STARTERS

1. 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)

2. 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

3. 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

4. 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

5. 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

6. Homework Book Page 172 – 174

7. 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)

8. Note 9: Volume of Prisms 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 Volume = Area of cross section × Length

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

10. Cylinder – A Circular Prism Volume = Area of cross section × Length V = πr2× h = π(1.2cm)2 × 8cm 1.2 cm = 36.19 cm 3 (4 sf) 8 cm Volume (cylinder) = πr2× h

11. Homework Book Page 172 – 174

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

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

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

15. 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

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

17. Example 2: 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

18. Homework Book Page 182

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

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

21. 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? Volume (sphere) = 4/3πr3 = 268.08 cm3 Volume (cylinder) = πr2 x h 28 cm = 1005.31 cm3 Radius = 4 cm Total volume = 268.08 + 1005.31 20 cm = 1273.39 cm3

22. Homework Book Page 183

23. 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

24. Note 12: Liquid Volume (Capacity) Metric system – Weight/volume conversions for water. 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)

25. Example: 600 ml = $ 0.83/0.6 L 2 = $ 1.383 / L 3 1 L = $ 1.39/ L 2 L = $ 2.76/ 2 L 1 = $ 1.38 / L

26. Example 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

27. Homework Book Page 185

28. Starter 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

29. Note 13: Time 60 seconds 60 mins = 60 x 60 secs = 3600 seconds 24 hours = 24 x 60 x 60 seconds = 86 400 seconds 0630 hours 6:30 am 24 hour time is represented using 4 digits 12 hour clock times are followed by am or pm Equivalent times (in seconds) 1 minute = 1 hour = 1 day =

30. 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?

31. Homework Book Page 185

32. Starter 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 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 = 292 cm2

33. 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.

34. 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 To find the upper limit, add 5 to the nearest significant place. To find the lower limit, minus 5 to the nearest significant place.

35. 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

36. Exercise Questions 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° 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°