vii

Chapter 1 More about Coordinates .................................................... 2 1.1 Areas of figures in a coordinate plane 2

1.2 Distance and gradient 8

1.3 Mid-point of a line segment 14

1.4 Parallel and perpendicular lines 16

1.5 Combined problems in a coordinate plane 24

1.6 Equations of straight lines 30

Chapter 2 Laws of Indices ................................................................ 39 2.1 Basic index operations 39

2.2 Negative exponents 43

2.3 Indices and roots 46

Chapter 3 Polynomials ...................................................................... 51 3.1 Addition and subtraction of polynomials 52

3.2 Multiplication of polynomials 55

3.3 Definition of algebraic fractions 57

3.4 Multiplication and division of algebraic fractions 62

Chapter 4 Change of Subject............................................................ 68 4.1 Revision of equations 68

4.2 Formulas and substituting values into formulas 71

4.3 Change of subject 73

Contents

viii

Chapter 5 Factorization of Polynomials ........................................... 79 5.1 Factorization by finding common factors 79

5.2 Factorization of quadratics using the difference of

two squares and perfect squares 82

5.3 Factorization of quadratics using the cross method 85

Chapter 6 Surds and Rationalization ............................................... 90 6.1 Definition of surds 90

6.2 Operations on surds 91

6.3 Rationalization 96

Chapter 7 Angle Properties .............................................................101 7.1 Angle properties with intersecting lines 101

7.2 Angle properties with parallel lines 109

Chapter 8 Angles in Polygons .........................................................118 8.1 Interior angles in polygons 118

8.2 Exterior angles in polygons 125

Chapter 9 Congruent Triangles ...................................................... 131 9.1 Identifying congruent triangles 131

9.2 How to show two triangles are congruent 135

ix

Chapter 10 Similar Triangles ............................................................ 142 10.1 Identifying similar triangles 142

10.2 How to show two triangles are similar 145

Chapter 11 Properties of Quadrilaterals .......................................... 153

11.1 Properties of parallelograms 153

11.2 Properties of rectangles and rhombuses 160

11.3 Properties of squares and kites 165

Chapter 12 Graphical Presentation of Statistics ............................. 170

Chapter 13 Sets and Venn Diagrams ............................................... 184 13.1 Sets 184

13.2 Venn diagrams 193

Answers .............................................................................................. 201

Index ................................................................................................... 234

2

IBMYP Mathematics STUDY GUIDE Level 3

Chapter 1More about Coordinates1.1 Areas of figures in a coordinate planeWe can find the area of a polygon in a coordinate plane if we know the coordinates of

each vertex. The following examples illustrate different methods of finding the area.

Example

Find the area of ∆ABC in the figure below:y

x2

6

4A B

C

O

Solution:

We need to find the base and height of ∆ABC:

Length of base = = − =4 2 2AB

Length of height = = − =6 0 6OC

Therefore, the area of ∆ABC

= ( )( )

= × ×

=

1212

2 6

6

AB OC

sq. units

The coordinate plane has uses far outside the classroom.

It is used to give locations on maps, to chart data

in statistics and to create designs for buildings and

clothing, among others.

Fun FactF

3

Chapter 1 – More about Coordinates

Sometimes it is necessary to break down or divide the figure into separate parts:

Example

Find the area of the quadrilateral ABCD in the figure below:

y

x0

−

−

( 1,3)

(3, 1)

(6,0)

B(3,4)A

C

D

Solution:

y

x0

−

−

( 1,3)

(3, 1)

(6,0)

B(3,4)A

C

DH

K

We can divide the quadrilateral into the two triangles ABC and ADC by joining AC.

AC is the common base of both triangles. BH is the height of ∆ABC where the

coordinates of H are 3 3,( ) . DK is the height of ∆ACD where the coordinates of K

are 3 0,( ) .

Area of quadrilateral ABCD

= ∆ ∆

= × × + × ×

= × × + × ×

= +=

area of + area of 1 12 21 15 4 5 32 210 7.517.5 sq. units

ABC ACD

AC BH AC DK

4

IBMYP Mathematics STUDY GUIDE Level 3

Example

Find the area of the triangle OYZ in the figure below:

y

xO

(2,5)Y

(6,3)Z

Solution:

First, surround the triangle OYZ by the rectangle OABC:

A B

C

y

xO

(2,5)Y

(6,3)Z

Referring to the above figure, we can divide rectangle OABC into four triangles:

∆ ∆ ∆ ∆, , andOAY YBZ OCZ OYZ.

Area of rectangle OABC

= ×= ×=5 630

OA OC

Area of ∆OAY

= × ×

= × ×

=

121 5 225

OA AY

Sometimes, instead of breaking a shape up, we surround it by a simpler shape, as in

this example.

5

Chapter 1 – More about Coordinates

Area of ∆YBZ

= × ×

= × ×

=

121 4 224

YB BZ

Area of ∆OCZ

= × ×

= × ×

=

121 6 329

OC CZ

Therefore, the area of ∆OYZ

= − ∆ − ∆ − ∆= − −

area of area of area of area of OABC OAY YBZ OCZ30 5 44 912

−= sq. units

Exercise 1.1

Section A

For questions 1 – 10, find the area of the given figures.

1. y

x1

2

20

1

3

2. y

x1

2

20

3

1

3 4

(1,3)

6

IBMYP Mathematics STUDY GUIDE Level 3

3. y

x5

10

100−10 −5

15

5

4. y

x2 40−2

2

−2

4

−4

( 2,4) (4,4)

( 4, 2) (2, 2)

−

− − −

5. y

x20−4−6

2

−4

−2−2

(2,2)

( 6, 4) ( 2, 4)− − − −

6. y

x50−10−15−20

5

10

−5−5

( 20,10)− ( 10,10)−

(5, 5)−( 15, 5)− −

7. y

x0 10 15

5

10

−5−5

55

( 3,8) (15,8)−

(7, 5)− (12, 5)−

8. y

x20−4 −2

2

−2

4

4

( 3.6, 2)− − (4.8, 2)−

9. y

x10 200

10

30−10

20

−10

(28,20)( 4,20)−

( 14, 10)− − (13, 10)−

10. y

x0 2 3−1

−1

−2

11.4

1

(2.4, 1.6)−(0.4, 1.6)−

7

Chapter 1 – More about Coordinates

Section B

For questions 1 – 10, find the area of the given figures.

1. y

x0

1

1

2

2−1−1

( 1,1)

(1,2)

(2,1)−

2. y

x0 2

1

−1−2 1

2

−1

3. y

x50−10

10

−5−15

5

−5

( 10,10)−

4. y

x2 40 6−2

−4

−6

2

−2

(2, 2)−

5. y

x

1

1

2

3

2 3 4−1−10

(3,3)

( 1,1) (4,1)−

6. y

x0 4

2

−2−4 2

4

−2

(4,4)( 2,6)−

( 4,2)−

(2, 2)−

7. y

x

10

100−10

5

−5

−10

1515

−5 5

8. y

x5 100−10 −5

5

−5

10

−10

1515(5,12)

( 10,9)−

( 17,5)−

( 14, 4)− − (10, 8)−