G20Geometry & MeasuresFoundation & HigherGrades 4–7

Pythagoras' Theorem

Q: Can I use Pythagoras to check whether a triangle is right-angled?

Yes. If a^2 + b^2 = c^2 exactly (with c the longest side), the triangle is right-angled. If not, it is not.

In any right-angled triangle, the square on the longest side equals the sum of the squares on the other two: $a^{2} + b^{2} = c^{2}$ . The longest side, $c$ , is the $hypotenuse$ — always opposite the right angle. With it you can find any missing side from the other two, which is why it turns up in coordinates, 3D shapes, trigonometry and circle problems.

40 min Video by Maths Genie AQA GCSE Maths

The big picture

The theorem is literally about squares. Build a square on each side of a 3-4-5 triangle: the areas are 9, 16 and 25, and $9 + 16 = 25$ . That is true for every right-angled triangle, and only for right-angled ones.

There are only two cases. Finding the hypotenuse: square, $add$ , square root. Finding a shorter side: square, $subtract$ , square root. A quick sense check — the hypotenuse must be the longest side — tells you which one you need.

Pythagoras and TrigonometryCurated by Kepler, not produced by us. Watch it, then read the notes below.

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What you'll learn

Identify the hypotenuse of a right-angled triangle
Find the hypotenuse and the shorter sides using $a^{2} + b^{2} = c^{2}$
Apply Pythagoras in context and to find distances between coordinates
Use Pythagoras in 3D shapes (Higher)

Finding the hypotenuse

Label the hypotenuse $c$ (opposite the right angle). Square the two shorter sides, add them, and square root the total.

a^{2} + b^{2} = c^{2}

, where

c

is the hypotenuse.

c^{2} = 7^{2} + 1 1^{2} = 49 + 121 = 170

c = 170 = 13.038 \dots

Answer

13.0

Your turnSame method, new numbers — try it before moving on.

A right-angled triangle has shorter sides 9 cm and 12 cm. Find the hypotenuse, in cm.

Your answer

Finding a shorter side

Rearrange: $a^{2} = c^{2} - b^{2}$ . Square the hypotenuse, subtract the square of the known shorter side, then square root.

a^{2} = 1 7^{2} - 8^{2} = 289 - 64 = 225

a = 15

Answer

15

Your turnSame method, new numbers — try it before moving on.

A right-angled triangle has hypotenuse 20 cm and one side 12.5 cm. Find the other side to 1 decimal place.

Your answer

Tip — Check that your hypotenuse is the longest side. If a "shorter" side comes out longer than the hypotenuse, you added when you should have subtracted.

Problems and coordinates

Look for the hidden right angle: a ladder against a wall, a diagonal across a rectangle, the height of an isosceles triangle, or the direct distance between two points.

The distance between $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is the hypotenuse of a triangle with sides equal to the horizontal and vertical changes.

1The ladder is the hypotenuse:

h^{2} = 5^{2} - 1. 4^{2} = 25 - 1.96 = 23.04

h = 4.8

Answer

4.8

Your turnSame method, new numbers — try it before moving on.

Find the distance between the points $(- 2, 3)$ and $(4, - 5)$ .

Your answer

Your turnSame method, new numbers — try it before moving on.

An equilateral triangle has sides of 6 cm. Find its perpendicular height, to 1 decimal place.

Your answer

Pythagoras in 3D (Higher)

Find a right-angled triangle inside the solid. Often you need two steps: first a diagonal across the base, then a triangle using that diagonal and the height.

For a cuboid with edges $a$ , $b$ , $c$ , the space diagonal is $d = a^{2} + b^{2} + c^{2}$ .

1Base diagonal:

6^{2} + 4^{2} = 52

2Space diagonal:

52 + 3^{2} = 61 = 7.81 \dots

Answer

7.8

Your turnSame method, new numbers — try it before moving on.

A box is 12 cm by 4 cm by 3 cm. What is the length of the longest straight rod that fits inside it, in cm?

Your answer

Keep it exact between steps

$a^{2} + b^{2} + c^{2}$ is just Pythagoras applied twice. Squaring the base diagonal undoes its square root, so $52$ becomes $52$ inside the second step — which is why keeping exact values avoids rounding errors.

Think like an examiner

Write $c^{2} = a^{2} + b^{2}$ with the numbers substituted — it earns the method mark.
Do not round between steps; keep surds or full calculator values until the end.
"Show that" questions often use Pythagoras to prove a triangle is right-angled: check $a^{2} + b^{2} = c^{2}$ exactly.

Pythagoras' theorem

a^{2} + b^{2} = c^{2}

c

is the hypotenuse.

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Distance between two points.

d = a^{2} + b^{2} + c^{2}

Space diagonal of a cuboid.

Watch out for these

Adding the squares when finding a shorter side.

Shorter side: subtract from the hypotenuse squared.

c = 7 + 11 = 18

Square first:

c = 49 + 121

Forgetting to square root at the end:

c = 170

c = 170 = 13.0

Using Pythagoras in a triangle with no right angle.

Use the sine or cosine rule for non-right-angled triangles.

Stretch yourself

A triangle has sides $2 n + 1$ , $2 n^{2} + 2 n$ and $2 n^{2} + 2 n + 1$ , where $n$ is a positive integer. Show that the triangle is always right-angled.

Hint — Check whether the square of the longest side equals the sum of the squares of the other two.

Questions students ask

How do I spot the hypotenuse?

It is opposite the right angle and is always the longest side.

What are Pythagorean triples?

Whole-number sets that fit the theorem, such as 3-4-5, 5-12-13, 8-15-17 and their multiples like 6-8-10. Recognising them saves time on non-calculator questions.

Can I use Pythagoras to check whether a triangle is right-angled?

Yes. If

a^{2} + b^{2} = c^{2}

exactly (with

c

the longest side), the triangle is right-angled. If not, it is not.

Key takeaways

Only for right-angled triangles; $c$ is the hypotenuse.
Hypotenuse: square, add, square root.
Shorter side: square, subtract, square root.
Look for hidden right angles in problems and coordinates.
In 3D, use two triangles or $a^{2} + b^{2} + c^{2}$ .

Exam-style questions

Practise it the way it is examined

3 original questions · 10 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 43 marks

A rectangle is 24 cm long and 10 cm wide. Work out the length of its diagonal.[3]

Question 2Grade 53 marks

Point $A$ is $(1, 7)$ and point $B$ is $(9, 1)$ .

Work out the length of $A B$ .[3]

Question 3Grade 74 marks

A square-based pyramid has base edges of 8 cm. Its sloping edges are each 9 cm long. The apex is directly above the centre of the base.

Work out the perpendicular height of the pyramid. Give your answer to 3 significant figures.[4]

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Pythagoras' Theorem

Finding the hypotenuse

Finding a shorter side

Problems and coordinates

Pythagoras in 3D (Higher)

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Skill drill

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Related topics

Pythagoras' Theorem

Finding the hypotenuse

Finding a shorter side

Problems and coordinates

Pythagoras in 3D (Higher)

Practise it the way it is examined

Skill drill

Practise this topic

Related topics