KS4 Mathematics
-
Upload
sebastien-romano -
Category
Documents
-
view
27 -
download
5
description
Transcript of KS4 Mathematics
© Boardworks Ltd 2005 1 of 51
S8 Measures
KS4 Mathematics
© Boardworks Ltd 2005 2 of 51
Contents
A
A
A
A
AS8.1 Converting units
S8 Measures
S8.3 Calculations involving bounds
S8.4 Compound measures
S8.5 Bearings
S8.2 Accuracy in measurement
© Boardworks Ltd 2005 3 of 51
Converting units
It is important to be able to convert between a variety of units quickly and accurately.
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.
1) When using a formula
2) When comparing measurementsFor example, suppose one mother gives the weight of her baby in pounds and another mother gives the weight of her baby in kilograms.
We usually use metric units for calculations.
How can we compare the babies’ weights?To compare the weights we convert them to the same unit.
© Boardworks Ltd 2005 4 of 51
Metric units
The metric system of measurement is based on powers of ten and uses the following prefixes:
These prefixes are then followed by a base unit.
The base unit for length is metre.metre.
The base unit for mass is gram.gram.
The base unit for capacity is litre.litre.
Kilo-Centi-Milli-Micro-
meaning one thousandmeaning one hundredthmeaning one thousandthmeaning one millionth
© Boardworks Ltd 2005 5 of 51
Metric units of length, mass and capacity
You should know the following metric conversions for length, mass and capacity (liquid volume):
LengthLength
1 km = 1000 m
1 m = 100 cm
1 m = 1000 mm
1 cm = 10 mm
MassMass
1 tonne = 1000 kg
1 kg = 1000 g
1 g = 1000 mg
Capacity and Volume
Capacity and Volume
1 litre = 1000 ml
1 cl = 10 ml
1 m3 = 1000 litres
1 cm3 = 1 ml
© Boardworks Ltd 2005 6 of 51
Metric units of area and volume
The metric units for area are mm2, cm2, m2 and km2.
1 m2 = 100 cm × 100 cm
1 m2 = 10,000 cm21 m2 = 10,000 cm2
1 cm2 = 10 mm × 10 mm
1 cm2 = 100 mm21 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 cm31 m3 = 1,000,000 cm3
1 cm3 = 10 mm × 10 mm × 10 mm
1 cm3 = 1000 mm31 cm3 = 1000 mm3
© Boardworks Ltd 2005 7 of 51
Converting metric units
To convert from a larger metric unit to a smaller one we need to _______ by 10, 100, or 1000.
To convert from a larger metric unit to a smaller one we need to _______ by 10, 100, or 1000.multiply
For example,
0.43 kg = 0.43 × 1000 g
Convert 0.43 kg to grams
= 430 g
To convert from a smaller metric unit to a larger one we need to _______ by 10, 100, or 1000.
To convert from a smaller metric unit to a larger one we need to _______ by 10, 100, or 1000.divide
For example,
7.6 cm = 7.6 ÷ 100 m
Convert 7.6 cm to metres
= 0.076 m
© Boardworks Ltd 2005 8 of 51
Converting metric units
© Boardworks Ltd 2005 9 of 51
Ordering units
© Boardworks Ltd 2005 10 of 51
Imperial units
Imperial units are still frequently used and you should be aware of the following imperial conversions:
1 foot = ___ inches12
1 yard = ___ feet3
1 pound = ___ ounces16
1 stone = ___ pounds14
1 gallon = ___ pints8
© Boardworks Ltd 2005 11 of 51
Converting imperial units
We can convert between metric and imperial units using the following approximate conversions:
5 miles is about ___ kilometres8
1 foot is about ___ centimetres30
1 inch is about ___ centimetres2.5
1 kilogram is about ___ pounds2.2
1 gallon is about ___ litres4.5
1 litre is about ____ pints1.75
© Boardworks Ltd 2005 12 of 51
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 kilograms1 stone is about 6.4 kilograms
© Boardworks Ltd 2005 13 of 51
Metric and imperial conversions
© Boardworks Ltd 2005 14 of 51
Spider diagram
© Boardworks Ltd 2005 15 of 51
Contents
A
A
A
A
A
S8.2 Accuracy in measurement
S8.1 Converting units
S8.3 Calculations involving bounds
S8.4 Compound measures
S8.5 Bearings
S8 Measures
© Boardworks Ltd 2005 16 of 51
Continuous measurements
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:
13 14 15 13 14 15 13 14 15
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.
© Boardworks Ltd 2005 17 of 51
Accuracy in measurement
Remember, any measurement given to the nearest whole unit could be up to half a unit longer or shorter.
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 cm14.15 cm ≤ l < 14.25 cm
© Boardworks Ltd 2005 18 of 51
Upper and lower bounds
When we give a range for a measurement as in:
14.15 cm ≤ length < 14.25 cm
this value is called the lower bound …this value is called the lower bound …
… and this value is called the upper 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.
This is an inequality
© Boardworks Ltd 2005 19 of 51
Upper and lower bounds
© Boardworks Ltd 2005 20 of 51
Contents
A
A
A
A
A
S8.3 Calculations involving bounds
S8.1 Converting units
S8.2 Accuracy in measurement
S8.4 Compound measures
S8.5 Bearings
S8 Measures
© Boardworks Ltd 2005 21 of 51
Adding measures
The following triangle has sides of length 3 cm, 4 cm and 5 cm.
3 cm5 cm
4 cm
What is the range of possible lengths for the perimeter?
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
© Boardworks Ltd 2005 22 of 51
Adding measures
The following triangle has sides of length 3 cm, 4 cm and 5 cm.
3 cm5 cm
4 cm
What is the range of possible lengths for the perimeter?
The range of possible values for the perimeter is
Notice that the more lengths that are added together the greater the error on either side.
10.5 cm ≤ perimeter < 13.5 cm10.5 cm ≤ perimeter < 13.5 cm
© Boardworks Ltd 2005 23 of 51
Adding measures
When a calculation involves adding two or more measurements together:
The lower bound is found by
adding the lower bounds togetheradding the lower bounds together
The upper bound is found by
adding the upper bounds togetheradding the upper bounds together
© Boardworks Ltd 2005 24 of 51
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
© Boardworks Ltd 2005 25 of 51
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
© Boardworks Ltd 2005 26 of 51
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 cm110 cm ≤ length < 130 cm
© Boardworks Ltd 2005 27 of 51
Calculations involving bounds
What is the range of possible values for the angle a?
The smallest a could be is 180° – (35.5° + 78.5°) =
The range of possible values for a is therefore,
66° ≤ a < 68° 66° ≤ a < 68°
The angles in the following diagram are rounded to the nearest degree:
66°
35° 78°a
The largest a could be is 180° – (34.5° + 77.5°) = 68°
This angle could be between 34.5° and 35.5°.
This angle could be between 77.5° and 78.5°.
© Boardworks Ltd 2005 28 of 51
Subtracting measures
When a calculation involves subtracting two measurements:
The lower bound is found by:
subtracting the upper bound from the lower boundsubtracting the upper bound from the lower bound
The upper bound is found by:
subtracting the lower bound from the upper boundsubtracting the lower bound from the upper bound
© Boardworks Ltd 2005 29 of 51
Calculations involving bounds
The dimensions of a small tile are given as 15 mm by 18 mm.
15 mm
18 mm
What is the smallest possible area the tile could have?
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
© Boardworks Ltd 2005 30 of 51
Calculations involving bounds
The dimensions of a small tile are given as 15 mm by 18 mm.
15 mm
18 mm
What is the largest possible area the tile could have?
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
© Boardworks Ltd 2005 31 of 51
Calculations involving bounds
The dimensions of a small tile are given as 15 mm by 18 mm.
15 mm
18 mm
What is the range of possible values for the area?
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 mm2253.75 mm2 ≤ area < 286.75 mm2
© Boardworks Ltd 2005 32 of 51
Multiplying measures
When a calculation involves multiplying two or more measurements together:
The lower bound is found by:
multiplying the lower bounds togethermultiplying the lower bounds together
The upper bound is found by:
multiplying the upper bounds togethermultiplying the upper bounds together
© Boardworks Ltd 2005 33 of 51
Calculations involving bounds
A boy runs 200 metres in 27.8 seconds.
What is his greatest possible average speed to 2 decimal places?
The distance is given to the nearest metre and the time is given to the nearest tenth of a second.
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.
greatest possible average speed
=greatest distance
shortest time=
200.5 m27.75 s
= 7.23 m/s
© Boardworks Ltd 2005 34 of 51
Dividing measures
When a calculation involves dividing two measurements:
The lower bound is found by:
dividing the lower bound by the upper bounddividing the lower bound by the upper bound
The upper bound is found by:
dividing the upper bound by the lower bounddividing the upper bound by the lower bound
© Boardworks Ltd 2005 35 of 51
Contents
A
A
A
A
A
S8.4 Compound measures
S8.1 Converting units
S8.2 Accuracy in measurement
S8.5 Bearings
S8.3 Calculations involving bounds
S8 Measures
© Boardworks Ltd 2005 36 of 51
Compound measures
Which is heavier: a kilogram of cotton wool or a kilogram of rocks?
© Boardworks Ltd 2005 37 of 51
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.
© Boardworks Ltd 2005 38 of 51
Compound measures
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
Which is heavier: 1 cm3 of cotton wool or 1 cm3 of rock?
For example, if a block of metal has a mass of 760 g and a volume of 80 cm3:
Density =76080
= 9.5 g/cm3
© Boardworks Ltd 2005 39 of 51
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.
Speed is usually measured in km/h, m/s or mph.
Speed =distance travelled
time taken
© Boardworks Ltd 2005 40 of 51
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
Average speed =1560300
= 5.2 m/s
© Boardworks Ltd 2005 41 of 51
Common compound measures
Commonly used compound measures include:
Fuel consumptionMeasured in km/l or
mpg.
PressureMeasured in N/m2 or
N/cm2.
SpeedMeasured in m/s, km/h
or mph.
DensityMeasured in g/cm3,
kg/m3 or kg/l.mass
volume
distancetime
forcesurface area
distancevolume
© Boardworks Ltd 2005 42 of 51
Pairs – compound measures
© Boardworks Ltd 2005 43 of 51
Calculating densities
© Boardworks Ltd 2005 44 of 51
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 kmSo 1 mile = 1.6 km45 miles ≈ 45 × 1.6 km
= 72 km= 72,000 m
1 hour = 60 minutes= 3600 seconds
© Boardworks Ltd 2005 45 of 51
Converting compound units
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 ≈72,0003600
m/s
= 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.We can divide the speed in mph by 9 and multiply it by 4.
© Boardworks Ltd 2005 46 of 51
Contents
A
A
A
A
A
S8.5 Bearings
S8.1 Converting units
S8.2 Accuracy in measurement
S8.4 Compound measures
S8.3 Calculations involving bounds
S8 Measures
© Boardworks Ltd 2005 47 of 51
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 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°.
N
75°
P
Bearings are always measured clockwise from North and are written as three figures.
© Boardworks Ltd 2005 48 of 51
Compass points
N
EW
S
NE
SESW
NW
000°
045°
090°
135°
180°
225°
270°
315°
© Boardworks Ltd 2005 49 of 51
Measuring bearings
© Boardworks Ltd 2005 50 of 51
Back bearings
The bearing from point A to point B is 105º.What is the bearing from point B to point A?
A
N
105º
B
N
?
The angle from B to A is
105º + 180º = 285º
This is called a back bearing.105º
180°
© Boardworks Ltd 2005 51 of 51
Investigating back bearings