Pythagoras' theorem | Pearson Edexcel GCSE Maths Notes

Visual model

Pythagoras works only in right-angled triangles

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

a

b

c

longest sideright angle

Find the right angle first.

The hypotenuse is opposite the right angle.

Add for the hypotenuse, subtract for a shorter side.

Gold-standard guide

22 mins

What you will learn

I can label the hypotenuse and shorter sides of any right-angled triangle.

I can use a² + b² = c² to find the hypotenuse.

I can rearrange Pythagoras' theorem to find a shorter side.

I can apply Pythagoras in real-world and geometric contexts.

I can test whether a triangle is right-angled using a² + b² = c².

Useful before you start

Squaring numbers (on a calculator and by hand)Square roots (using the √ key)Substitution into formulasIdentifying right anglesRounding and significant figures

Core knowledge

Know the rule, then use it

These are the short notes. Read each one, then check you can use it in the worked example below.

Method

Hypotenuse

The longest side — always opposite the right angle; labelled c in the formula.

Step 1

Right-angled triangle

A triangle with exactly one 90° angle — Pythagoras only applies here.

Step 2

Square (of a number)

A number multiplied by itself (e.g. 7² = 49).

Step 3

Square root

The inverse of squaring — √49 = 7; the final step in every Pythagoras calculation.

Step 4

Pythagorean triple

Three integers satisfying a² + b² = c² exactly (e.g. 3, 4, 5 or 5, 12, 13).

Step 5

Exact form

Leaving an answer as a surd (e.g. 5√2) rather than a rounded decimal.

Pythagoras' theorem: a² + b² = c²

a and b are the two shorter sides; c is the hypotenuse (opposite the right angle).Always identify which side is the hypotenuse before substituting.

Finding the hypotenuse: c = \sqrt{a² + b²}

Square both shorter sides, add the results, then take the positive square root.

Finding a shorter side: a = \sqrt{c² − b²}

Square the hypotenuse, subtract the square of the known shorter side, then take the positive square root.Subtract — never add — when finding a shorter side.

Worked example

A ladder of length 10 m leans against a vertical wall. The foot of the ladder is 4 m from the base of the wall. How high up the wall does the ladder reach? Give your answer to 1 decimal place.

Draw a right-angled triangle: the ladder is the hypotenuse (10 m), the base is one shorter side (4 m), and the height on the wall is the unknown shorter side h.

Apply Pythagoras: h² + 4² = 10².

h² + 16 = 100.

h² = 84.

h = √84 ≈ 9.165... ≈ 9.2 m (to 1 d.p.).

Final answer

9.2 m

Question ladder

Build up to the hardest questions

Do them in order. If you miss a step, read the solution, then redo the question without looking.

Diagnosticrecall

In a right-angled triangle, which side is the hypotenuse?

short sideshort side

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

1 mark1 minpyth-d1

Show solution

Worked solution

1.Look for the 90° angle.
2.The side directly opposite that angle is the hypotenuse.

Final answer

The side opposite the right angle (the longest side).

Mark points

B1: 'opposite the right angle' or 'the longest side' (both answers acceptable; the geometric definition is preferred).

Watch out

Choosing the side that looks longest in a particular diagram orientation, without considering which angle it is opposite.Rotated triangles frequently catch students out.

Easyprocedure

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

short sideshort side

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

2 marks2 minspyth-e1

Show solution

Worked solution

1.Apply the theorem: c² = 9² + 12².
2.c² = 81 + 144 = 225.
3.c = √225 = 15 cm.

Final answer

15 cm

Mark points

M1: correct substitution into a² + b² = c² giving c² = 225 (or equivalent correct sum of squares).
A1: c = 15 cm (accept 15 without units if units were not given in the question, but units should be encouraged).

Watch out

Forgetting the square root at the end and writing c = 225 cm (a common final-step error under time pressure).

Mediumprocedure

A right-angled triangle has hypotenuse 20 cm and one shorter side 12 cm. Find the other shorter side, giving your answer to 1 decimal place.

hypotenuseknown side

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

3 marks3 minspyth-m1

Show solution

Worked solution

1.Let the unknown side be a.
2.Use a² + 12² = 20².
3.a² + 144 = 400.
4.a² = 256.
5.a = √256 = 16 cm.
6.(No rounding needed — this is a Pythagorean triple: 12, 16, 20 = 4 × (3, 4, 5).)

Final answer

16 cm

Mark points

M1: correct rearrangement: a² = 20² − 12² seen (accept a² = 400 − 144).
M1: obtains a² = 256.
A1: a = 16 cm.

Watch out

Adding 12² to 20² rather than subtracting (confusing finding a shorter side with finding the hypotenuse), giving a² = 544 and a ≈ 23.3 cm.

Hardreasoning

ABCD is a rectangle with AB = 8 cm and BC = 6 cm. Point M is the midpoint of CD. Find the length AM, giving your answer in surd form.

midpoint

4\text{ cm}

6\text{ cm}

AM

4 marks4 minspyth-h1

Show solution

Worked solution

1.In the rectangle, AB = CD = 8 cm and BC = AD = 6 cm.
2.M is the midpoint of CD, so CM = DM = 4 cm.
3.Draw triangle ABM (or consider right triangle with one vertex at A).
4.Better: use coordinates. Let A = (0, 0), B = (8, 0), C = (8, 6), D = (0, 6).
5.M = midpoint of CD = midpoint of (8,6) and (0,6) = (4, 6).
6.AM = √((4−0)² + (6−0)²) = $\sqrt{16 + 36}$ = √52 = 2√13.

Final answer

AM = 2√13 cm

Mark points

g. horizontal distance 4 and vertical distance 6 seen.
M1: correctly substitutes into Pythagoras: AM² = 4² + 6² = 52.
M1: takes square root to give AM = √52.
A1: simplifies correctly to 2√13.

Watch out

Using AB = 8 as the horizontal distance rather than half of AB = 4, or forgetting to simplify √52 into surd form.

Very Hardproblem solving

A right-angled triangle has sides in the ratio 3:4:5, scaled so that its area is 96 cm². Find the length of its hypotenuse.

short sideshort side

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

5 marks5 minspyth-vh1

Show solution

Worked solution

1.Let the sides be 3k, 4k and 5k for some positive k.
2.Check: (3k)² + (4k)² = 9k² + 16k² = 25k² = (5k)². ✓ So 5k is the hypotenuse.
3.The two shorter sides form the right angle, so they are the base and height.
4.Area = ½ × base × height = ½ × 3k × 4k = 6k².
5.6k² = 96 → k² = 16 → k = 4.
6.Hypotenuse = 5k = 5 × 4 = 20 cm.

Final answer

20 cm

Mark points

M1: recognises ratio form 3k:4k:5k or equivalent scaled triple.
M1: forms area equation ½ × 3k × 4k = 96.
M1: solves k² = 16 to obtain k = 4.
M1: identifies hypotenuse as 5k.
A1: 20 cm.

Watch out

Using the area formula with the hypotenuse as the base (treating all three sides as if they could be base and height), which requires an additional step to find the perpendicular height.

Grade 9 stretchproblem solving

An equilateral triangle has side length 10 cm. Find the exact area of the triangle, leaving your answer in surd form.

10\text{ cm}

5\text{ cm}

h

6 marks6 minspyth-g9

Show solution

Worked solution

1.Divide the equilateral triangle into two congruent right-angled triangles by dropping a perpendicular from the apex to the base.
2.Each right-angled triangle has hypotenuse 10 cm and base 5 cm.
3.Height h: h² + 5² = 10² → h² = 75 → h = √75 = 5√3.
4.Area of equilateral triangle = ½ × base × height = ½ × 10 × 5√3 = 25√3.

Final answer

25√3 cm²

Mark points

M1: identifies that the perpendicular bisector creates a right-angled triangle with hypotenuse 10 and base 5.
M1: correct Pythagoras application → h² = 75.
M1: correctly simplifies √75 to 5√3.
M1: correct area formula ½ × 10 × 5√3.
A1: 25√3 cm² (exact surd form required).
A1: correct method throughout with all steps shown.

Watch out

g.66) rather than leaving it in surd form, then giving an approximate decimal area instead of the required exact answer.

Exam Techniqueexam trap

A student is given a right-angled triangle with sides 7 cm, x cm and 25 cm. They write x² = 7² + 25² = 49 + 625 = 674, so x = √674 ≈ 25.96 cm. Without doing any calculation, explain why this answer must be wrong, then find x correctly.

hypotenuseknown side

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

3 marks3 minspyth-et1

Show solution

Worked solution

1.The answer x ≈ 25.96 cannot be right because x would then be longer than 25 cm.
2.But the hypotenuse is the longest side. If 25 cm is the hypotenuse, then x must be shorter than 25 cm.
3.The student incorrectly added both known values as if finding a hypotenuse, but 25 is the hypotenuse.
4.Correct method: x² + 7² = 25² → x² = 625 − 49 = 576 → x = 24 cm.

Final answer

x = 24 cm. The error was using the addition formula (treating 25 cm as a shorter side) rather than the subtraction formula for a shorter side.

Mark points

B1: explains that the result exceeds the hypotenuse, which is geometrically impossible.
M1: correct Pythagoras rearrangement: x² = 25² − 7² = 576.
A1: x = 24 cm.

Watch out

Accepting an answer that is longer than the hypotenuse without questioning it — always check that any shorter side is genuinely shorter than the hypotenuse.

Timed checkpoint

12 mins - 9 marks

Switch between skills

Set a timer and attempt all four questions before opening any answers. This is closer to the way skills appear in a real paper.

1Pythagoras' theorem - 2 marksFind the hypotenuse of a right triangle with legs 5 cm and 12 cm.Mark answer

Answer

13 cm

2Angles, lines and polygons - 2 marksThe exterior angle of a regular polygon is 24°. How many sides?Mark answer

Answer

3Properties of shapes - 2 marksName all 2D shapes with equal diagonals that bisect each other.Mark answer

Answer

Rectangle, square

4Perimeter, area and volume - 3 marksFind the perimeter of a rectangle with length 13 cm and width 8 cm.Mark answer

Answer

42 cm

Mastery check

I can recall Pythagoras' theorem and correctly identify which side is the hypotenuse before writing the formula.
I can calculate both the hypotenuse and a shorter side, choosing to add or subtract the squares correctly at each step.
I can apply Pythagoras' theorem to real-life contexts (ladders, distances, coordinates) and interpret the answer with appropriate units.
I can work with exact surd answers rather than rounding, including simplifying surds such as √75 = 5√3.
I can apply Pythagoras in multi-step unfamiliar settings — such as an equilateral triangle or a rectangle with an internal line — by decomposing the problem into right-angled triangles.

Ask Nova Bot

Explain Pythagoras' theorem with a visual proof — ideally using the classic square-on-each-side diagram — so I truly understand why a² + b² = c².I keep mixing up when to add and when to subtract the squares. Give me a one-sentence rule I can remember in an exam.Quiz me on Pythagoras: give me five questions of increasing difficulty, one involving an exact surd answer, and mark my responses.What are the two most common errors on Pythagoras GCSE questions, and how does checking my answer help me catch them?Show me a real-world scenario where Pythagoras is used — for example in architecture, navigation or sport — and explain the calculation.

Get help with anything that still feels tricky.

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