Pythagoras Theorem

Pythagoras was a Greek mathematician who lived over 2500 years ago
He is most famous for Pythagoras’ Theorem, which includes the important formula for right-angled triangles

The longest side in a right-angled triangle is called the hypotenuse
- The hypotenuse will always be the side opposite the right angle
If we label the hypotenuse c, and label the other two sides a and b then Pythagoras’ Theorem tells us that:
- a² + b² = c²
  Where a, b, and c are the lengths of the three sides

Right Angled Triangle abc, IGCSE & GCSE Maths revision notes

To find the length of the hypotenuse use $c = \sqrt{a^{2} + b^{2}}$
- This is just a rearrangement of the formula a² + b² = c², to make c the subject
- Note that when finding the hypotenuse you add inside the square root
To find the length of one of the other sides use $a = \sqrt{c^{2} - b^{2}}$ or $b = \sqrt{c^{2} - a^{2}}$
- This is a rearrangement of the formula a² + b² = c², to make either a or b the subject
- Note that when finding one of the shorter sides you subtract inside the square root

‘Reality check’ your answers, if the hypotenuse ends up being shorter than another side in your answer, you’ve made a mistake somewhere
If your calculator gives you a ‘Math ERROR’ result, you probably subtracted the wrong way around in the square root
In questions with multiple steps, don’t round off (to 1 decimal place, 3 significant figures, etc) until the very end

In the following diagram:
$A B = 12 cm$
$A C$ is a straight line, with $A D = 9 cm$ and $A C = 22 cm$

Back to Back Right Angled Triangles, IGCSE & GCSE Maths revision notes

Find $x$ , the length of side $B C .$ Give your answer to 1 decimal place.

To find $x$ , we first need to find the length of $B D$ using triangle $A B D$
Note that $B D$ is a shorter side

$B D = \sqrt{12^{2} - 9^{2}} = \sqrt{63} = 7.93725 . . .$

It is best to leave rounding until the very end, use $\sqrt{63}$ in subsequent working

Now we can find $x$ using triangle $B C D$

$D C = 22 - 9 = 13 cm$

$x = \sqrt{B D^{2} + D C^{2}} = \sqrt{{(\sqrt{63})}^{2} + 13^{2}} = \sqrt{63 + 169}$

$x = \sqrt{232} = 15.23154621 . . .$

Round to 1 decimal place

$x = 15.2 cm$

Pythagoras Theorem (CIE IGCSE Maths: Extended)