**Table of Contents**

** Unit 1 | Algebra**

Page 1 | Expressions and Formulae

Page 3| Solving Linear Equations

Page 4| Expanding and Factorising

Page 5| Factorising Quadratics and expanding double brackets

Page 6| Patterns and Sequences

Page 7| Simultaneous Equations

Page 8| Changing the subject of a Formula

Page 9| Adding , subtracting algebraic formulas

** Unit 2 |Graphs**

Page 1 | Straight line graphs

Page 2 | Graphs of Quadratic functions

** Unit 3 |Geometry and Measure **

Page 2 | Symmetry

Page 3 | Coordinates

Page 4 | Perimeter, Area, Volume

Page 6 | Measurement

Page 7 | Trigonometry

Page 8 | Pythagoras

Page 9 | Angles

Page 10 | Shapes

Page 11| Time

Page 12 | Locus

**Unit 4 | Numbers**

Page 1 | Speed, Distance and time

Page 2 | Rounding and estimating

Page 3 | Ratio and proportion

Page 4 | Factors, Multiples and primes

Page 5 | Powers and roots

Page 7 | Positive and negative numbers

Page 8 | Basic operations

Page 9 | Fractions

Page 10 | Percentages

** Unit 5 | Statistics and Probability **

Page 1 | Sampling data (MA)

Page 2 | Recording and representing data

Page 3 | Mean median range and mode

Page 4 | Standard deviation

**Unit 4 | Calculus **

*Pythagoras’ Theorem*

L.O To find the length of an unknown side using Pythagoras

This is the method of finding the length of different sides in a **right angled triangle.**

Firstly, you should be comfortable with labelling the: **adjacent**, **opposite** and **hypotenuse** of any right angle triangle, we will cover this more on the Trigonometry section.

All you have to do is follow one simple rule:

**Let’s try an example:**

Here we have a right angle triangle with side lengths of 3cm,4cm and 5cm.

Pythagoras’ Theorem uses the square of the length of each side.

Now if we look at the areas of each square:

The area of the square on the side length of 3cm is 9cm

The area of the square on the side length of 4cm is 16cm

The area of the square on the side length of 5cm (the hypotenuse) is 25cm

**Example 1:**

Work out the length of the hypotenuse of this triangle

Always state Pythagoras’ Theorem first

**Example 2:**

Work out the length of the Opposite side of this triangle to 3 s.f

Now we have to rearrange Pythagoras’ Theorem

However the question asks the answer to 3 significant figures so make sure your answer is __13.9__

**Example 3:**

Work out the length of the adjacent side to 3 s.f.

Step 1: Rearrange in the same way as you did in the example above but this time subtract b^{2}.