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 $\textbf{hypotenuse}$ — 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 Cognito AQA GCSE Maths

Before you start

Square positive numbers, find a square root and distinguish length from area.

Try this first. If you need the answer, revisit the prerequisite before continuing.

Starting check: A right-angled triangle is rotated on the page. How do you identify its hypotenuse?

Show answer and reasoning

Find the right angle. The side opposite it is the hypotenuse, regardless of the drawing’s orientation; it is the longest side.

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, $\textbf{add}$ , square root. Finding a shorter side: square, $\textbf{subtract}$ , square root. A quick sense check — the hypotenuse must be the longest side — tells you which one you need.

GCSE Maths - Pythagoras' Theorem And How To Use It (2026/27 exams)Curated by Kepler, not produced by us. Watch it, then read the notes below.

Open on YouTube

Watch, pause, try

Video by Cognito. The video introduces right-angled triangles. The application below adds a length comparison and a reasonableness check.

Viewing focus: Locate the right angle and name the opposite side before choosing whether to add or subtract squares.

After watching, attempt these questions before opening the answers. You can also use the written lesson below.

After watching 1: The shorter sides are 9 cm and 12 cm. Find the hypotenuse.

Show answer and reasoning

c² = 9² + 12² = 225, so c = 15 cm.

After watching 2: The hypotenuse is 13 m and one shorter side is 5 m. Find the other shorter side.

Show answer and reasoning

b² = 13² − 5² = 144, so b = 12 m. Adding would produce a side longer than the hypotenuse.

KEPLER VISUAL LAB

Explore the idea

Predict what will change, then use the controls to test your explanation.

Vertical side a3 cmHorizontal side b4 cm

3² + 4² = 25 cm², so c = √25 = 5 cm. The hypotenuse is opposite the right angle and is longer than either shorter side.

On a narrow screen, scroll the diagram sideways to read every label.

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.

Example 1A right-angled triangle has shorter sides 7 cm and 11 cm. Find the hypotenuse to 1 d.p.

c^2 = 7^2 + 11^2 = 49 + 121 = 170

c = \sqrt{170} = 13.038\ldots

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

Show worked solution

Solution

1 $c^2 = 81 + 144 = 225$ .
2 $c = \sqrt{225} = 15$ cm.

Answer: $15$ cm

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.

Example 1A right-angled triangle has hypotenuse 17 cm and one side 8 cm. Find the other side.

a^2 = 17^2 - 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

Show hint

The hypotenuse is given, so subtract.

Show worked solution

Solution

1 $a^2 = 400 - 156.25 = 243.75$ .
2 $a = \sqrt{243.75} = 15.61\ldots$ , so $15.6$ cm.

Answer: $15.6$ cm

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.

Example 1A 5 m ladder leans against a wall with its foot 1.4 m from the wall. How high up the wall does it reach?

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

Show worked solution

Solution

1Horizontal change $6$ , vertical change $8$ .
2 $d = \sqrt{36 + 64} = \sqrt{100} = 10$ .

Answer: $10$

Example 2An isosceles triangle has sides 10 cm, 10 cm and 12 cm. Find its area.

1The height splits the base into two 6 cm halves:

h^2 = 10^2 - 6^2 = 64

, so

h = 8

2Area

= \frac{1}{2} \times 12 \times 8

Answer

48

cm²

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

Show worked solution

Solution

1Half the base is 3 cm: $h^2 = 36 - 9 = 27$ .
2 $h = \sqrt{27} = 5.196\ldots$ , so $5.2$ cm.

Answer: $5.2$ cm

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 $d = \sqrt{a^2 + b^2 + c^2}$ .

Example 1A cuboid is 6 cm by 4 cm by 3 cm. Find the length of its longest diagonal, to 1 d.p.

1Base diagonal:

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

2Space diagonal:

\sqrt{52 + 3^2} = \sqrt{61} = 7.81\ldots

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

Show worked solution

Solution

1 $d = \sqrt{12^2 + 4^2 + 3^2} = \sqrt{144 + 16 + 9} = \sqrt{169}$ .
2 $= 13$ cm.

Answer: $13$ cm

Keep it exact between steps

$\sqrt{a^2 + b^2 + c^2}$ is just Pythagoras applied twice. Squaring the base diagonal undoes its square root, so $\sqrt{52}$ 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 = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Distance between two points.

d = \sqrt{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 = \sqrt{49 + 121}

Forgetting to square root at the end:

c = 170

c = \sqrt{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 $2n + 1$ , $2n^2 + 2n$ and $2n^2 + 2n + 1$ , 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.

Reveal solution

Solution

(2n + 1)^2 + (2n^2 + 2n)^2 = 4n^2 + 4n + 1 + 4n^4 + 8n^3 + 4n^2 = 4n^4 + 8n^3 + 8n^2 + 4n + 1

. And

(2n^2 + 2n + 1)^2 = 4n^4 + 8n^3 + 8n^2 + 4n + 1

. They are equal, so by the converse of Pythagoras the triangle is right-angled. (

n = 1

gives 3, 4, 5;

n = 2

gives 5, 12, 13.)

\blacksquare

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 $\sqrt{a^2 + b^2 + c^2}$ .

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]
Show mark scheme
- M1 $24^2 + 10^2$ or $576 + 100$
- M1 $\sqrt{676}$
- A1 $26$ cm
Answer: $26$ cm

Question 2Grade 53 marks

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

Work out the length of $AB$ .[3]
Show mark scheme
- M1Horizontal and vertical differences 8 and 6
- M1 $\sqrt{8^2 + 6^2}$
- A1 $10$
Answer: $10$

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]
Show mark scheme
- M1Base diagonal $= \sqrt{8^2 + 8^2} = \sqrt{128}$
- M1Half diagonal $= \sqrt{32}$ (5.656…)
- M1 $h^2 = 9^2 - 32 = 49$
- A1 $h = 7.00$ cm
Answer: $7.00$ cm

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Complete the method

Some of the setup is done. Finish it on paper, then compare your reasoning.

Your turn: A 10 m ladder reaches a point 8 m above level ground. Complete: distance from wall squared = ___ − ___. Find that distance.

Show a hint

The ladder is the hypotenuse.

Show answer and reasoning

10² − 8² = 36. The foot is 6 m from the wall, assuming a vertical wall and a right angle at the ground.

Spot and repair the mistake

Identify the first incorrect step and explain how to fix it.

Diagnose it: A student claims a triangle with sides 6, 8 and 11 is right-angled because 6 + 8 exceeds 11. Is that enough?

Show answer and reasoning

No. The inequality checks whether the lengths can form a triangle. For a right angle, test the longest side: 6² + 8² = 100, while 11² = 121. They differ, so it is not right-angled.

Apply it in an exam

Write a full response before revealing the solution. These are Kepler-authored exam-style questions; the suggested marks are a self-check guide.

Question 1 · 4 marks: A rectangular lawn is 18 m by 24 m. A gardener walks diagonally between opposite corners rather than along two edges. How much shorter is the diagonal route?

Show answer and reasoning

Diagonal = √(18² + 24²) = √900 = 30 m. The edge route is 18 + 24 = 42 m, so the saving is 12 m.

Identify the diagonal as the hypotenuse.
Use the sum of squared side lengths.
Find a diagonal length of 30 m.
Compare with the 42 m edge route to obtain 12 m.

Explore further

AQA 8300 specification (opens in a new tab)
Check the assessed content, tier requirements and official course documents.
Corbettmaths: Pythagoras practice (opens in a new tab)
More right-triangle calculations and questions that test whether a triangle is right-angled.

Recall and check later

Return with the notes closed. Answer from memory, check the reasoning, then practise anything you missed.

Recall 1: Tomorrow: a right-angled triangle has hypotenuse 17 cm and a shorter side 8 cm. Find the third side.

Show answer and reasoning

√(17² − 8²) = √225 = 15 cm.

Recall 2: In a week: why is the answer to a squared-length calculation not yet a length?

Show answer and reasoning

Squaring gives an area-sized quantity in square units. Take the positive square root to recover the physical length.

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

Hypotenuse of a triangle with shorter sides 6 and 8?
- $10$
- $14$
- $100$
- $7$
Show answer and explanation
Answer: $10$
$\sqrt{36 + 64}$ .
Which side is the hypotenuse?
- Opposite the right angle
- Next to the right angle
- The shortest side
- The base
Show answer and explanation
Answer: Opposite the right angle
It is always the longest side.
Hypotenuse 13, one side 5. The other side is…
- $12$
- $8$
- $13.9$
- $18$
Show answer and explanation
Answer: $12$
$\sqrt{169 - 25}$ .
Shorter sides 4 cm and 7 cm. Hypotenuse (1 d.p.)?
- 8.1 cm
- 11.0 cm
- 5.7 cm
- 65.0 cm
Show answer and explanation
Answer: 8.1 cm
$\sqrt{65} = 8.06$ .
Is a triangle with sides 7, 24, 25 right-angled?
- Yes
- No
- Only if drawn accurately
- Cannot tell
Show answer and explanation
Answer: Yes
$49 + 576 = 625 = 25^2$ .
Distance between $(0, 0)$ and $(5, 12)$ ?
- $13$
- $17$
- $7$
- $169$
Show answer and explanation
Answer: $13$
$\sqrt{25 + 144}$ .
A square has diagonal 10 cm. Its side length is…
- $\sqrt{50}$ cm
- 5 cm
- $\sqrt{10}$ cm
- 20 cm
Show answer and explanation
Answer: $\sqrt{50}$ cm
$2s^2 = 100$ .
Space diagonal of a 2 × 3 × 6 cuboid?
- $7$
- $11$
- $49$
- $\sqrt{13}$
Show answer and explanation
Answer: $7$
$\sqrt{4 + 9 + 36}$ .

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