Measurement

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

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

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 3789600cm to km

Homework Book Page 159-160

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?

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

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

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

Examples: Change 74m/s into km/h 74m/s = 3600x74m/h = 266400m/h = 266.4km/h

Homework Book Page 162-163

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?

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

Homework Book Page 164 - 166

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

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)

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° - 120 ° = 240 ° Arc Length = x/360 x 2πr = 240/360 x 2 x π x 6m = 25.1m (1dp) Perimeter = 2 x 6m + 25.1m = 37.1m

Homework Book Page 167 - 169

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.

Note 5: Area Area is measured in square units.

Examples converting units:

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

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

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)

Homework Book Page 170 – 171

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?

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.

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

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

Homework Book Page 172 – 174

STARTERS Find the area of Trapezium = 750 Rectangle = 1000 Half Circle = 628.3 Area = 750 + 1000 - 628.3 = 1121.7 cm2

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

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

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

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

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

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)

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

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 = 24 + 24 + 120 + 72 + 96 = 336 cm2

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 0.82 = 2.01m2 Bottom circular end = 2.01m2 Curved surface = 2 x  x 0.8 x 3 = 15.08m2 Total surface area = 2.01 + 2.01 + 15.08 = 19.1m2

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 = 4071.5m2

Homework Book Page 172 – 174

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)

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

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

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 = 36.19 cm 3 (4 sf) 8 cm

Homework Book Page 172 – 174

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

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

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

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

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

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

Homework Book Page 182

Starter Vwith peel = 4/3 π (45)3 Vno peel = 4/3 π (40)3 mm. Vwith peel = 4/3 π (45)3 = 381 704 mm3 Vno peel = 4/3 π (40)3 = 268 083 mm3 Vpeel = Vwith peel – Vno peel = 113 621 mm3 = 114 000 mm3

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

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 = 268.08 cm3 Volume (cylinder) = πr2 x h = 1005.31 cm3 Radius = 4 cm Total volume = 268.08 + 1005.31 = 1273.39 cm3

Homework Book Page 183

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 = 1.263 cm3 × 19.3g/cm3 = 24.38 cm3 Length of ‘opened up’ ring = π × 1.6 cm Value = 24.38 g x $59.70 = 5.0265 cm = $1456 Volume of ‘opened up’ ring = ½ × π × 0.42 × 5.0265 cm = 1.263 cm3

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

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

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 = 41580 cm3 42 cm Vwater = 4/5 (41580) = 33264 cm3 = 33 L 55 cm

Homework Book Page 185

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 = 41580 cm3 42 cm Vwater = 4/5 x 41580 = 33264 cm3 = 33 L 55 cm

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 = 86 400 seconds 24 hour time is represented using 4 digits 12 hour clock times are followed by am or pm 0630 hours 6:30 am

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

Homework Book Page 185

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 = 908-616 = 292 cm2

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.

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 17 + 0.5 = 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

Example 2: At the Otago vs Canterbury game at the stadium it was reported that 23500 people attended. Give the limits of accuracy for the number of people attending the game? The upper limit is 23500 + 50 = 23550 The lower limit is 23500 – 50 = 23450 Therefore the limits of accuracy are: 23450 ≤ People < 23550

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 ≤ 397.5 mm 3.5 seconds ≤ x ≤ 4.5 seconds 45 g ≤ x ≤ 55 g 5885 kg ≤ x ≤ 5895 kg Homework pg 162-163 815 cm ≤ x ≤ 825 cm 91.5 kg ≤ x ≤ 92.5 kg 89.05° ≤ x ≤ 89.15° 72