GCSE Maths · Geometry and measures

Pythagoras' theorem GCSE Questions and Worked Answers

In a right-angled triangle, the squares of the two shorter sides add to the square of the hypotenuse: a² + b² = c². Add to find c²; subtract to find a shorter side squared; then take the positive square root.

Foundation & Higher6 worked examples10 original questions

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Start with the meaning

What you need to know about pythagoras' theorem

A rectangular field is 6 m across and 8 m long. Walking across and then along makes a right-angled route; a straight diagonal is shorter. The diagonal and the two edges form a right-angled triangle. The side opposite the 90° corner is called the hypotenuse, and it is the longest side. Pythagoras' theorem connects its length to the other two lengths.

See the idea first

The relationship is between square areas

A square built on a 6-unit side has area 6² = 36 square units. One on an 8-unit side has area 64. For a right-angled triangle, these add to the area of the square on the longest side: 36 + 64 = 100. The side of that square is 10 because 10 × 10 = 100.

A 6–8–10 right-angled triangle. The small square marks 90°; the sloping side is opposite that corner.

From problem to method

Typical problems you need to be able to solve

These are the main kinds of question in this topic. Each card first states the problem, then explains how to solve it.

Find the hypotenuse

What the problem asks: The perpendicular sides are 5 cm and 12 cm. Find the longest side.

How to solve it: Its square is 5² + 12² = 169. The positive square root gives 13 cm.

Find a shorter side

What the problem asks: The hypotenuse is 13 cm and one shorter side is 5 cm. Find the other.

How to solve it: The unknown square plus 25 equals 169, so subtract: 169 − 25 = 144. The side is 12 cm.

Find a diagonal or coordinate distance

What the problem asks: Two points differ by 3 horizontally and 4 vertically. Find their straight-line distance.

How to solve it: Horizontal and vertical changes form perpendicular sides of a triangle. The distance is √(3² + 4²) = 5 units.

Check whether a triangle is right-angled

What the problem asks: Are sides 7 cm, 24 cm and 25 cm a right-angled triangle?

How to solve it: Test the longest side: 7² + 24² = 625 = 25². The converse of Pythagoras confirms a right angle opposite the 25 cm side.

A reliable routine

For one missing side in a right-angled triangle

This method uses the square-area relationship a² + b² = c², where c must be opposite the right angle. It is not valid for a triangle without a right angle. If the hypotenuse is known, rearranging the equality requires subtraction rather than addition.

Identify the right angle and label the opposite side c.
Write a² + b² = c² and insert the two known lengths in matching units.
Add the two shorter-side squares, or subtract a known shorter-side square from c².
Take the positive square root. Round only at the end and check that c is the longest side.

Check: Squaring a length gives an area-like quantity; taking the square root returns a length. Do not report the squared total as the final side length.

Fully worked

Pythagoras' theorem GCSE worked examples

Each solution explains the clue, why the method fits and how to check the result.

Example 1

Find the longest side

3 marks

Question

A right-angled triangle has perpendicular sides 9 cm and 12 cm. Find its hypotenuse.

c^2=9^2+12^2

c^2=81+144=225

c=\sqrt{225}=15

The hypotenuse is 15 cm, longer than either given side.

Example 2

Find a shorter side

3 marks

Question

A right-angled triangle has hypotenuse 17 cm and another side 8 cm. Find the third side.

Let the missing shorter side be a.

a^2+8^2=17^2

a^2=289-64=225

a=15

The missing side is 15 cm. Adding would incorrectly produce a side longer than the hypotenuse.

Example 3

A ladder

4 marks

Question

A 6 m ladder rests against a vertical wall on level ground. Its foot is 1.5 m from the wall. Find the vertical height reached, to 2 decimal places.

The wall and ground meet at 90°. The ladder is the hypotenuse.

h^2+1.5^2=6^2

h^2=36-2.25=33.75

h=\sqrt{33.75}\approx5.81

The height is 5.81 m. Keep the root unrounded until the last step.

Example 4

Coordinate distance

3 marks

Question

Find the distance between A(−2, 1) and B(4, 9).

The horizontal change is 4 − (−2) = 6; the vertical change is 9 − 1 = 8.

d^2=6^2+8^2=100

d=10

The distance is 10 units.

Example 5

Test a triangle

3 marks

Question

Is a triangle with sides 6 cm, 8 cm and 11 cm right-angled? Show a calculation.

The longest side is 11 cm.

6^2+8^2=36+64=100

11^2=121

Since 100 ≠ 121, it is not right-angled. Having one long side is not enough.

Example 6

A cuboid diagonal

Higher only4 marks

Question

A cuboid measures 3 cm by 4 cm by 12 cm. Find its space diagonal, joining opposite vertices.

First find the base diagonal d.

d^2=3^2+4^2=25

The base diagonal is perpendicular to the vertical height. Apply Pythagoras again to the space diagonal D.

D^2=d^2+12^2

D^2=25+144=169

D=13

The space diagonal is 13 cm.

10 original questions · total 30 marks

Pythagoras' theorem GCSE exam-style questions

Try each question before opening its fully worked answer. The difficulty rises through the set.

Before you startAllow about 35 minutes · Show your working. Give exact answers unless a question asks you to round, and include units where needed. · answers start collapsed

Identify the side

1 mark

In a right-angled triangle ABC, angle B is 90°. Which side is the hypotenuse?

Show worked answer

AC is opposite the right angle at B, so AC is the hypotenuse. AB and BC meet at the right angle.

Add squares

3 marks

The perpendicular sides are 5 cm and 12 cm. Find the hypotenuse.

Show worked answer

c^2=5^2+12^2=169

c=\sqrt{169}=13

The hypotenuse is 13 cm.

Subtract squares

3 marks

The hypotenuse is 25 cm and one shorter side is 7 cm. Find the other side.

Show worked answer

a^2+7^2=25^2

a^2=625-49=576

a=24

The missing side is 24 cm.

Round at the end

3 marks

Perpendicular sides are 4 cm and 7 cm. Find the hypotenuse to 3 significant figures.

Show worked answer

c^2=4^2+7^2=65

c=\sqrt{65}\approx8.06

The length is 8.06 cm.

Rectangle diagonal

3 marks

A rectangle is 8 m by 15 m. Find its diagonal.

Show worked answer

The rectangle's corner is a right angle.

d^2=8^2+15^2=289

d=17

The diagonal is 17 m.

Ladder height

3 marks

A 10 m ladder has its foot 6 m from a vertical wall on level ground. How high does it reach?

Show worked answer

h^2+6^2=10^2

h^2=100-36=64

h=8

It reaches 8 m vertically.

Distance between points

3 marks

Find the distance between (1, −2) and (7, 6).

Show worked answer

Changes are 6 horizontally and 8 vertically.

d^2=6^2+8^2=100

d=10

The distance is 10 units.

Check the converse

3 marks

Show that sides 8 cm, 15 cm and 17 cm form a right-angled triangle.

Show worked answer

8^2+15^2=64+225=289

17^2=289

The squares match, so the triangle is right-angled opposite its longest side, 17 cm.

Isosceles height

Harder4 marks

An isosceles triangle has equal sides 13 cm and base 10 cm. Find its perpendicular height.

Show worked answer

The altitude from the vertex bisects the base into two 5 cm lengths. In either right-angled half,

h^2+5^2=13^2

$h^2=169-25=144$

h=12

The height is 12 cm. Do not use the full 10 cm base in the half-triangle.

Space diagonal

Higher only4 marks

Find the space diagonal of a cuboid measuring 2 cm by 3 cm by 6 cm.

Show worked answer

First find the base diagonal d: $d^2=2^2+3^2=13$ That diagonal and the height form a second right-angled triangle. $D^2=d^2+6^2$ $D^2=13+36=49$

D=7

The space diagonal is 7 cm.

Examiner-style feedback

Common pythagoras' theorem mistakes

Using the wrong hypotenuse

It is opposite the right angle, not necessarily the sloping side in every orientation.

Always adding

Finding a shorter side requires subtracting its partner's square from the hypotenuse square.

Missing the square root

If c² = 169, then c = 13 for a positive length, not 169.

30-second recap

Choose, carry out, check

Translate the wording before you reach for a rule.

Check for a right angle.
The hypotenuse is opposite 90°.
Add or subtract squares according to the missing side.
Take a positive root and retain units.

Quick answers

Pythagoras' theorem FAQ

Can I use Pythagoras for any triangle?

No. The equality applies to right-angled triangles. A non-right triangle may need another relationship, such as the cosine rule.

Why is there no negative length answer?

The equation a² = 25 has roots ±5, but a physical side length is positive, so the length is 5.

Build connected skills

What to revise next

Understand the angles

Angle rules

Identify right angles and triangle relationships.

Revise angle rules

Use an angle and a side

Trigonometry

Choose sine, cosine or tangent for right-triangle problems.

Revise trigonometry

Content standards

Curriculum and rights review

G20: Pythagoras in 2D across tiers, with labelled Higher 3D applications. All side lengths are positive. Questions, suggested marks, solutions and diagrams are original practice material, not official past-paper questions.

Official specification references

AQA GCSE Maths subject content