There is a rule connecting the side lengths of right-angled triangles. It is called Pythagoras' theorem, and it is true for all triangles with a right-angle.
Look at the diagrams below. The areas of the squares are marked inside them (the diagrams are drawn to different scales). How are the squares' areas related to each other?
Did you spot it?
If you add the areas of the smallest two squares you get the area of the largest square.
In any right-angled triangle, the square of the longest side is the sum of the squares of the other two sides. This can be written in the formula:
a2 + b2 = c2
Where c is the longest side.
That is Pythagoras' theorem.
In your exam, you may be required to work out other mathematical problems using Pythagoras' theorem. This is easier than it sounds.
For instance, if you know two sides of a right-angled triangle, you can find the third like this:
Trigonometry can be used to calculate the lengths of sides and sizes of angles in right-angled triangles.
The sides of the right-angled triangles are given special names - thehypotenuse, the opposite and the adjacent.
The hypotenuse is the longest side and is always opposite the right angle. The opposite and adjacent sides relate to the angle under consideration.
Click on the correct side of the triangle on the diagram below.
There are three formulae involved in trigonometry:
sin = opposite / hypotenuse
cos = adjacent / hypotenuse
tan = opposite / adjacent
Which formula you use will depend on the information given in the question.
There are a couple of ways to help you remember which formula to use. Remember SOHCAHTOA (it sounds like 'Sockatoa')
You often need to use Pythagoras' theorem more than once in each problem as we create extra right-angled triangles in the 3D shapes.
This cuboid has side lengths of 2cm, 3cm and 6cm.
Work out the length of the diagonal AF.
First use Pythagoras' theorem in triangle ABC to find length AC.
AC2 = 62 + 22
AC = √40
You do not need to find the root as we will need to square it in the following step. Next we use Pythagoras' theorem in triangle ACF to find length AF.
AF2 = AC2 + CF2
AF2 = 40 + 32
AF = √49
AF = 7cm
Sometimes you will be asked to find 2 unknown values by solving 2 equations at the same time. These types of equations are called simultaneous equations.
The first step is to try to eliminate one of the unknowns.
Solve these simultaneous equations and find the values of x and y.
Add the two equations to eliminate the ys:
So the answers are x = 3 and y = 1
Sometimes you will need to multiply one of the equations before you can add or subtract
You will most likely get a few rearranging questions on the exam paper. Or you may want to know the area of a circle and need to find the radius. To do this, we rearrange the formula to make the radius the subject.
The area of a circle (A) is πr2. So:
A = πr2
We will now rearrange the formula to make 'r' the subject.
A = πr2
Start by dividing both sides by π.
Then take the square root of both sides.
Standard index form is also known as standard form. It is very useful when writing very big or very small numbers.
In standard form, a number is always written as: A × 10 n
A is always between 1 and 10. n tells us how many places to move the decimal point.
Example : Write 15 000 000 in standard index form.
15 000 000 = 1.5 × 10 000 000
This can be rewritten as:
1.5 × 10 × 10 × 10 × 10 × 10 × 10 × 10
= 1. 5 × 10 7
You can convert from standard form to ordinary numbers, and back again. Have a look at these examples:
3 x 104 = 3 × 10 000 = 30 000 (Since 104 = 10 × 10 × 10 × 10 = 10 000)
2 850 000 = 2.85 × 1 000 000 = 2.85 × 106 Make the first number between 1 and 10.
0.000467 = 4.67 × 0.0001 = 4.67 × 10 -4