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Finding one side of a right-angled triangle when you know the other two

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.

Using Pythagoras theorem to solve problems

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:

1. square (multiply by itself) the lengths you know
2. add or subtract them
3. find the square root

The three formulae: sin, cos, tan

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.

Example

This cuboid has side lengths of 2cm, 3cm and 6cm.

Work out the length of the diagonal AF.

Solution

First use Pythagoras' theorem in triangle ABC to find length AC.

1. AC² = 6² + 2²

2. 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.

1. AF² = AC² + CF²

2. AF² = 40 + 3²

3. AF = √49

4. AF = 7cm

Simultaneous equations are two equations with two unknowns. They are called simultaneous because they must both be solved at the same time.

The first step is to try to eliminate one of the unknowns.

Example

Solve these simultaneous equations and find the values of x and y.

• Equation 1: 2x + y = 7
• Equation 2: 3x - y = 8

Add the two equations to eliminate the ys:

• 2x + y = 7
• 3x - y = 8
• ------------
• 5x = 15
• x = 3

• Now you can put x = 3 in either of the equations.
• Substitute x = 3 into the equation 2x + y = 7:
• 6 + y = 7
• y = 1

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.

 or 

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.

Solution

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