STARTERS Calculate the length if the area is 60cm 2 A rotating irrigation jet waters an area of 2600m 2. 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 = 24cm 2 Back triangle = ½ x 6 x 8 = 24cm 2 Left rectangle = 10 x 12 = 120cm 2 Bottom rectangle = 6 x 12 = 72cm 2 Right rectangle = 8 x 12 = 96cm 2 Total surface area = = 336 cm 2

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

Example 3: Calculate the surface area of this sphere SA= 4  r 2 = 4 x  x 18 2 = m 2 The formula for the surface area of a sphere is: SA = 4  r 2

Homework Book Page 172 – 174

STARTERS Calculate the length if the area is 60cm 2 A rotating irrigation jet waters an area of 2600m 2. If you did not want to get wet, how far would you have to stand from the jet? (to the nearest metre)

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

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 = (½b×h) × L = ½ × 4cm × 5cm × 7cm = 70 cm 3

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 = cm 3 (4 sf)

Homework Book Page 172 – 174

Starter 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 m 3 × 5 = 3120 m 3 /hr = 52 m 3 /min

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

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 = 21.2 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

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

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

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

Homework Book Page 182

Starter V with peel = 4 / 3 π (45) 3 = mm 3 mm. V no peel = 4 / 3 π (40) 3 = mm 3 V peel = V with peel – V no peel = mm 3 = mm 3

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 Radius = 4 cm Volume (sphere) = 4 / 3 πr 3 Volume (cylinder) = πr 2 x h = cm 3 = cm 3 Total volume = = cm 3

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. 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 = cm Volume of ‘opened up’ ring = ½ × π × × cm = cm 3 Mass = cm 3 × 19.3g/cm 3 = cm 3 Value = g x $59.70 = $1456

Note 12: 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.

Example: 600 ml = $ 0.83/0.6 L = $ / L 1 L = $ 1.39/ L 2 L = $ 2.76/ 2 L = $ 1.38 / L

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

Homework Book Page 185

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

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

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? HoursMinutes 8:23am :15pm415 Total time752

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 = 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 = = 292 cm 2

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 = The lower limit is 17 – 0.5 = Therefore the limits of accuracy are 16.5 km ≤ Bluff < 17.5 km 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 16.5 km 17.5 km

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 = The lower limit is – 50 = Therefore the limits of accuracy are: ≤ People < 23550

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 mm ≤ x ≤ 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°