1 / 63

1.5 Measurement

1.5 Measurement. AS 90130 Internal (3 credits). Calculate the area of the following shapes. 2 cm. A = 9 cm 2. 3 cm. 4 cm. 6 cm. A = 12.6 cm 2. 4 cm. 5 cm. 5 cm. 3 cm. A = 45 cm 2. 2 rugby fields side by side. 5 mL. 8L. 3760m. 750m 2. 55m 2. 1600cm 3. 60W. 2m. 750mL.

ilyssa
Download Presentation

1.5 Measurement

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 1.5 Measurement AS 90130 Internal (3 credits)

  2. Calculate the area of the following shapes. 2 cm A = 9 cm2 3 cm 4 cm 6 cm A = 12.6 cm2 4 cm 5 cm 5 cm 3 cm A = 45 cm2

  3. 2 rugby fields side by side

  4. 5 mL 8L 3760m 750m2 55m2 1600cm3 60W 2m 750mL 80kg 1.3kg 40L -22°C 1500m 160kPa 2000cm2 37.4° 1m 50g 9.85s 180mm

  5. Try These – Metric Conversions 55000 5.5 ha = _______m2 25 m = ________ cm 345 m = _______km 4321 mm = _______m 0.12 m = ________ mm 237 cm = _________ m 998 km = ____________ cm 199500 cm = ________ km 2500 5700 5.7 L = ______mL 3400kg = ______T 0.42 T = _______kg 4300 m2 = ______ha 3 m2 = ________cm2 400000mm2 ____m2 • 4500cm3 _______ L .345 3.4 4.321 420 120 0.430 2.37 30000 99 800 000 0.4 1.995 4.5 IWB Fundamentals Ex. 8.02 pg 220

  6. Starter 12 ha of land is subdivided in 650 m2 lots. How many lots of this size can be created? 120000 12 ha = _______ m2 184 lots can be created. 120000 m2 = 184.6 650 m2 A teaspoon holds 5 cm3. How many teaspoons are needed to measure out exactly 1 m3 of water? 5 5 cm3 = ____ mL 1000 L= 200 000 teaspoons .005 L .005 = ____ L 1000 1 m3 = ____ L

  7. 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.

  8. 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 17.5 km 16.5 km

  9. 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

  10. e.g. At the Otagovs 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

  11. Exercise Questions 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°

  12. Note 2: Areas 2 mm

  13. Note 2: Areas 5 cm

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

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

  16. 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

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

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

  19. 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. 2x = 4.8 m Calculate the height (x) and diameter (2x) Length of steel required = 21x + 2(2πx) x = 33.57 x Amount of steel to purchase = $350 $4.35 33.57 x = 80.46 Hint – there are 21 uprights and 2 circles = 80.46 m x = 2.40 m

  20. 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.

  21. e.g. Find the area of this shape Area of compound shape: = area of Rectangle + area of Triangle + area of semi-circle 2 cm 5 cm 4 cm Area = b × h + ½ b × h + ½ π × r2 = 4 × 5 + ½ 4 × 2 + 0.5 π × (2)2 = 30.3 cm2 (1 dp)

  22. 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 100 m d = ? Acircle + Arectangle = Atotal 3183.1 + 6366.2 = 9549.3 m2 (1dp)

  23. 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

  24. Note 6: Area of a Sector/ Arc Length πr2 Recall: Area of a circle = • The sector is part of a circle • Area of a sector = x° × πr2 360° x° r e.g. 35° Area of a sector = 35° × π× (6 cm)2 360° = 11.0 cm2 6 cm

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

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

  27. 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. C = 2πr 25.4 cm = 2π × r e.g. The circumference of a circle is 25.4 cm. Calculate the radius. r = 25.4 cm 2π r = 4.0 cm (1 dp)

  28. e.g. The area of a circle is 35.6 cm2 Calculate the diameter. A = πr2 35.6 = π× r2 35.6 = r2 π r = 35.6 π r = 3.37 cm Diameter = 6.7 cm (1 dp) √ IWBFundamentals Ex9.04pg 258-259

  29. 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. 2.5 m 18 m 2 m 24 m

  30. 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. Area courtyard = (24 ×18) – 4(π×1.252) – 22 = 408.37 m2 2.5 m No. of moulds = 408.37 m2/ (.25 × .15) 18 m = 10 890 2 m Total cost = 10890 × $ 0.83 ≈ $9100 24 m

  31. 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 a a SA (cube) = 6a2

  32. Surface Area Calculate the surface area of this triangular prism 6.4 cm 5 How many faces are there? 2 Triangles: 2 × (1/2 b × h) = 2 × 0.5 × 5cm × 4 cm = 20 cm2 Left Rectangle: b × h = 7 cm × 4 cm = 28 cm2 Right Rectangle: b × h = 7 cm × 6.4 cm = 44.8 cm2 4 cm Base Rectangle: b × h = 7 cm × 5 cm = 35 cm2 7 cm Total surface Area = 127.8 cm2 5 cm

  33. Surface Area SA (cone)= the curved surface + the circular base = π r l + π r2 SA (sphere) =4πr2 l r r

  34. Surface Area (cylinder) SA (cylinder) = area of rectangle + 2 ×(area of circle) = 2πr × h + 2 ×(πr2) = 2πrh + 2πr2 r SA (cylinder) = 2πr (h + r) Surface area depends on whether the ends are open or not. Open cylinders are called pipes. h C = 2πr Textbook Ex 12.04 pg 156-157 Ex 12.05 pg 157-160 IWBFundamentals Ex 9.06 pg 271-273 Ex 9.07 pg 275-279

  35. Starter A gardener wishes to sow grass seed on his land. 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? 15 m 7.0 m 9.2 m 7.6 m 4.1 m 22.0 m Total Area = 22×15 – (4.1×9.2) – (7.0×7.6) = 239.08 m2 239.08 m2 / 26 = 9.2 1kg bags = 239 m2 (3 sf) 9 x $8.95 + 1 x $5.95 = $86.50

  36. 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

  37. Examples: • Volume = (b × h) × L = 4 m × 4 m × 10 m = 160 m3 • Volume = (1/2 b×h) × L = ½× 4cm ×5cm × 10 cm = 70 cm3 10 m 4 m 4 cm 7 cm 5 cm

  38. Cylinder – A circular Prism Volume (cylinder) = πr2× h • 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

  39. 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. Volume = ½ × 13 × 12 × 8 = 624 m3 12 m 8.0 m 624 m3 × 5 = 3120 m3/hr 13 m 3 × = 52 m3/min

  40. Starter B 10cm 1 cm A 8 cm 10cm 10cm 125 cm 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 cm3 B = 1000 cm3

  41. 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 Strips per roll – 8 m ÷ 2.5 m = 3 / roll 4.2 m Long Wall – 9 m ÷ 0.5 m = 18 strips across 18 ÷3 = 6 rolls per side = 12 rolls

  42. The walls of a lounge are to be wallpapered. The room’s dimensions are depicted below. Merit 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. Short Wall – 4.2 m ÷ 0.5 m 2.5 m 9 m = 8.4 (9) strips across 4.2 m 9 ÷ 3 (strips per roll) = 3 rolls per side = 6 rolls + 12 rolls (long) 18 x $34.95= $629.10 = 18

  43. 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

  44. Volume of Cones 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 = 21.21 cm3 (4 sf) 1.5 cm

  45. 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 IWBFundamentals Ex 10.02 pg 296-299 Ex 13.02 pg 165-167

  46. 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? V =4/3 πr3 V = 4/3 π(75 m)3 V = 1767146 m3 = 1.77 x 106 kL = 1.77 x 109 L

  47. 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? 3.2 m Possible models Half cylinder Trapezoidal Prism 1.6 m Volume = ½ x π (1.6)2 x 245 = 985.2 m3 IWBFundamentals Ex 10.03 pg 301-304 * diagram not to scale

  48. Starter 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

  49. Note 11: 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.

More Related