e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity 9 Sequences and sets
a
A B f
e d
ca
b k
h g
j i
Review b Look in the region that is contained within the overlap of both circles. This region
contains the set {a, h}. So A ∩ B = {a, h}.
c n(A ∩ B) = 2 as there are two elements in the set A ∩ B.
d A ∪ B = set of elements of A or B or both = {a, b, c, d, e, g, h, j}.
e n(A ∪ B) = 8
f A ∩ B’ = set of all elements that are both in set A and not in set B = {c, e, j}
Exercise 9.9 1 Use the given Venn diagram to answer the
following questions.
A 48 B 14
a List the elements of A and B 6
b List the elements of A ∩ B. 18 12 16 10
c List the elements of A ∪ B.
24 20 2
22
Review 2 Use the given Venn diagram to answer the P e Q
following questions. a f hj
a List the elements that belong to: cb gi
i P ii Q d
b List the elements that belong to both P and Q.
c List the elements that belong to:
i neither P nor Q
ii P but not Q.
3 Draw a Venn diagram to show the following sets and write each element in its correct space.
a The universal set is {a ,b, c, d, e, f, g, h}.
A = {b, c, f, g} and B = {a, b, c, d, f}.
b ℰ = {whole numbers from 20 to 36 inclusive}.
A = {multiples of four} and B = {numbers greater than 29}.
Review 4 The universal set is: {students in a class}. V S
V = {students who like volleyball}. 10 x 86
S = {students who play soccer}.
There are 30 students in the class.
The Venn diagram shows numbers of students.
a Find the value of x.
b How many students like volleyball?
c How many students in the class do not play
soccer?
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5 Copy the Venn diagram and shade the region B
which represents the subset A ∩ B′.
Review Set builder notation
So far the contents of a set have either been given as a list of the elements or described by a rule
(in words) that defines whether or not something is a member of the set. We can also describe
sets using set builder notation. Set builder notation is a way of describing the elements of a set
using the properties that each of the elements must have.
For example:
A = {x : x is a natural number}
This means:
A = { x : x is a natural number }
Set A is all values such each value of x
the set of (x) that is a natural
number
Review In other words, this is the set: A = {1, 2, 3, 4, … }
Sometimes the set builder notation contains restrictions.
For example, B = {{x : x is a letter of the alphabet, x is a vowel}
In this case, set B = {a, e, i, o, u}
Here is another example:
A = {integers greater than zero but less than 20}.
In set builder notation this is:
A = {x : x is an integer, 0 < x < 20}
This is read as: ‘A is the set of all x such that x is an integer and x is greater than zero but less
than 20’.
The following examples should help you to get used to the way in which this notation is used.
Review Worked example 8
List the members of the set C if: C = {x: x ∈ primes, 10 < x < 20}.
Read the set as: ‘C is the set of all x such that x is a member of the set of primes and x is
greater than 10 but less than 20’.
The prime numbers greater than 10 but less than 20 are 11, 13, 17 and 19.
So, C = {11, 13, 17, 19}
192 Unit 3: Number
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Worked example 9
Express the following set in set builder notation:
D = {right-angled triangles}.
So, D = {x : x is a triangle, x has a If D is the set of all right-angled triangles
right-angle} then D is the set of all x such that x is a
triangle and x is right-angled.
Review As you can see from this last example, set builder notation can sometimes force you to write
more, but this isn’t always the case, as you will see in the following exercise.
Exercise 9.10 1 Describe each of these sets using set builder notation.
Set builder notation is very useful a square numbers less than 101
when it isn’t possible to list all b days of the week
the members of set because the c integers less than 0
set is infinite. For example, all the d whole numbers between 2 and 10
numbers less than −3 or all whole e months of the year containing 30 days
numbers greater than 1000.
2 Express each of the following in set builder notation.
Review a {2, 3, 4, 5, 6, 7, 8}
b {a, e, i, o, u}
c {n, i, c, h, o, l, a, s} E
d {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}
e {1, 2, 3, 4, 6, 9, 12, 18, 36}
3 List the members of each of the following sets.
a {x : x is an integer, 40 < x < 50}
b {x : x is a regular polygon and x has no more than six sides}
c {x : x is a multiple of 3, 16 < x < 32}
4 Describe each set in words and say why it’s not possible to list all the members of each set.
a A = {x, y : y = 2x + 4}
b B = {x : x3 is negative}
5 If A = {x : x is a multiple of three} and B = {y : y is a multiple of five}, express A ∩ B
in set builder notation.
6 ℰ = {y : y is positive, y is an integer less than 18}.
A = {w : w > 5} and B = {x : x 5}.
a List the members of the set:
Review i A∩B ii A′ iii A′ ∩ B iv A ∩ B′ v (A ∩ B′)′
b What is A ∪ B?
c List the members of the set in part (b).
Unit 3: Number 193
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ReviewSummary Are you able to …? E
E
Do you know the following? • continue sequences
E • describe a rule for continuing a sequence
• A sequence is the elements of a set arranged in a E • find the nth term of a sequence
particular order, connected by a rule. • use the nth term to find later terms
E • find out whether or not a specific number is in a
• A term is a value (element) of a sequence.
• If the position of a term in a sequence is given the letter sequence
n then a rule can be found to work out the value of the • generate sequences from shape patterns
nth term. • find a formula for the number of shapes used in a
• A rational number is a number that can be written as a pattern
fraction.
• write a recurring decimal as a fraction in its lowest terms
• An irrational number has a decimal part that continues • describe a set in words
forever without repeating. • find the complement of a set
• represent the members of set using a Venn diagram
• A set is a list or collection of objects that share a • solve problems using a Venn diagram
characteristic. • describe a set using set builder notation.
Review• An element is a member of a set.
• A set that contains no elements is called the
empty set (∅).
• A universal set (ℰ) contains all the possible elements
appropriate to a particular problem.
• The complement of a set is the elements that are not in
the set (′).
• The elements of two sets can be combined (without
repeats) to form the union of the two sets (∪).
• The elements that two sets have in common is called the
intersection of the two sets (∩).
• The elements of a subset that are all contained within a
larger set are a proper subset (⊆).
• If it is possible for a subset to be equal to the larger set,
then it is not a proper subset (⊂).
• A Venn diagram is a pictorial method of showing sets.
• A shorthand way of describing the elements of a set is
called set builder notation.
Review
194 Unit 3: Number
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Exam-style questions
1 Pattern 1 Pattern 2 Pattern 3
The first three patterns in a sequence are shown above.
a Copy and complete the table.
Review Pattern number (n) 1 2 3 4
Number of dots (d) 5
b Find a formula for the number of dots, d, in the nth pattern.
c Find the number of dots in the 60th pattern.
d Find the number of the pattern that has 89 dots.
2 The diagram below shows a sequence of patterns made from dots and lines.
1 dot 2 dots 3 dots
a Draw the next pattern in the sequence.
b Copy and complete the table for the numbers of dots and lines.
Review Dots 12345 6
Lines 4 7 10
c How many lines are in the pattern with 99 dots?
d How many lines are in the pattern with n dots?
e Complete the following statement:
There are 85 lines in the pattern with . . . dots.
Review Past paper questions [1]
[1]
1 a Here are the first four terms of a sequence: [1]
27 23 19 15
i Write down the next term in the sequence. [2]
ii Explain how you worked out your answer to part (a)(i).
b The nth term of a different sequence is 4n – 2.
Write down the first three terms of this sequence.
c Here are the first four terms of another sequence:
–1 2 5 8
Write down the nth term of this sequence.
[Cambridge IGCSE Mathematics 0580 Paper 11 Q23 October/November 2013]
Unit 3: Number 195
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AB A B
A′ ∪ B A′ ∩ B′
[Cambridge IGCSE Mathematics 0580 Paper 22 Q1 May/June 2013]
Review 3 The first five terms of a sequence are shown below. [2]
13 9 5 1 –3
Find the nth term of this sequence.
4 Shade the required region in each of the Venn diagrams. [Cambridge IGCSE Mathematics 0580 Paper 22 Q3 May/June 2013]
[2] E
PQ
AB
R
A′ (P ∩ R) ∪ Q
Review [Cambridge IGCSE Mathematics 0580 Paper 23 Q9 October/November 2012]
5 Shade the region required in each Venn diagrams. [2]
AB AB
(A ∪ B)′ A′ ∩ B
[Cambridge IGCSE Mathematics 0580 Paper 22 Q4 October/November 2014]
6 The Venn diagram shows the number of students who study French (F), Spanish (S) and Arabic (A).
Review FS
7 45
1
23
8 0
A
a Find n(A ∪ (F ∩ S)). [1]
b On the Venn diagram, shade the region F' ∩ S.
[1]
[Cambridge IGCSE Mathematics 0580 Paper 22 Q6 October/November 2015]
196 Unit 3: Number
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Layer 2
Layer 3
Review The diagrams show layers of white and grey cubes.
Khadega places these layers on top of each other to make a tower.
a Complete the table for towers with 5 and 6 layers.
Number of layers 1 234 5 6
Total number of white cubes 0 1 6 15
Total number of grey cubes 1 5 9 13 [4]
Total number of cubes 1 6 15 28 [2]
[1]
Review b i Find, in terms of n, the total number of grey cubes in a tower with n layers.
ii Find the total number of grey cubes in a tower with 60 layers. [2]
iii Khadega has plenty of white cubes but only 200 grey cubes.
How many layers are there in the highest tower that she can build?
. [Cambridge IGCSE Mathematics 0580 Paper 42 Q9 (a) & (b) October/November 2014]
8 Write the recurring decimal 0.36 as a fraction.
E
Give your answer in its simplest form.
. [3]
[0.36 means 0.3666…]
[Cambridge IGCSE Mathematics 0580 Paper 22 Q12 May/June 2016]
Review
Unit 3: Number 197
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University Copy Copy Copy Copy
Chapter 10: Straight lines and
quadratic equations
UniversitCyoPprye-ssC-aRmebrviidegwe
Review Key words
• Equation of a line
• Gradient
• y-intercept
• Constant
• x-intercept
• Line segment
• Midpoint
• Expand
• Constant term
• Quadratic expression
• Factorisation
• Difference between two
squares
• Quadratic equation
UniversitCyoPprye-ssC-aRmebrviidegwe
Review In this chapter you
will learn how to:
UniversitCyoPprye-ssC-aRmebrviidegwe
EXTENDED • construct a table of values Geoff wishes he had paid more attention when his teacher talked about negative and positive gradients and
and plot points to draw rates of change.
graphs
On 4 October 1957, the first artificial satellite, Sputnik, was launched. This satellite orbited
• find the gradient of a the Earth but many satellites that do experiments to study the upper atmosphere fly on short,
straight line graph sub-orbital flights. The flight path can be described with a quadratic equation, so scientists
know where the rocket will be when it deploys its parachute and so they know where to recover
• recognise and determine the instruments. The same equation can be used to describe any thrown projectile including a
the equation of a line baseball!
EXTENDEDReview • determine the equation of a
line parallel to a given line
• calculate the gradient of a
line using co-ordinates of
points on the line
• find the gradient of parallel
and perpendicular lines
• find the length of a
line segment and the
co-ordinates of its midpoint
• expand products of
algebraic expressions
• factorise quadratic
expressions
• solve quadratic equations
by factorisation
- Cambridge
ess - Review
198 Unit 3: Algebra
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RECAP
You should already be familiar with the following algebra and graph work:
Table of values and straight line graphs (Stage 9 Mathematics)
A table of values gives a set of ordered pairs (x, y) that you can use to plot graphs on a coordinate grid.
x −1 0 1 2 y
y 3 4 56 10
8
(−1, 3), (0, 4), (1, 5) and (2, 6) are all points on the graph. 6
Plot them and draw a line through them. 4
Review Equations in the form of y = mx + c (Year 9 Mathematics) –10 –8 –6 –4 2 x
The standard equation of a straight line graph is y = mx + c 2 4 6 8 10
• m is the gradient (or steepness) of the graph –2 0
• c is the point where the graph crosses the y-axis (the y-intercept) –2
–4
–6
Gradient of a straight line (Year 9 Mathematics) –8
–10
Gradient (m) = rise = change in y - values
run change in x - values y
10
Drawing a straight line graph (Year 9 Mathematics) 8
6
You can use the equation of a graph to find the gradient and 4
y-intercept and use these to draw the graph. 2 + gradient 1 up, 3 right
Review For example y 1 x − 2
3
Expand expressions to remove brackets (Chapter 2) –10 –8 –6 –4 –2 0 246 x
To expand 3x(2x − 4) you multiply the term outside the bracket –2 8 10
by each term inside the first bracket
–4 y-intercept
3x (2x − 4) = 3x × 2x − 3x × 4 –6
= 6x2 − 12x –8
–10
Review
Unit 3: Algebra 199
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10.1 Straight lines
Review REWIND Using equations to plot lines
You will recognise that the formulae Mr Keele owns a boat hire company. If Mr Keele makes a flat charge of $40 and then another $15
used to describe nth terms in per hour of hire, you can find a formula for the total cost $y after a hire time of x hours.
chapter 9 are very similar to the Total cost = flat charge + total charge for all hours
equations used in this chapter. y = 40 + 15 × x
or (rearranging)
y = 15x + 40
Now think about the total cost for a range of different hire times:
one hour: cost = 15 × 1 + 40 = $55
two hours: cost = 15 × 2 + 40 = $70
three hours: cost = 15 × 3 + 40 = $85
and so on.
If you put these values into a table (with some more added) you can then plot a graph of the total
cost against the number of hire hours:
Number of hours (x) 123456789
Total cost (y) 55 70 85 100 115 130 145 160 175
LINK y Costs for hiring Mr Keele’s boats
200
Review Equations of motion, in
physics, o en include terms 150
that are squared. To solve Total
some problems relating to cost ($)
physical problems, therefore,
physicists o en need to solve 100
quadratic equations.
50
0 x
2 4 6 8 10
Number of hours
Review The graph shows the total cost of the boat hire (plotted on the vertical axis) against the number
of hire hours (on the horizontal axis). Notice that the points all lie on a straight line.
The formula y = 15x + 40 tells you how the y co-ordinates of all points on the line are related to
the x co-ordinates. This formula is called an equation of the line.
The following worked examples show you how some more lines can be drawn from given
equations.
200 Unit 3: Algebra
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Worked example 1
A straight line has equation y = 2x + 3. Construct a table of values for x and y and draw the
line on a labelled pair of axes. Use integer values of x from −3 to 2.
Substituting the values −3, −2, −1, 0, 1 and 2 into the equation gives the values in the
following table:
x −3 −2 −1 0 1 2
y −3 −1 1 3 5 7
Notice that the y-values range from −3 to 7, so your y-axis should allow for this.
Review Graph of y = 2x + 3
y
7
6
5
4
3
2 x
12 3
1
0
–4 –3 –2 –1
–1
–2
–3
Review Worked example 2
Draw the line with equation y = − x + 3 for x-values between −2 and 5 inclusive.
The table for this line would be:
x −2 −1 0 1 2 3 4 5
y 5 4 3 2 1 0 −1 −2
Graph of y = – x + 3
y
5
4
3
Review 2
1
–2 –1 0 x
–1 1 2 34 5
–2
Unit 3: Algebra 201
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Before drawing your axes, always To draw a graph from its equation:
check that you know the range of
y-values that you need to use. • draw up a table of values and fill in the x and y co-ordinates of at least three points (although
you may be given more)
• draw up and label your set of axes for the range of y-values you have worked out
• plot each point on the number plane
• draw a straight line to join the points (use a ruler).
Exercise 10.1 1 Make a table for x-values from −3 to 3 for each of the following equations.
Plot the co-ordinates on separate pairs of axes and draw the lines.
Review a y = 3x + 2 b y=x+2 c y = 2x − 1 d y = 5x − 4
e y = −2x + 1 f y = −x − 2 g y=6−x h y = 3x + 1
i y = 1x +1 j y = 4x k y = −3
n x−y=2 o y=x 2
2 l y = −1 − x
m x+y=4
p y = −x
2 Plot the lines y = 2x, y = 2x + 1, y = 2x − 3 and y = 2x + 2 on the same pair of axes.
Use x-values from −3 to 3. What do you notice about the lines that you have drawn?
3 For each of the following equations, draw up a table of x-values for −3, 0 and 3.
Complete the table of values and plot the graphs on the same set of axes.
a y=x+2 b y = −x + 2 c y=x−2 d y=−x−2
4 Use your graphs from question 3 above to answer these questions.
Review a Where do the graphs cut the x-axis?
b Which graphs slope up to the right?
c Which graphs slope down to the right?
d Which graphs cut the y-axis at (0, 2)?
e Which graphs cut the y-axis at (0, −2)?
f Does the point (3, 3) lie on any of the graphs? If so, which?
g Which graphs are parallel to each other?
h Compare the equations of graphs that are parallel to each other. How are they similar?
How are they different?
Gradient
The gradient of a line tells you how steep the line is. For every one unit moved to the right, the
gradient will tell you how much the line moves up (or down). When graphs are parallel to each
other, they have the same gradient.
Vertical and horizontal lines
Look at the two lines shown in the following diagram:
y
5
4
3 x=3
Review 2
1
–5 –4 –3 –2 ––110 x
1 2 34 5
y = –2 –2
–3
–4
–5
202 Unit 3: Algebra
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Every point on the vertical line has x co-ordinate = 3. So the equation of the line is simply x = 3.
Every point on the horizontal line has y co-ordinate = −2. So the equation of this line is y = −2.
All vertical lines are of the form: x = a number.
All horizontal lines are of the form: y = a number.
The gradient of a horizontal line is zero (it does not move up or down when you move to
the right).
Review Exercise 10.2 1 Write down the equation of each line shown in the diagram.
y
7
(d)
6
5
(a)
4
3 (c)
(e)
(b)
2
1
Review –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7x
–1
–2
–3
–4
–5
(f)
–6
–7
Review 2 Draw the following graphs on the same set of axes without plotting points or drawing up a
table of values.
a y=3 b x=3 c y = −1 d x = −1
h x = −7
e y = −3 f y=4 g x=1
2 2
i a graph parallel to the x-axis which cuts the y-axis at (0, 4)
j a graph parallel to the y-axis which goes through the point (−2, 0)
Unit 3: Algebra 203
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FAST FORWARD Lines that are neither vertical nor horizontal
You will deal with gradient as a rate y
of change when you work with 10
kinematic graphs in Chapter 21.
8
(a)
6
4
A
2
Review –4 –2 0 x
–2 246
–4
–6
Review –8
(b)
–10
The diagram shows two different lines. If you take a point A on the line and then move to the
right then, on graph (a) you need to move up to return to the line, and on graph (b) you need to
move down.
The gradient of a line measures how steep the line is and is calculated by dividing the change in
the y co-ordinate by the change in the x co-ordinate:
gradient = y-change
x-increase
Another good way of remembering For graph (a): the y-change is 8 and the x-increase is 2, so the gradient is 8 = 4
the gradient formula is 2
gradient = ’rise’. The ‘run’ must For graph (b): the y-change is −9 (negative because you need to move down to return to the line)
’run’ and the x-increase is 4, so the gradient is −9 = −2.25.
always be to the right (increase x). 4
It is essential that you think about x-increases only. Whether the y-change is positive or negative
Review tells you what the sign of the gradient will be.
204 Unit 3: Algebra
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Worked example 3
Calculate the gradient of each line. Leave your answer as a whole number or fraction in its
lowest terms.
y y
a 12 b3
10 2
8
1
6
Review 4 –2 0 x
246
–1
2
–2 0 x
246
–2
a Notice that the graph passes through the points (2, 4) and (4, 10).
gradient = y -change = 10 − 4 = 6 =3
x-increase 4−2 2
b Notice that the graph passes through the points (2, 1) and (4, 0).
gradient = y -change = 0 −1 = − 1
x-increase 4−2 2
Review E
Worked example 4
Calculate the gradient of the line that passes through the points (3, 5) and (7, 17).
(7, 17) Think about where the points would
be, in relation to each other, on a pair
of axes. You don’t need to draw this
accurately but the diagram will give
you an idea of how it may appear.
17 – 5 = 12
Review (3, 5) 7 – 3 = 4 17 − 5
y -change 7−3 12
gradient = x-increase = = 4 = 3
Unit 3: Algebra 205
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Exercise 10.3 1 Calculate the gradient of each line. Leave your answers as a fraction in its lowest terms.
ay by cy
6 6 6
4 44
2 2 2
–2 0 x –2 0 x –2 0 x
–2 246 –2 246 –2 246
d y e y fy
6 12 6
Review 4 10 4
2 82
–6 –4 –2 0 x 6x
–2 –4 0 2 4 6
4 –4
2
–6 –4 –2 0 x
2
–2
gy i y
6 hy 2
6
4 4 –6 –4 –2 0 x
2 2 –2 2
Review –2 0 x –2 0 x –4
–2 246 –2 246
–6
Think carefully about whether you 2 Calculate the gradient of the line that passes through both points in each case. E
expect the gradient to be positive Leave your answer as a whole number or a fraction in its lowest terms.
or negative.
a A (1, 2) and B (3, 8) b A (0, 6) and B (3, 9)
c A (2, −1) and B (4, 3) d A (3, 2) and B (7, −10)
e A (−1, −4) and B (−3, 2) f A (3, −5) and B (7, 12)
ReviewThink carefully about the problem Applying your skills
and what mathematics you need to 3 If the car climbs 60 m vertically how far must the car have travelled horizontally?
do to find the solution.
2
15
GRADIENT
vertical
distance
horizontal d.istance
206 Unit 3: Algebra
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Finding the equation of a line yc y
4 y = –2x + 4 4
Look at the three lines shown below.
2 2
a yb
4
y = 3x + 2
2
– 4 –2 0 x x x
–2 2 4 –4 –2 0 2 4 –4 –2 0 24
Review –4 –2 –2
y = 1 x−3
–4
2
–4
Check for yourself that the lines have the following gradients:
• gradient of line (a) = 3
• gradient of line (b) = −2
• gradient 1
of line (c) = 2
REWIND Notice that the gradient of each line is equal to the coefficient of x in the equation and that the
You met the coefficient in point at which the line crosses the y-axis (known as the y-intercept) has a y co-ordinate that is
chapter 2.
equal to the constant term.
In fact this is always true when y is the subject of the equation: c
y = mx +
y is the subject gradient y-intercept
of the equation
Review
In summary:
• equations of a straight line graphs can be written in the form of y = mx + c
• c (the constant term) tells you where the graph cuts the y-axis (the y-intercept)
• m (the coefficient of x) is the gradient of the graph; a negative value means the graph slopes
down the to the right, a positive value means it slopes up to the right. The higher the value of
m, the steeper the gradient of the graph
• graphs which have the same gradient are parallel to each other (therefore graphs that are
parallel have the same gradient).
Worked example 5
Find the gradient and y-intercept of the lines given by each of the following equations.
a y = 3x + 4 b y = 5 − 3x c y = 1 x + 9 d x+y=8
e 3x + 2y = 6 2
Review a y = 3x + 4 The coefficient of x is 3.
The constant term is 4.
Gradient = 3
y-intercept = 4
b y = 5 − 3x Re-write the equation as y = −3x + 5.
The coefficient of x is −3.
Gradient = −3
y-intercept = 5 The constant term is 5.
Unit 3: Algebra 207
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c y = 1 x + 9
2
Gradient = 1 The gradient can be a fraction.
2
y-intercept = 9
d x+y=8 Subtracting x from both sides, so that y is the
subject, gives y = −x + 8.
Gradient = −1
y-intercept = 8
e 3x + 2y = 6 Make y the subject of the equation.
Gradient = −3
Review 2 3x + 2y = 6
y-intercept = 3 2y = 3x + 6
y = −3 x + 6
22
y = −3 x +3
2
Worked example 6
Find the equation of each line shown in the diagrams.
ay by
6 2
5
4 1
3 –1 0 x
–1 123
Review 2
1
– 2 –1–10 x –2
12
–2
a Gradient = 6 and the y-intercept = −1 Gradient = 6 = 6
So the equation is y = 6x − 1 1
Graph crosses y-axis at −1
You should always label your axes x b Gradient = −3 and the y-intercept = 1 Gradient = −1 5 = −3
and y when drawing graphs – even 4 24
when they are sketches.
Graph crosses y-axis at 1.
So the equation is y = − 3 x + 1
4
Review Exercise 10.4 1 Find the gradient and y-intercept of the lines with the following equations. Sketch
the graph in each case, taking care to show where the graph cuts the y-axis.
Look carefully at your sketches for
answers 1(d) and 1(g). If you draw a y = 4x − 5 b y = 2x + 3 c y = −3x − 2 d y = −x + 3
them onto the same axes you will
see that they are parallel. These lines e y = 1 x + 2 f y = 6 − 1 x g x+y=4 h x + 2y = 4
have the same gradient but they cut 3 4 k x = y +2 l 2x − 3y = −9
the y-axis at different places. If two
or more lines are parallel, they will i x+ y =3 j x = 4y − 2 4
have the same gradient. 2
208 Unit 3: Algebra
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2 Rearrange each equation so that it is in the form y = mx + c and then find the
gradient and y-intercept of each graph.
a 2y = x − 4 b 2x + y − 1 = 0 c x= y −2 d 2x − y − 5 = 0
2
e 2x − y + 5 = 0 f x + 3y − 6 = 0 g 4y = 12x − 8 h 4x + y = 2
i y =x+2 j y = 2x − 4 k x − 4 y = 12 l −y = 4x −2
2 3 2 3
3 Find the equation (in the form of y = mx + c) of a line which has:
a a gradient of 2 and a y-intercept of 3
b a gradient of −3 and a y-intercept of −2
c a gradient of 3 and a y-intercept of −1
o−f232aannddaayg-rinadteirecnetpotfa−t (430,
Review d a gradient of −0.5)
e a y-intercept
f a y-intercept of −3 and a gradient of 4
8
g a y-intercept of −0.75 and a gradient of 0.75
h a y-intercept of −2 and a gradient of 0
i a gradient of 0 and a y-intercept of 4
4 Find an equation for each line.
ay by cy
2 7 9
1 6 8
–4 –3– 2 ––110 x 5 7
1234
4 6
–2 3 5
–3 2 4
–4 1 3
–5 –4 –3–2 ––110 x 2
1234
–6 1
–7 –2 –4 –3– 2 ––110 x
1234
–8 –3
Review
–9 –4 –2
dy ey fy
9 9 9
8
7 88
6
5 77
4
3 66
2
1 55
0x
–4 –3– 2 ––11 0 1 2 3 4 44
–2
33
22
11
–4 –3– 2 ––110 x –4 –3–2 ––110 x
1234 1234
–2 –2
gy hy i7
5 7 6
4 6 5
3 5 4
2 4 3
Review 1 3 2
–4 –3– 2 ––110 x 2 1
1234
1
x
–2 –4 –3– 2 ––110 x –4 –3– 2 ––110 1234
1234
–3
–2
–4 –2
–5 –3 –3
–6 –4 –4
Unit 3: Algebra 209
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5 Find the equation of the line which passes through both points in each case. E
E
a A (2, 3) and B (4, 11) b A (4, 5) and B (8, −7)
c A (−1, −3) and B (4, 6) d A (3, −5) and B (7, 12)
6 Write down the equation of a line that is parallel to:
a y = −3x b y = 2x − 3 c y=x +4
2
d y = −x − 2 e x=8 f y = −6
7 Which of the following lines are parallel to y = 1 x?
2
a y = 1 x +1 b y = 2x c y + 1 = 1 x d 2y + x = −6 e y = 2x − 4
2 2
Review 8 Find the equation of a line parallel to y = 2x + 4 which:
a has a y-intercept of −2
b passes through the origin
c passes through the point (0, −4)
d has a y-intercept of 1
2
9 A graph has the equation 3y − 2x = 9.
a Write down the equation of one other graph that is parallel to this one.
b Write down the equation of one other graph that crosses the y-axis at the same
point as this one.
c Write down the equation of a line that passes through the y-axis at the same point
as this one and which is parallel to the x-axis.
ReviewIf the product of the gradients of Parallel and perpendicular lines
two lines is equal to –1, it follows
that the lines are perpendicular to You have already seen that parallel lines have the same gradient and that lines with the
each other. same gradient are parallel.
Perpendicular lines meet at right angles. The product of the gradients is −1.
So, m1 × m2 = −1, where m is the gradient of each line.
The sketch shows two perpendicular graphs.
y
10
8
1 2 6 y ϭ 3x Ϫ 4
3 4
y ϭϪ x ϩ
2
Review –6 –4 –2 0 2 4 6 x
–2 8
–4
–6
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y = − 1 x + 2 has a gradient of − 1 E
33
y = 3x – 4 has a gradient of 3
The product of the gradients is − 1 × 3 = −1.
3
Worked example 7
Given that y = 2 x + 2, determine the equation of the straight line that is:
3
a perpendicular to this line and which passes through the origin
b perpendicular to this line and which passes through the point (−3, 1).
Review a y = mx + c
m = −3 The gradient is the negative reciprocal of 2
2 3
c=0 3
2
The equation of the line is y = − x.
b y = −3x + c Using m = − 3 from part (a) above.
2 2
x = − 3 and y = 1
1 = − 3(–3) + c Substitute the values of x and y for the given point to
2 solve for c.
1=9 +c
2
c = −3 1
2
Review
y = − 3x − 3 1
2 2
Exercise 10.5 1 A line perpendicular to y = x + 3 passes through (1, 3). What is the equation of the line?
5
2 Show that the line through the points A(6, 0) and B(0, 12) is:
a perpendicular to the line through P(8, 10) and Q(4, 8) –13 )
b perpendicular to the line through M(–4, –8) and N(–1, 2
3 Given A(0, 0) and B(1, 3), find the equation of the line perpendicular to AB with a
y-intercept of 5.
Review 4 Find the equation of the following lines:
a perpendicular to 2x – y – 1 = 0 and passing through (2, –1)
2
b perpendicular to 2x + 2y = 5 and passing through (1, –2)
5 Line A joins the points (6, 0) and (0, 12) and Line B joins the points (8, 10) and (4, 8).
Determine the gradient of each line and state whether A is perpendicular to B.
6 Line MN joins points (7, 4) and (2, 5). Find the equation of AB, the perpendicular bisector of MN.
7 Show that points A(–3, 6), B(–12, –4) and C(8, –5) could not be the vertices of a rectangle ABCD.
Unit 3: Algebra 211
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Intersection with the x–axis
So far only the y-intercept has been found, either from the graph or from the equation. There is,
of course, an x-intercept too. The following sketch shows the line with equation y = 3x − 6.
y
10
8
6
y = 3x− 6
4
Review 2 x
–10 –8 –6 –4 –2 0 2 4 6 8 10
–2
–4
–6
–8
–10
Notice that the line crosses the x-axis at the point where x = 2 and, importantly, y = 0. In fact, all
points on the x-axis have y co-ordinate = 0. If you substitute y = 0 into the equation of the line:
FAST FORWARD y = 3x − 6 (putting y = 0)
0 = 3x − 6 (add 6 to both sides)
You will need to understand this 3x = 6 (dividing both sides by 3)
method when solving simultaneous x=2
equations in chapter 14.
Review this is exactly the answer that you found from the graph.
You can also find the y-intercept by putting x = 0. The following worked examples show
calculations for finding both the x- and y-intercepts.
Worked example 8
Find the x- and y-intercepts for each of the following lines. Sketch the graph in each case.
a y = 6x − 12 b y = −x + 3 c 2x + 5y = 20
a y = 6x − 12 y
10
x = 0 ⇒ y = −12
y = 0 ⇒ 6 x − 12 = 0 8 y = 6x – 12
6
⇒ x=2
4
2
–10 –8 –6 –4 –2–20 x
2 4 6 8 10
Review –4
–6
–8
–10
–12
212 Unit 3: Algebra
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b y = −x + 3 y
x=0⇒y =3 10
y =0⇒ x+3=0
⇒ x=3 8
c 2x + 5y = 20 y = –x + 3 6
x = 0 ⇒ 5y = 20
⇒y=4 4
y = 0 ⇒ 2x = 20
⇒ x = 10 2
–10 –8 –6– 4 ––220 x
2 4 6 8 10
–4
–6
–8
–10
–12
Review y
10
8 2x + 5y = 20
6
4 x
2 2 4 6 8 10
–10 –8 –6– 4 ––220
–4
–6
–8
–10
–12
Review Exercise 10.6 1 Find the x- and y-intercepts for each of the following lines. Sketch the graph in each case.
a y = −5x + 10 b y = x −1 c y = −3x + 6 d y = 4x + 2
3
e y = 3x + 1 g y = 2x − 3 h y = 2x −1
i y= x −2 f y = −x + 2 k −2 + y = x 3
j y = 2x +1 l −y = 4x −2
4 4
5 3
2 For each equation, find c, if the given point lies on the graph.
a y = 3x + c (1, 5) b y = 6x + c (1, 2)
c y = −2x + c (−3, −3) d y = 3x + c (4, −5)
4
e y = 1x + c (−2, 3) f y = c − 1x (−4, 5)
2 2
g y = c + 4x (−1, −6) h 2x + c = y (3, 4)
3
Review
Unit 3: Algebra 213
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Finding the length of a straight line segment E
FAST FORWARD Although lines are infinitely long, usually just a part of a line is considered. Any section
of a line joining two points is called a line segment.
Pythagoras’ theorem is covered
in more detail in chapter 11. If you know the co-ordinates of the end points of a line segment you can use Pythagoras’
Remember though, that in any right- theorem to calculate the length of the line segment.
angled triangle the square on the
hypotenuse is equal to the sum of Worked example 9
the squares on the other two sides.
We write this as a2 + b2 = c2. Find the distance between the points (1, 1) and (7, 9)
Review y
10
(7, 9)
8
6a
48
2
(1, 1) 6 x
0 2 4 6 8 10
a2 = 82 + 62 a2 = b2 + c2 (Pythagoras’ theorem)
a2 = 64 + 36 Work out each expression.
a2 = 100 Undo the square by taking the
∴a = 100 square root of both sides.
a = 10 units
Review Worked example 10
Given that A(3, 6) and B(7, 3), find the length of AB.
y
10
8 B(7, 3)
6 A(3, 6)
43
2 C(3, 3) 4
0x
0 2 4 6 8 10
AB2 = AC 2 + CB 2 a2 = b2 + c2 (Pythagoras’ theorem)
AB2 = 32 + 42 Work out each expression.
Review = 9 + 16
= 25
∴AB = 25 = 5 units
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Midpoints E
It is possible to find the co-ordinates of the midpoint of the line segment (i.e. the point
that is exactly halfway between the two original points).
Consider the following line segment and the points A(3, 4) and B(5, 10).
y B(5,10)
10
8
6
Review 4
A(3, 4)
2
x
0 12345
FAST FORWARD If you add both x co-ordinates and then divide by two you get (3 5) = 8 = 4.
In chapter 12 you will learn about 22
the mean of two or more numbers.
The midpoint uses the mean of the If you add both y co-ordinates and then divide by two you get (4 1 ) = 14 = 7.
x co-ordinates and the mean of the 22
y co-ordinates.
This gives a new point with co-ordinates (4, 7). This point is exactly half way between A and B.
Exercise 10.7
Review 1 Find the length and the co-ordinates of the midpoint of the line segment joining each
REWIND pair of points.
Check that you remember how to
deal with negative numbers when a (3, 6) and (9, 12) b (4, 10) and (2, 6) c (8, 3) and (4, 7)
adding. d (5, 8) and (4, 11) e (4, 7) and (1, 3) f (12, 3) and (11, 4)
g (−1, 2) and (3, 5) h (4, −1) and (5, 5) i (−2, −4) and (−3, 7)
2 Use the graph to find the length and the midpoint of each line segment.
y
8
CA
6 E
D
4
2 B
4
F H G
–8O –6 –4
Review –2 0 I2 x
N M 6 K8
L
–2
–4
–6 J
P
–8
Unit 3: Algebra 215
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3 Find the distance from the origin to point (−3, −5). E
4 Which of the points A(5, 6) or B(5, 3) is closer to point C(−3, 2)?
5 Which is further from the origin, A(4, 2) or B(−3, −4)?
6 Triangle ABC has its vertices at points A(0, 0), B(4, −5) and C(−3, −3). Find the
length of each side.
7 The midpoint of the line segment joining (10, a) and (4, 3) is (7, 5). What is the
value of a?
8 The midpoint of line segment DE is (−4, 3). If point D has the co-ordinates (−2, 8),
what are the co-ordinates of E?
Review x 10.2 Quadratic (and other) expressions
x x2
3 The diagram shows a rectangle of length (x + 3) cm and width (x + 5) cm that has been divided
5 5x 3x into smaller rectangles.
15 The area of the whole rectangle is equal to the sum of the smaller areas, so the area of whole
rectangle = (x + 3) × (x + 5).
The sum of smaller rectangle areas: x2 + 3x + 5x + 15 = x2 + 8x 15.
This means that (x + 3) × (x + 5) = x2 + 8x + 15 and this is true for all values of x.
Notice what happens if you multiply every term in the second bracket by every term in
the first:
(x + 3)( x + 5) (x + 3)( x + 5) (x + 3)( x + 5) (x + 3)( x + 5)
x2 5x 3x 15
Review Notice that the four terms in boxes are exactly the same as the four smaller areas that were
calculated before.
Another way to show this calculation is to use a grid: x3
x x2 3x
5 5x 15
Review You will notice that this is almost the same as the areas method above but it can also be used
when the constants are negative, as you will see in the worked examples shortly.
When you remove the brackets and re-write the algebraic expression you are expanding or
multiplying out the brackets. The resulting algebraic expression contains an x2 term, an
x term and a constant term. This is called a quadratic expression.
The following worked example shows these two methods and a third method for expanding
pairs of brackets. You should try each method when working through the next exercise and
decide which you find easiest, though you will begin to notice that they are all, in fact, the same.
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Worked example 11
Expand and simplify: b (x − 7)(x + 6) c (2x − 1)(x + 9)
a (x + 2)(x + 9)
a x+2 x+9 x2 + 9 x + 2x + 18 In this version of the method
= x2 + 11x + 18 you will notice that the arrows
have not been included and
You need to choose which method the multiplication ‘arcs’ have
works best for you but ensure that
you show all the appropriate stages been arranged so that they
of working clearly.
are symmetrical and easier to
LINK
remember.
Quadratic expressions
Review and formulae are useful b x –7 x2 − 7 x + 6 x − 42 The grid method with a negative
for modelling situations
that involve movement, x x2 –7x = x2 − x − 42 value.
including acceleration,
stopping distances, velocity +6 6x –42
and distance travelled
(displacement). These c (2x − 1)(x + 9) 2x2 + 18 x − 9 A third method that you can
situations are studied in = x2 + 17 x − 9 remember using the mnemonic
Physics but they also have Firsts: 2x × x = 2x2 ‘FOIL’ which stands for First,
real life applications in Outsides: 2x × 9 = 18x Outside, Inside, Last. This means
situations such as road or Insides: −1 × x = − x that you multiply the first term
plane accident investigations. Lasts: −1 × 9 = −9 in each bracket together then
the ‘outside’ pair together (i.e.
Review the first term and last term), the
‘inside’ pair together (i.e. the
second term and third term) and
the ‘last’ pair together (i.e. the
second term in each bracket).
The product of more than two sets of brackets E
You can multiply in steps to expand three (or more) sets of brackets. Your answer might
contain terms with powers of 3 (cubic expressions).
Worked example 12
Expand and simplify (3x + 2)(2x + 1)(x − 1)
Review (3x + 2)(2x + 1)(x − 1) Expand the first two brackets.
= (6x2 + 4x + 3x + 2)(x − 1) Collect like terms.
= (6x2 + 7x + 2)(x − 1)
= 6x3 + 7x2 + 2x − 6x2 − 7x − 2 Multiply each term in the first
= 6x3 + x2 − 5x − 2 bracket by each term in the second.
Collect like terms to simplify.
Unit 3: Algebra 217
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Exercise 10.8 1 Expand and simplify each of the following.
REWIND a (x + 3)(x + 1) b (x + 6)(x + 4) c (x + 9)(x + 10)
You will need to remember how to d (x + 3)(x + 12)
multiply fractions. This was covered g (x + 4)(x − 7) e (x + 1)(x + 1) f (x + 5)(x + 4)
in chapter 5. j (x − 9)(x + 8)
m (y + 3)(y − 14) h (x − 3)(x + 8) i (x − 1)(x + 1)
REWIND p (h − 3)(h − 3)
Refer to chapter 2 to remind you k (x − 6)(x − 7) l (x −13)(x + 4)
how to multiply different powers of
the same number together. n (z + 8)(z − 8) o (t + 17)(t − 4)
q (g − 1)(g + )
r (d + 3 )(d − 3 )
2 4
2 Find the following products.
a (4 − x)(3 − x) b (3 − 2x)(1 + 3x) c (3m − 7)(2m − 1)
d (2x + 1)(3 − 4x) e (4a − 2b)(2a + b) f (2m − n)(−3n − 4m)
Review g x + 1 x + 1 h 2x + 1 x − 1 i (2x2 4y)( y x2 )
2 4 3 2
l (3x − 3)(5 + 2x)
j (7 − 9b)(4b + 6) k (x + y)(2y2 4x3 )
3 Expand and simplify each of the following.
a (2x + 3)(x + 3) b (3y + 7)(y + 1) c (7z + 1)(z + 2)
d (t + 5)(4t − 3) e (2w − 7)(w − 8) f (4g − 1)(4g + 1)
g (8x − 1)(9x + 4) h (20c − 3)(18c − 4) i (2m − 4)(3 − m)
4 Expand and simplify each of the following. E
a (3x2 + 1)(2x + 3) b (5x2 − 1)(3x2 − 3) c (3x2 − y)(2x + y)
5 Expand and simplify.
a (5x + 2)(3x − 3)(x + 2)
b (x − 5)(x − 5)(x + 5)
c (4x − 1)(x + 1)(3x − 2)
d (x + 4)(2x + 4)(2x + 4)
Review e (2x − 3)(3x − 2)(2x − 1)
f (3x − 2)2 (2x − 1)
g (x + 2)3
h (2x − 2)3
i (x2y2 + x2)(xy + x)(xy − x)
j 1 + x 1 − x2 1 − x
3 2 9 4 3 2
6 The volume of a cuboid can be found using the formula V = lbh, where l is the length,
b is the breadth and h is the height. A cuboid has length 2x + 1 m, breadth (x − 2) m
and height (x − 2) m. 2
a Write an expression for the volume of the cuboid in factor form.
b Expand the expression.
c Determine the volume of the cuboid when x = 2.2 m.
Review
218 Unit 3: Algebra
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Squaring a binomial E
(x + y)2 means (x + y)(x + y)
To find the product, you can use the method you learned earlier.
(x + y)(x + y) = x2 + xy + xy + y2 = x2 + 2xy + y2
However, if you think about this, you should be able to solve these kinds of expansions by
inspection. Look at the answer. Can you see that:
• the first term is the square of the first term (x2)
• the middle term is twice the product of the middle terms (2xy)
• the last term is the square of the last term (y2)?
Review Exercise 10.9 1 Find the square of each binomial. Try to do this by inspection first and then check
your answers.
a (x − y)2 b (a b)2 c (2x 3y)2 d (3x 2y)2
e (x + 2y)2 f ( y x )2 2 ( )g x2 y2 2 h (2 + )3 2
i (−2x − 4 y )2
1 − 1 3x y 2 1 b 2
2x 4 y 4 2 2
j k − l a
m (−ab − )2 ( )n 3x2y 1 2 2x y 2 p [−( − )]2
3
o + 4
2 Simplify.
a (x 2)2 − (x − 4)2 b (x + 2)(x − 2) − (3 − x)(5 + x)
c ( y + 2x)2 + (2x − y)(− y + 2x) d 1 (3 2) x + 2
2 3
e 3(x + 2)(2x + 0.6)
Review g (x + 4)(x − 5) − 2(x −1)2 f ( 2x − y)( 2x + y) − (4x − y)2
h (2x y)2 + (x − 2y)(x + 2y) − (x + 4 y)2
i −2x(x + 1)2 − (x − 5)(−3x) j (3 2x)2 − 5(5x + 2)
3 Evaluate each expression when x = 4.
a (x + 7)(x − 7) − x2 3) b x2 − (x − 3)(x 3)
d (x + )2
c (3 2x)2 − (2x + 3)(2 f (2x 3)2 − 4(x + 1)(2 3x)
e (x2 + 3)(x 4)
Review Factorising quadratic expressions
Look again at the expansion of (x + 2)(x + 9), which gave x2 + 11x + 18:
2 + 9 = 11
( x + 2)( x + 9) = x2+ 11x + 18
2 × 9 = 18
Here the two numbers add to give the coefficient of x in the final expression and the two
numbers multiply to give the constant term.
Unit 3: Algebra 219
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This works whenever there is just one x in each bracket. E
Worked example 13
Expand and simplify: a (x + 6)(x + 12) b (x + 4)(x − 13)
a (x + 6)(x + 12) = x2 + 18x + 72 6 + 12 = 18 and 6 × 12 = 72 so this gives 18x
and 72.
b (x + 4)(x − 13) = x2 − 9x − 52 4 + −13 = −9 and 4 × −13 = −52 so this gives
−9x and −52.
Review If you use the method in worked example 11 and work backwards you can see how to put
a quadratic expression back into brackets. Note that the coefficient of x2 in the quadratic
REWIND expression must be 1 for this to work.
1 × 72 and 6 × 12 are the factor
pairs of 72. You learned about Consider the expression x2 + 18x + 72 and suppose that you want to write it in the form
factor pairs in chapter 1. (x + a)(x + b).
List the factor pairs of 12. From the worked example you know that a + b = 18 and a × b = 72.
(If you spot which pair of numbers
works straight away then you don’t Now 72 = 1 × 72 but these two numbers don’t add up to give 18.
need to write out all the other
factor pairs.) However, 72 = 6 × 12 and 6 + 12 = 18.
So, x2 + 18x + 72 = (x + 6)(x + 12).
The process of putting a quadratic expression back into brackets like this is called factorisation.
Worked example 14
Review Factorise completely: b x2 − 6 x − 16 c x2 − 8 x + 15
a x2 + 7 x + 12
a You need two numbers that multiply to give
12 and add to give 7.
12 = 1 × 12
12 = 2 × 6 These don’t add to give 7.
12 = 3 × 4 and 3 + 4 = 7
So, x2 + 7 x + 12 = ( x + 3)( x + 4) These don’t add to give 7.
These multiply to give 12 and add to give 7.
b −8 × 2 = −16 and −8 + 2 = −6 You need two numbers that multiply to give
So, x2 − 6 x − 16 = ( x − 8)( x + 2). −16 and add to give −6. Since they multiply to
give a negative answer, one of the numbers
must be negative and the other must be
positive. (Since they add to give a negative, the
Review larger of the two numbers must be negative.)
c −5 × −3 = 15 and −5 + −3 = −8 You need two numbers that multiply to give
So, x2 8 x + 15 = ( x − 3)( x − 5). 15 and add to give −8. Since they multiply to
give a positive value but add to give a negative
then both must be negative.
220 Unit 3: Algebra
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Exercise 10.10 1 Factorise each of the following. E
a x2 + 14x + 24
When looking for your pair of d x2 + 12x + 35 b x2 + 3x + 2 c x2 + 7x + 12
integers, think about the factors g x2 + 11x + 30 e x2 + 12 + 27 f x2 + 7x + 6
of the constant term first. Then j x2 + 8x + 7 h x2 + 10x + 16 i x2 + 11x + 10
choose the pair which adds up to k x2 + 24x + 80 l x2 + 13x + 42
the x term in the right way. 2 Factorise each of the following. b x2 − 9x + 20 c x2 − 7x + 12
a x2 − 8x + 12 e x2 −12x + 32 f x2 −14x + 49
Review d x2 − 6x + 8 h x2 − 7x −18 i x2 − 4x − 32
g x2 − 8x − 20 l x2 + 8x − 33 m x2 + 10x − 24
k x2 + x − 6 b p2 + 8p − 84 c w2 − 24w + 144
e v2 + 20v + 75 f x2 −100
3 Factorise each of the following.
a y2 + 7 y −170
d t2 + 16t − 36
Difference between two squares
The very last question in the previous exercise was a special kind of quadratic.
To factorise x2 − 100 you must notice that x2 − 100 = x2 + 0x − 100.
Now, proceeding as in worked example 12:
10 × −10 = −100 and −10 + 10 = 0 so, x2 + 0x −100 = (x −10)(x + 10).
Now think about a more general case in which you try to factorise x2 a2.
Notice that x2 − a2 = x2 + 0x − a2.
Since a × −a = −a and a + −a = 0, this leads to: x2 − a2 = (x − a)(x + a).
You must remember this special case. This kind of expression is called a difference
between two squares.
Review Worked example 15
Factorise the following using the difference between two squares:
a x2 − 49 b x2 − 1 c 16y 2 25w 2
4
a 49 = 72 Use the formula for the difference
x2 − 49 = x2 − 72 between two squares: x2 − a2 =
(x − a)(x + a).
= ( x − 7)( x 7)
You know that 49 = 7 so you can
write 49 as 72. This gives you a2.
Substitute 72 into the formula.
b 21 2 = 1 1 is 1 so you can rewrite
4 42
Review x2 − 1 = x2 − 21 2 1 as 21 2 and substitute it into the
4 4
= x − 21 )( x + 1 formula for the difference between
2
( ) two squares.
Unit 3: Algebra 221
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E
c (4y )2 = 4y × 4y 16y 2 The 16y 2 (4y )2.
and 25w2 = (5w)2
(5w )2 = 5w × 5w 25w2 Substitute in (4y)2 and (5w)2.
16y 2 − 25w 2 = (4y )2 (5w )2
= (4y − 5w )(4y 5w )
Exercise 10.11 1 Factorise each of the following.
ReviewFrom question (l) you should a x2 − 36 b p2 − 81 c w2 − 16 d q2 − 9
notice that the numbers given are e k2 − 400 f t2 −121 g x2 y2 h 81h2 16g 2
not square. Try taking a common i 16 p2 36q2 j 144s2 c2 k 64h2 49g 2 l 27x2 48 y2
factor out first.
m 200q2 98 p2 n 20d2 125e2 o x4 − y4 p xy2 − x3
2 Factorise and simplify 362 − 352 without using a calculator.
3 Factorise and simplify (6 1 )2 − (5 3 )2 without using a calculator.
4 4
Using factors to solve quadratic equations
You can now use the factorisation method to solve some quadratic equations.
A quadratic equation is an equation of the form ax2 + bx + c = 0 . The method is
illustrated in the following worked examples.
Worked example 16
Solve each of the following equations for x.
Review a x2 − 3x = 0 b x2 − 7 x + 12 0 c x2 + 6 x − 4 = 12
d x2 − 8 x + 16 0
a Notice that both terms of the left-hand side are multiples of x so you can
use common factorisation.
x2 − 3x = 0
x( x − 3) = 0
Now the key point:
If two or more quantities multiply to give zero, then at least one of the
quantities must be zero.
So either x = 0 or x − 3 = 0 ⇒ x = 3.
Check: 02 − 3 × 0 = 0 (this works).
32 − 3 × 3 = 9 − 9 = 0 (this also works).
In fact both x = 0 and x = 3 are solutions.
Review b Use the factorisation method of worked example 12 on the left-hand side
of the equation.
x2 − 7 x + 12 0
( x − 4)( x − 3) = 0
Therefore either x − 4 = 0 ⇒ x = 4
or x − 3 = 0 ⇒ x = 3.
Again, there are two possible values of x.
222 Unit 3: Algebra
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E
Review When solving quadratic equations c x2 + 6 x − 4 = 12 ⇒ 2
they should be rearranged so 2 = ⇒ x2 + 6 x − 16 0 (subtract 12 from both sides)
that a zero appears on one side,
i.e. so that they are in the form Factorising, you get (x + 8)(x − 2) = 0
ax2 + bx + c = 0 So either x + 8 = 0 ⇒ x = −8
There are still two solutions here, or x − 2 = 0 ⇒ x = 2.
but they are identical.
d Factorising, x2 − 8 x + 16 0
( x − 4)( x − 4) = 0
So either x − 4 = 0 ⇒ x = 4
or x − 4 = 0 ⇒ x = 4
Of course these are both the same thing, so the only solution is x = 4.
Exercise 10.12 1 Solve the following equations by factorisation.
a x2 − 9x = 0 b x2 + 7x = 0 c x2 − 21x = 0
d x2 − 9x + 20 0 e x2 + 8x + 7 = 0 f x2 + x − 6 0
g x2 + 3x + 2 = 0 h x2 + 11x + 10 = 0 i x2 − 7x + 12 0
j x2 − 8x + 12 0 k x2 −100 = 0 l t2 + 16t − 36 = 0
m y2 + 7 y −170 = 0 n p2 + 8p − 84 0 o w2 − 24w + 144 = 0
ReviewSummary
Do you know the following? Are you able to …?
• The equation of a line tells you how the x- and • draw a line from its equation by drawing a table and
y co-ordinates are related for all points that sit on the line. plotting points
• The gradient of a line is a measure of its steepness. • find the gradient, x-intercept and y-intercept from the
• The x- and y-intercepts are where the line crosses the equation of a line
x- and y-axes respectively. • calculate the gradient of a line from its graph
• find the equation of a line if you know its gradient and
• The value of m in y = mx + c is the gradient of the line. y-intercept
• The value of c in y = mx + c is the y-intercept.
• The x-intercept can be found by substituting y = 0 and • find the equation of a vertical or horizontal line
• calculate the gradient of a line from the E
solving for x.
co-ordinates of two points on the line
• The y-intercept can be found by substituting x = 0 and
solving for y. • find the length of a line segment and the
co-ordinates of its midpoint
• Two lines with the same gradient are parallel.
• The gradients of two perpendicular lines will multiply E • expand double brackets
Review • expand three or more sets of brackets
to give −1. • factorise a quadratic expression E
• There is more than one way to expand brackets. • factorise an expression that is the difference
• Some quadratic expressions can be factorised to solve between two squares
quadratic equations. • solve a quadratic equation by factorising.
• Quadratic equations usually have two solutions, though
these solutions may be equal to one another.
Unit 3: Algebra 223
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Exam-style questions
1 Expand and simplify each of the following.
a (x + 2)(x + 18) b (2x + 3)(2x − 3) c (4y2 − 3)(3y2 +1)
iii d2 − 196
Review 2 a Factorise each of the following. ii y2 − 13y + 42 iii d2 − 196 = 0
i 12x2 6x ii y2 − 13y + 30 = −12
b Solve the following equations.
i 12x2 − 6x = 0
Past paper questions
1y
5
L4
3
2
1 x
B 123456
–6 –5 –4 –3 –2 –1–10
–2
–3
–4
Review a On the grid mark the point (5, 1). Label it A. [1]
b Write down the co-ordinates of the point B. [1]
c Find the gradient of the line L. [2]
[Cambridge IGCSE Mathematics 0580 Paper 13 Q19 October/November 2012]
Review
224 Unit 3: Algebra
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P
NOT TO
SCALE
l
0x
Review The equation of the line l in the diagram is y = 5 − x. [1]
a The line cuts the y-axis at P. [1]
Write down the co-ordinates of P.
b Write down the gradient of the line l.
[Cambridge IGCSE Mathematics 0580 Paper 22 Q5 May/June 2014]
3 Factorise
9w2 − 100,
[Cambridge IGCSE Mathematics 0580 Paper 22 Q15 (a) October/November 2015]
4 Factorise [2]
mp + np − 6mq − 6nq.
[Cambridge IGCSE Mathematics 0580 Paper 22 Q15(b) October/November 2015]
Review
Review
Unit 3: Algebra 225
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University Copy Copy Copy Copy
Chapter 11: Pythagoras’ theorem and
similar shapes
UniversitCyoPprye-ssC-aRmebrviidegwe
Review Key words
• Right angle
• Hypotenuse
• Similar
• Corresponding sides
• Corresponding angles
• Scale factor of lengths
• Scale factor of volumes
• Scale factor of areas
• Congruent
• Included side
• Included angle
UniversitCyoPprye-ssC-aRmebrviidegwe
Review In this chapter you
will learn how to:
UniversitCyoPprye-ssC-aRmebrviidegwe
EXTENDEDRevieEXTENDEDw • use Pythagoras’ theorem One man – Pythagoras of Samos – is usually credited with the discovery of the Pythagorean theorem, but
to find unknown sides of a there is evidence to suggest that an entire group of religious mathematicians would have been involved.
right-angled triangles
Right-angled triangles appear in many real-life situations, including architecture, engineering
• learn how to use and nature. Many modern buildings have their sections manufactured off-site and so it is
Pythagoras’ theorem to important that builders are able to accurately position the foundations on to which the parts will
solve problems sit so that all the pieces will fit smoothly together.
• decide whether or not Many properties of right-angled triangles were first used in ancient times and the study of these
triangles are mathematically properties remains one of the most significant and important areas of Mathematics.
similar
• use properties of similar
triangles to solve problems
• find unknown lengths in
similar figures
• use the relationship
between sides and areas
of similar figures to find
missing values
• recognise similar solids
• calculate the volume and
surface area of similar solids
• decide whether or not
shapes are congruent.
• use the basic conditions for
congruency in triangles
- Cambridge
ess - Review
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RECAP
You should already be familiar with the following number and shape work:
Squares and square roots (Chapter 1)
To square a number, multiply it by itself. 72 = 7 × 7 = 49.
You can also use the square function on your calculator x2 . 121 = 11.
To find the square root of a number use the square root function on your calculator
Pythagoras’ theorem (Stage 9 Mathematics)
Pythagoras’ theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the
two shorter sides.
Review For this triangle:
hypotenuse
c
a
b
a2 + b 2 = c 2
The hypotenuse is the longest side and it is always opposite the right angle.
11.1 Pythagoras’ theorem
Review Centuries before the theorem of right-angled triangles was credited to Pythagoras, the Egyptians
knew that if they tied knots in a rope at regular intervals, as in the diagram on the left, then they
would produce a perfect right angle.
In some situations you may be given a right-angled triangle and then asked to calculate the
length of an unknown side. You can do this by using Pythagoras’ theorem if you know the
lengths of the other two sides.
c = hypotenuse a Learning the rules
b Pythagoras’ theorem describes the relationship between the sides of a right-angled triangle.
The longest side – the side that does not touch the right angle – is known as the hypotenuse.
Tip
For this triangle, Pythagoras’ theorem states that: a2 + b2 = c2
You will be expected to In words this means that the square on the hypotenuse is equal to the sum of the squares on
remember Pythagoras' the other two sides. Notice that the square of the hypotenuse is the subject of the equation. This
theorem. should help you to remember where to place each number.
Worked example 1
Find the value of x in each of the following triangles, giving your answer to one decimal place.
Review ab x cm 8 cm
x cm 17 cm
5 cm
3 cm
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Review a a2 + b2 = c2 Notice that the final answer needs to be
32 + 52 = x2 rounded.
9 + 25 = x2
⇒ x2 = 34 Notice that a shorter side needs to be found so,
x = 34 = 5.8309. . . after writing the Pythagoras formula in the usual
way, the formula has to be rearranged to make
≈ 5.8 cm (1dp) x2 the subject.
b a2 + b2 = c2
82 + x2 = 172
64 + x2 = 289
x2 = 289 − 64
x2 = 225
x = 225 = 15cm (1dp)
Checking for right-angled triangles
You can also use the theorem to determine if a triangle is right-angled or not. Substitute the
values of a, b and c of the triangle into the formula and check to see if it fits. If a2 + b2 does not
equal c2 then it is not a right-angled triangle.
Worked example 2
Notice here the theorem is written Use Pythagoras’ theorem to decide whether or not the triangle shown below is right-angled.
as c2 = a2 + b2; you will see it
Reviewwritten like this or like a2 + b2 = c2 3.1 m 4.2 m Check to see if Pythagoras’ theorem is satisfied:
in different places but it means the 5.3 m c2 = a2 + b2
same thing. 3.12 + 4.22 = 27.25
5.32 = 28.09 ≠ 27.25
The symbol ‘≠’ means ‘does not Pythagoras’ theorem is not satisfied, so the triangle is
equal’. not right-angled.
Exercise 11.1 For all the questions in this exercise, give your final answer correct to three significant figures
where appropriate.
REWIND 1 Find the length of the hypotenuse in each of the following triangles.
You will notice that some of your a bc
answers need to be rounded. Many
of the square roots you need to take 6 cm x cm 12 cm 1.2 cm
produce irrational numbers. These
were mentioned in chapter 9. h cm
Review 6 cm 2.3 cm
8 cm
d e tm
6m
1.5 cm 4m
p cm
0.6 cm
228 Unit 3: Shape, space and measures
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2 Find the values of the unknown lengths in each of the following triangles.
a 8m b c
3m y cm 11 cm
x m 4.3 cm 14 cm
2.3 cm t cm
Review d e a cm
pm 13 m 10 cm
5m 8 cm
3 Find the values of the unknown lengths in each of the following triangles.
a b c
2.3 cm
6 cm h cm 6 cm
x cm 4 cm
Review 1.6 cm
y cm 4.2 cm
d e k cm f
p km 3 km 9 cm
h cm
8 km 12 cm
6 cm
8 cm
g h 12 m
8m dm
Review fm
6m 3m 4m
3m
Unit 3: Shape, space and measures 229
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4 Use Pythagoras’ theorem to help you decide which of the following triangles are right-angled.
a 6 cm b c
12 cm
8 cm 10 cm 12 cm 14 cm
6 cm
5 cm
d 3.6 km e
6 km 24 cm
25 cm
Review 4.8 km
7 cm
Applications of Pythagoras’ theorem
This section looks at how Pythagoras’ theorem can be used to solve real-life problems. In each case
look carefully for right-angled triangles and draw them separately to make the working clear.
Worked example 3
1.85 m 1.6 m The diagram shows a bookcase that has fallen against a
wall. If the bookcase is 1.85 m tall, and it now touches
the wall at a point 1.6 m above the ground, calculate the
distance of the foot of the bookcase from the wall. Give your
answer to 2 decimal places.
ReviewIt is usually useful to draw the 1.85 m 1.6 m Apply Pythagoras’ theorem: Think what triangle the
triangle that you are going to use xm situation would make and
as part of your working. a2 + b2 = c2 then draw it. Label each side
x2 + 1.62 = 1.852 and substitute the correct
sides into the formula.
x2 = 1.852 − 1.62
= 3.4225 − 2.56
= 0.8625
x = 0.8625 = 0.93m (2dp)
Worked example 4
It can be helpful to draw diagrams Find the distance between the points A(3, 5) and B(−3, 7).
when you are given co-ordinates.
y AB = 7 − 5 = 2 units Difference between
8 y-co-ordinates.
B(–3, 7) AC = 3 − −3 = 6 units Difference between
6 x-co-ordinates.
Review A C(3, 5) BC 2 = 22 + 62 Apply Pythagoras’
4 = 4 + 36 theorem.
x = 40
2 246
So BC = 40
–6 –4 –2 0 = 6.32 units (3sf)
–2
230 Unit 3: Shape, space and measures
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Exercise 11.2 1 B The size of a television screen is its longest
21.6 inches diagonal. The diagram shows the length
You generally won't be told to and breadth of a television set. Find the
use Pythagoras' theorem to solve distance AB.
problems. Always check for right-
angled triangles in the context of A 48.6 inches
the problem to see if you can use 2
the theorem to solve it. The diagram shows a ladder that is leaning against a wall. Find the
length of the ladder.
3m
Review 0.4 m
3 Sarah stands at the corner of a rectangular field. If the field measures 180 m by 210 m, how far
would Sarah need to walk to reach the opposite corner in a straight line?
4 2m 2m The diagram shows the side view of a shed.
Calculate the height of the shed.
height
2.4 m
Review 3.2 m The diagram shows a bridge that can be lifted
5 AB to allow ships to pass below. What is the
distance AB when the bridge is lifted to the
6m position shown in the diagram?
(Note that the bridge divides exactly in half
when it lifts open.)
86 m
6 Find the distance between the points A and B with co-ordinates:
a A(3, 2) B(5, 7)
b A(5, 8) B(6, 11)
c A(−3, 1) B(4, 8)
d A(−2, −3) B(−7, 6)
7 The diagonals of a square are 15 cm. Find the perimeter of the square.
Review 11.2 Understanding similar triangles
Two mathematically similar objects have exactly the same shape and proportions, but may be
different in size.
When one of the shapes is enlarged to produce the second shape, each part of the original will
correspond to a particular part of the new shape. For triangles, corresponding sides join the
same angles.
Unit 3: Shape, space and measures 231
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All of the following are true for similar triangles:
Corresponding angles are equal.
Review B E ‘Internal’ ratios of sides are the same for both
CD triangles. For example:
A
B AB = DE
BC EF
A
F
E Ratios of corresponding sides are equal:
CD
AB = BC = AC
DE EF DF
F
Review If any of these things are true about two triangles, then all of them will be true for both triangles.
Worked example 5
Explain why the two triangles shown in E
the diagram are similar and work out x
and y. B
27° 8 m ym
6m
18 m
108°
REWIND A xm 108° 45°
C D 9m F
You learned in chapter 2 that
the angle sum in a triangle is Angle ACB = 180° − 27° − 108° = 45°
always 180°.
Angle FED = 180° − 45° − 108° = 27°
Review So both triangles have exactly the same three angles and are, therefore, similar.
Since the triangles are similar: DE = EF = DF
AB BC AC
y 18
So : 8 = 6 = 3 ⇒ y = 24 m
and: 9 = 18 = 3 ⇒ = 3m
x 6
232 Unit 3: Shape, space and measures
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Worked example 6 1.2 m 1.8 m F C
A B E 1.2 m
The diagram shows a tent that has been
attached to the ground using ropes AB 0.9 m HD
and CD. ABF and DCF are straight lines. G F
Find the height of the tent.
Consider triangles ABG and AEF:
Review 1.2 m + 1.8 m = 3 m height
B
1.2 m
0.9 m
A GA E
Angle BAG = FAE Common to both triangles.
Angle AGB = AEF = 90°
Angle ABG = AFE BG and FE are both vertical, hence parallel lines. Angles correspond.
Therefore triangle ABG is similar to triangle AEF.
So: height = 3 ⇒ height = 0 9 × 3 = 2 25m
0.9 1.2 12
Review Exercise 11.3 1 For each of the following decide whether or not the triangles are similar in shape. Each
decision should be explained fully.
Always look for corresponding
sides (sides that join the ab
same angles).
58° 10 m
3m
5m 6m
4m 8m
63° 59° 63°
cd
5 cm 15 cm 22 cm
30° 7 cm
83°
6 cm 18 cm
69°
30°
Review e f 27 km 9 km 7 km
49° 77° 18 km
54° 49°
6 km
21 km
Unit 3: Shape, space and measures 233
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g h 12.3 cm
7m 16.2 cm
8 m 60 m 5.4 cm 3.2 cm
4.1 cm
44 m
9.6 cm
10 m 48 m
iA B jA
triangles triangles
C ABC and CDE ABE and ACD
EB
Review DC
ED
2 The pairs of triangles in this question are similar. Calculate the unknown (lettered) length in
each case.
a 9 cm b
6 cm
x cm 8 cm 24 cm
8 cm
y cm 15 cm
c d a cm
Review 12 m 16 m 7 cm 28 cm
9m pm 3 cm
e f 3 cm
b cm 7 cm 12 cm
2.1 m
1.6 m
4m
c cm
3 The diagram shows triangle ABC. If AC is B
parallel to EF, find the length of AC.
5.1 cm
Review
EF
3.6 cm
7.3 cm
AC
234 Unit 3: Shape, space and measures
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4 In the diagram AB is parallel to DE. A 6.84 cm B E
Explain why triangle ABC is mathematically
similar to triangle CDE and find the length 4.21 cm
of CE. C
5 The diagram shows a part of a children’s E 7.32 cm D
climbing frame. Find the length of BC. A
Review 1.73 m
E 0.82 m B
D 2.23 m C
6 Swimmer A and boat B, shown in the diagram, are 80 m apart, and boat B is 1200 m
from the lighthouse C. The height of the boat is 12 m and the swimmer can just see
the top of the lighthouse at the top of the boat’s mast when his head lies at sea level.
What is the height of the lighthouse?
A 80 m 12 m C
B 1200 m
Review 7 The diagram shows a circular cone that has been r cm r cm
filled to a depth of 18 cm. Find the radius r of the top 12 cm
of the cone.
24 cm
18 cm
8 The diagram shows a step ladder that is held in place E
by an 80 cm piece of wire. Find x.
x cm
80 cm
30 cm
Review 120 cm
Unit 3: Shape, space and measures 235
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11. 3 Understanding similar shapes
In the previous section you worked with similar triangles, but any shapes can be similar. A shape
is similar if the ratio of corresponding sides is equal and the corresponding angles are equal.
Similar shapes are therefore identical in shape, but they differ in size.
You can use the ratio of corresponding sides to find unknown sides of similar shapes just as you
did with similar triangles.
Worked example 7
LINK Ahmed has two rectangular flags. One measures 1000 mm by 500 mm, the other
measures 500 mm by 350 mm. Are the flags similar in shape?
When trying to understand
Review how molecules fit together, 1000 = 2 and 500 = 1 43 (Work out the ratio of corresponding sides.)
chemists will need to have a 500 350
very strong understanding of
shape and space. 1000 ≠ 500
500 350
The ratio of corresponding sides is not equal, therefore the shapes are not similar.
Worked example 8
Given that the two shapes in the diagram are 8m 20 m
mathematically similar, find the unknown length x. xm
12 m
Review Using the ratios of corresponding sides: x = 20 = 2 5
12 8
⇒ x = 12 × 2.5 = 30m
Exercise 11.4 1 Establish whether each pair of shapes is similar or not. Show your working.
a2 by
6x
5
4
c 5 4 d 45
4 3 60
80
60
Review e f 80°
6 9
8 60°
12
236 Unit 3: Shape, space and measures
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a 2 In each part of this question the two shapes given are mathematically similar to one another.
5 cm Calculate the unknown lengths in each case.
3 cm
b
c
15 cm 11 cm
7.28 m x cm
p cm 22 cm 7 cm y cm
d
3.62 cm
Review 1.64 cm 10.3 cm 11.6 cm
y cm
8.4 cm
e 50 20 40 f 8 10 28
40 y x y 21 80
x y
g 25 x 40 x
12 h 120 6.9
267
Review 15
24
Area of similar shapes E
Each pair of shapes below is similar:
2.3
10
5
4 8 Area = 5.29 Area = 47.61
Area = 10 Area = 40
The ratio that compares the
Review measurements of two similar Scale factor = 10= 2 Scale factor = 6.9 = 3
shapes is called the scale factor. 5 2.3
Area factor = 40 = 4 Area factor = 47.61 = 9
10 5.29
If you look at the diagrams and the dimensions you can see that there is a relationship
between the corresponding sides of similar figures and the areas of the figures.
In similar figures where the ratio of corresponding sides is a : b, the ratio of areas is a2 : b2.
In other words, scale factor of areas = (scale factor of lengths)2
Unit 3: Shape, space and measures 237
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Worked example 9 E
These two rectangles are similar. What is the ratio of the smaller area to the larger?
18 21
Ratio of sides = 18 : 21
Ratio of areas = (18)2 : (21)2
= 324 : 441
= 36 : 49
Review Worked example 10
Similar rectangles ABCD and MNOP have lengths in the ratio 3 : 5. If rectangle
ABCD has area of 900 cm2, find the area of MNOP.
Area MNOP = 52
Area ABCD 32
Area MNOP = 25
900 cm2 9
Area MNOP = 25 ×900
9
= 2500 cm2
The area of MNOP is 2500 cm2.
Review Worked example 11
The shapes below are similar. Given that the area of ABCD = 48 cm2 and the area
of PQRS = 108 cm2, find the diagonal AC in ABCD.
P
A
18
D BS Q
C
R
Let the length of the diagonal be x cm.
48 = x2
108 182
Review 48 = x2
108 324
48 × 324 = x2
108
x2 = 144
x = 12
Diagonal AC is 12 cm long.
238 Unit 3: Shape, space and measures
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Exercise 11.5 1 In each part of this question, the two figures are similar. The area of one figure is given. E
Find the area of the other.
ab
15 m
7 m Area =17.0 m2
20 cm 30 cm
Area =187.5 cm2
Review d 25 cm
c 15 cm
80 m 50 m
Area = 4000 m2
Area = 135 cm2 xm
2 In each part of this question the areas of the two similar figures are given.
Find the length of the side marked x in each.
ab
16.5 m
Review 32 cm
x cm
Area = 592 cm2 Area = 333 cm2 Area = 272.25 m2 Area = 900 m2
22.5 cm
c d
2 cm x cm
x cm
Area = 4.4 cm2 Area = 6.875 cm2 Area = 135 cm2 Area = 303.75 cm2
Review 3 Clarissa is making a pattern using a cut out regular pentagon. How will the area of the
pentagon be affected if she:
a doubles the lengths of the sides?
b trebles the lengths of the sides?
c halves the lengths of the sides?
4 If the areas of two similar quadrilaterals are in the ratio 64 : 9, what is the ratio of
matching sides?
Unit 3: Shape, space and measures 239
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Similar solids E
Three-dimensional shapes (solids) can also be similar.
Similar solids have the same shape, their corresponding angles are equal and all corresponding
linear measures (edges, diameters, radii, heights and slant heights) are in the same ratio. As with
similar two-dimensional shapes, the ratio that compares the measurements on the two shapes is
called the scale factor.
Volume and surface area of similar solids
Review
The following table shows the side length and volume of each of the cubes above.
×5
×2
Length of side (units) 2 4 10
Volume (units3) 2×2×2=8 4 × 4 × 4 = 64 10 × 10 × 10 = 1000
×8
Review × 125
Sometimes you are given the scale Notice that when the side length is multiplied by 2 the volume is multiplied by 23 = 8
factor of areas or volumes rather Here, the scale factor of lengths is 2 and the scale factor of volumes is 23.
than starting with the scale factor of Also, when the side length is multiplied by 5 the volume is multiplied by 53 = 125.
lengths. Use square roots or cube This time the scale factor of lengths is 5 and the scale factor of volumes is 53.
roots to get back to the scale factor In fact this pattern follows in the general case:
of lengths as your starting point.
scale factor of volumes = (scale factor of lengths)3
By considering the surface areas of the cubes you will also be able to see that the rule from
page 221 is still true:
scale factor of areas = (scale factor of lengths)2
Review In summary, if two solids (A and B) are similar:
• the ratio of their volumes is equal to the cube of the ratio of corresponding linear measures a 3
b
(edges, diameter, radii, heights and slant heights). In other words: Volume A ÷ Volume B =
• the ratio of their surface areas is equal to the square of the ratio of corresponding linear
a 2
measures. In other words: Surface area A ÷ Surface area B = b
The following worked examples show how these scale factors can be used.
240 Unit 3: Shape, space and measures
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