Pythagoras' theorem is one of the most frequently tested topics in GCSE Maths and one of the most reliable sources of marks on the exam. It appears on both Foundation and Higher papers across AQA, Edexcel and OCR, sometimes as a standalone question, sometimes embedded inside a larger problem involving coordinates, trigonometry or 3D shapes. This topic guide gives you the theorem, the step-by-step method for finding both longer and shorter sides, worked examples at both tiers (including 3D Pythagoras for Higher), common pitfalls and practice questions to sharpen your skills. For a deeper exploration, read our dedicated Pythagoras' theorem blog post, and for a broader formula reference check our GCSE Maths formulas guide.
What Is Pythagoras' Theorem?
Pythagoras' theorem states that in any right-angled triangle:
$$a^2 + b^2 = c^2$$
where c is the hypotenuse — the longest side, always opposite the right angle — and a and b are the other two sides.
This formula is not given on the AQA, Edexcel or OCR formula sheets. You must memorise it.
Identifying the Hypotenuse
- It is always opposite the right angle (marked with a small square).
- It is always the longest side.
- Mislabelling the hypotenuse is the single most common error in Pythagoras questions.
When to Use Pythagoras
Use Pythagoras' theorem whenever you have a right-angled triangle and need to find a missing side. You need to know two sides to find the third.
Step-by-Step Method
Finding the Hypotenuse (Longest Side)
- Label the hypotenuse c and the other sides a and b.
- Square both known sides: a² and b².
- Add them: a² + b² = c².
- Take the square root: c = √(a² + b²).
Finding a Shorter Side
- Label the hypotenuse c (the known longest side).
- Rearrange: a² = c² − b².
- Substitute and subtract.
- Take the square root.
3D Pythagoras (Higher)
For three-dimensional problems, apply Pythagoras twice:
- Use two sides to find a diagonal across a face.
- Use that diagonal with the third dimension to find the space diagonal.
The direct formula for a cuboid with dimensions l, w, h: $$d = \sqrt{l^2 + w^2 + h^2}$$
Worked Example 1 — Foundation Level
A right-angled triangle has legs of 5 cm and 12 cm. Find the hypotenuse.
Step 1: c² = a² + b² = 5² + 12² = 25 + 144 = 169.
Step 2: c = √169 = 13 cm.
Finding a Shorter Side
A right-angled triangle has hypotenuse 15 cm and one leg of 9 cm. Find the other leg.
Step 1: a² = c² − b² = 15² − 9² = 225 − 81 = 144.
Step 2: a = √144 = 12 cm.
Worked Example 2 — Higher Level
A cuboid measures 6 cm by 4 cm by 3 cm. Find the length of the space diagonal AG. Give your answer to 1 decimal place.
Step 1: Find the diagonal across the base (let us call it AC). $$AC = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52}$$
Step 2: Use AC and the height to find AG. $$AG = \sqrt{AC^2 + 3^2} = \sqrt{52 + 9} = \sqrt{61} = 7.8 \text{ cm (1 d.p.)}$$
Or in one step: AG = √(6² + 4² + 3²) = √(36 + 16 + 9) = √61 = 7.8 cm.
Distance Between Two Points
Find the distance between A(2, 3) and B(8, 11).
Step 1: Horizontal distance = 8 − 2 = 6. Vertical distance = 11 − 3 = 8.
Step 2: Distance = √(6² + 8²) = √(36 + 64) = √100 = 10 units.
Common Mistakes
- Labelling the wrong side as the hypotenuse. Always check which side is opposite the right angle.
- Adding when you should subtract. Finding the hypotenuse → add. Finding a shorter side → subtract. Get this the wrong way round and the answer is completely wrong.
- Forgetting to square root at the end. The formula gives you c² — you must take the square root to find c.
- Squaring incorrectly. 5² = 25, not 10. Take care with basic arithmetic.
- Applying Pythagoras to non-right-angled triangles. The theorem only works for right-angled triangles. If the triangle does not have a 90° angle, you need the cosine rule instead.
Exam Tips
- Draw a diagram if one is not provided. Label the right angle and the hypotenuse. This takes seconds and prevents errors.
- Show every step — write the formula, show the substitution, show the addition or subtraction, then the square root. Each step can earn a method mark.
- Use exact values where possible. If the question says "give an exact answer", leave as √52 rather than rounding.
- In coordinate geometry, the distance formula is just Pythagoras applied to the horizontal and vertical differences.
- For 3D questions, sketch the triangle you are using and label the sides clearly.
- Pythagorean triples — 3, 4, 5 and 5, 12, 13 and 8, 15, 17 — appear frequently. Recognising them speeds up your work.
Practice Questions
Question 1 (Foundation) A right-angled triangle has legs 8 cm and 15 cm. Find the hypotenuse.
Question 2 (Foundation) A ladder leans against a wall. The ladder is 5 m long and its base is 3 m from the wall. How high up the wall does it reach?
Question 3 (Higher) Find the distance between the points (−1, 4) and (5, −4).
Question 4 (Higher) A cone has a base radius of 7 cm and a slant height of 25 cm. Find the perpendicular height of the cone.
Question 5 (Higher) Find the length of the space diagonal of a cuboid with dimensions 8 cm, 9 cm and 12 cm.
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Related Topics
- Trigonometry (SOHCAHTOA) — the next step after Pythagoras for right-angled triangles
- Coordinate Geometry — distance between points uses Pythagoras
- Surface Area — slant heights of cones often require Pythagoras
- 3D Shapes — space diagonals use 3D Pythagoras
- Circle Theorems — some proofs involve Pythagoras
Summary
Pythagoras' theorem states that a² + b² = c² in any right-angled triangle, where c is the hypotenuse. To find the hypotenuse, add the squares of the other two sides and take the square root. To find a shorter side, subtract and then square root. For 3D problems on Higher, apply the theorem twice or use d = √(l² + w² + h²). Always identify the hypotenuse correctly, show every step of working, and remember that the theorem only applies to right-angled triangles. It is one of the most tested topics in GCSE Maths and a dependable source of marks.