276563235-Pearson-International-Mathematics-for-the-MYP-Year-2

Name Diagram Properties

Scalene • No sides are equal.
triangle • All angles are different.

Acute-angled • All angles are acute (less than 90°).
triangle

Right-angled • One angle is a right angle (90°).
triangle

Obtuse-angled • One angle is an obtuse angle
triangle (greater than 90°).

• The longest side is opposite the
largest angle.

A:07E Constructing Quadrilaterals

Often you can construct a quadrilateral in several different ways.

worked examples C 3·8 cm
D
To construct this quadrilateral, use the following steps.
Step 1 Draw a 60° angle with your protractor.
Step 2 Mark off the sides AB (5 cm) and AD (3·2 cm).
Step 3 Draw an 80° angle with its vertex at B.
Step 4 Mark off the side BC (3·8 cm).
Step 5 Join CD.
3·2 cm 60° 5 cm 80°
3·2 cmA B

1 2345

CC

DDD 3·8 cm D

60° 60° 80° A BA B
A A 5 cm B A B

26 INTERNATIONAL MATHEMATICS 2

A:07F Properties of Quadrilaterals

Name Diagram Properties
• One pair of opposite sides are
1 Trapezium
parallel.
2 Parallelogram
• Two pairs of sides are parallel.
3 Rhombus xx • Opposite sides are equal.
4 Rectangle xx • Opposite angles are equal.
• Diagonals bisect one another.
5 Square
A rhombus has all the properties of a
6 Kite parallelogram and …
• All sides are equal.
• Diagonals bisect each other at right

angles.
• Diagonals bisect the angles through

which they pass.

A rectangle has all the properties of a
parallelogram and …
• All angles are right angles.
• Diagonals are equal.

A square has all the properties of a
rhombus and a rectangle.

• Two pairs of adjacent sides equal.
• Diagonals are perpendicular.
• One diagonal is an axis of symmetry.

27APPENDIX A

A:07G Finding the Size of an Angle

Diagram Rule Example

triangle Find the value of a.
b° a° + b° + c° = 180° 70° a + 50 + 70 = 180

a° c° a + 120 = 180
∴ a = 60
b° c° quadrilateral
a° a° + b° + c° + d° = 360° 50° a°

d° 55° Find the value of b.
b° b + 55 + 115 + 110 = 360

110° 115° b + 280 = 360
∴ b = 80

a° b° isosceles triangle x° 3 cm Find the value of x.
b° a° = b° 72° 3 cm This is an isosceles
triangle since two
a° equilateral triangle sides are equal.
c° a° = b° = c° ∴ x = 72

a° b° straight angle y° Find the value of y.
a° + b° = 180° 135° This is an equilateral
y° triangle as all sides
are equal.
∴ All angles are equal.
3y = 180

y = 60

Find the value of y.
135 + y = 180

∴ y = 45

angles at a point 70° Find the value of m.
a° + b° + c° = 360° m° 30° m + 70 + 30 + 40 = 360

a° b° 40° m + 140 = 360
c° ∴ m = 220

A:07H The Angle Sum of a Quadrilateral

The sum of the angles of any quadrilateral is 360° (or one revolution).

The angle sums of other polygons can be found by multiplying the number that is two less than the
number of sides by 180°:
angle sum = (n − 2) × 180°, where n is the number of sides,
eg angle sum of a hexagon = (6 − 2) × 180°

= 720°

28 INTERNATIONAL MATHEMATICS 2

A:08 | Solid Shapes see

A:08A Know the Names of Solids 1:08

The ID cards at the front of this book (pages xv to xx) should be used often to help you remember
the language of Mathematics.

A:08B Prisms

(1) (2) (3) (4)

Prisms

All prisms have a special pair of parallel faces. These faces are the only two faces that need not be
rectangular in shape. If the prism is ‘sliced’ parallel to these faces, the same shape always results.
This shape is called the cross-section.

A prism is named according to the shape of its cross-section.

Name this solid. worked example

Solution ■ The prisms above are:
1 cube
The shape of the cross- 2 triangular
section of this solid is a 3 hexagonal
hexagon so it is called a 4 pentagonal.
hexagonal prism.

29APPENDIX A

A:08C Pyramids (3) (4)

(1) (2)

Pyramids

All pyramids have one face that need not be triangular. This face is used to name the pyramid.
A right pyramid has its point above the centre of its base.

A pyramid is named according to the shape of its non-triangular face. If all of its faces
are triangles the pyramid is a triangular pyramid.

Name this solid. worked example ■ The pyramids
above are:
Solution ■ An oblique 1 square
pyramid does 2 triangular
The solid is a pyramid. not have its 3 hexagonal
The non-triangular face point above 4 pentagonal.
is a square. Hence the the centre of
solid is a square pyramid. its base.

Euler’s theorem: For any convex polyhedra, F + V = E + 2 where
F = number of faces, V = number of vertices and E = number of edges.

A:08D Nets of Solids F = 5, V = 6, E = 9

Solid shapes can be made from plane shapes. This can be done by drawing the net of the solid on a
piece of paper. The net shows how the faces of the solid are joined to each other. When these faces
are folded along their edges, the solid is formed.

worked example

Draw the net of a square pyramid on a sheet of
square grid paper.

Solution

A square pyramid has four
identical triangular faces
joined to a square base, so:
Step 1 Draw a square base.
Step 2 Add the four adjoining

triangles.

30 INTERNATIONAL MATHEMATICS 2

A:08E Drawing Pictures of Solids

Below, a rectangular prism has been drawn using two different projections.
Rectangular prism

Use grids like these
to practise drawing solids.

1 Using square grid

2 Using triangular grid

worked example

Draw the front view, top view and right Solution
side view of the solid pictured below.

top

front front right side

A:09 | Measurement

A:09A Metric Units and Equivalents see

The Metric Units ID card on page xv and the Metric Equivalents table on page xiv should be used 1:09
to help you remember this essential information.

A:09B Measuring Lengths

worked example

Write down each measurement in centimetres, giving your An interval is
answers correct to the nearest millimetre. part of a line. It
has a definite length.
a bcd

cm 1 2 3 4 5678 9

Solution b 3·8 cm (ie 38 mm)
a 1·5 cm (ie 15 mm) d 8·4 cm (ie 84 mm)
c 6·1 cm (ie 61 mm)

31APPENDIX A

A:09C Conversions in the Metric System

When converting a measurement from one unit to another, you need to know the meanings of
metric prefixes.

kilo … hecto … deca … deci … centi … milli …

1000 100 10 1 --1--- ----1---- -----1------
10 100 1000

The value of each column is ten times the value of the column on its right.

To convert a unit into a larger unit you divide.
To convert a unit into a smaller unit you multiply.

worked examples

1 3000 mm = 3000 ÷ 10 cm 2 2500 mL = 2500 ÷ 1000 L 3 630 mg = 630 ÷ 1000 g

= 300 cm = 2·5 L = 0·63 g

4 7 km = 7 × 1000 m 5 7·8 kg = 7·8 × 1000 g 6 2·5 m = 2·5 × 100 cm

= 7000 m = 7800 g = 250 cm

7 An interval is 8·4 cm long. The interval must be divided into twelve equal parts. How many

millimetres would be in each part?

The length 8·4 cm = 8·4 × 10 mm

= 84 mm

∴ the length of each part = 84 ÷ 12
= 7 mm

A:09D Perimeter

The perimeter of a figure is the sum of the side lengths.
It is the distance around the figure.

Find the perimeter 8m worked examples ■ Perimeter of a square:
of each figure P = 4 × s (or P = 4s)
1 9m Solutions
s
5m P = (5 + 10 + 8 + 9) m
= 32 m

The perimeter is 32 metres.

10 m

32 INTERNATIONAL MATHEMATICS 2

2 For a square, all sides are equal. ■ Perimeter of a
P = (5·2 + 5·2 + 5·2 + 5·2) m rectangle:
5·2 m P=2×l+2×b
= (4 × 5·2) m (or P = 2l + 2b)
3 = 20·8 m
The perimeter is 20·8 metres. l
2·6 cm
1·2 cm For a rectangle, opposite sides b
are equal.
P = (2 × 1·2 + 2 × 2·6) cm The formulae
make the working
= (2·4 + 5·2) cm
= 7·6 cm easier.
The perimeter is 7·6 centimetres.

A:09E Time

You should be familiar with both conventional and digital time.

worked examples

1a b c d

12 12 12 12
93 3
93 93
6 6
6
25 minutes A quarter to 5
past 1 Half past 10 10 minutes to 7
d
2a AM b c
PM
PM AM

15 minutes 36 minutes past 20 minutes 55 minutes
past 10 8 (after noon) past 11 past 7

(before noon) (before noon) (after noon)

33APPENDIX A

3
A 24-hour clock uses a 4-digit number, the first two digits indicating
the hour past midnight and the second two indicating the minutes.

a bc d

20 minutes past 30 minutes past 12 noon 57 minutes past
5 (before noon) 5 (after noon) 11 (after noon)

■ The Days in a Month Jan
To work out how many days are in each month, Feb
we can use the knuckles on two hands.
The knuckle stands for a month of 31 days; Mar
the lower skin joining two knuckles stands for a Apr
month of 30 days, with the exception of February
which has 28 days, and 29 days each leap year. May
June

July
Aug
Sep
Oct
Nov
Dec
31 31 31 31 31
28 31 31
30 30
30 30

or

29

A:09F Distance, Speed and Time

If speed does not change (or we are referring to average speed), the following formulae can be used.

• Distance = Speed × Time
• Speed = Distance ÷ Time
• Time = Distance ÷ Speed
We can write these formulae using letters instead of words:
D = S × T, S = D ÷ T, T = D ÷ S

worked examples

1 Rajiv ran at a speed of 5 metres per second for 20 seconds. How far did he run?

2 Taya walked at a constant speed for 50 seconds. During this time she travelled 150 metres.
What was her speed?

3 A train travelling with a speed of 30 kilometres per hour travelled a distance of
120 kilometres. How long did it take?

Solutions 2 Time = 50 s. Distance = 150 m.
Find the speed.
1 Speed = 5 m/s. Time = 20 s. Speed = Distance ÷ Time
Find the distance. S=D÷T
Distance = Speed × Time ∴ S = 150 ÷ 50 m/s
D=S×T = 3 m/s
∴ D = 5 × 20 m
= 100 m ∴ Taya’s speed was 3 m/s.

∴ Rajiv ran 100 metres.

34 INTERNATIONAL MATHEMATICS 2

3 Speed = 30 km/h. Distance = 120 km. km/h means
Find the time. kilometres
Time = Distance ÷ Speed per hour.
T=D÷S
= 120 ÷ 30
= 4h

∴ the train took 4 hours.

A:09G Area 1 cm One square
centimetre
When we measure the area of a shape, we are measuring
the amount of space there is inside that shape. We try to (1 cm2).
work out how many square units would cover the shape.
One square
Area of a rectangle metre
(1 m2).
• Area = length × breadth l 100 cm
A=l×b

b 1 m2
100 cm

1 m2 = 10 000 cm2

worked examples

1 Find the area of the rectangle below. 100 m

Solution

3m A=l×b 100 m 1 ha
7·5 m = 7·5 × 3
One
= 22·5 m2 hectare.

The area is 22·5 m2. 1 ha = 10 000 cm2

2 Find the area of the figure below. Note: i We must make all units the same,
so 1 m = 100 cm.
1 m (100 cm)
ii We divide the figure into two rectangles.
30 cm 1
Solution
50 cm 2
30 cm A = Area 1 + Area 2
= (100 × 30) + (50 × 30)
= 3000 + 1500
= 4500 cm2

The area of the figure is 4500 cm2.

35APPENDIX A

Area of a square • Area of a square
s
A square is, of course, a rectangle
with its length equal to its breadth, s
so if a square has sides of s cm, then A = s2
the area is equal to s × s = s2 cm2.

worked example

Find the area of a square that has sides of length 7 mm.

7 mm A = s2
=7×7
= 49 mm2

The area of the square is 49 mm2.

Area of a triangle

Area of a triangle = 1-- (base × height)

2

A = b-----×-----h- or A = 1-- bh h h h
2 b b b
2

Find the area of each triangle. worked examples
12
3

5 cm 8m 5 cm
6m
12 cm 7 cm
A = b-----×-----h-
A = -b----×-----h- 2 A = -b----×-----h-
2 2
= 6-----×-----8-
= -1---2----×-----5-- 2 = -7----×-----5-
2 2
= 24 m2
= 30 cm2 = 17·5 cm2

36 INTERNATIONAL MATHEMATICS 2

A:09H Volume One cubic
centimetre (1 cm3).
When we measure the space an object occupies, we are
measuring its volume. We try to work out how many 1 cm
cubic units it would take to fill the object. 1 cm 1 cm

Irregular solids
We may be able to find the volume by counting cubes.

worked example One cubic 1 m3 = 100 ϫ 100 ϫ 100

metre = 1 000 000 cm3

Find the volume of the solid below.

3 cm 3 cm There are 5 cm3 in each layer. 100 cm
3 cm
There are 3 layers. 100 cm 100 cm
∴ volume = 5 × 3

= 15 cm3
∴ the volume of the solid is
15 cm3.

Rectangular prisms
To find the volume of a rectangular prism, use the following rule:

V = l × b × h or simply V = lbh h
b
l

worked examples

Find the volume of each rectangular prism. 3 cm ■ Notice:
12 The number of cubes
in each layer is the
2 cm same as the area of
the base rectangle in
3 cm square units.

7 cm 6 cm 2 cm

V = lbh V = lbh
=7×3×2 =6×2×3
= 42 cm3 = 36 cm3

∴ volume is 42 cm3. ∴ volume is 36 cm3.

37APPENDIX A

3

20 m V = lbh
= 11 × 9 × 20
9m = 1980 cm3
11 m
∴ volume is 1980 cm3.

Summary: Perimeter, area and volume 8 P = 24 units
1 Perimeter: distance around 4 A = 12 units2
V = 48 units3
4 8
4
2 Area: number of squares inside
3

3 Volume: number of cubes inside

3
4

4

A:09I Capacity

When we measure the amount of a liquid that a container can hold, we are measuring its capacity.

We might say that an amount of a liquid
is so many cupfuls or spoonfuls, but as
with the volume of solids, we need some
standard units.

The basic unit in the metric system for capacity is the litre (L).

This means that:

1000 millilitres (mL) = 1 litre (L)

1000 litres (L) = 1 kilolitre (kL)

There is a link between the capacity of a liquid and the volume of a solid.

A container with a volume of 1 cube centimetre would hold 1 millilitre of liquid.
∴ 1 mL = 1 cm3 1 L = 1 dm3 1 L = 1000 cm3

38 INTERNATIONAL MATHEMATICS 2

A:09J Mass

Mass is the amount of matter that makes up an object. We can measure the mass of an object by
measuring its weight at sea level or by comparing it with a known mass.

To get a measure of an object’s mass, we need some standard units. The metric units for mass are:

100 milligrams (mg) = 1 gram (g) The base unit
is the kilogram.
1000 grams (g) = 1 kilogram (kg)

1000 kilograms (kg) = 1 tonne (t)

This small packet of dried
peas has a mass of 50 g.

This bag
of potatoes
has a mass
of 20 kg.

The combined mass of a container and its contents is its gross mass.
The mass of the contents only is its net mass.

Calculate the missing mass. worked examples 3 Net mass = 800 g
Container’s mass = 170 g
1 Gross mass = 500 g 2 Gross mass = 1·2 kg Gross mass = g
Net mass = 400 g Container’s mass = 250 g
Container’s mass = g Net mass = g Gross mass
Net mass = Net + Container
Container’s mass = Gross − Container = 800 g + 170 g
= Gross − Net = 1200 g − 250 g = 970 g
= 500 g − 400 g = 950 g
= 100 g

A:10 | Directed Numbers

A:10A Directed Numbers ... –4, –3, –2, –1, 0, see
1, 2, 3, 4, ...
Directed numbers have both size and direction. 1:10

−20 is read as ‘negative 20’ or ‘minus 20’.
+20 is read as ‘positive 20’ or just ‘20’.
Note: +20 can be written more simply as 20.

■ If a directed number is a whole
number, it is called an integer.

39APPENDIX A

A:10B Addition and Subtraction

When adding or subtracting, a negative can be thought of as a loss and a positive can be thought
of as a gain.

1 −7 + 11 A loss of 7 and worked examples A loss of 9 and
=4 a gain of 11. a gain of 2.
2 −9 + 2
= −7

3 2 − 13 A gain of 2 and 4 −7 − 5 A loss of 7 and
= −11 a loss of 13. = −12 a loss of 5.

A:10C Signs Side by Side

When two signs occur side by side (with no numeral between them):
• two like signs give a plus
• two unlike signs give a minus.

worked examples

1 6 − −10 2 3 + (+7) 3 10 − (3 − 9) 4 3 + (−7 + 11)
= 6 + 10 =3+7 = 10 − −6 = 3 + (+4)
= 16 = 10 = 10 + 6 =3+4
= 16 =7

A:10D Multiplication and Division

For multiplication and division:
• two like signs give a plus
• two unlike signs give a minus.

worked examples

1 −4 × −3 2 −0·2 × −3 3 −4 × 14 4 7 × (−1·1)
= 12 = 0·6 = −56 = −7·7

5 −35 ÷ (−5) 6 60 ÷ −6 7 -–---2---1- 8 -–---1---·--8--
=7 = −10 –3 2
=7
= −0·9

9 −8 + 6 × −3 10 (2 − 20) ÷ 3
= −8 + −18 = −18 ÷ 3
= −8 − 18 = −6
= −26

40 INTERNATIONAL MATHEMATICS 2

A:11 | The Number Plane

A:11A The Number Line Extended see

To graph a number on the number line, place a large dot at that position on the number line. 1:11

worked examples ■ A set of numbers is a
collection of numbers.
1 Graph the set of numbers {−2, 0, 1·5, 4}.

–3 –2 –1 0 1 2 3 4 5

2 List the set of numbers graphed below.

0 0·5 1 1·5

The set is {−0·3, 0·2, 0·6, 1}.

A:11B The Number Plane

(2, 5) is called an ordered pair or the coordinates of a point.
The first number is the reading on the x-axis (how far right or left).
The second number is the reading on the y-axis (how far up or down).

(–3, 1) is 3 to y
the left and 1 up. 4

3 (4, 3) is above
‘4’ on the x-axis
2 and across from
1 ‘3’ on the y-axis.

–5 –4 –3 –2 –1 0 12 3 4x
–1
(–5, –4) is (3, 0) is
5 to the (0, 0) is called –2 3 to the right
left and 4 the origin. It is –3
0 to the right and and 0 up.
down.
0 up. (2, –3) is
2 to the right
–4 and 3 down.

41APPENDIX A

A:11C Mastery Test

Do this test again and again until you get all the questions right.

y Answers
4 2 Name this ax.is
9 What letter
is at (4, –4)?
EF 1 x-axis 6 (−4, 0)
2 y-axis 7 (0, −2)
8 How long 3 3 origin, 8 3 units
is CD? D
(0, 0) 9 A
C 2 4 Give coordinates 4 (1, 3) 10 F
5 (−5, −2)
1 1 Name this ax.is

–5 –4 –3 –2 –1 1 234 5x
6 Give coords. –1 3 Give name
10 What letter
5 Give coords. –2 and coords. is at (–2, 4)?
B –3 7 Give coords.
–4 Check your
A answers when
you finish the

test.

A:12 | Algebra

see A:12A Describing Number Patterns

A symbol can be used to describe the rule used to obtain a number pattern. Symbols that represent
numbers are called pronumerals. The most commonly used pronumerals are the letters of the
1:12 alphabet.

A pronumeral takes the place of a numeral.

worked examples

1 In each example below, t represents the number of triangles, and m the number of matches
used. Find the rule to describe each pattern and use it to complete the table of values given.
a

, , , ...

3 6 9

42 INTERNATIONAL MATHEMATICS 2

The number of matches used is always three times the number of triangles.
m=3×t

t 12345678
m 3 6 9 12 15 18 21 24

b

,, , , ...

35 7 9

The number of matches used is always two times the number of triangles plus one.
m = (2 × t) + 1

t 12345678

m 3 5 7 9 11 13 15 17

2 At Joe’s fruit shop, apples cost 25 cents each. If C stands for the cost in cents and n stands
for the number of apples, write down a rule to give the cost of each lot of apples bought,
and use it to complete the table of values.
Cost of apples = (number of apples) × (25 cents)
C = n × 25

n 1 2 3 4 5 10 30 100

C 25 50 75 100 125 250 750 2500

A:12B Algebraic Expressions

If an expression contains one or more pronumerals, it is an algebraic expression.
eg 4 + a, 9 × y, p ÷ q, 5 + x × y

Note:
• 1 × m = 1m = m

The numeral 1 does not have to be written in front of a pronumeral.
• 4 × y = 4y

This means ‘4 times y’ or ‘4 lots of y’ or ‘y + y + y + y’.
• f × 5 is written as 5f, not f5.
• a × b = ab = ba

Since 5 × 11 = 11 × 5, a × b = b × a
• x ÷ 5 = -x- , a ÷ 12 = --a--- .

5 12
The fraction is usually used instead of the division sign.

43APPENDIX A

worked examples

Simplify: 2 a+3×y 3 6 × (a + 7) 4 5×a÷7
1 5×k+2 = a + 3y = 6(a + 7) = -5---a-
7
= 5k + 2

Rewrite these expressions showing all multiplication and division signs.

5 3a + 8 6 5p − 6q 7 4(x + 2) 8 -a----+-----7-
=3×a+8 =5×p−6×q = 4 × (x + 2) 3

= (a + 7) ÷ 3

A:12C Substitution

You will often be asked to evaluate an expression using values given for the pronumerals.

worked examples

1 Given x = 3: b 2(x + 5) = 2 × (3 + 5) c 5x2 = 5 × 32
a 6x = 6 × 3 =2×8 =5×9
= 18 = 16 = 45

Note: The multiplication signs must be written in after the values have been substituted for
pronumerals.

2 If a = 2 and b = 5 then: b -1---0---a- = 1----0----×-----2-- c 4a(b − a) = 4 × 2 × (5 − 2)
a 3a + 7b = 3 × 2 + 7 × 5 b5 =4×2×3
= 6 + 35 = 20 ÷ 5 = 24
= 41 =4

3 If the number of matches, m, needed to make t triangles is given by m = 2t + 1, how many
matches would be needed to make 100 triangles?
m = 2t + 1
= 2 × 100 + 1 (when t = 100)
= 200 + 1
= 201
So 201 matches would be needed to make 100 triangles.

A:12D Finding the Rule y ϭ 3x ϩ 4
x 12345
Rules like y = 3x + 4 have a difference of 3 between y 7 10 13 16 19
successive values of y as x increases by 1 each time.
3333
Rules like y = 7x + 4 have a difference of 7 between
successive values of y as x increases by 1 each time.

44 INTERNATIONAL MATHEMATICS 2

worked example x 12345
y 11 15 19 23 27
Discover the rule connecting x and y. If the differences
( ) are the same, the rule will be like: y = x + 4444
As the differences are all 4,

y = 4x +
From the table we see that when x = 1, y = 11.

So 11 = 4 × 1 +
∴ =7
So the rule is y = 4x + 7.

A:12E Simplifying Algebraic Expressions

If terms have identical pronumeral parts, then they are called like terms.

Like terms may be added or subtracted.

1 7a + 5a = 12a worked examples 3 8x − x = 7x
4 5m + 7m − 10m = 2m
2 3pq − 10pq = −7pq 6 7p + 2q + 3p + q
7 6a + 7b − 2a + 5b +5b 5 5x + 2y + 7y = 5x + 9y = 10p + 3q
= 6a −2a +7b
= 4a + 12b ■ The plus or minus 8 7a2 − 4a + 2a2
sign always belongs to = 9a2 − 4a
the term after it!

When multiplying algebraic terms, multiply numbers first and then pronumerals. When dividing,
write the division as a fraction and reduce the fraction to its lowest terms.

worked examples

1 3 × 5a 2 7m × 3p 3 a × 4b 4 −3k × −5
= 15a = 21mp = 4ab = 15k

5 12t ÷ 3 6 6m ÷ 2a 7 15r ÷ 10 8 3ab × 7a
= 4-1---2---t = 36----m--- = 3-1---5---r- = 21aba
13 12a 2 10 = 21aab
= -3---m--- = -3---r- = 21a2b
= 4t a 2

A:12F Grouping Symbols

To ‘expand’ an expression, we write it without grouping symbols.

• a(b + c) = ab + ac ■ • Parentheses: ( )
• a(b − c) = ab − ac • Brackets: [ ]

45APPENDIX A

worked examples

1 5(x + 2) 2 9(2a − 3) 3 m(m + 7)
=5×x+5×2 = 9 × 2a − 9 × 3 =m×m+m×7
= 5x + 10 = 18a − 27 = m2 + 7m

A:12G Index Notation ■ The plural of ‘index’ is
‘indices’.
In a4, ‘a’ is called the base, ‘4’ is called the index.

worked examples

4 2 53 = 5 × 5 × 5

1 a×a×a×a=a

3 p × p × p × p × p × p × p = p7 (since the p is used as a factor 7 times).

4 m × m × m × b × b × b × b × b = m3b5 5 5 × t × t × 7 × t = 35t3

A:12H Algebraic Sentences or Equations

To ‘solve’ an equation, we must find the value of the pronumeral that will make the equation true.

worked examples

1 x+5=9 2 12 − a = 5 3 6m = 42
Ask: ‘What number Ask: ‘12 minus what Ask: ‘6 times what
plus 5 is equal to 4?’ number is equal to 5?’ number is equal to 42?’
x=4 a=7 m=7

4 -x- = 21 5 m+7=2 6 2x + 5 = 13
3
Ask: ‘What number Ask: ‘What number ‘What plus 5 gives 13?’
plus 7 is equal to 2?’ 2x = 8
divided by 3 is equal
m = −5 ‘What times 2 gives 8?’
to 21?’ x=4
x = 63

A:12I Problem Solving Using Algebra

We can use equations to solve problems.

worked example

The sum of two consecutive numbers is 91. What are the numbers? Consecutive numbers

Solution follow one after the other,
eg 29, 30, 31, ...

Let the numbers be x and x + 1.

Then x + (x + 1) = 91

2x + 1 = 91

2x = 90

x = 45

∴ x + 1 = 46 ■ Using algebra is a

The numbers are 45 and 46. problem-solving strategy.

46 INTERNATIONAL MATHEMATICS 2

APPENDIX B

Challenge B:01 | Finite differences challenge

Here we are investigating patterns in the differences columns of equations like y = 5x2 + 2x + 1. B:01
1 y = x2 + 4

xy Differences Step 1
15 Make sure that the values of x increase one at a time.
28 3 Find the values of y.
3 13 2
4 Step 2
5 5 Find the differences between the values of y, and then
6 the first column of boxes. Put the answer in the box to
the right each time.

Step 3
Has a pattern emerged? Is there a connection to the
original equation?

2 y = 2x2 + x 3 y = 3x2 − 1

xy Differences xy Differences
1 1
2 2
3 3
4 4
5 5
6 6

4 y = 4x2 + 2x 5 y = 5x2 − x + 1

xy Differences xy Differences
1 1
2 2
3 3
4 4
5 5
6 6

47APPENDIX B

APPENDIX C
C:01 | Multiplying with Indices

prep quiz Simplify these expressions.

C:01 1 42 2 23 3 92 4 33 ÷ 32 5 24 × 22 6 (23)2

Rewrite these expressions using index notation.

7 m×m×m×m×m 8 a×a×b×b×b

9 5×a×a×6×a×a 10 3 × y × z × 4 × y × y × z

This Prep Quiz should have reminded you of how indices can be used to simplify expressions in
algebra.

n6 means 6 n’s multiplied together.
n6 = n × n × n × n × n × n

worked examples

1 x5 × x7 = x5 + 7 2 n6 × n = n6 × n1
= x12 = n7

■ Rule 1 3 6a2 × 3a3 = 18a2 + 3 4 m6n3 × m2n5
When multiplying powers = 18a5 = m6 + 2n3 + 5
of x, add indices: = m8n8
xm × xn = xm + n
5 3xy4 × 5x2y3 = 15x1 + 2y4 + 3
= 15x3y7

48 INTERNATIONAL MATHEMATICS 2

Exercise C:01

1 Simplify these products.

a x3 × x2 b m5 × m2 c n7 × n3 d q3 × q7
g x6 × x2 h y × y4
e a × a3 f p6 × p k n5 × n7 l x4 × x6
o m6 × m3 p t3 × t4
i t×t j w4 × w4 s x × x6 t q10 × q10

m h7 × h4 n f2 × f2 c x2 × 4x2 d q7 × 7q2
g 7a3 × 3a2 h 5m × 10m2
q w2 × w10 r p7 × p7 k 3x × 6x4 l 8p2 × p4
o 5n3 × 7n7 p 3x4 × 4x3
2 Simplify: b 5p2 × p3 s 10b3 × 12b2 t 8y7 × 7y8
f 4x2 × 2x4
a 2x2 × x5 j 7t × 8t3 c a4b3 × b3 d p5 × pq4
e 3p2 × 2p3 n 6m2 × 6m2 g x4y3 × x5y2 h tw3 × t2w
i 6a4 × 4a r 9a4 × 7a5 k mn4 × m4n l a2b3 × b2a
m w5 × 3w3 o ab2 × a4b2 p m3n2 × m10n
q 3p3 × 5p s tw × tw4 t a10b4 × a10b5

3 Simplify: b p2q3 × q2 c 4xy4 × 3x3 d 7t2w4 × 2t5
f a2b7 × a4b g 6a2b × a4b3 h 2x2y5 × 7xy
a a4 × a2b j am4 × a4 k p4q5 × 5pq2 l 3m2n5 × 2mn
e m4n3 × m2n5 n x4y5 × xy o 4a × 5b × 2ab3 p 6p2q4 × 2pq × 4p
i x2y × x3 r p4q5 × pq
m k2l5 × l4k2
q a2b × a4b

4 Simplify each product.

a 3a2b × 4a4 b 5p2q3 × 3q2

e 6x4y2 × 3x5y4 f 3px4 × 8p4x

i 8m2n4 × 4m5n6 j 6a4b × a5b4

m 2ab2 × 3a × 4b n 7a3b4 × 2a × 3b2

C:02 | Dividing with Indices

worked examples

■ Rule 2 1 x7 ÷ x3 = x7 – 3 2 a4 ÷ a = a4 ÷ a1
When dividing powers of = x4 = a3
x, subtract indices:
xm ÷ xn = xm – n 3 10p4 ÷ 5p2 = 2p4 – 2 4 a5b4 ÷ a2b3 = a5 – 2b4 – 3
= 2p2 = a3b1
= a3b
5 -2---4---a---6- = 4a6 – 4
6a4 = 4a2 6 -a---4--m-----5- = a4 – 1m5 – 4
am4 = a3m

49APPENDIX C

Exercise C:02

1 Simplify these divisions.

a x5 ÷ x2 b a7 ÷ a4 c m4 ÷ m2 d n5 ÷ n3
g y6 ÷ y5 h k7 ÷ k6
e a4 ÷ a f y5 ÷ y k x10 ÷ x8 l m12 ÷ m7
o a6 ÷ a2 p l9 ÷ l6
i z6 ÷ z4 j b9 ÷ b7 s t16 ÷ t4 t w14 ÷ w2

m t8 ÷ t7 n w6 ÷ w c 7k4 ÷ k d 8y7 ÷ y5
g 16n8 ÷ 4n6 h 24b9 ÷ 6b5
q x12 ÷ x6 r m15 ÷ m10 k 8l6 ÷ 2l3 l 12n10 ÷ 3n7
o 7m6 ÷ 7m4 p 14w6 ÷ 14w5
2 Simplify: b 6m4 ÷ m2 s 24y8 ÷ 12y4 t 36p12 ÷ 12p4
f 20a4 ÷ 4a3
a 5x7 ÷ x3 j 18y7 ÷ 9y2 c m-----8- d -t--1--0
e 10m6 ÷ 2m2 n 5a4 ÷ 5a3 m6 t2
i 12x6 ÷ 4x5 r 25n15 ÷ 5n3
m 15t7 ÷ 15t g 6----a---4- h 1----2---t--9-
q 30a10 ÷ 10a5 3a2 4t7

3 Simplify: b a----1--0 k 1----0---m-----5 l 2----4---n---1---0
a7 2m4 6n6
a -x---7
x5 f 1----0---m-----6 c y6z4 ÷ z3 d t6w5 ÷ t4
m5 g m6n4 ÷ m5n h k4m2 ÷ km
e -6---m-----7 k -t--7--w-----4 l f--7---h---5-
m5 j -1---8---a---7-
9a w3 f5
i -1---5---n---7- o p----6--q---7- p c---1--0---d---7-
3n5
q3p3 dc8
4 Simplify these divisions. s 24p6q3 ÷ 6q2 t 8t4w5 ÷ t3w2
w 6----x---y---4- x 1----8---a---7---b---2
a a7b5 ÷ a5 b m7n6 ÷ m4
3xy2 6a2b2
e p6q4 ÷ p4q3 f a8b4 ÷ a6b2

i -a---4--x---7- j m-----1--0---n---7-
x4 m8

m -m----6---n---1--0- n a----6--b---8-
m4n7 ab5

q 6a4b3 ÷ a3b2 r 10m7n3 ÷ 5m6

u -1---0---m-----6--n----5 v -1---2---a---4---b---7
m3n2 6a3

C:03 | Power of a Power

worked examples

■ Rule 3 1 (x5)4 = x5 × 4 2 (3x2)3 = 33 × x2 × 3
For a power of a power, = x20 = 27x6
multiply the powers:
(xm)n = xmn 3 (2x)5 = 25 × x1 × 5 4 (4m3n2)2 = 42 × m3 × 2n2 × 2
= 32x5 = 16m6n4

50 INTERNATIONAL MATHEMATICS 2

Exercise C:03

1 Simplify these expressions.

a (a2)4 b (x3)4 c (m2)5 d (n3)2
g (x3)3 h (w7)3
e (y7)4 f (t4)3 k (h3)5 l (b2)2
o (w3)4 p (x10)10
i (x7)2 j (y4)5
c (5m3)2 d (2x4)3
m (n9)3 n (m2)10 g (3p2)3 h (4q2)3
k (6n)3 l (10a5)3
2 Simplify: b (3a4)2
f (2y)5 c (a4x2)4 d (pq5)2
a (2x3)2 j (7p3)2 g (k4m2)3 h (x4y3)3
e (4n5)2 k (10mn4)2 l (5xy4)2
i (10a3)2 o (2p4q2)4 p (5a3b2)3

3 Simplify: b (m4n2)3
f (t2w2)4
a (a2b)2 j (3x2y3)2
e (mn4)2 n (10m4n3)3
i (2a2b)2
m (4p2q3)2

C:04 | Miscellaneous Exercises on Indices

worked examples

■ Index Laws Now try this set of mixed exercises using one or
• am × an = am + n
Remember these more of the index laws.
laws! ie when multiplying,
add the indices. 1 x7 × x5 ÷ x6 2 15a7 ÷ 5a4 × 2a2
• am ÷ an = am – n
ie when dividing, = x7 + 5 – 6 = 3a3 × 2a2
subtract the indices. = x6 = 6a5
• (am)n = amn
ie for a power of a 3 -2---m-----4---×-----6----m----2- 4 -(--2---m-----2---)--3----×-----3---m-----5
power, multiply the 4m3 4m × m3
indices.
• a0 = 1 = 1----2---m-----6 = 8----m-----6---×-----3----m----5-
ie any number to the 4m3 4m4
power of zero equals 1.
= 3m3 = 2----4---m-----1--1-
4m4

= 6m7

Exercise C:04

1 Use the index laws to simplify each expression fully.

AB C
p6q × p2q3
a x6 × x5 a10 ÷ a5 20m2 ÷ 10m
q6 ÷ q6
b (a5)3 (3m)3 (5p2)2

c 6n2 × 3n3 p7 × p

d 9p4q ÷ 3p3 a4b3 × ab

51APPENDIX C

e 48tw4 ÷ 12tw 8m4 × m2n 10x4y × 3xy4

f -a---1--2 -1---8---q---6- -3---2---a---7---b---3
a4 9q3 8a6b3
8x6 ÷ 8x5 (2pq2)5
g (4a4b)2 a4 × (a2)3 (p4)3 × p6
h (x3)2 × x5 (t5)2 ÷ t4 x10 ÷ (x2)4
i (a4)3 ÷ a10 (3x4)2 × 4x2 (2m3)2 × (3m)2
j (2p2)3 ÷ 4p5
-(--2---m-----4---)--2- -(--2---n----3--)---3
k (---3---a----4--)--2- m6 4n7
3a5
2m4 × 3(m2)3 (4x4)2 ÷ 8(x3)2
l 4(a4)2 × 5a3 3a4 × 2b2 5ab × 2b × 3a
m p2q × p3 × q2 20m4n3 ÷ 5m2n × 2mn 24t4 ÷ 8t ÷ 3
n 10xy4 × 2x5 ÷ 4x2y2

2 Check out this ■ Looking at the division 53 ÷ 53 in two ways:

interesting fact! First, a number divided by Also, using the index laws,

itself must be equal to 1. 53 ÷ 53 = 53 – 3

ie 53 ÷ 53 = 1 = 50

This means that 50 = 1.

Actually, any number with an index of zero is equal to 1: a0 = 1

If this is true, find the answers to these.

a 30 b 70 c 80 d 1230
h (5pq2)0
e x0 f p0 g (2a)0 l 25m0
p (9x2)0 × (3x0)2
i 2a0 j 5x0 k 3p0

m 5a0 × (5a)0 n (2x)0 × 4x0 o 3x2 × 4x0

C:05 | Large Numbers on a Calculator

inve stigation Investigation C:05 | Large numbers on a calculator
C:05
1 A five-cent coin is placed on the first square of a chess
board. Two five-cent coins are placed on the second
square, four on the next, eight on the next and so on
until there is a pile on each square.

My calculator tells me that on the 40th square there
would be 5·497558139 11 coins.

• What does this mean?
• What is this pile worth?
• How high is this pile of coins?
• Work out some questions of your own and use a

calculator to solve them.

2 If light travels at 300 000 km/s, how far would light travel in one year? This distance is called
a light year.

52 INTERNATIONAL MATHEMATICS 2

APPENDIX D
D:01 | Non-numerical Geometrical

Reasoning (Extension)

AD Copy the diagram. prep quiz
1 Mark ∠ABC with ‘X’
2 Mark ∠BAC with ‘•’. D:01

In the diagram, which angle is equal to:

B CE 3 ∠ABC? 4 ∠BAC?

5 Which angle is adjacent to ∠DCE?

6 Which two adjacent angles make ∠ACE?

7 If a + b = 180, then 180 − a = . . . .? 8 If a + c = 180, then c = . . . .?

9 If b = 180 − a and c = 180 − a 10 If a = b and b = c, what can we say

then b = . . . .? about a and c?

Many problems in geometry are non-numerical. In these problems, the reasoning process becomes
more involved. As there are no numbers involved, pronumerals are used to represent unknown
quantities. With the use of pronumerals, the reasoning will involve algebraic skills learned in other
parts of the course.

We have already seen examples of reasoning involving non-numerical problems in Chapter 7.
In sections 7:04 and 7:05, reasoning was applied to non-numerical problems to prove that the
angles of a triangle add to 180° and the angles of a quadrilateral add to 360°. These were excellent
examples of reasoning applied to non-numerical exercises.

Because exercises do not involve specific numbers, the results we obtain will be true irrespective of
the numbers used. The results obtained are called generalisations or, more commonly, proofs.

worked examples

1 In the diagram, prove that x = y. 2 In the diagram, prove that ∠ABD = ∠CEF.

AC A

G y° E F H D B
x° E C
F
BD

3 In the diagram, prove that: A
a ∠ABD = ∠DAC
b ∠BAD = ∠ACD

B DC

53APPENDIX D

Solutions 2 ∠ABD = ∠BDE (alt. ∠s, AC // DF)
∠CEF = ∠BDE (corresp. ∠s DB // EC)
1 ∠CFE = x° (vert. opp. ∠s)
∠CFE = y° (corresp. ∠s, AB // CD) ∴ ∠ABD = ∠CEF (both equal to ∠BDE)
∴ x = y (both equal to ∠CFE)
I’ll have to be
3A wide awake for

y° this work!
w°

x° z°
B DC

Let ∠ABD = x°, ∠BAD = y°, ∠DAC = w° and ∠ACD = z°.

a x + y = 90 (comp. ∠s in ∆ABD) b x + y = 90 (comp. ∠s in ∆ABD)
w + y = 90 (comp. ∠s) x + z = 90 (comp. ∆ABC)
∴x=w ∴y=z

∴ ∠ABD = ∠DAC ∴ ∠BAD = ∠ACD

Exercise D:01

1 a Prove that x = y. b Prove that a = b. These won’t stump you if you
remember your work on angles
A E x° B DF
A and parallel lines.

C F D a° X
y° Y b°

B
CE

2 In each of the following, prove that x = y.

a bX

y° B A x° B C
A

x° y°
CD DE

Y

3 ABCD is a parallelogram. A B
Prove that: a° y°
a x=y
b a=b x° b°
D C
54 INTERNATIONAL MATHEMATICS 2

This is an important result.

The opposite angles of a parallelogram are equal. x
x

4 a Prove that b Prove that c Prove that
∠FAB = ∠ECD. a = b = c = 90°. ∠AEC = 90°.

E BD A B BD
a° b°
F

c° E
C
G D p°p° q°
C A q°

C

A

5 a Prove that a = b. b Prove that c Prove that
∠ABC = ∠ADC. ∠BCA = ∠ABD.
A
b° A A
D a° a° D
BD
BC
a°x° x° b° b°
B C C

55APPENDIX D

APPENDIX E

fun spot Fun Spot E:01 | The Tower of Hanoi

E:01 This is a puzzle that consists of a series of discs
which vary in size and which are stacked to form
a tower. The object of the puzzle is to move the
discs to form the tower on one of the other two pegs.
Two rules, however, must be obeyed:
• Only one disc at a time may be moved.
• A larger disc must never be placed on

top of a smaller disc.

Of course, the puzzle gets harder the more discs you use. For 3 discs the tower can be moved
in 7 moves, like so:

123

45 6 7

It can be done in 15 moves with 4 discs. You could use
7 discs, however, would take at least some coins.
127 moves!

56 INTERNATIONAL MATHEMATICS 2

APPENDIX F
F:01 | Bisecting Intervals and

Angles

Bisect means to cut in half. In Year 7 Watch me bisect this person
you would have bisected an interval with my ruler and compasses!
(or found the halfway point) using a
ruler. To bisect (or halve) an angle you
would have used a protractor.

The ancient Greeks knew how to bisect
intervals and angles using compasses and
a ruler. The steps that need to be followed
to do this are given below in diagrammatic
form with an explanation following.

Bisecting an Interval 3 4 That’s
BA C easy!
12
E
A BA
BA B

D

• If we join the points A, C, B and D we get a rhombus.
Step 1 The interval AB.
Step 2 With the point of the compasses at A, draw an arc above and below the interval.
Step 3 With the point of the compasses at B, draw two arcs to cut the arcs already drawn at C and D.
Step 4 Joint C to D. The line CD is called the perpendicular bisector of AB, while E is the

midpoint of AB.

57APPENDIX F

Bisecting an Angle 3 45

12 A A FA F

AA EE E
E

D

BC B DC B DC B DC B C

Step 1 The angle ABC. Work in
pencil so
Step 2 With the point of the compasses at the you can
vertex B, draw an arc to cut both arms erase easily.
of the angle at D and E.

Step 3 From D draw an arc.

Step 4 From E draw an arc to cut the arc drawn
in step 3 at F.

Step 5 Join B to F.

Exercise F:01 (Practical)

You will need: compasses, ruler, protractor, pencil, eraser
1 Draw three intervals similar to the ones shown below and construct the perpendicular bisector

of each.
a bc

2 a Draw any triangle ABC and bisect the three sides.
b The bisectors should meet at a point called the circumcentre. What do you notice about the
distances of the points A, B and C from the circumcentre?
c Use the circumcentre to draw a circle (called a circumcircle) that passes through the points
A, B and C.

3 a Draw a circle that will pass through the points •Y •X
X, Y and Z. (You will need to trace the positions •Z
of X, Y and Z in your book and use the method
used in Question 2.)

b Mark any three points on a page (they cannot be
in a straight line) and construct a circle that passes
through these points.

58 INTERNATIONAL MATHEMATICS 2

4 C
B
Draw any quadrilateral ABCD and by constructing the
A perpendicular bisectors, locate the midpoint of each side.
What shape is obtained by joining the midpoints?

D

5a B Draw any circle and mark the centre, O.
C Draw any two chords AB and CD.
A Bisect the chords AB and CD. What do you notice?

O

D

b Using what you have learnt in Part a, describe how you could find the centre of a circle.

6 a Construct the isosceles triangle ABC shown. C
b Bisect the base AB.
c Does the perpendicular bisector pass through C?

To find the position of 7 cm
C, remember that it is
7 cm from A and 7 cm A 5 cm B

from B, then use
your compasses,

7 a Draw an arc of a circle. Choose two points on the arc and join
O A them to the centre O as shown in the diagram.

b What type of triangle is ∆AOB? Why?
c Bisect AB.
d Does the perpendicular bisector pass through O?

B

8 Trace the angles below and bisect them. c
ab

59APPENDIX F

9A a Use your compass to copy ∆ABC.
B b Bisect ∠ABC.
c Bisect ∠BCA.
d Bisect ∠BAC.
e The bisectors meet at a point called the incentre.

Use the incentre to draw the incircle (the circle
that touches the three sides of the triangle on the
C inside).

10 a Use compasses and protractor to construct a square.
b Bisect one of the angles of the square.
c Does the bisector also bisect the opposite angle?
d Does the diagonal of the square bisect the angles through which it passes?

11 a Construct the rectangle, parallelogram and rhombus shown. C D

D C DC

3 cm 3 cm 4 cm

3 cm

A 4 cm B 60° 120° 50° A
A 4 cm B B 4 cm

b Bisect ∠ABC in each figure.
c In which figure does the bisector bisect the opposite angle?
d In which figure does the diagonal bisect the angles through which it passes?

12 Draw an interval BC and mark a point A on it.

a Bisect the straight angle BAC. •
b Check the size of the bisected angle with a B AC

protractor. What size should it be?

60 INTERNATIONAL MATHEMATICS 2

F:02 | Constructing angles of

60°, 120°, 30° and 90°

1 What does ‘bisect’ mean? 2 1-- of 60° 3 -1- of 180° 4 1-- of 90° prep quiz
5 Find the value of x.
2 2 2 F:02
x°
6 What type of triangle is ∆ABC? C
8
7 What is the size of ∠CAB?
D
A 2 cm B

9A D 10 A

D

A BC BC 60° C
B
The diagram shows the The diagram shows the
bisection of the straight bisection of the right The diagram shows the
angle ABC. What is the angle ABC. What is the bisection of a 60° angle.
size of ∠ABD? size of ∠ABD? What is the size of
∠ABD?

In this section we will learn how to construct some angles with special sizes.

Question 6 in the Prep Quiz should have reminded you how to construct an equilateral triangle.
When we remember that the angles in an equilateral triangle are all 60° in size, it means that all
we have to do to construct a 60° angle is to construct an equilateral triangle.

This is shown in the series of diagrams below.

Constructing a 60° Angle 4

123

CC ■ The arcs
have AB as
60° radius.
A B A BA B A B

Step 1 Draw the arm AB.
Step 2 With the point of the compasses at A, draw an arc that passes through B.
Step 3 With the point of the compasses at B, draw an arc that passes through A. Draw this arc

so that it cuts the arc drawn in step 2 at C.
Step 4 Join A to C. The angle CAB is a 60° angle (since ∆CAB is equilateral).
Once you know how to construct a 60° angle you can also construct 30° and 120° angles.

61APPENDIX F

Constructing a 30° Angle

To construct a 30° angle, first construct a 60° angle and then bisect this angle. In practice, this
procedure can be shortened, as shown below.

1 2 3 45

CD CD

AB AB AB AB AB

Step 1 With the point of the compasses at A draw an arc. The point where the arc cuts the arm is B.
Step 2 With the point of the compasses at B, draw an arc to cut the arc already drawn at C.
Step 3 With the point of the compasses still at B, draw another arc as shown.
Step 4 With the point of the compasses at C, draw an arc to cut the arc drawn in step 3 at D.
Step 5 Join D to A. The angle DAB is a 30° angle.

Constructing a 120° Angle To construct a 120° angle, we remember that a 60° angle and
a 120° angle, when they are adjacent, form a straight angle.
D Therefore, if we wish to construct a 120° angle, we can construct
a 60° angle and then extend one of its arms. This is shown in the
120° 60° diagrams below.
AB C

The dotted

section is only

drawn lightly. 1 2 3 4 5

120°
A A A AA

62 INTERNATIONAL MATHEMATICS 2

Constructing a 90° Angle Don’t forget–you could do it by
bisecting a straight angle.
To construct a 90° angle, we can either bisect
a straight angle or follow the steps shown in 5
the following diagrams.

1234

A AB AB AB AB

worked examples

Construct the following figures using compasses and ruler only.

1C 2 D3 D 25 mm C

32 mm
E
35 mm

30° 60°
A 25 mm B 25 mm
2·5 cm 25 mm 25 mm
2·5 cm
60°
2·5 cm
B 3 cm A C A 25 mm B

Solutions 2 Construct 60° 3 4 Join BC. Just follow the frames.
C Each new piece of work
1 angle. Measure to C
is shown in colour.
1 Draw AB find C.

3 cm long.

B 3 cm A B 3 cm A B 3 cm A B 3 cm A

2 2 Measure 25 mm 3 Construct 60° angle and 4 Construct 30° angle and
measure 35 mm to locate E.
1 Draw a line and from B to locate measure 32 mm to D
A and C. locate D. 35 mm D
mark the point B.
25 mm 25 mm25 mm 60° E
B AB C A 25 mm B 25 mm C 30°
32 mm
A 25 mm B 25 mm C

3 2 Set compasses 3 With compasses at B and then D, 4 Join CD and CB.

1 Construct 90° angle to 25 mm and draw arcs to meet at C.

at A.

D locate B and D. DC DC

A A 25 mm B AB AB

63APPENDIX F

Exercise F:02 c

1 Construct each of the following 60° angles in your book.
a b 60°

60° 60°

2 Construct each of the following 30° angles in your book. c 30°
ab

30°
30°

3 Construct each of the following angles in your book. c 120°
ab

120°

4 a Construct the triangle ABC shown. A
b What should the size of ∠BAC be? Check with your protractor.
c Measure the lengths AB and AC to the nearest millimetre.

60° 30°

B 4 cm C

5C a Construct the triangle ABC shown.
b What should the size of ∠ACB be? Check with your

protractor.
c Measure the lengths AC and BC to the nearest millimetre.

120° 30°

A 5 cm B

64 INTERNATIONAL MATHEMATICS 2

6 a Construct the trapezium shown. AD
b Measure ∠ADC. What should it be?
c Measure AD and DC to the nearest millimetre. 25 mm

120°

B 44 mm C

7 Construct the following figures:

aD bDC cD

20 mm 3 cm 3 cm 22 mm

B 30 mm C 60° 60° 60° 43 mm B
A B A 30°

5 cm 32 mm

C

Check your accuracy by measuring the length of CD in each figure.

8 Construct the following figures. L To construct a 45° angle,
a Eb just bisect a
90° angle.

45° 60° 4 cm 45°
C 3 cm D M N

F:03 | Constructing Perpendicular and

Parallel Lines

A line is perpendicular to another line when it intersects or meets it at right Perpendicular lines
angles. You need to be able to construct a perpendicular to a line from: form right angles.

1 a point on the line, 2 a point not on the line, 65APPENDIX F
eg perpendicular to eg perpendicular to
AB through C. AB through C.

• BA B
AC

C•

The first of these constructions can be done by either:
a constructing a right angle at C, or
b bisecting the straight angle ACB.

The second of these constructions is shown below.

Constructing the Perpendicular from C to AB

1 2 3 4 5
C C C C
C

A BA BA B A BA B

To construct a pair of parallel lines, we need to remember that when C E
parallel lines are cut by a transversal, the corresponding angles formed A D
are equal (see diagram on the right).
AS B
Therefore, to make CD parallel to AB we must make ∠ECD equal to CP Q
∠CAB. That is, we must make a copy of ∠CAB.

To copy an angle we follow the steps shown below.

Copying an Angle

1 2 3 Here the arcs 4
A A A drawn from D B
and R have
B CP E E radius DE.
R
QB D R D
CP QB CP Q

This construction can also be used to draw a line parallel to a given line through a given point.

Drawing a Line Parallel to AB through P

1 2 3 4 5
A A A A AS

P PP P P
B BB B B

We have constructed equal
corresponding angles, so the
lines AB and SP are parallel.

66 INTERNATIONAL MATHEMATICS 2

Exercise F:03

1 Draw figures similar to those below and then construct the perpendicular through C.

a b c

• •C
C

•C

d •C e f

•C
•C

2 Construct the following figures. Measurements are in mm for b and c. B
aC b C c 18

F 25 C
2 cm
25

50
28 37
A 1 cm D 4 cm B D 35 E
B 40 F 15 D

A 20 E

A

3 Construct the following figures. C2 cm In each case, D will
aC b D have to be located by
60°
4 cm B constructing the
perpendicular to AD
B
120° through C.

3 cm
60°

A 5 cm
DA

4 Draw three angles in your book. Copy them using the construction for copying an angle.

67APPENDIX F

5 Use your compasses only to copy the figures given.

aC b D C
B
Use me to transfer

the side lengths.

A

AB

6 Draw three diagrams similar to those given below, then construct a line through C parallel to
AB in each.
a b Bc

•C A

A B C• •C
A B

7 Draw a line in your book. Construct a line parallel to this line which is 3 cm away from it.

8 Construct the following figures.

aC Db C D c D 2 cm C

4 cm 4 cm
3 cm

60° 60° A 45°
A 4 cm B 5 cm
A 5 cm B
B

68 INTERNATIONAL MATHEMATICS 2

F:04 | Constructing Regular Polygons

in a Circle

A regular polygon is a plane figure that has all its sides equal in length and all its angles equal in
size. The equilateral triangle and the square are the simplest regular polygons.

Complete: prep qu

1 How many vertices does 3 How many vertices does F:04 iz

the triangle have? the square have?

2 How many sides does 4 How many sides does

the triangle have? the square have?

5 If a polygon has six sides, how many vertices will it have?

6 How many degrees are there in an angle of revolution?

7 360° ÷ 3 8 360° ÷ 4 9 360° ÷ 6 10 360° ÷ 8

To construct a regular polygon in a circle, we have to space the vertices of the polygon equally
around the circle. We do this by drawing equal-sized angles from the centre. The size of the angle
is found by dividing 360° by the number of vertices.

worked example

Construct a regular hexagon in a circle of radius 2 cm. CB

Solution 60°

Draw a circle of radius 2 cm. Since a hexagon D 60° 60° A
has six vertices, we must construct six equal 60° 60°
angles at the centre.
60°
The size of the angles is 360° ÷ 6 = 60°.
EF

A hexagon has 6 sides.
360° Ϭ 6 = 60°.
Therefore I would

need six 60° angles.

Exercise F:04 A

You will need: compasses, protractor, ruler, pencil, eraser. 120°
1 a Make an accurate full-sized drawing of the figure shown. O 120°

b Check the accuracy of your construction by measuring to 2 cm
see if ∆ABC is equilateral. B

C

69APPENDIX F

2 a Copy the figure shown, making sure that the arms of the angles are longer than 3 cm.

120° This is
O 120° another
way to do
question 1.

b With O as centre, draw a circle of radius 3 cm to cut the three arms of the angles at
A, B and C.

c Join ABC. What type of figure is it?

3 Construct a square in a circle by copying the diagram below.

D

Forgotten the names of
the plane shapes?

AO C

B

4 Construct a regular hexagon inside a circle of radius 3 cm.

5 a How many vertices does a regular pentagon have?
b What is the size of the angle needed to space the vertices equally around the circle?
c Construct a regular pentagon inside a circle of radius 2 cm.

6 a How many vertices does a regular octagon have?
b What is the size of the angle needed to space the vertices equally around the circle?
c Construct a regular octagon inside a circle of radius 2 cm.

7 Construct a regular dodecagon inside a circle of radius 3 cm. A dodecagon has 12 sides.

70 INTERNATIONAL MATHEMATICS 2

8 Construct a regular decagon inside a circle of radius 3 cm. A decagon has 10 sides.

9 Construct a regular polygon with 18 sides inside a circle of radius 3 cm.

10 Construct each of the following figures. c
ab

1 cm

1 cm 1 cm 2 cm

11 C Sometimes we need to construct a regular polygon given the length of
its side. To do this we can work backwards. Consider the diagram on
120° 120° the left. Notice that each of the triangles formed by joining the vertices
O to the centre of the circle is isosceles.

120° Since in an isosceles triangle the two base angles are equal, we can
AB write the following equation to describe each triangle.

2x + 120 = 180 O
2x = 60 120°
x = 30
x° x°
A B

Hence, to construct this equilateral triangle given its side length, we draw the side length and
then draw 30° angles from each end of the line. The point of intersection gives the centre of the
circle. The rest of the figure is then completed in the normal way.
a

O Find the value of x for the regular pentagon and then use this value to
construct a regular pentagon with a side length of 3 cm.
72
x°

b

Find the value of x for the regular dodecagon and then use this value
to construct a regular dodecagon with a side length of 2 cm.

x°
30°

71APPENDIX F

APPENDIX G
G:01 | The Minimum Conditions for

Congruent Triangles

prep quiz From the triangles given, find which triangle is congruent to:

G:01 1 ∆ABC 2 ∆DEF

AD T

P

B CE FQ RU V

M RQ

J ∆XYZ ≡ ∆PRQ X
P
∆JKL ≡ ∆MNO NO

KL YZ

3 List the pairs of corresponding angles. 5 List the pairs of corresponding angles.

4 List the pairs of corresponding sides. 6 List the pairs of corresponding sides.

7 If two triangles are congruent, are they the same size?

A D 8 Is ∆ABC ≡ ∆DEF?

60° 30° 60° 30° 9 Are the angles of ∆ABC the same sizes as the angles of
B CE ∆DEF?
F 10 If the angles of one triangle are the same as the angles

of another triangle, are the triangles definitely congruent?

Two triangles are congruent if everything about them is the same. In the following exercise we will
investigate the smallest amount of information that we need to know about two triangles before we
can say they are congruent.

For example, Questions 7 to 10 in the Prep Quiz showed that two triangles having their angles
equal is not a minimum condition for congruence. It is possible to have two triangles with the same
angles which are not congruent, so we would need to know more about two triangles than the fact
that their angles are equal if we wanted to be sure they were congruent.

Exercise G:01 (Practical)

You will need: paper, scissors, compasses, pencils, protractor.
1 Copy one of the following triangles onto paper and cut it out.

72 INTERNATIONAL MATHEMATICS 2

C
F

A BD E

a Do the triangles have their sides equal in length?
b By superimposing the cut-out triangle over the other triangle, see if the triangles are

congruent.
c Do the triangles have the same-sized angles?

2 Construct one of the triangles below and cut it out.

F

A

5 cm 6 cm
4 cm
4 cm

B 6 cm C D 5 cm E Use your compasses
and ruler to construct
a Do the corresponding sides have equal lengths? a triangle.
b Are the triangles congruent?
c Are the corresponding angles equal?

In Questions 1 and 2, the triangles with corresponding sides of equal length were
congruent. We say that three pairs of sides equal is a minimum condition for congruent
triangles. This is abbreviated to SSS.

3 The diagram shows a triangle with sides AB and AC of given lengths. The angle CAB is allowed
to vary in size so that C moves on a circle with radius AC and centre A.
a What happens to the length of BC as ∠BAC gets
bigger?
C b If you fix ∠BAC at a certain size (say 30°), is it
possible to get two different lengths for BC?

3 cm

A 2 cm B I see! The Yes, and for each angle size, there is only
angle size one corresponding length.
determines the
length of BC.

73APPENDIX G

4 Construct the triangle shown, on a piece of paper. 4 cm C
a Measure BC to the nearest millimetre. 5 cm
b Is it possible to get more than one triangle from 60° B
this information? A
c Cut your triangle out and compare it with the
triangles of other students in your class. D
Are they all congruent? 120°

5 Construct one of the triangles below and cut it out.

A E 5 cm

3 cm
3 cm

120° 5 cm C F
B

a Is AC = EF?
b By superimposing, find if ∆ABC is congruent to ∆DEF.

In Questions 3, 4 and 5 we found that when two sides and the angle between them
in one triangle are equal to two sides and the angle between them in the other triangle,
then the triangles are congruent. It is important that the angle is included by
(ie between) the two sides. This is a minimum condition for congruent triangles, and is
abbreviated to SAS.

6 In each of the following triangles, the angles match and one side in each has the same length.
Construct each triangle and cut it out.

70° You could need this
60° 50° in Question 6.

I 50° II III
60°

5 cm 70° 50° 60° 70°
5 cm 5 cm

60° 70° 50° ■ If two sides are
opposite equal
IV V VI angles, they are
50° 70° 70° 60° corresponding
50° sides.
5 cm 5 cm 5 cm
60°

a Which triangles are congruent?
b For each pair of congruent triangles, how could you describe the position of the 5 cm side?

74 INTERNATIONAL MATHEMATICS 2

7 Construct or copy each triangle on paper, and cut it out. L
40°
D
A 60°
80°

60° 40° 80° 4 cm 40°
B 4 cm C E F

80° 4 cm 60°
M N

a Which angle is the 4 cm side opposite in: X
60°
i ∆ABC? ii ∆DEF? iii ∆LMN?

b Which of the triangles shown is congruent to

∆XYZ on the right?

c How could you describe the position of the

4 cm side in each of the congruent triangles?

40° 4 cm 80°
Y Z

Questions 6 and 7 have shown us that if the angles of one triangle are equal to the
angles of another triangle, and a side in one is equal to a side in the same position of
the other, then the triangles are congruent. This is the third minimum condition, and is
abbreviated to AAS.

8 The fourth set of minimum conditions is restricted to The hypotenuse is the
right-angled triangles only. longest side. For this
Copy one of the triangles below and cut it out. reason, Pythagoras’
Does your cut-out match both triangles?
Theorem says
C c2 = a2 + b2

D

5 cm 4 cm

A 4 cm BE 5 cm F

a Write down the pairs of corresponding sides.
b Are the triangles congruent?

75APPENDIX G