KS4 Mathematics S8 Measures.

KS4 Mathematics S8 Measures

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

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. Explain why it is important to be able to convert between a variety of units. Explain that metric units are the easiest to work with because they are based on powers of ten like the number system. It is also the only system of measurement that is recognized around the world. To compare the weights we convert them to the same unit. How can we compare the babies’ weights? We usually use metric units for calculations.

Metric units The metric system of measurement is based on powers of ten and uses the following prefixes: Kilo- meaning one thousand Centi- meaning one hundredth Milli- meaning one thousandth Micro- meaning one millionth These prefixes are then followed by a base unit. Ask pupils why they think powers of ten are used for the metric system of measurement. Discuss the fact that it is easy to convert between units by multiplying or dividing by a power of ten. The base unit for length is metre. metre. The base unit for mass is gram. gram. The base unit for capacity is litre. litre.

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

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 = cm2 1 cm2 = 100 mm2 The metric units for volume are mm3, cm3, m3 and km3. 1 m3 = 100 cm × 100 cm × 100 cm You may like to mention that we can also measure area in hectares. 1 hectare is 100 m × 100 m = 10000m2. 1 m3 = cm3 1 cm3 = 10 mm × 10 mm × 10 mm 1 cm3 = 1000 mm3

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 Pupils often believe that to convert from a smaller unit to a larger unit we should multiply to make it larger and that to convert from a larger unit to a smaller unit we should divide to make it smaller. Stress that to convert from a smaller unit to a larger unit we actually have to divide because there are fewer larger units for each smaller unit. To convert from a larger unit to a smaller unit we have to multiply because there are more smaller units in each larger unit. If pupils are in doubt they should check that their answers make sense. For example, 23 cm cannot possibly be equal to 2300 m. For example, Convert 7.6 cm to metres 7.6 cm = 7.6 ÷ 100 m = m

Converting metric units
For each problem ask pupils what we have to multiply or divide by to complete the conversion before asking for the solution.

Ordering units Each of the amount shown could be converted to the same units to enable them to be ordered.

Imperial units Imperial units are still frequently used and you should be aware of the following imperial conversions: 1 foot = ___ inches 12 1 yard = ___ feet 3 1 pound = ___ ounces 16 Tell pupils that until relatively recently, we used the imperial system of measurement in the UK. We still use imperial units in many everyday situations. Ask pupils to give examples of instances where imperial units of length are commonly used. For example, in British cars the speedometer measures speed in miles per hour. Also, most people still give their heights in feet and inches and their weight in stones and pounds. 1 stone = ___ pounds 14 1 gallon = ___ pints 8

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

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. Explain carefully why 14 pounds must be divided by 2.2. 14 pounds ≈ 14 ÷ 2.2 kilograms = 6.4 kilograms 1 stone is about 6.4 kilograms

Metric and imperial conversions

Spider diagram Use proportional reasoning to complete the spider diagram.

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

Continuous measurements
The continuous nature of measurements means that they can never be exact. There is always ant 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: 13 14 15 13 14 15 13 14 15 Discuss the meaning of continuous. Explain that for any two values on a continuous scale, there will always be another value between them. This means that between any two values there are infinitely many other values. Ask pupils to imagine repeatedly zooming in to a continuous scale. Explain that the accuracy of a measurement usually depends on the size of the object and the measuring instrument being used. Stress that the pencil given in the example cannot be exactly 14.5 cm because it would then be rounded up to 15 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.

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. We can assume that a measurement given as 14.2 has been given to 1 decimal place. Point out that if this measurement had been given as 14.20, we would assume that it had been given to 2 decimal places. 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 < cm

Upper and lower bounds When we give a range for a measurement as in:
This is an inequality 14.15 cm ≤ length < cm this value is called the lower bound … … and this value is called the upper bound. The length could be equal to cm so we use a greater than or equal to symbol. We can read the inequality as “the length is between cm and cm, but doesn’t include cm”. Explain that if the length is strictly less than cm, it can be any value up to but not including cm If the length was equal to cm however, it would have been rounded up to 14.3 cm. The length is therefore “strictly less than” cm and so we use the < symbol.

Upper and lower bounds Stress that when we illustrate an inequality on a number line, the value indicated by the empty circle is not included.

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

What is the range of possible lengths for the perimeter?
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 = = 10.5 cm The most the lengths could be is 3.5 cm, 4.5 cm and 5.5 cm. The largest possible perimeter = = 13.5 cm

What is the range of possible lengths for the perimeter?
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 < cm Notice that the more lengths that are added together the greater the error on either side.

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

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 remaining piece? The original piece of wood could be between 165 cm and 175 cm. Discuss which two values would give the smallest possible length when subtracted. 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

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 remaining piece? The original piece of wood could be between 165 cm and 175 cm. Discuss which two values would give the largest possible length when subtracted. 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

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 remaining piece? The original piece of wood could be between 165 cm and 175 cm. Discuss which two values would give the smallest possible length when subtracted. 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

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 35° 78° What is the range of possible values for the angle a? The smallest a could be is 180° – (35.5° °) = 66° Explain that to find the lower bound for a we have to add together the upper bounds for the two given angles and subtract them from 180°. To find the upper bound for a we have to add together the lower bounds for the two given angles and subtract them from 180°. The largest a could be is 180° – (34.5° °) = 68° The range of possible values for a is therefore, 66° ≤ a < 68°

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: Stress that to find the smallest possible value we have to take the biggest number from the smallest number. To find the biggest possible value we have to take the smallest number from the biggest number. subtracting the lower bound from the upper bound

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. Explain that when we perform a calculation involving two of more measurements the errors in measurements are combined to make a larger margin of error. In this case, if we had multiplied 15 mm by 18 mm we would have got 270 mm2. This is mm2 more than mm2. To calculate the smallest possible area we multiply these values together. Smallest possible area = 14.5 × 17.5 = mm2

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. If we had multiplied 15 mm by 18 mm we would have got 270 mm2. This is mm2 less than mm2. To calculate the largest possible area we multiply these values together. Largest possible area = 15.5 × 18.5 = mm2

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 mm2 and the largest value for the area is mm2 The range of possible values for the area is therefore, mm2 ≤ area < mm2

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

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 m and m. Remind pupils that to calculate average speed we have to divide the distance by the time taken. Explain why, to find the greatest possible average speed we have to divide the greatest possible distance by the shortest possible time. Ask pupils to calculate the boys least possible speed. The time taken could be between s and s. greatest possible average speed = greatest distance shortest time = 200.5 m 27.75 s = 7.23 m/s

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

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

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. Explain that the ‘/’ in g/cm3 is read as ‘per’ and means ‘for every’ or ‘in every’. A density of 1.2 g/cm3 means that every 1 cm3 of the material has a mass of 1.2 grams.

Compound measures 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: Density = mass volume For example, if a block of metal has a mass of 760 g and a volume of 80 cm3 : 760 80 = Density = 9.5 g/cm3

Compound measures 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. When we use this formula we assume that the person or object is travelling at constant speed. Speed = distance travelled time taken Speed is usually measured in km/h, m/s or mph.

Average speed 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: Average speed = total distance travelled total time taken For example, if the man runs 1560 metres in 300 seconds 1560 300 Average speed = = 5.2 m/s

Common compound measures
Commonly used compound measures include: Density mass volume Measured in g/cm3, kg/m3 or kg/l . Speed distance time Measured in m/s, km/h or mph . Pressure force surface area Measured in N/m2 or N/cm2 . Give examples where necessary. For example, the fuel consumption of a car depends on how much petrol it uses over a given distance. Discuss the equivalent formulae for each one. For example, If density is mass divided by volume, then mass is density times volume and volume is mass divided by density. Fuel consumption distance volume Measured in km/l or mpg .

Pairs – compound measures
Ask pupils to come to the board and choose two cards that show units that can be used to measure the same rate. Ask them to tell you what the matching unit could be used to measure. For example, m2/cm3 and cm2/ml could be used to measure the amount of surface covered per unit volume of paint. Some of the units shown may by less familiar such as m/s2 and ft/s2 for measuring acceleration. Another way of expressing this unit is m/s/s or ft/s/s. Point out that when there is a ‘p’ instead of a ‘/’ we are usually referring to imperial units. For example, mph for miles per hour or mpg for miles per gallon. Also point out that volume can be measured in cm3, m3 or in litres, l.

Calculating densities
Pupils must find the missing measurements by multiplying or dividing. Establish that if density is mass divided by volume, then mass is density times volume and volume is mass divided by density. When the table is complete ask pupils to tell you if any of the materials in the table would float on water. The density of water is 1 g/cm3.

Converting compound units
What is 45 mph in m/s? When we convert compound unit 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 Explain that because we are converting between metric and imperial units the conversion is approximate. = 72 km = m 1 hour = 60 minutes = 3600 seconds

Converting compound units
What is 45 mph in m/s? Travelling 45 miles in 1 hour is equivalent to travelling m in 3600 seconds. 72 000 3600 m/s Therefore, mph ≈ = 20 m/s How could we use this answer to convert any speed given in mph to a speed in m/s? Explain that because we are converting between metric and imperial units the conversion is approximate. Pupils do not need to know how to convert between mph and m/s using 9 mph = 4 m/s. However, it is interesting to see how this conversion is derived. Ask pupils how we could convert m/s to mph using the same conversion (divide by 4 and multiply by 9). Link this to work on proportion. 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.

Bearings 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 075°. 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°. N Bearings are always measured clockwise from North and are written as three figures. 75° P

Compass points 000° N 315° 045° NW NE 270° W E 090° SW SE 225° 135° S
Revise the basic compass points. These can be remembered using ‘Naughty Elephants Squirt Water’. Revise the points NE, SE, SW and NW, pointing out that N or S always comes before E and W. Ask pupils to tell you how many degrees there are between each compass point (45°). Ask questions such as, A ship is sailing due southwest. What bearing is it sailing on? Reveal the directions of the compass using bearings in orange. You may like to mention that in addition to these points we can also have NNE (north by north east), ENE (east by north east), ESE (east by south east), SSE (south by south east), SSW (south by south west), WSW (west by south west), WNW (west by north west) and NNW (north by north west). As an extension exercise ask pupils to draw these compass points using a ruler and a protractor and give the bearing of each one. There will be 22.5º between each compass point. SW SE 225° 135° S 180°

Measuring bearings Use this activity to demonstrate how to find the bearing from one numbered place on the map to another. Also, ask pupils to use the scale given on the map to estimate the distance between two points. Ask, for example, What is the approximate bearing from 6 to 5? Approximately how many metres are there between 1 and 3? What point is on a bearing of 255° from 6? Find the bearing from 7 to 9. Ask pupils if they can give you the bearing from 9 to 7 without measuring.

Back bearings The bearing from point A to point B is 105º.
What is the bearing from point B to point A? N The angle from B to A is 105º + 180º = N 285º 105º A This is called a back bearing. 105º ? Explain that the north lines are parallel and so we can work out he corresponding angle of 105°. The angle from B to A is therefore 105° + 180° = 285°. B 180°

Investigating back bearings
If a given bearing is less than 180º, we find the reciprocal or back bearing by adding 180°. If a given bearing is more than 180º, we find the back bearing by subtracting 180°.

KS4 Mathematics S8 Measures.

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