Igcse Math Chapters 1 to 4

8/19/2019 Igcse Math Chapters 1 to 4

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Amal Language Schools Mr. Walid M. Gadalla

Chapter 1 Algebra Page 1

Algebraic Manipulation

Aims: At the end of this module you should be able to:- Evaluate expressions by substituting numerical values for letters. Collect like terms. Multiply a single term over a bracket. Expand the product of two simple linear expressions. Expand the product of two linear expressions. Take out single common factors. Understand the concept of a quadratic expression and be able to factorise such expressions. Manipulate algebraic fractions where the numerator and/or the denominator can be numeric,

linear or quadratic.

Simplifying Expressions

Drill 1

1.

Expansion of Brackets

Remember

a b c a b a c where , ,a b c

a b c a b a c where , ,a b c

Drill 1

1.

2.

3.

4.


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Multiplying Two Polynomials

Note that

To simplify an expression, we have to :

1. Rearrange the expression in descending order

2. Expand the brackets.

3. Add like terms.

1) Direct multiplication of two binomials

Method A: Horizontally:(x+2y) ( a+b )= x(a+b) +2y( a+b)=ax+bx+2ay+2by

Method B ( The FOIL method)

Multiply 2 3 2x y z Solution

2 3 2 6 4 3 2x y z xz x yz y

Note that: If the terms of the second binomial differ in ordering of the terms of the first binomial,

we have to rearrange it first.

Example : (x-3y) ( y+2x) = (x-3y)(2x+y) we use the FOIL method If the terms of multiplicand and the multiplier are like , then

add the product of Outer terms and the product of Inner terms.

Example: 2 2 4 2 2 4 23 1 5 3 15 9 5 3 15 4 3a a a a a a a

2) Special Cases of the product of two binomials:

The Square of a binomial “Perfect square trinomial” Rule:

2 2 2

2a b a ab b

2 2 2

2a b a ab b

Remark:

2 2 2

2a b a ab b

Example: Expand each of the following: 2(2a+3b) =…………………….

2

3x y =…………………….

The product of the difference of two squares: Rule:

2 2a b a b a b

2 2a b a b a b

Note that: The result 2 2a b is called the “difference of two squares”.

Example: Expand each of the following:

3 3x y x y =………………….

3 2 3 2b a b a =………………

Product of

First terms

Product of

Outer terms

Product of

Inner terms

Product of

Last terms

Product of

First termTwice the

Product of the two terms

Product of

Last term


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Factorisation of expressions

Remember: “2 is a factor of 10” means “10 is divisible by 2”. “ a is a factor of b” means “ b is divisible by a“. The Factorisation process is the inverse of the expanding of multiplication. The prime expression is divisible by 1 and itself only.

H.C.F “ Highest Common Factor ”.

Note that: The complete factorisation means to factorise into prime factors.

Example (1) : Factorise :a) 12 b) 18 c) 250

Answer: 12=………… 18=……… 250=……….………

Example (2): Find the H.C.F of :a) 12 and 18 b) 63 and 105

Answer: H.C.F of 12 and 18=……… H.C.F of 63 and 105=…...

The H.C.F for any algebraic terms is a term which has common variables have the leastpower found.

Example (3): Find the H.C.F for the following algebraic terms: 3a and 5a Answer:……………..

324a and 36a Answer:……………..

2 6 3 4 2 224 ,36 and 54 .m n m n m n Answer:……………..

( ) and ( )x a b y a b Answer:……………..

Factorising by extracting the H.C.F: Remark: To factorise by extracting the H.C.F:

(1) Find the H.C.F of the terms of the expression. (2) Divide the given expression by the H.C.F. (3) The product of the H.C.F by the quotient is the required form.

Example (4) : Factorise : 8 6x y =…………………….

12 15 3a b =…………………….

ab ac ad =…………………….

Factorising quadratic expression in the form of 2x bx c :

Remark: x bx c is called a quadratic trinomial.

Note that:

When you factorise the trinomial you must:

(1) Arrange its three terms in descending order according to the powers of the givensymbol.

(2) Take the H.C.F. ( If exists ).


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The first Case: c is positive Rule: To factorise the expression 2x bx c

If the absolute term ( c ) is positive ,then the signs of m and n must be the same as the sign of b. moreover the sum of m and n is equal to b.

Example (1): 2 11 24 3 8x x x x

Example (2):

211 24 3 8x x x x

Example : Factorise each of the following expressions:

a) 2 5 6x x b) 10 16x x c) 2 210 9x xy y

The second Case: c is negative Rule: To factorise the expression 2x bx c

If the absolute term ( c ) is negative ,then the signs of m and n must be differentand the greater number of m or n has the same sign of the middle term ( bx ).

Example (1): 2 5 24 8 3x x x x Example (2): 2 5 24 8 3x x x x

Example : Factorise each of the following expressions:

a) 2 5 6x x b) 2 49 36x x c) 2 5 6x x d) 2 384 2 2a a a

Factorising quadratic expressions in the form of 2

, 1ax bx c a

1) Factorise each of the following expressions: a) 22 6x x c) 26 3 3x x

b)

2

3 2 8a a d)3 2

2 10x x x

Factorization of the difference of two squares Rule: 2 2a b a b a b

The difference of two squares is equal to the product of their sum by their difference. Note that: 2 2a b has no factors The sum of two squares has no factors in .


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Remark: Always check whether the factors you get can be factorised again or not. Don’t forget to check for the existence of the H.C.F.

Example (1) : Factorise each of the following:

a) 2 28 144a b b) 3 25a a

Algebraic fractions

1. Expressing a fraction in its simplest form

Drill 21.

2.

3.

4.

5.


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Expressions and Formulae

Aims: At the end of this module you should be able to:- Understand that a letter may represent an unknown number or a variable. Use correct notational conventions for algebraic expressions and formulae. Substitute positive and negative integers, decimals and fractions for words and letters in

expressions and formulae. Use formulae from mathematics and other real life contexts expressed initially in words or

diagrammatic form and converting to letters and symbols.

Understand the process of manipulating formulae to change the subject where the subject mayappear twice or a power of the subject occurs.

Using Formulae

Drill 1

Construct and Use Simple Formulae

Drill 2

1.


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2.

Substitution into Formulae

Drill 3 1.

More Complex Formulae

Drill 4

1.

2.


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Changing the Subject

Drill 5

1.

2.

Further Change of Subject

You may need to apply a number of rules or principles at the same time.

These steps are useful in manipulating a formula:

• Clear the formula of any fractions

• Eliminate any square or other roots

• Expand any brackets

• Rearrange the terms so that those containing the new subject are isolated on one side of

the formula

• Factorise this side so that it has the format (new subject) x (expression)

• Divide both sides by (expression) so the formula becomes (new subject) = ...

Drill 6 1. Given that ab+ c = d(b + 2), express b in terms of a, c and d.

(This means 'make b the subject of the formula'.)

2. Make u the subject of the formula1 1 2

v u r


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3. Make b the subject of the formulaa b

r a b

4. Given that RF R r

, express R in terms of F and r .

5. Make a the subject of the formula 2 a

T g

6.


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Proportion

Aims: At the end of this module you should be able to:- Set up problems involving direct or inverse proportion and relate algebraic solutions to graphical

representation of the equations

Direct Proportion

Drill 1

Inverse Proportion


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Drill 2

1.

Functional and Graphical Representations

Example:


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Linear Equations

Aims: At the end of this module you should be able to:- Solve linear equations with integer or fractional coefficients in one unknown in which the

unknown appears on either side or both sides of the equation Set up simple linear equations from data given

Simple Equations

Drill 1

1.

2.

Solving Equations

Drill 2

1.

2.


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3.

4.

5.


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Simultaneous Linear Equations

Aims: At the end of this module you should be able to:- Calculate the exact solution of two simple simultaneous equations in two unknowns. Calculate the exact solution of two simultaneous equations in two unknowns Interpret the equations as lines and the common solution as the point of intersection

Simultaneous Linear Equations

Drill 1

1.

2.

3.


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Quadratic Equations

Aims: At the end of this module you should be able to:- Solve quadratic equations by factorisation. Solve quadratic equations by using the quadratic formula. Form and solve quadratic equations from data given in a context. Solve simultaneous equations in two unknowns, one equation being linear and the other equation

being quadratic.

Solving Quadratic Equations by Factorisation

Drill 1

1.

2.

3.

4.


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5.

Solving Quadratic Equations Using the Formula

Drill 2

1.

2.

3.


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Algebraic Fractions and Quadratic Equations This section deals with finding the solutions of equations such as

Drill 3

1.

2.

Solving two equations in two variables, one of 1st degree

and the other of 2nd degree

Drill 4

1. x -y = 1

x2 + y2 = 13.


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2.

x-y = 3

xy + 2 = 0.

3.

2

2

2 32

x y

x y


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Inequalities

Aims: At the end of this module you should be able to:- Understand and use the symbols >,


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Drill 11.

Solution of Linear Inequalities

Drill 21.

2.

3.

4.


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Inequalities Involving Quadratic Terms


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Drill 31.

2.

Graphical Approach to Inequalities


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Drill 41.

2. By shading the unwanted region, show the region that represents the inequality

3 x - 5y 15.

Dealing With More Than One Inequality


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Drill 5 1. Given that x and y are whole numbers, find the pairs of values ( x, y) that satisfy all the

inequalities x + y 4, y- 2x 2, y > 0.

2.


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Sequences

Aims: At the end of this module you should be able to:- Generate terms of a sequence using term-to-term and position-to-term definitions of the sequence. Find subsequent terms of an integer sequence. Use linear expressions to describe the nth term of an arithmetic sequence.

Simple Number Patterns

Drill 11.

2.

Recognising Number Patterns


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Drill 2

1.

2.

Extending Number Patterns


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Drill 3

1.

Formulae and Number Patterns

Drill 4

1.

2.


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Module 2- Function notation

Aims: At the end of this module you should be able to:- Understand the concept that a function is a mapping between elements of two sets Use function notations of the form f(x) = … and f : x … Understand the terms domain and range and which parts of a domain may need to be excluded Understand and use the notations composite function fg and inverse function f -1

FunctionsIf X and Y are 2 non-empty sets. Any relation R from X to Y is called a function if:

Each element of X appears one and only one as an element of Y, Such that x R y

Remarks:

The set X is called the Domain.The set Y is called the Co-Domain.

The Range:

The set of all images in the Co-Domain which have correspondingelements in the domain.

Example: 1

Let X = {x: x Z, -1 x 4} and f(x) = 2x - 3 1. Find the Domain and then deduce the Range.

Determining the domain of a function

Rules:

1. If f(x) is a polynomial, then its domain is R.

2. If f (x) is a Fractional function, then its Domain = R - { the Zeroes of the denominator}.

3. If f (x) = g x , then its domain is the set of all real numbers. where g(x) 0.

4. If F(x) =1

( ) g x, then its domain is the set of all real numbers such that. g(x) > 0.

Drill 1Find the excluded value(s) of the domain of the following:

1. f(x) = -4x

2. f(x) =

3

2 3x

3. f(x) =2

1

9 25x

4. f(x) = 35

5. f(x) =2

1

7x

6. f(x) = 3x

7. f(x) = 1 x

8. f(x) = 1

2x

9. f(x) =

10. f(x) =

11. f(x) =

12. f(x) =

13. f(x) =

14. f(x) =

15. f(x) =


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Function notationConsider the familiar y=x+2. This can also be written as f (x) = x + 2.

Example: g(x) = x2 - 2x- 3 and h(x) = 4x + 1.

Bear in mind that the following are all the same function:

2 3

2 3

2 3

x x

t t

y y

Drill 21. Given f ( x) = x

2 - 3 x and g( x) = 4x- 6, find the value of:

a) f (6) b) f (-3) c) g( 1

2 ) d) g(6)

2. Given 2: 9h x x

a) Write down the expression for h(x): 29h x x

b) Find the image of:(i) 0 (ii) 3

3. Given the functions f(x) = x2 and g( x) = x + 2, a) Solve the equation f ( x) = g( x).

b) solve the equation 4g(x) = g(x) - 3.

Composite functionsYou can think of a composite function as a function of a function.It is the result of applying one function to a number and then applying another function to the result.

Consider the two functions: f(x) = 2x+ 1 and g(x) = x2 f (4) = 9 and g(9) = 81

This can be written as g[ f (4)] =81, but it is normally shortened to g f (4).

Remember:

g f (x) stands for g[ f (x)].Thus g f (x) is a composite function in which f is applied first and g second.

Drill 3Given the functions f (x) = x2-2x and g(x) = 3-x , find the values of:a) g f (4) =

b) f g(4) =

c) ff (-1) =

d) gg(100) =


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Flow diagramsThe steps taken to work out the value of any function f (x) can be shown on a flow diagram.For example, the function f (x) = 2x + 5 can be represented as:

......

.... ..... ..........x

g( x) = 2( x+ 5) can be represented as:

......

.... ..... ..........

x

Notice that these flow diagrams show the same operations but the order is different.

Inverse of a functionThe inverse of a function is the function that will do the opposite of f or, in other words, undo the effects of f

For example, if f maps 4 onto 13, then the inverse of f will map 13 onto 4.In general, if f is applied to a number and the inverse of f is applied to the result, you will get back to thenumber you started with.

In simple cases, you can find the inverse of a function by inspection.

For example, the inverse of 7x x must be 7x x because subtraction is the opposite of addition,

and to undo +7 you have to subtract 7.

Similarly, the inverse of 2x x is2

x

x because to undo 2 you have to

divide by 2.

The inverse of the function f is denoted by 1 f .

Hence, if f ( x) = x + 7, then f -1( x) = …….

And if g( x) = 2 x, then g-1

(x) = …….

Not all functions have an inverse function.

Consider 2x x . This is a function because for every value of x there is only one value of x2. However, the

inverse is not a function because a positive number has two square roots (one negative and one positive).

Note

When a graph of y = f(x) has been drawn, then f(1) requires the y value when x = 1.

1 1 f requires the x value to be read from the graph when y = 1.

This is because 1 1 f = x implies f(x) =1.

Finding the inverse of a functionThere are two methods of finding the inverse:Method 1 - Using a flow diagram:In this method, you draw a flow diagram for the function and then obtain the inverse by reversing the flowand 'undoing' the operations in the boxes.

Example Find the inverse of f (x) = 3x- 4.

1

4

1

: ...... ......

: ...... ......

let be input:

......+........... ......

.............

.......

x

f

f

x

x

f x

input output

output input


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Example

Given g(x) = 5 - 2 x, find 1 g x .

Drill 4

1.

Method 2 - Reversing the mapping:

In this method, you use the fact that if f maps x onto y, then 1 f maps y onto x. To find 1 f you have to find

a value of x that corresponds to a given value of y.

Example Find the inverse of the function f(x) = 3x - 4.

Example:

Given that g(x) = 5 - 2x, find 1 g x .

1

5

1

: ...... ......

: ...... ......

let be input:

......+.....

...... ................... .......

....... ......

x

g

g

x

x

g x

input output

output input


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Graphs

Aims: At the end of this module you should be able to:- Interpret information presented in a range of linear and non-linear graphs understand and use conventions for rectangular Cartesian coordinates Plot points (x , y) in any of the four quadrants of a graph Locate points with given coordinates Determine the coordinates of points identified by geometrical information Determine the coordinates of the midpoint of a line segment given the coordinates of the two end

points Draw and interpret straight line conversion graphs Understand the concept of a gradient of a straight line Recognise that equations of the form y = mx + c are straight line graphs Generate points and plot graphs of linear and quadratic functions plot and draw graphs with equation: y = Ax3 + Bx2 + Cx + D in which

(i) The constants are integers and some could be zero(ii) The letters x and y can be replaced with any other two letters

Or:

3 2

2

E F y Ax Bx Cx D

x xin which

(i) The constants are numerical and at least three of them are zero.

(ii) The letters x and y can be replaced with any other two letters Find the gradients of non-linear graphs Find the intersection points of two graphs, one linear (y1) and one non-linear (y2), and recognise

that the solutions correspond to the solutions of y2 - y1 = 0. Calculate a gradient of a straight line given two coordinates Recognise that equations of the form y = mx+c are straight line graphs with gradient m and

intercept on the y- axis at the point (0, c) find the equation of a straight line parallel to a given line


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Drill 2

1.


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2.

3. Find the gradient of the line that joins the points with coordinates (-2,4) and (4,1).

4. Find the coordinates of the midpoint of AB where A(3,-2) and B(2,3).

5. If the coordinates of the midpoint of AB is (2,4) and A(6,5), then find the coordinate of B.


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Drill 3

1.


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Drill 4


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Drill 6

:


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Drill 7


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Drill 8


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Calculus

Aims: At the end of this module you should be able to:- Understand the concept of a variable rate of change Differentiate integer powers of x Determine gradients, rates of change, turning points (maxima and minima) by differentiation and

relate these to graphs Distinguish between maxima and minima by considering the general shape of the graph Apply calculus to linear kinematics and to other simple practical problems


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Stationary points


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The geometrical meaning of the rate of change

The gradient of the tangent of the function [y = f (x)] = The rate of change = dy

dx

Exercise:

1. Find the gradient of the tangent of each of the following curves at the given point:

a. y = 6 + 3x 4 – x 2 at the point (-1, 2)

b. y = (2x – 5)(x 2-1) at the point (2, -3)

c. y =1

x

x

at the point (1, 2)

2. Find the points on the curve Y = 2x3 – 3x

2 – 12x + 5 at which the tangent on the curve is

parallel to the X – axis.


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4. If p(x) = – 12

4

x

, then find the value of x, which makes p x = – 8

Maxima and Minima


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: 1.


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Drill

1. For a particle moving in a straight line, its displacement s m from a point O on the line is

given by 2 4 5s t t , where t is the time in seconds from the start. Find:

(i) The initial distance of the particle from O.(ii) Its initial velocity.(iii) The time when the particle comes to instantaneous rest.

(iv) At what time (s) after the start it passes through O. The distance travelled in the first 4seconds.


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Chapter 2 Numbers & the number system Page 60

Powers and RootsAims: At the end of this module you should be able to:- Calculate squares, square roots, cubes and cube roots Use index notation and index laws for multiplication and division of positive integer powers Express integers as the product of powers of prime factors. Understand the meaning of surds Manipulate surds, including rationalising the denominator where the denominator is a pure surd Use index laws to simplify and evaluate numerical expressions involving integer, fractional and

negative powers Evaluate Highest Common Factors (HCF) and Lowest Common Multiples (LCM)

Squares, Cubes, Square Roots and Cube Roots

Drill 1

1.


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Index Notation

Drill 2

1.

2.

3.


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Factors

A factor of a number will divide exactly into it.

Drill 2

1.

2.

Prime Factors

Drill 3

1.

2.


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Further Index Notation

Drill 5

1.

2.


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Surds

Surds are numbers left in 'square root form' (or 'cube root form' etc). They are therefore irrationalnumbers. The reason we leave them as surds is because in decimal form they would go on foreverand so this is a very clumsy way of writing them.

Multiplication of Surds

5 × 15 = 75 (= 15× 5)

= 25 × 3

= 5 3

(1 + 3)(2 - 8) [The brackets are expanded as usual]

= 2 - 8 + 2 3 - 24

= 2 - 2 2 + 2 3 - 2 6

Addition and Subtraction of Surds

Adding and subtracting surds are simple- however we need the numbers being square rooted (orcube rooted etc) to be the same.

4 7 - 2 7 = 2 7.

5 2 + 8 2 = 13 2

Note: 5√2 + 3√3 cannot be manipulated because the surds are different (one is2

and one is 3 ).

However, if the number in the square root sign isn't prime, we might be able to split it up in order tosimplify an expression.

Example

12 + 27

12 = 3 × 4. So 12 = (3 × 4) = 3 × 4 = 2 × 3.

Simplify

Similarly, 27 = 3 3 .

Hence 12 + 27 = 2 3 + 3 3 = 5 3

Rationalising the Denominator

It is untidy to have a fraction which has a surd denominator. This can be 'tidied up' by multiplyingthe top and bottom of the fraction by a particular expression. This is known as rationalising thedenominator, since surds are irrational numbers and so you are changing the denominator from an

irrational to a rational number.

http://www.mathsrevision.net/gcse/pages.php?page=4http://www.mathsrevision.net/gcse/pages.php?page=4


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Drill 6

1. Rationalise the denominator of:

a) 1

2

b) 1 2

1 2

2. Simplify :

a. 20 45 6 5

b. 3

5 2

c. 5 7 4 28 3 63 7

d. 3 27

53

e. 54 12

f. 3 3 3128 10 25


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Set Language and NotationAims: At the end of this module you should be able to:- Understand the definition of a set of numbers Use the set notation , and Understand the concept of the Universal Set and the Null Set and the symbols for these sets Understand sets defined in algebraic terms understand and use subsets Understand and use the complement of a set Use Venn diagrams to represent sets and the number of elements in sets Use the notation n(A) for the number of elements in the set A Use sets in practical situations

Sets

A set is a well-defined collection of objects that usually have some connection with each other. Sets can be described in words.

For example: set A is a set of the oceans of the world; set B contains natural numbers lessthan or equal to 10.

Sets can also be listed between curly brackets {} or braces. For example:

A = {Indian, Atlantic, Pacific, Arctic, Antarctic}B = {1, 2, 3, 4, 5, 6, 7, 8, 9}

Objects that belong to a set are called elements and are indicated by the symbol. means “is anelement of ”.In the examples above, we can say Atlantic ….. or 2 ….. means “is not an element of “.

Again using our examples, we can say CairoA and 11 B.The sets described and listed above are finite sets - they have a fixed number of elements.Sets that do not have a fixed number of elements are infinite sets.

The set of natural numbers greater than 10 is an example of an infinite set. This can be listed as {10,11, 12, 13, ...}.A set may also have no elements. Such a set is called an empty set.

The symbols {} or indicate an empty set.An example of an empty set would be women over 6 m tall.

Notice that The number of elements in an empty set is 0 but {0} is not an empty set - it is a setcontaining one element 0.

Set builder notation

This is a method of defining a set when it is difficult to list all the elements or when the elementsare unknown.

Examples1. The infinite set of even natural numbers can be listed like this:

{2; 4; 6; 8; 10;...}. This set can be described in set builder notation like

this: : , is an even numberx x x .You read this as 'the set of all elements x such that x is an element of the set of natural numbers andis even'.

2. The finite set of natural numbers between 9 and 90 can be listed as{10; 11; 12; 13; ...; 89} or written in set builder notation

as : ,9 90x x x .


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You read this as 'the set of elements x such that x is an element of the set of natural numbers and xis bigger than 9 and smaller than 90'.

Relationships between sets

Equal setsSets that contain exactly the same elements are said to be equal. Consider these sets:

A = The set of letters in the word END = {E, N, D}B = The set of letters in the word DEN = {D, E, N}We can say A = B. The order of the elements in the set does not matter.

SubsetsIf every element of set A is also an element of set B, then A is a subset of B.This is written as A B. means “is a proper subset of ”.means 'not a proper subset of.

Proper subsets always contain fewer elements than the set itself.

The proper subsets of {D, E, N} are:{…} {….} {….} {….,….} {….,….} {…., …}

Trivial subsets of {D, E, N}are {} (the empty set) and{…, …, ….}.

Notice that

If a set has n elements, it will have 2n subsets.

For example, a set with 3 elements will have 23 subsets. That is 2x2x2 = 8 subsets.The set of elements from which to select to form subsets is called the universal set.

The symbol E is used to denote the universal set.

Notice thatYou should remember that the universal set could change from problem to problem.

Example 1. Find a universal set for each of the following sets.

the set of people in your class that have long hair the set of vowels {2,4,6,8} {goats, sheep, cattle}

2. If E is the set of students at your school, define five subsets of E .

Intersection and union of sets

The intersection of two sets refers to elements that are found in both sets.

For example:A = {a, b, c, d }B = {c, d, e, f }

The intersection of these two sets is{………..} You write this as A B .

Remember that

A B B A

. When two sets have no elements in common, they are called disjoint sets.

The intersection of disjoint sets is the empty set or .


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The elements of two or more sets can be combined to make a new set. This is called the unionof the sets.

For example:A ={1,2, 3, 4}B = {4, 5, 6}C = {1, 2, 3, 4, 5, 6}

Set C is the union of set A and set B. We write this as A B C .

Remembern(A) means the number of elements in set A. In the union of sets, when n(A) + n(B) = n(C) then Aand B were disjoint sets.When n(A) + n(B) n(C) then the sets were not disjoint; in other words

A B .

Example

List the set which is the intersection of the two sets.{1, 2, 3, 4, 5, 6} and {4, 5, 8, 9, 10}

…………………..……………………………..…………………..……………….. Example

Write down the union of the following sets.A = {a, b, c} and B = {d, e, f }

…………………..……………………………..…………………..………………..

The complement of a set

The complement of a set A is the set of those elements that are in the universal set but are not inA. The complement of set A is written as A'.

For example, if E = {1, 2, 3, 4, 5} and A = {2, 4, 5}, then all the members of E that are not in A

make the subset {….,….}.This subset is the complement of A, so A' = {….,….}. Notice that

A set and its complement are disjoint. ' A A .

The union of a set and its complement is the universal set. ' A A E

Venn diagrams

Sketches used to illustrate sets and the relationships between them are called Venn diagrams. Youneed to understand the basics of Venn diagrams before you can use them to help you solve

problems involving sets.ExampleUse the given Venn diagram to answer the following questions.a) List the elements of A and B.

b) List the elements in A B .

c) List the elements in A B .

……………………………………………………….


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PercentageAims: At the end of this module you should be able to:- Understand that 'percentage' means 'number of parts per 100'. Express a given number as a percentage of another number. Express a percentage as a fraction and as a decimal. Convert simple fractions of a whole to percentages of the whole and vice versa. Understand the multiplicative nature of percentages as operators. Solve simple percentage problems, including percentage increase and percentage decrease. Use reverse percentages.

Drill 1

1.

2.

3.

4.


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Percentages of Quantities

Drill 2

1.

2.

Quantities as Percentages

Drill 3

1.

2.

More Complex Percentages

Drill 3

1.

2.


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Percentage Increase and Decrease

Drill 4

1.

2.

Calculation of selling price

Drill 5

1. A shopkeeper buys an article for $20 and sells it at a profit of 30%. For how much does hesell it?

2. Find the selling price of an article that is bought for $400 and sold at a loss of 10%.

Reverse Percentage ProblemsYou've seen how to calculate the percentage profit, if you know what the cost price and the sellingprice of an article are.Now you will find out how to calculate the cost price, if you know the selling price and percentageprofit.It is important to remember that profit is expressed as a percentage of the cost price. It is not apercentage of the selling price.

Drill 6 1. Find the cost price of an article sold at $360 with a profit of 20%.

2. If a shopkeeper sells an article for $440 and loses 12% on the sale, find his cost price.


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Simple interest The money you pay to live in somebody else's house is called rent.

The money you pay if you hire a car is called rental

The money you pay to the post office to use one of their post boxes is also called rental.

If you borrow money from a bank, you also have to pay for the use of the money borrowed.

This money you pay is called interest.

If you deposit (or invest) money in a bank, the bank will pay you interest. The letter I stands

for interest.

The interest depends on the sum of money borrowed or invested The initial amount of money

borrowed or invested is called the principal amount. The letter P stands for the principal

amount.

The interest also depends on the length of time for which the money is borrowed or invested. T

is the letter used for time. Time is usually measured in years.

Finally, the interest depends on the rate of interest.This is expressed as a percentage per year and is denoted by R.

For example, if a bank's interest rate is 5% per annum, it will pay R(5) per year for everyR(100) invested.

So, to summarise, you invest an amount of money (P) in the bank.

The bank pays you interest (I) at a fix rate (R) per annum100

PTRI .

For any number of years (T) you have kept your money in the bank, you can calculate the finalamount (A) of money. The final amount will be equal to the principal plus interest (A = P + I).

Examples 1. $500 is invested at 10% per annum “per year” simple interest. How much interest is

earned in 3 years?

2. How long will it take for $250 invested at the rate of 8% per annum simple interest to

amount to $310?

3. A farmer gets a loan of $8 000 and clears the loan at the end of 5 years by paying $12 000.

What rate percentage of simple interest did the farmer have to pay per annum?


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Compound Interest and Depreciation

Every year, if the money is left sitting in the bank account, the amount of interest paid wouldincrease each year.

This phenomenon is known as compound interest.

i.e:

no of years100 + %change

× original value100

Drill 7

1.

2.


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Ratio and ProportionAims: At the end of this module you should be able to:- Use ratio notation including reduction to simplest form and its various links to fraction notation. Divide a quantity in a given ratio. Use the process of proportionality to evaluate unknown quantities. Calculate an unknown quantity from quantities that vary in direct proportion. Solve word problems about ratio and proportion.

Simple Ratios

If the ratio of one length to another is 1 : 2, this means that the second length is twice as large as thefirst.

If a boy has 5 sweets and a girl has 3, the ratio of the boy's sweets to the girl's sweets is 5 : 3 . The

boy has5

3 times more sweets as the girl, and the girl has

3

5 as many sweets as the boy. Ratios

behave like fractions and can be simplified.

Drill 1

1.

2.

Proportion and Ratio

Drill 1

1.



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2.

Map ScalesThe scale of a map is usually given as a ratio in the form of 1: n. If a map has a scale of 1 : 50 000,this means that 1 unit on the map is actually 50 000 units across the land.

So 1cm on the map is 50 000cm along the ground (= 0.5km). So 1cm on the map is equivalent tohalf a kilometre in real life.For 1 : 25 000, 1 unit on the map is the same as 25 000 units on the land.So 1 inch on the map is 25 000 inches across the land, or 1cm on the map is 25 000 cm in real life.You can manipulate these ratios if necessary.

Drill 2

1.

2.

Proportional Division

Drill 3

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2.


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Degree of AccuracyAims: At the end of this module you should be able to:- Round integers to a given power of 10. Round to a given number of significant figures or decimal places. Identify upper and lower bounds where values are given to a degree of accuracy. Use estimation to evaluate approximations to numerical calculations. Solve problems using upper and lower bounds where values are given to a degree of accuracy.

Rounding Numbers

Decimal Places (dp)

Often you are asked to write an answer to a given number of decimal places (be careful to read thequestion properly!).

What you need to do:

1. Count the number of decimal places you need. 2. Look at the next digit. If it’s 4 or below just write down the answer with the right amount ofdecimal places. If it’s 5 or above write down the number but put your last decimal place up by one.

For example: 2.3635 to two decimal places

For example: 53.586 to two decimal places

What if the last digit is a 9?

A 9 goes up to a 10 so you need to put a zero in the last column and add one to the previousnumber.

For example: 8.6397 to three decimal places

Tips!

If you are not told how many places to write just be sensible!

Generally, you should go to one more place than the numbers used in the question.

With angles, no more than one decimal place should be used unless told otherwise.


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Significant Figures ( sf )

These involve all digits, not just decimal places. Zeros are only “significant” if they separate twoother non-zero digits!

What you need to do:

1. Start counting at the first non-zero digit until you have the number of digits that you need.

2. Look at the next digit. If it’s a 4 or below just write the number down leaving the last digit thesame. If it’s a 5 or above put the last digit up by one.

3. If you are rounding whole numbers (i.e. to the left of the decimal point) put zeros in all theother columns after your last digit until you reach the decimal point.

Example: 12 736 to three significant figures is …….. Example: 6530 to one significant figure is ………. Example: 0.576 to two significant figures is …….

Tip!

In real situations, use common sense to decide on your accuracy. E.g. Length of a back gardenwould not be written as 8.5632 metres. It would be more sensible to write 8.6 metres!

Example: 0.00256023164, rounded off to 5 decimal places (d.p.) is …………... You writedown the 5 numbers after the decimal point.

To round the number to 5 significant figures, you write down 5 numbers. However, you donot count any zeros at the beginning. So to 5 s.f. (significant figures), the number is0.0025602 (5 numbers after the first non-zero number appears).

From what I have just said, if you rounded 4.909 to 2 decimal places, the answer would be………., because the next number is a 9.

So 3.486 to 3 s.f. is 3.490.0096 to 3d.p. is …………..… (This is because you add 1 to the 9, making it 10.

Estimating

Arithmetic Operations

You should always do a quick estimate in your head when doing arithmetic so you can see if youranswer is reasonable.

Sometimes an exam question will test your ability to do this!

Generally, you should round each number involved to one significant figure and then it’s easy toestimate by using the single digits and moving the point around.

Let’s have a look at one:

936 27 this is difficult to do in your head but if we round both numbers to one significant figureit becomes 900 x 30.

Now this is easy to do in your head by doing 93 = 27 then moving the point 3 times (putting three

noughts on!) giving the answer 27 000 which is a good estimate of the real answer 25 272.

Here’s some more!


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45 72 becomes 50 70 which is ……..

317 23 becomes 300 20 which is …….

Check these are reasonable estimates of the real answers!

Here’s a more difficult one:

Tip!

If you can’t perform your estimate in your head then it’s too complicated - think again!

Error

It is important to remember that most measurement is approximate.

If you say your garden is 8 metres long you are rounding to the nearest metre and it could beanything from 7.5 to 8.5 metres long.

Upper and Lower Bounds

The real value can be as much as half the rounded unit above or below the value given.

So, if you are given 5.4cm the upper bound is 5.45cm and the lower bound is 5.35cm.

For 6.0kg you need to go 0.05kg either way so the upper bound is 6.05kg and the lower bound is5.95kg.

Note that: Sometimes you will be asked the upper and lower bounds of the area.

Example

If the side of a square field is given as 90m, correct to the nearest 10m:The smallest value the actual length could be is 85m (since this is the lowest value which, to thenearest 10m, would be rounded up to 90m). The largest value is 95m.

Using inequalities, ...... length < ......

Less than or equal to because

5.35 would round to 5.4

Less than, not equal to, because

5.45 rounds to 5.5

5.35 5.45x


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The area will be smallest when the side of the square is 85m. In this case, the area will be 7725m².The largest possible area is 9025m² (when the length of the sides are 95m).

Drill 1

1. In a race, Nomatyala ran 100 m in 15.3 seconds. The distance is correct to the nearest

metre and the time is correct to one decimal place. Write down the lower and upper

bounds of:

a) The actual distance Nomatyala ran

b) The actual time taken.

2.

The length of a piece of thread is 4.5 m to the nearest 10 cm. The actual length of thethread is L cm. Find the range of possible values for L.

Maximum and Minimum Values

For calculations you must use the upper or lower bounds of each measurement depending on whatcalculation you are doing.

Addition - For the maximum use the upper bound of each measurement, for the minimum use thelower bound of each measurement.

For example:

If a piece of wood measuring 15cm is joined to another piece measuring12cm you can see the maximum and minimum values of the addition .

Subtraction - For the maximum you need the biggest difference between the two measurements i.e.the upper bound of the first number and the lower bound of the second and for the minimum it’s theother way round.

For example:

David and Steven were given seeds to plant in Biology and decidedto see whose would grow the highest. After two weeks they

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measured them to the nearest centimetre and David’s had grown to 11cm whereas Steven’s hadgrown to 15cm. What are the maximum and minimum values of Steven’s victory?

Examples 1. What are the upper and lower bounds of the sum of the measurements 8 cm and 4 cm, each of

which is correct to the nearest centimetre?

2. Two lengths are given correct to 3 significant figures as 2.63 m and 4.75 m.

Find the upper and lower bounds of the sum of these measurements.

3. What are the upper and lower bounds of the difference between the measurements 8 cm and4 cm, each begin correct to the nearest centimetre?

Multiplication - Same as for Addition

Division - Same as for Subtraction

Examples1. The dimensions of a rectangle are 32.6 cm and 20.8 cm correct to 3 significant figures.

Calculate the lower and upper bounds for the area.The length lies within the range …….. cm to ……. cm. The breadth lies within the range …… cm to ……. cm.

Hence the lower bound for the area is …..… x ….… = 675.4125 cm2

and the upper bound of the area is ……. x …….. = 680.7525 cm2.

So the area lies between 675.4125 cm2 and 680.7525 cm2, or 675 cm2 and 681 cm2 correct to 3significant figures.

Maximum Value Minimum Value


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Standard FormAims: At the end of this module you should be able to:- Express numbers in the form a × 10n where n is an integer and 1 ≤ a < 10 Solve problems involving standard form

Standard Form

Drill 1

1.

2.

3.


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Applying NumberAims: At the end of this module you should be able to:- Use and apply number in everyday personal, domestic or community life. Carry out calculations using standard units of mass, length, area, volume and capacity. Understand and carry out calculations using time. Carry out calculations using money, including converting between currencies.

MeasurementToday most countries in the world use a decimal system of measurement. Decimal units ofmeasurements are also called SI (System International) units.The table below will help you to remember what units are used to measure length, mass, capacity,area and volume.This table shows you only the commonly used units. However, there are some other units that wedo not use often.

Measure Units used Equivalent to ...

Length:how long (or tall) something is Millimetres (mm) 10 mm = 1 cmCentimetres (cm) 100 cm = 1 m

Metres (m) 1 000 m = 1 km

Kilometres (km) 1 km = 1 000 000 mm

Mass:The amount ofmaterial in an object,sometimes incorrectlycalled weight

Milligrams (mg) 1 000 mg = 1 g

Grams (g) 1 000 g = 1 kg

Kilograms (kg) 1 000 kg = 1 t

Tonnes (t) 1 t = 1 000 000 g

Capacity:

The inside volume of a container,how much it holds

Millilitres (ml) 10 ml = 1 cl

Centilitres (cl) 100 cl = 1 l Litres (l) 1 l = 1 000 ml

Area :The amount of space taken up by aflat(two-dimensional) shape, alwaysmeasured in square Units

Square millimetre (mm2) 100 mm2 = 1 cm2

Square centimetre (cm2) 10 000 cm2 = 1 m2

Square metre (m2) 1 000 000 m2 = 1 km2

Square kilometre (km2) 1 km2 = 100 ha

Hectare (ha) 1 ha = 10 000 m2

Volume:The amount of space taken up by athree- dimensional object, always

measured in cubic units

Cubic millimetre (mm3) 1 000 mm3 = 1 cm3

Cubic centimetre (cm3) 1 000 000 cm3 = 1 m3

Cubic metre (m3) 1 m3 = 1 000 t

Millilitre (ml) 1 cm3 = 1 ml

Remember For square units, each place counts 100 or 102. For cube units, each place counts 1 000 or 10 3

Converting units of measurements Sometimes you will need to change the units of a measurement.For example, you may want to change metres to kilometres or centimetres to millimetres. To change to a bigger unit, you have to divide. To change to a smaller unit, you have to multiply.

The amount by which you divide or multiply will depend on how many places you are moving -the diagram below will help you to see how this works. Each place is worth a power of 10, so ifyou move one place, you will multiply or divide by 10; two places, by 100; three places, by 1


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000 and so on.

Prefix kilo hecto deca standard units (m, g, l) deci centi milli

Remember When you change from a large unit (km) to a smaller unit (m), you multiply.

Remember When you change from a small unit (m) to a larger unit (km), you divide.

Examples

1. Express 5 km in metres.

1 km =……..m So 5 km = 5 …… m = …….m

2. Express 3.2 cm in millimetres.

1 cm = …… mm So 3.2 cm = 3.2 ….. mm = ….. mm

3. Express 1 425 m in kilometres. 1 000 m = 1 km1 425 m = (1 425 1 000) km = 1.425 km

4. Express 2 m 37 cm in centimetres.Only the 2 m must be changed into centimetres.

1 m =….. cm So 2 m = 2 100 cm= …….cm2 m 37 cm = ……..cm + …..cm = ……….cm .

5. Express 2 000 000 cm2 in m2.

10 000 cm2 = l m2 2 000 000 cm2 = (2 000 000 10 000) m

2 = 200 m2

Money

Working with money is the same as working with decimal fractions, because most money amountsare given as decimals. Remember, though, that when you work with money, you need to include theunits ($ or cents) in your answers.

Notice that

Many countries use the dollar as their main currency. When we write US$, we are referring to thecurrency of the USA, and not to all dollars.In 2002, 11 countries in Europe changed to a common currency called the Euro (€) 1€ = 100 cents.


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Foreign currency

The money a country uses is called its currency. Each country has its own currency and mostcurrencies work on a decimal system (100 small units are equal to 1 main unit). This table showsyou the currency units of a few different countries.

Foreign exchange

When you change one currency for another, it is called foreign exchange. The rate of exchangedetermines how much of one currency you will get for another. Exchange rates can change daily.The daily rates are published in the press and displayed at banks. When you are asked to convertfrom one currency to another, you will be given a rate of exchange to work with.

Country Main unit Smaller unit

USA Dollar ($) = 100 cents

Japan Yen (¥) = 100 sen

UK Pound (£) — 100 pence

Germany Euro( €) = 100 cents

France Euro( € ) = 100 cents

Time

You have already learnt how to tell the time and you should know how to readand write time using the 12-hour and 24-hour system.The clock dial on the right shows you the times from 1 to 12 (a.m. and p.m.times). The outside dial shows what the times after 12 p.m. are in the 24-hoursystem.

Remember

Always bear in mind that time is written in hours and minutes and that there are 60 minutes in anhour. This is very important when calculating time - if you put 1.5 hours into your calculator, it willassume the number is decimal and work with parts of 100.

Working with time intervals

Drill 1

1. Sara and John left home at 2.15 p.m. Sara returned at 2.50 p.m. and John returned at 3.05

p.m. How long was each person away from home?

2. A journey takes from 0535 to 1820. How long is this?

Remember You cannot subtract 15 minutes from 5 minutes, so carry one whole hour over to make 65 minutes.

3. How much time passes from 1935 on Monday to 0355 on Tuesday?


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Chapter 3 Shape, Space and Measurement 88


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Angles associated with parallel lines

Lines AB and CD are parallel to one another (hence the »on the lines).

a and d are known as vertically opposite angles.

Vertically opposite angles are equal. (b and c, e and h, f

and g are also vertically opposite).

g and c are corresponding angles.

Corresponding angles are equal. (h and d, f and b, e and a are also corresponding).

d and e are alternate angles.

Alternate angles are equal. (c and f are also alternate). Alternate angles form a 'Z' shape and are

sometimes called 'Z angles'.

a and b are adjacent angles.

Adjacent angles add up to 180 degrees. (d and c, c and a, d and b, f and e, e and g, h and g, h and f

are also adjacent).

d and f are interior angles.

These add up to 180 degrees (e and c are also interior).

Any two angles that add up to 180 degrees are known as supplementary angles.

Exercise


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Exterior angles

The exterior angles of a shape are the angles you get if you extend the

sides. The exterior angles of a hexagon are shown:

A polygon is a shape with straight sides. All of the exterior angles of a polygon add up to 360°. because if you put them all together they form the angle all the way round

a point:

Therefore if you have a regular polygon (in other words, where all the

sides are the same length and all the angles are the same),

Each of the exterior angles will have size 360 ÷ the number of sides. So,

For example, each of the exterior angles of a hexagon are 360/6 = 60°.

Exercise


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Module 2-Polygons Aims: At the end of this module you should be able to:- Recognise and give the names of polygons. Understand and use the term quadrilateral and the angle sum property of quadrilaterals. Understand and use the properties of the parallelogram, rectangle, square, rhombus, trapezium

and kite. Understand the term regular polygon and calculate interior and exterior angles of regular

polygons. Understand and use the angle sum of polygons. Understand congruence as meaning the same shape and size. Understand that two or more polygons with the same shape and size are said to be congruent to

each other.

The geometric properties of quadrilaterals


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Rule:

For a polygon with n sides, the sum of the interior angles is ( - 2) 180n

The measure of each angle of a regular polygon with n sides is ( - 2) 180n

n

Exercise


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Symmetry Aims: At the end of this module you should be able to:- Recognise line and rotational symmetry Identify any lines of symmetry and the order of rotational symmetry of a given two-dimensional

figure.

Two-dimensional symmetry When two sides of a shape or object are identical (each side is a mirror image of the other), the shape or

object is symmetrical.There are two kinds of symmetry in flat shapes:

1. Line symmetry2. Rotational symmetry.

Line symmetryLook at the drawing of the African mask on the right. The drawing is said to have l ine symmetry and the dashed line is called thedrawing's line of symmetry.

The li ne of symmetry is sometimes called a 'mirror line'.

RotationalsymmetryThe shape onthe left can beturned (orrotated),keeping its

centre point P in a fixed position. The shape can be turned so that X is in position X, Y or Z, and the shape will still look the same. We say that the

shape has rotational symmetry . In this case, it fits onto itself three times whenrotated through 360° (one full revolution). We therefore say it has rotational

symmetry of order 3 . Order of rotational symmetry = the number of times a shape looks the same when rotated through 360°.

Hint: There is another way of working out the order of rotational symmetry. If the smallest angle through which

the shape can be rotated and still look the same is A°, then the order of rotational symmetry =360°

A°


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Measures Aims: At the end of this module you should be able to:- Interpret scales on a range of measuring instruments. Calculate time intervals in terms of the 24-hour and 12-hour clock. Make sensible estimates of a range of measures. Understand angle measure including three-figure bearings. Measure an angle to the nearest degree. Understand and use the relationship between average speed, distance and time.

Discrete and Continuous Measures


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2.

3.


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Drill

1

2


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Speed, Distance and Time

The following is a basic but important formula which applies when speed is constant (in other

words the speed doesn't change):

Remember, when using any formula, the units must all be consistent.

For example: speed could be measured in m/s, distance in metres and time in seconds.

If speed does change, the average (mean) speed can be calculated:

Average speed = total distance travelled

total time taken

Units

In calculations, units must be consistent, so if the units in the question are not all the same (e.g. m/s,

m and s or km/h, km and h), change the units before starting, as above.

The following is an example of how to change the units:

Drill

1. Change 15km/h into m/s.

2. If a car travels at a speed of 10m/s for 3 minutes, how far will it travel?

Velocity and Acceleration

Velocity is the speed of a particle and its direction of motion (therefore velocity is a vector quantity,

whereas speed is a scalar quantity).

When the velocity (speed) of a moving object is increasing we say that the object isaccelerating.

If the velocity decreases it is said to be decelerating. Acceleration is therefore the rate of change of velocity (change in velocity /time) and is

measured in m/s².

Example: A car starts from rest and within 10 sec is travelling at 10m/s. What is its acceleration?



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ConstructionAims: At the end of this module you should be able to:- Measure and draw lines to the nearest millimetre. Construct triangles and other two-dimensional shapes using a combination of a ruler, protractor

and compasses. Solve problems using scale drawings. Use straight edge and compasses to:

(i) Construct the perpendicular bisector of a line segment.(ii) Construct the bisector of an angle.

Scale Drawings

1.1.


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Constructing Triangles and Other ShapesDrawing triangles

Drawing parallels and Perpendiculars


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Drill


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Module 6-Circle PropertiesAims: At the end of this module you should be able to:- Recognise the terms centre, radius, chord, diameter, circumference, tangent, arc, sector and

segment of a circle. Understand chord and tangent properties of circles. understand and use the internal and external intersecting chord properties recognise the term cyclic quadrilateral

understand and use angle properties of the circle including angle subtended by an arc at the centre of a circle is twice the angle subtended at any point on the

remaining part of the circumference angle subtended at the circumference by a diameter is a right angle angles in the same segment are equal the sum of the opposite angles of a cyclic quadrilateral is 180° the alternate segment theorem.

Angles and Circles 1


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Angles and Circles 2

Drill 1


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Circles and Tangents


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Trigonometry

And PythAgorAs’ theorem Aims: At the end of this module you should be able to:- Understand and use Pythagoras' theorem in two dimensions Understand and use sine, cosine and tangent of acute angles to determine lengths and angles of a

right-angled triangle,

Apply trigonometrical methods to solve problems in two dimensions. Understand and use sine, cosine and tangent of obtuse angles Understand and use angles of elevation and depression Understand and use the sine and cosine rules for any triangle Use Pythagoras’ theorem in 3 dimensions Understand and use the formula ½bc sin A for the area of a triangle Apply trigonometrical methods to solve problems in 3 dimensions including finding the angle

between a line and a plane

Pythagoras' Theorem


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Further Work with Pythagoras' Theorem


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Sine, Cosine and Tangent

Finding Lengths in Right Angled Triangles


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Finding Angles in Right Angled Triangles


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Mixed problems with trigonometry


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ine and cosine rules


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Using Pythagoras' Theorem and Trigonometry inThree Dimensions


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Angles and Planes


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MensurationAims: At the end of this module you should be able to:- Convert measurements within the metric system to include linear, area and volume units. Find the perimeter of shapes made from triangles and rectangles. Find the area of simple shapes using the formulae for the areas of triangles and rectangles. Find the area of parallelograms and trapezia. Find circumferences and areas of circles using relevant formulae. Find the surface area of simple shapes using the area formulae for triangles and rectangles. Find the volume of right prisms, including cuboids and cylinders, using an appropriate formula. Understand the terms face, edge and vertex in the context of a three-dimensional solid. Find perimeters and areas of sectors of circles. Find the surface area and/or volume of a sphere and a right circular cone using relevant formulae. Convert between volume measures.

Squares, Rectangles and Triangles

1.

2.

1.

Drill 1

2.

Drill 1


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2.

3. .

4.


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Area and Circumference of circle

Drill 6

2

2.

1.


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Volumes of Cubes, Cuboids, Cylinders and Prisms

3.

6.


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Areas of Parallelograms, Trapeziums, Kites and

Rhombuses

10.

11.


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Surface Area

Exercises

1.

2.


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Volumes, Areas and Lengths


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SimilarityAims: At the end of this module you should be able to:- Understand and use the geometrical properties that similar figures have corresponding lengths in

the same ratio but corresponding angles remain unchanged. Understand that areas of similar figures are in the ratio of the square of corresponding sides. Understand that volumes of similar figures are in the ratio of the cube of corresponding sides. Use areas and volumes of similar figures in solving problems.

Similarity


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1


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VectorsAims: At the end of this module you should be able to:- Understand that a vector has both magnitude and direction. Understand and use vector notation. Multiply vectors by scalar quantities. Add and subtract vectors. Calculate the modulus (magnitude) of a vector. Find the resultant of two or more vectors. Apply vector methods for simple geometrical proofs.

Vectors and Scalars


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Vectors and Geometry


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The magnitude of a vectorThe length of a vector is called the magnitude or modulus of the vector.

The magnitude of vector AB is written as AB .

And the magnitude of a is |a |.

Rule

2 2If AB = , then | | =x

AB x yy

Examples

1. If a=5

12

find |a|

2. If A is point (-1, -2) and B is (5, 6), find |AB|.

Note that

If a and b are parallel vectors (parallel means pointing in the same direction), then a will be a scalar

multiple of b and vice-versa. So there will be a constant k with a = k b

Example

If a =

5

3

and b =

2

1

, find the magnitude of their resultant.

The resultant of two or more vectors is another word for their sum.

The resultant therefore is....

....

.

The magnitude of this is …………………

Remark

The addition and subtraction of vectors can be shown diagrammatically.

To find a + b, draw a and then draw b at the end of a.

The resultant is the line between the start of a and the end of b.

To find a - b, find -b (see above) and add this to a.

Example

.... ....,

.... ....

.... .... ..........

.... .... ....

| | ...... ..... ......

OA OB

AB OB

AB

Igcse Math Chapters 1 to 4

Documents

Transcript of Igcse Math Chapters 1 to 4