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KS4 Mathematics

KS4 Mathematics. S8 Measures. S8 Measures. Contents. A. S8.2 Accuracy in measurement. A. S8.1 Converting units. S8.3 Calculations involving bounds. A. S8.4 Compound measures. A. S8.5 Bearings. A. Converting units.

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KS4 Mathematics

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  1. KS4 Mathematics S8 Measures

  2. S8 Measures Contents • A S8.2 Accuracy in measurement • A S8.1 Converting units S8.3 Calculations involving bounds • A S8.4 Compound measures • A S8.5 Bearings • A

  3. Converting units It is important to be able to convert between a variety of units quickly and accurately. 1) When using a formula For example, if we are using a formula to find the volume of an object we must make sure that all the lengths are written using the same units before using the formula. 2) When comparing measurements For example, suppose one mother gives the weight of her baby in pounds and another mother gives the weight of her baby in kilograms. How can we compare the babies’ weights? To compare the weights we convert them to the same unit. We usually use metric units for calculations.

  4. Metric units Kilo- The base unit for length is metre. The metric system of measurement is based on powers of ten and uses the following prefixes: meaning one thousand Centi- meaning one hundredth Milli- meaning one thousandth Micro- meaning one millionth These prefixes are then followed by a base unit. metre. The base unit for mass is gram. gram. The base unit for capacity is litre. litre.

  5. Metric units of length, mass and capacity Length Mass Capacity and Volume 1 km = 1000 m 1 tonne = 1000 kg 1 litre = 1000 ml 1 m = 100 cm 1 cl = 10 ml 1 kg = 1000 g 1 m = 1000 mm 1 m3 = 1000 litres 1 g = 1000 mg 1 cm = 10 mm 1 cm3 = 1 ml You should know the following metric conversions for length, mass and capacity (liquid volume):

  6. Metric units of area and volume The metric units for area are mm2, cm2, m2 and km2. 1 m2 = 100 cm × 100 cm 1 cm2 = 10 mm × 10 mm 1 m2 = 10,000 cm2 1 cm2 = 100 mm2 The metric units for volume are mm3, cm3, m3 and km3. 1 m3 = 100 cm × 100 cm × 100 cm 1 m3 = 1,000,000 cm3 1 cm3 = 10 mm × 10 mm × 10 mm 1 cm3 = 1000 mm3

  7. Converting metric units To convert from a larger metric unit to a smaller one we need to _______ by 10, 100, or 1000. multiply For example, Convert 0.43 kg to grams 0.43 kg = 0.43 × 1000 g = 430 g To convert from a smaller metric unit to a larger one we need to _______ by 10, 100, or 1000. divide For example, Convert 7.6 cm to metres 7.6 cm = 7.6 ÷ 100 m = 0.076 m

  8. Converting metric units

  9. Ordering units

  10. Imperial units 1 foot = ___ inches Imperial units are still frequently used and you should be aware of the following imperial conversions: 12 1 yard = ___ feet 3 1 pound = ___ ounces 16 1 stone = ___ pounds 14 1 gallon = ___ pints 8

  11. Converting imperial units 5 miles is about ___ kilometres We can convert between metric and imperial units using the following approximate conversions: 8 1 foot is about ___ centimetres 30 1 inch is about ___ centimetres 2.5 1 kilogram is about ___ pounds 2.2 1 gallon is about ___ litres 4.5 1 litre is about ____ pints 1.75

  12. Converting imperial units 1 kilogram is about 2.2 pounds. About how many kilograms are there in 1 stone? 1 stone = 14 pounds Every 2.2 pounds is worth about 1 kilogram. This means that we have to divide 14 pounds by 2.2 to find the equivalent number of kilograms. 14 pounds ≈ 14 ÷ 2.2 kilograms = 6.4 kilograms 1 stone is about 6.4 kilograms

  13. Metric and imperial conversions

  14. Spider diagram

  15. S8 Measures Contents S8.1 Converting units • A • A S8.2 Accuracy in measurement S8.3 Calculations involving bounds • A S8.4 Compound measures • A S8.5 Bearings • A

  16. Continuous measurements 13 13 13 14 14 14 15 15 15 The continuous nature of measurements means that they can never be exact. There is always an element of rounding involved. If we were measuring the length of a pencil, for example, we would measure to the nearest cm or mm. These three pencils all measure 14 cm to the nearest cm: The length of a pencil given as 14 cm to the nearest cm can be any length between 13.5 cm and 14.5 cm.

  17. Accuracy in measurement Remember, any measurement given to the nearest whole unit could be up to half a unit longer or shorter. Suppose we are told that a pencil measures 14.2 cm. Can we assume that this measurement is exact? No, although this measurement has been given to a higher degree of accuracy it is not exact. What is the shortest and longest length it could be? The length l of a pencil given as 14.2 cm, to the nearest 0.1 cm, could be anywhere in the range 14.15 cm ≤ l <14.25 cm

  18. Upper and lower bounds When we give a range for a measurement as in: This is an inequality 14.15 cm ≤ length <14.25 cm this value is called the lower bound … … and this value is called the upper bound. The length could be equal to 14.15 cm so we use a greater than or equal to symbol. If the length was equal to 14.25 cm however, it would have been rounded up to 14.3 cm. The length is therefore “strictly less than” 14.25 cm and so we use the < symbol.

  19. Upper and lower bounds

  20. S8 Measures Contents S8.1 Converting units • A S8.2 Accuracy in measurement • A S8.3 Calculations involving bounds • A S8.4 Compound measures • A S8.5 Bearings • A

  21. Adding measures The following triangle has sides of length 3 cm, 4 cm and 5 cm. 5 cm 3 cm What is the range of possible lengths for the perimeter? 4 cm The least the lengths could be is 2.5 cm, 3.5 cm and 4.5 cm. The smallest possible perimeter = 2.5 + 3.5 + 4.5 = 10.5 cm The most the lengths could be is 3.5 cm, 4.5 cm and 5.5 cm. The largest possible perimeter = 3.5 + 4.5 + 5.5 = 13.5 cm

  22. Adding measures The following triangle has sides of length 3 cm, 4 cm and 5 cm. 5 cm 3 cm What is the range of possible lengths for the perimeter? 4 cm The range of possible values for the perimeter is 10.5 cm≤perimeter < 13.5 cm Notice that the more lengths that are added together the greater the error on either side.

  23. Adding measures When a calculation involves adding two or more measurements together: The lower bound is found by adding the lower bounds together The upper bound is found by adding the upper bounds together

  24. Calculations involving bounds A piece of wood measuring 170 cm has a piece of length 50 cm cut off of it. These are given to the nearest 10 cm. What is the range of possible sizes for the remaining piece? The original piece of wood could be between 165 cm and 175 cm. The piece of wood that is cut off could be between 45 cm and 55 cm. The smallest possible size of the remaining piece is: 165 – 55 = 110 cm

  25. Calculations involving bounds A piece of wood measuring 170 cm has a piece of length 50 cm cut off of it. These are given to the nearest 10 cm. What is the range of possible sizes for the remaining piece? The original piece of wood could be between 165 cm and 175 cm. The piece of wood that is cut off could be between 45 cm and 55 cm. The largest possible size of the remaining piece is: 175 – 45 = 130 cm

  26. Calculations involving bounds A piece of wood measuring 170 cm has a piece of length 50 cm cut off of it. These are given to the nearest 10 cm. What is the range of possible sizes for the remaining piece? The original piece of wood could be between 165 cm and 175 cm. The piece of wood that is cut off could be between 45 cm and 55 cm. The range of possible sizes for the remaining piece is: 110 cm≤length < 130 cm

  27. Calculations involving bounds The angles in the following diagram are rounded to the nearest degree: This angle could be between 34.5° and 35.5°. This angle could be between 77.5° and 78.5°. a 78° 35° What is the range of possible values for the angle a? The smallest a could be is 180° – (35.5° + 78.5°) = 66° The largest a could be is 180° – (34.5° + 77.5°) = 68° The range of possible values for a is therefore, 66°≤a< 68°

  28. Subtracting measures When a calculation involves subtracting two measurements: The lower bound is found by: subtracting the upper bound from the lower bound The upper bound is found by: subtracting the lower bound from the upper bound

  29. Calculations involving bounds The dimensions of a small tile are given as 15 mm by 18 mm. 18 mm What is the smallest possible area the tile could have? 15 mm The smallest values for the length and the width of the tile are 14.5 mm by 17.5 mm. To calculate the smallest possible area we multiply these values together. Smallest possible area = 14.5 × 17.5 = 253.75 mm2

  30. Calculations involving bounds The dimensions of a small tile are given as 15 mm by 18 mm. 18 mm What is the largest possible area the tile could have? 15 mm The largest values for the length and the width of the tile are 15.5 mm by 18.5 mm. To calculate the largest possible area we multiply these values together. Largest possible area = 15.5 × 18.5 = 286.75 mm2

  31. Calculations involving bounds The dimensions of a small tile are given as 15 mm by 18 mm. 18 mm What is the range of possible values for the area? 15 mm The smallest value for the area is 253.75 mm2 and the largest value for the area is 286.75 mm2 The range of possible values for the area is therefore, 253.75 mm2 ≤area < 286.75 mm2

  32. Multiplying measures When a calculation involves multiplying two or more measurements together: The lower bound is found by: multiplying the lower bounds together The upper bound is found by: multiplying the upper bounds together

  33. Calculations involving bounds greatest possible average speed greatest distance 200.5 m = = = shortest time 27.75 s A boy runs 200 metres in 27.8 seconds. The distance is given to the nearest metre and the time is given to the nearest tenth of a second. What is his greatest possible average speed to 2 decimal places? The distance could be between 195.5 m and 200.5 m. The time taken could be between 27.75 s and 27.85 s. 7.23 m/s

  34. Dividing measures When a calculation involves dividing two measurements: The lower bound is found by: dividing the lower bound by the upper bound The upper bound is found by: dividing the upper bound by the lower bound

  35. S8 Measures Contents S8.1 Converting units • A S8.2 Accuracy in measurement • A S8.4 Compound measures S8.3 Calculations involving bounds • A • A S8.5 Bearings • A

  36. Compound measures Which is heavier: a kilogram of cotton wool or a kilogram of rocks?

  37. Compound measures Of course, a kilogram of rocks weighs the same as a kilogram of cotton wool – they both weigh one kilogram! If you thought that the rocks were heavier, you were probably thinking of their density rather than their weight. Density is an example of a compound measure. It is a measure of the mass of an object per unit volume. Density is usually measured in g/cm3 or kg/m3. It can also be measured in kg/l. A density of 1.2 g/cm3 means that every 1 cm3 of the material has a mass of 1.2 grams.

  38. Compound measures mass Density = volume 760 = 80 Which is heavier: 1 cm3 of cotton wool or 1 cm3 of rock? This is a more sensible question. Rock is denser than cotton wool and so 1 cm3 of rock is heavier than 1 cm3 of cotton wool. We can find the density of a given material using the following formula: For example, if a block of metal has a mass of 760 g and a volume of 80 cm3: Density = 9.5 g/cm3

  39. Compound measures distance travelled Speed = time taken We use compound measures when we are comparing how one measurement changes with another. When one measurement changes in direct proportion with another it is said to change at a constant rate. For example, suppose a man is running around a race track. The total distance he has run changes with time. The rate at which he runs is called his speed. Speed is usually measured in km/h, m/s or mph.

  40. Average speed total distance travelled Average speed = total time taken 1560 300 In many situations the speed is not constant. For example, the man running around the track will probably speed up or slow down as he runs. We can still calculate his average speed using the following formula: For example, if the man runs 1560 metres in 300 seconds Average speed = = 5.2 m/s

  41. Common compound measures distance time force mass surface area volume distance volume Commonly used compound measures include: Density Measured in g/cm3, kg/m3 or kg/l. Speed Measured in m/s, km/h or mph. Pressure Measured in N/m2 or N/cm2. Fuel consumption Measured in km/l or mpg.

  42. Pairs – compound measures

  43. Calculating densities

  44. Converting compound units What is 45 mph in m/s? When we convert compound units we usually have to do it in several steps, especially if both units are being changed. 45 mph stands for 45 miles per hour. We have to change the 45 miles into metres and 1 hour into seconds. 5 miles = 8 km So 1 mile = 1.6 km 45 miles ≈ 45 × 1.6 km = 72 km = 72,000 m 1 hour = 60 minutes = 3600 seconds

  45. Converting compound units m/s 72,000 3600 What is 45 mph in m/s? Travelling 45 miles in 1 hour is equivalent to travelling 72 000 m in 3600 seconds. Therefore, 45 mph ≈ = 20 m/s How could we use this answer to convert any speed given in mph to a speed in m/s? 45 mph ≈ 20 m/s Dividing by 5 we have, 9 mph ≈ 4 m/s We can divide the speed in mph by 9 and multiply it by 4.

  46. S8 Measures Contents S8.1 Converting units • A S8.2 Accuracy in measurement • A S8.5 Bearings S8.3 Calculations involving bounds • A S8.4 Compound measures • A • A

  47. Bearings N 75° P Bearings are a measure of direction taken from North. If you were travelling North you would be travelling on a bearing of 000°. If you were travelling from the point P in the direction shown by the arrow then you would be travelling on a bearing of 000°. If you were travelling from the point P in the direction shown by the arrow then you would be travelling on a bearing of 075°. Bearings are always measured clockwise from North and are written as three figures.

  48. Compass points 000° N 045° 315° NW NE E 270° W 090° SW SE 225° 135° S 180°

  49. Measuring bearings

  50. Back bearings N N 105º A B The bearing from point A to point B is 105º. What is the bearing from point B to point A? The angle from B to A is 105º + 180º = 285º This is called a back bearing. ? 105º 180°

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