you have to stand from the jet? (to the nearest metre)

38. Note 8: Surface Area The surface area of a solid is the sum of the
areas of all its faces .
Example 1: Calculate the surface area of this triangular prism
The prism has 5 faces
2 triangles
3 rectangles

39. Example 1: Calculate the surface area of this triangular prism
Front triangle = ½ x 6 x 8 = 24cm2
Back triangle = ½ x 6 x 8 = 24cm2
Left rectangle = 10 x 12 = 120cm2
Bottom rectangle = 6 x 12 = 72cm2
Right rectangle = 8 x 12 = 96cm2
Total surface area =
= 336 cm2

40. Example 2: Calculate the surface area of this cylinder
The cylinder has 3 faces:
2 circular ends
A curved rectangular face
Radius = 1.6 ÷ 2 = 0.8m
Top circular end = r2
=  x = 2.01m2
Bottom circular end = 2.01m2
Curved surface = 2 x  x 0.8 x = 15.08m2
Total surface area =
= 19.1m2

41. Example 3: Calculate the surface area of this sphere
The formula for the surface area of a sphere is:
SA = 4r2
Example 3: Calculate the surface area of this sphere
SA = 4r2
= 4 x  x 182
= m2

42. Homework Book
Page 172 – 174

43. STARTERS Calculate the length if the area is 60cm2
A rotating irrigation jet waters an area of 2600m2. If you did not want to get wet, how far would you have to stand from the jet? (to the nearest metre)

44. Volume = Area of cross section × Length
Note 9: Volume of Prisms
Volume = Area of cross section × Length
The volume of a solid figure is the amount of space it occupies. It is measured in cubic centimetres, cm3 or cubic metres, m3
Area L

45. Examples: Volume = (b × h) × L = 4 m × 4 m × 10 m = 160 m3
4 cm
5 cm
7 cm
10 m
4 m
Volume = (½b×h) × L
= ½ × 4cm × 5cm × 7cm
= 70 cm3

46. Cylinder – A Circular Prism
Volume = Area of cross section × Length
Volume (cylinder) = πr2× h
V = πr2× h
= π(1.2cm)2 × 8cm
1.2 cm
= cm 3 (4 sf)
8 cm

47. Homework Book
Page 172 – 174

48. Starter = 624 m3 624 m3 × 5 = 3120 m3/hr = 52 m3/min
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.
Volume = ½ × 13 × 12 × 8
= 624 m3
12 m
8.0 m
624 m3 × 5 = 3120 m3/hr
13 m
= 52 m3/min

49. Note 10: Volume of Pyramids, cones & Spheres
V = 1/3 A × h
A = area of base
h = perpendicular height
Apex
8 m
(altitude)
V = 1/3 A × h
= 1/3 (5m×4m)×8m
= 53.3 m3
5 m
4 m

50. Volume of Cones = 1/3 × π×(1.5cm)2 × 9cm
A cone is a pyramid on a circular base
V = 1/3 × πr2× h A = πr2 (area of base) h = perpendicular height
Vertex
9 cm
V = 1/3 × πr2× h
= 1/3 × π×(1.5cm)2 × 9cm
= 21.2 cm3 (4 sf)
1.5 cm

51. Spheres A sphere is a perfectly round ball.
It has only one measurement: the radius, r.
The volume of a sphere is:
V = 4/3πr3

52. Example 1: Calculate the volume of the pyramid below V = 1/3 A x h
= 1/3 x 5m x 4m x 6m
= 40m3

53. Example 2: V = 4/3πr3 V = 4/3 × π(11.5cm)3 = 6371 cm 3
Calculate the volume of a football with a diameter of 23cm
V = 4/3πr3
Radius = 23 ÷ 2 = 11.5cm
V = 4/3 × π(11.5cm)3
= 6371 cm 3

54. Homework Book
Page 182

55. Starter Vwith peel = 4/3 π (45)3 Vno peel = 4/3 π (40)3
mm.
Vwith peel = 4/3 π (45)3
= mm3
Vno peel = 4/3 π (40)3
= mm3
Vpeel = Vwith peel – Vno peel
= mm3
= mm3

56. Note 11: Compound Volume
Divide compound solids into solids, such as, prisms and cylinders and add or subtract as for area compounds.

57. Example:
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 volume of the time capsule?
28 cm
20 cm
Volume (sphere) = 4/3πr3
= cm3
Volume (cylinder) = πr2 x h
= cm3
Radius =
4 cm
Total volume =
= cm3

58. Homework Book
Page 183

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.
Density of gold is 19.3 g/cm3
16 mm
Estimate the value of the ring if it was melted down and recovered. Assume gold is currently traded for $59.70/g
4 mm
Mass = cm3 × 19.3g/cm3
= cm3
Length of ‘opened up’ ring = π × 1.6 cm
Value = g x $59.70
= cm
= $1456
Volume of ‘opened up’ ring = ½ × π × 0.42 × cm
= cm3

60. Note 12: Liquid Volume (Capacity)
There are 2 ways in which we measure volume:
Solid shapes have volume measured in cubic units (cm3, m3 …)
Liquids have volume measured in litres or millilitres (mL)
Metric system – Weight/volume conversions for water.
Weight
Liquid Volume
Equivalent Solid Volume
1 gram
1 mL
1 cm3
1 kg
1 litre
1000 cm3

61. 2 3 1 Example: 600 ml = $ 0.83/0.6 L = $ 1.383 / L 1 L = $ 1.39/ L

62. Example = 33 L = 7122 cm2 Vtank = 55 × 42 × 18 = 41580 cm3
Calculate the area of glass required for the fish tank
S.A. = 2(55 × 42) + 2(18 × 42) + (55 x 18)
= 7122 cm2
Calculate the volume of water in the tank. Give your answer to the nearest litre
Vtank = 55 × 42 × 18
= cm3
42 cm
Vwater = 4/5 (41580)
= cm3
= 33 L
55 cm

63. Homework Book
Page 185

64. Starter = 33 L = 7122 cm2 Vtank = 55 × 42 × 18 = 41580 cm3
Calculate the area of glass required for the fish tank
S.A. = 2(55 × 42) + 2(18 × 42) + (55 x 18)
= 7122 cm2
Calculate the volume of water in the tank. Give your answer to the nearest litre
Vtank = 55 × 42 × 18
= cm3
42 cm
Vwater = 4/5 x 41580
= cm3
= 33 L
55 cm

65. Note 13: Time Equivalent times (in seconds) 1 minute = 1 hour =
1 day =
60 seconds
60 mins = 60 x 60 secs = 3600 seconds
24 hours = 24 x 60 x 60 seconds
= seconds
24 hour time is represented using 4 digits
12 hour clock times are followed by am or pm
0630 hours
6:30 am

66. Example:
If I arrive at school at 8:23am and leave at 4:15pm. How long in hours and minutes do I spend at school?
Hours
Minutes
8:23am - 12 3 37
12 - 4:15pm 4 15
Total time 7 52

67. Homework Book
Page 185

68. 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 = πr2
r = 14 cm
A = π (14)2
A = 616 cm2
Area of porthole = πr2 , r = 17 cm
(including frame) = 908 cm2
Area of frame =
= 292 cm2

69. Note 14: 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.

70. To find the upper limit, add 5 to the nearest significant place.
To find the lower limit, minus 5 to the nearest significant place.
Example 1:
The distance to Bluff on a signpost reads 17 km.
The upper limit is =
The lower limit is 17 – 0.5 =
Therefore the limits of accuracy are 16.5 km ≤ Bluff < 17.5 km
17.5 km
16.5 km

71. Example 2:
At the Otago vs Canterbury game at the stadium it was reported that people attended. Give the limits of accuracy for the number of people attending the game?
The upper limit is = 23550
The lower limit is – 50 = 23450
Therefore the limits of accuracy are:
23450 ≤ People < 23550

72. Exercise Questions 67.5 mm ≤ x ≤ 68.5 mm 396.5 mm ≤ x ≤ 397.5 mm
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 ≤ mm
3.5 seconds ≤ x ≤ 4.5 seconds
45 g ≤ x ≤ 55 g
5885 kg ≤ x ≤ 5895 kg
Homework pg
815 cm ≤ x ≤ 825 cm
91.5 kg ≤ x ≤ 92.5 kg
89.05° ≤ x ≤ 89.15° 72