Maths

Pythagoras

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.

# 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
3. find the square root

For these questions you will need to use the button on your calculator.

Question

Work out the missing measurements on the right-angled triangles below:

Triangle 1

• Square the lengths you know: 62 = 36, 82 = 64.
• To find the square of the longest side, add them together 36 + 64 = 100.
• Find the square root = 10

• ? = 10

Triangle 2

• Square the lengths you know 132 = 169, 52 = 25.
• You know the longest side, so subtract the shorter side from the longest 169 - 25 = 144.
• Find the square root = 12

• ? = 12

Triangle 3

• Square the lengths you know 102 = 100, 72 = 49.
• To find the longest side, add them together 100 + 49 = 149.
• Find the square root

= 12.206555

Give your answer to a sensible degree of accuracy, eg 2 decimal places (d.p.).

• ? = 12. 21 (2 d.p.).

Be careful to give your answer correct to the number of decimal places asked for. You may need to revise Rounding and estimating numbers.

Question

Calculate the missing side lengths (shown with letters). Give your answers to one decimal place (1dp).

Remember that if you use your calculator you should still show your working out. Writing out what you do will help you to see how it works, and show the examiner that you know the process.

Here is how to work out the answers:

• Triangle 1
• x2 = 212 + 202
• 212 + 202 = 841
• Find the square root of 841
• x = 29

• Triangle 2
• y2 = 3.22 + 5.52
• 3.22 + 5.52 = 40.49
• Find the square root of 40.49
• y = 6.4

• Triangle 3
• z2 = 192 - 162
• 192 - 162 = 105
• Find the square root of 105
• z = 10.2

# Using Pythagoras' theorem in more complicated diagrams

You now know how to use Pythagoras' theorem to find any side of a right-angled triangle. Sometimes you have to use it more than once in the same problem.

Question

How would you find S in this diagram, where a right-angled triangle has been split in half?

You cannot find S until you have found the length of one other side of the whole triangle - the side which is not split in half. We will call it 'x'. Note that x also forms one side of the smaller right-angled triangle.

Looking at this smaller triangle, you will see that

x2 = 72 - 32.

So x2 = 49 - 9 = 40

Looking at the bigger right-angled triangle, you will see that

S2 = x2 + 62.

Which means s2 = 40 + 36 = 76

So S =

By using a calculator, we can find out that S = 8.72 (to two decimal places).

Now try this question:

Question

How would you find c in this diagram?

The answer is 2. If you had problems, look at the solution.

Looking at the first triangle, you will see that

a2 = 12 + 12 = 2

Looking at the next triangle, you will see that

b2 = a2 + 12 = 2 + 1 = 3

Looking at the last triangle, you will see that

c2 = b2 + 12 = 3 + 1 = 4.

So c = 2

Back to Revision Bite