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:
For these questions you will need to use the button on your calculator.
Work out the missing measurements on the right-angled triangles below:
Find the square root = 10
Find the square root = 12
Find the square root
Give your answer to a sensible degree of accuracy, eg 2 decimal places (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.
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:
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.
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:
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