1MA1 · G20Geometry & MeasuresFoundation & HigherGrades 3–6

Pythagoras’ Theorem

In any right-angled triangle, the square on the longest side equals the sum of the squares on the other two: $a^{2} + b^{2} = c^{2}$ . That single relationship finds missing lengths in ladders against walls, diagonals of rectangles, distances on maps and coordinate grids — and, at Higher tier, lengths inside 3D shapes.

45 min Pythagoras & Trigonometry

The big picture

The longest side, opposite the right angle, is the $\textbf{hypotenuse}$ . Every Pythagoras question is one of two types: find the hypotenuse (add the squares) or find a shorter side (subtract the squares). Identifying the hypotenuse first decides which.

Pythagoras is the engine behind distance in coordinate geometry, the equation of a circle, vector magnitudes and 3D diagonals. Edexcel often hides it inside a larger problem, so spotting the right-angled triangle is the real skill.

What you'll learn

Use Pythagoras’ theorem to find the hypotenuse or a shorter side
Solve problems in context, including checking whether a triangle is right-angled
Find the distance between two points on a coordinate grid
Use Pythagoras in 3D shapes (Higher)

The theorem

For a right-angled triangle with hypotenuse $c$ and shorter sides $a$ and $b$ : $a^{2} + b^{2} = c^{2}$ .

$\textbf{Finding the hypotenuse}$ : square the two shorter sides, add, then square root.

$\textbf{Finding a shorter side}$ : square the hypotenuse, subtract the square of the known shorter side, then square root.

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

c is the hypotenuse, opposite the right angle.

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 = \sqrt{23.04}

Answer4.8 m

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

A 13 m ladder leans against a wall with its foot 5 m from the wall. How high up the wall does it reach, in metres?

Your answer

Show worked solution

Solution

1 $h^2 = 13^2 - 5^2 = 144$ .
2 $h = 12$ m.

Answer: $12$ m

Tip — Sense check: the hypotenuse must be the longest side. If your "shorter side" comes out longer than the hypotenuse, you added instead of subtracting.

Right-angled or not?

The converse of Pythagoras is also true: if the side lengths satisfy $a^{2} + b^{2} = c^{2}$ , the triangle has a right angle opposite $c$ .

Triples of whole numbers that work, like $3, 4, 5$ and $5, 12, 13$ , are called Pythagorean triples. Multiples also work: $6, 8, 10$ .

Example 1A triangle has sides 9 cm, 12 cm and 16 cm. Is it right-angled?

9^{2} + 12^{2} = 81 + 144 = 225

16^{2} = 256 \neq 225

AnswerNo — the squares of the two shorter sides do not add to the square of the longest.

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

Is a triangle with sides 8 cm, 15 cm and 17 cm right-angled? Type yes or no.

Your answer

Show worked solution

Solution

1 $8^2 + 15^2 = 64 + 225 = 289$ .
2 $17^2 = 289$ , so yes.

Answer: yes

Distance between points

The line between two points is the hypotenuse of a right-angled triangle whose other sides are the horizontal and vertical differences.

So the distance between $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ .

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

Distance between two points.

Example 1Find the distance between

(-2, 3)

and

(4, 11)

1Differences:

6

across and

8

up.

d = \sqrt{36 + 64}

Answer10 units

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

Find the distance between $(1, 2)$ and $(8, -3)$ , to 1 decimal place.

Your answer

To 1 decimal place.

Show worked solution

Solution

1Changes: 7 and $-5$ .
2 $\sqrt{49 + 25} = \sqrt{74} = 8.60\ldots$

Answer: $8.6$

Why this works

Squaring removes signs, so it does not matter which point you subtract from which — the distance is always positive.

Pythagoras in 3D

To find the diagonal of a cuboid, use Pythagoras twice: first the diagonal of the base, then the triangle formed by that diagonal and the vertical edge.

Combined, the space diagonal of a cuboid with sides $a$ , $b$ , $c$ is $\sqrt{a^{2} + b^{2} + c^{2}}$ .

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

Diagonal of a cuboid.

Example 1A box measures 6 cm by 4 cm by 3 cm. Will a 7.5 cm pencil fit inside, lying along the longest diagonal?

1Base diagonal:

\sqrt{36 + 16} = \sqrt{52}

2Space diagonal:

\sqrt{52 + 9} = \sqrt{61} = 7.81

cm.

AnswerYes — the diagonal is 7.81 cm, longer than 7.5 cm.

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

Find the length of the space diagonal of a 5 cm by 4 cm by 3 cm cuboid, to 2 decimal places.

Your answer

Show worked solution

Solution

1 $\sqrt{25 + 16 + 9} = \sqrt{50}$ .
2 $= 7.07$ cm.

Answer: $7.07$ cm

Tip — Sketch the right-angled triangle you are using inside the 3D shape. Keep $\sqrt{52}$ exact between steps to avoid rounding errors.

Think like an examiner

Identify the hypotenuse before substituting.
Show the squares and the sum or difference before square rooting.
Round only at the end, and give units.

Pythagoras

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

Hypotenuse.

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

Shorter side.

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

Cuboid diagonal.

Watch out for these

Adding squares when finding a shorter side.

Subtract from the hypotenuse squared.

Forgetting to square root at the end.

c^{2} = 100

means

c = 10

Using Pythagoras in a triangle without a right angle.

It only applies to right-angled triangles.

Writing

\sqrt{36 + 64} = 6 + 8

Add first:

\sqrt{100} = 10

Stretch yourself

An isosceles triangle has two sides of 13 cm and a base of 10 cm. Find its area.

Hint — The line from the apex to the midpoint of the base is perpendicular to the base, creating two right-angled triangles.

Reveal solution

SolutionThe perpendicular height splits the base into two halves of 5 cm. In either right-angled triangle,

h^{2} = 13^{2} - 5^{2} = 169 - 25 = 144

, so

h = 12

cm. Area

= \tfrac12 \times 10 \times 12 = 60

cm².

\blacksquare

Questions students ask

How do I know which side is the hypotenuse?

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

Does Pythagoras work for any triangle?

Only right-angled triangles. For others, use the cosine rule.

Should I leave answers as surds?

Only if the question asks for an exact answer. Otherwise round to the accuracy requested.

Key takeaways

$a^{2} + b^{2} = c^{2}$ for right-angled triangles, with $c$ the hypotenuse.
Add squares for the hypotenuse; subtract for a shorter side.
The converse tests whether a triangle is right-angled.
Distance between points uses the horizontal and vertical differences.
In 3D, apply Pythagoras twice: $d = \sqrt{a^{2} + b^{2} + c^{2}}$ .

Exam-style questions

Practise it the way it is examined

3 original questions · 7 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 32 marks

A rectangle has a diagonal of 13 cm and a width of 5 cm. Work out its length.[2]
Show mark scheme
- M1 $13^2 - 5^2$
- A1 $12$ cm
Answer: $12$ cm

Question 2Grade 53 marks

An isosceles triangle has sides 10 cm, 10 cm and 16 cm. Work out its area.[3]
Show mark scheme
- M1Height $= \sqrt{100 - 64} = 6$
- M1 $\frac{1}{2} \times 16 \times 6$
- A1 $48$ cm²
Answer: $48$ cm²

Question 3Grade 62 marks

Work out the distance between $(-3, 4)$ and $(5, -2)$ .[2]
Show mark scheme
- M1Differences 8 and 6
- A1 $10$
Answer: $10$

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