What is Pythagoras Theorem and how is it used in Geometry?

Pythagoras Theorem is a fundamental principle in Geometry that states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. This theorem is used to calculate the length of one side of a right-angled triangle when the lengths of the other two sides are known. It is a crucial concept in the Cambridge IGCSE Mathematics curriculum.

How can I apply Pythagoras Theorem to solve real-world problems?

Pythagoras Theorem can be applied to various real-world problems, such as determining the distance between two points on a plane, calculating the height of a building using a ladder, or finding the diagonal of a rectangular object. By identifying right-angled triangles in these scenarios, you can use the theorem to find unknown lengths, making it a practical tool in fields like architecture, engineering, and navigation.

What are the conditions necessary for using Pythagoras Theorem?

To use Pythagoras Theorem, the triangle in question must be a right-angled triangle, meaning it has one angle that is exactly 90 degrees. Additionally, you need to know the lengths of at least two sides of the triangle to calculate the third side. This theorem is only applicable in Euclidean geometry, which is the type of geometry typically studied in the Cambridge IGCSE Mathematics curriculum.

Can Pythagoras Theorem be used in non-right-angled triangles?

Pythagoras Theorem is specifically applicable to right-angled triangles. For non-right-angled triangles, other methods such as the Law of Cosines or the Law of Sines are used to find unknown sides or angles. These laws are extensions of Pythagorean principles and are also covered in the Cambridge IGCSE Mathematics curriculum to handle a wider range of geometric problems.

What are some common mistakes to avoid when using Pythagoras Theorem?

Common mistakes include applying the theorem to non-right-angled triangles, incorrectly identifying the hypotenuse, and arithmetic errors when squaring and adding the side lengths. It is important to ensure that the triangle is right-angled and to correctly label the sides before performing calculations. Double-checking your work can help avoid these errors, which is a good practice in any mathematical problem-solving, including in the Cambridge IGCSE Mathematics exams.

Why is "Identifying the wrong side as the hypotenuse" a common mistake in Pythagoras Theorem?

Why it happens: Students label the longest given side as c even when they're trying to find it. How to avoid it: The hypotenuse is opposite the right angle — find that vertex first. Source: 0580/22 — recurring

Why is "Adding when finding a leg (should subtract)" a common mistake in Pythagoras Theorem?

Why it happens: Memorising a^2 + b^2 = c^2 as a single line of arithmetic. How to avoid it: If finding a LEG, rearrange first: a^2 = c^2 - b^2.

Why is "Forgetting the final square root" a common mistake in Pythagoras Theorem?

Why it happens: Stopping at c^2 = 169 and writing c = 169. How to avoid it: Always: square the legs, sum, take square root.

Why is "Doing only one Pythagoras step in a 3D problem" a common mistake in Pythagoras Theorem?

Why it happens: Students compute the base diagonal but forget to combine with the height. How to avoid it: 3D body diagonal needs TWO Pythagoras applications (or the cuboid shortcut).

GeometryCambridge IGCSE Maths 0580 Extended

Pythagoras Theorem

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Pythagoras' Theorem — Cambridge IGCSE 0580 Maths Extended (2026)

$a^{2} + b^{2} = c^{2}$ . The relationship between the three sides of a right-angled triangle. Find any side from the other two, then check for right-angledness with the converse.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E4.3 — Apply Pythagoras' theorem to right-angled triangles in 2D and 3D.

The theorem and its uses

Square the two legs, sum, square root for the hypotenuse. Subtract for a leg.

Pythagoras. In any right-angled triangle with legs $a$ , $b$ and hypotenuse $c$ : $a^{2} + b^{2} = c^{2}.$

Find the hypotenuse. $c = \sqrt{a^{2} + b^{2}}.$

Worked. Right triangle with legs $3$ and $4$ . Find the hypotenuse.

$c = \sqrt{9 + 16} = \sqrt{25} = 5$ .

The hypotenuse c is the longest side, always opposite the right angle.

Find a leg. $a = \sqrt{c^{2} - b^{2}}.$

Worked. Right triangle with hypotenuse $13$ and one leg $5$ . Find the other leg.

$a = \sqrt{169 - 25} = \sqrt{144} = 12$ .

Always identify the hypotenuse first. It's the SIDE OPPOSITE the right angle, and it's always the LONGEST side.

Hypotenuse: opposite the right angle, longest side.
$c^{2} = a^{2} + b^{2}$ .
Find a leg: subtract from $c^{2}$ , then square root.
Both legs are interchangeable.

Pythagorean triples

Whole-number sets satisfying the equation. Memorise the common ones.

A Pythagorean triple is a set $(a, b, c)$ of positive integers with $a^{2} + b^{2} = c^{2}$ .

The classics:

$(3, 4, 5)$ .
$(5, 12, 13)$ .
$(8, 15, 17)$ .
$(7, 24, 25)$ .

Multiples are also triples. $(6, 8, 10)$ comes from doubling $(3, 4, 5)$ . $(9, 12, 15)$ from tripling, etc.

Why they help. Spotting a triple means you don't need a calculator — the answer is exact, no surds.

Worked. A right triangle has legs $9$ and $12$ . Find the hypotenuse.

Spot: $9$ and $12$ are $3 \times 3$ and $3 \times 4$ . So scaled $(3, 4, 5)$ → hypotenuse is $3 \times 5 = 15$ .
Check: $9^{2} + 12^{2} = 81 + 144 = 225 = 15^{2}$ ✓.

Memorise: $(3, 4, 5)$ , $(5, 12, 13)$ , $(8, 15, 17)$ .
Multiples count too.
Spotting a triple saves calculator time.

The converse — proving right-angledness

If $a^{2} + b^{2} = c^{2}$ in a triangle, the triangle MUST be right-angled.

Converse of Pythagoras. Given a triangle with sides $a, b, c$ (with $c$ the longest). If $a^{2} + b^{2} = c^{2}$ , then the triangle is right-angled at the vertex opposite $c$ .

Worked. Is the triangle with sides $5, 12, 13$ right-angled?

$5^{2} + 12^{2} = 25 + 144 = 169 = 13^{2}$ ✓.
Yes, right-angled at the vertex opposite the side of length $13$ .

Worked. Is the triangle with sides $4, 5, 6$ right-angled?

$4^{2} + 5^{2} = 16 + 25 = 41$ . $6^{2} = 36$ .
$41 \ne 36$ . Not right-angled.

Use case. Sometimes Cambridge gives you three sides and asks "Show that the triangle is right-angled" — the converse is the only proof.

Identify the LONGEST side as $c$ .
Check if $a^{2} + b^{2} = c^{2}$ .
Equal → right-angled. Not equal → not right-angled.
Right angle is OPPOSITE the longest side (the hypotenuse).

Pythagoras in 3D

For a cuboid space diagonal, use two Pythagoras steps. Or the 3D Pythagoras formula.

Two-step method. To find the space diagonal of a cuboid with dimensions $l, w, h$ :

Find the diagonal of the base: $d_{\text{base}} = \sqrt{l^{2} + w^{2}}$ .
Combine with the height: $d_{\text{space}} = \sqrt{d_{\text{base}}^{2} + h^{2}} = \sqrt{l^{2} + w^{2} + h^{2}}$ .

3D Pythagoras formula (which is just the two steps combined): $d = \sqrt{l^{2} + w^{2} + h^{2}}.$

Worked. A cuboid measures $3 \times 4 \times 12$ . Find the length of the space diagonal.

$d = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$ .

Find the base diagonal first, then combine it with the height for the space diagonal.

Tip. $3, 4, 12, 13$ is a popular cuboid Cambridge example because the answer is integer.

3D Pythagoras: $d = \sqrt{l^{2} + w^{2} + h^{2}}$ .
Or apply 2D Pythagoras twice.
Always identify the base diagonal first if going step by step.

How it’s examined

Pythagoras appears on every paper. Paper 2: 1-2 mark single-step questions. Paper 4: 3-5 mark questions combining 2D Pythagoras with trig, or applying 3D Pythagoras inside a cuboid problem. Examiner reports flag misidentifying the hypotenuse and using $a^{2} - b^{2}$ when $a^{2} + b^{2}$ is needed (or vice versa).

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E4.3); 0580/22 May/Jun 2024 — Q11 (Pythagoras); 0580/42 Oct/Nov 2024 — Q10 (3D Pythagoras); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

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Step-by-step worked examples — Pythagoras Theorem

Step-by-step solutions to past-paper-style questions on pythagoras theorem, written exactly the way a tutor would explain them at the board.

1Find the hypotenuse
Core• hypotenuse
Question
A right-angled triangle has legs $7$ cm and $24$ cm. Find the hypotenuse.
Step-by-step solution
1. Step 1
  $c^{2} = a^{2} + b^{2}$ .
  $c^{2} = 7^{2} + 24^{2} = 49 + 576 = 625$
2. Step 2
  Square root.
  $c = 25$
Answer
$25$ cm
2Find a missing leg
Core• Adapted from 0580/22 May/Jun 2024 Q3• leg
Question
The hypotenuse is $13$ cm and one leg is $5$ cm. Find the other leg.
Step-by-step solution
1. Step 1
  Rearrange.
  $b^{2} = 13^{2} - 5^{2} = 169 - 25 = 144$
2. Step 2
  Square root.
  $b = 12$
Answer
$12$ cm
3Pythagoras with decimals
Extended• decimals
Question
Find the hypotenuse if the legs are $4.5$ cm and $6$ cm. Give your answer to $3$ s.f.
Step-by-step solution
1. Step 1
  Apply Pythagoras.
  $c^{2} = 4.5^{2} + 6^{2} = 20.25 + 36 = 56.25$
2. Step 2
  Square root.
  $c = 7.5$
Answer
$7.50$ cm (3 s.f.)
4Distance between two points
Extended• coordinates
Question
Use Pythagoras to find the distance from $A(2, 3)$ to $B(11, 9)$ .
Step-by-step solution
1. Step 1
  Differences.
  $\Delta x = 9,\ \Delta y = 6$
2. Step 2
  Apply Pythagoras.
  $d^{2} = 81 + 36 = 117$
3. Step 3
  Square root.
  $d = \sqrt{117} \approx 10.8$
Answer
$\sqrt{117} \approx 10.8$ units (3 s.f.)
5Pythagoras in 3D
Extended• Adapted from 0580/42 Oct/Nov 2024 Q15• 3D
Question
A cuboid has edges $3$ , $4$ and $12$ cm. Find the length of the longest internal diagonal.
Step-by-step solution
1. Step 1
  First find the diagonal of the base.
  $d^{2} = 3^{2} + 4^{2} = 25 \implies d = 5$
2. Step 2
  Now use Pythagoras with this diagonal and the height as legs.
  $D^{2} = 5^{2} + 12^{2} = 169 \implies D = 13$
Answer
$13$ cm
Examiner tip
For a cuboid with edges $a, b, c$ the body diagonal is $\sqrt{a^{2} + b^{2} + c^{2}}$ . Useful shortcut.
6Ladder against a wall (real-world)
Extended• Adapted from 0580/22 Oct/Nov 2023 Q12• word problem, ladder
Question
A ladder of length $4.2$ m rests against a vertical wall. The foot of the ladder is $1.5$ m from the base of the wall. How high up the wall does the ladder reach? Give your answer to $3$ s.f.
Step-by-step solution
1. Step 1
  The ladder is the hypotenuse; the distance from the wall is one leg and the height up the wall is the unknown leg.
  $h^{2} = 4.2^{2} - 1.5^{2}$
2. Step 2
  Compute the squares.
  $h^{2} = 17.64 - 2.25 = 15.39$
3. Step 3
  Square root.
  $h = \sqrt{15.39} \approx 3.92\,\text{m}$
Answer
$h \approx 3.92$ m (3 s.f.)
Examiner tip
The 2023 examiner report flags candidates who add instead of subtract: the ladder is the longest side, so its square must be reduced, not increased, to find a leg.
7Find the height of an isosceles triangle
Extended• Adapted from 0580/42 Feb/Mar 2023 Q9• isosceles, height
Question
An isosceles triangle has two equal sides of $10$ cm and a base of $12$ cm. Find its vertical height. Give your answer in exact form.
Step-by-step solution
1. Step 1
  Drop a perpendicular from the apex to the base. It bisects the base, creating two congruent right-angled triangles each with hypotenuse $10$ and one leg $6$ .
2. Step 2
  Apply Pythagoras to find the height $h$ .
  $h^{2} = 10^{2} - 6^{2} = 100 - 36 = 64$
3. Step 3
  Square root.
  $h = 8\,\text{cm}$
Answer
$h = 8$ cm
Examiner tip
The perpendicular from the apex bisects the base only because the triangle is isosceles. Forgetting to halve the base (using $12$ instead of $6$ ) is the most common slip in this question type.
8Decide whether a triangle is right-angled (converse)
Core• Adapted from 0580/12 May/Jun 2022 Q7• converse
Question
A triangle has sides $9$ cm, $12$ cm and $15$ cm. Show that it is right-angled.
Step-by-step solution
1. Step 1
  The longest side ( $15$ ) is the candidate hypotenuse. Check whether the sum of squares of the other two equals its square.
  $9^{2} + 12^{2} = 81 + 144 = 225$
2. Step 2
  Compare with the square of the longest side.
  $15^{2} = 225$
3. Step 3
  Since $9^{2} + 12^{2} = 15^{2}$ , the converse of Pythagoras tells us the triangle is right-angled, with the right angle opposite the side of length $15$ .
Answer
Right-angled, since $9^{2} + 12^{2} = 15^{2} = 225$ .
Examiner tip
State the conclusion explicitly with reference to the converse — mark schemes reward the words 'right-angled triangle' alongside the numerical check.
9Multi-step: perimeter and cost
Challenge• Adapted from 0580/42 Oct/Nov 2022 Q4• multi-step, perimeter, cost
Question
A rectangular plot of land measures $24$ m by $7$ m. A straight diagonal path is built between two opposite corners. Fencing is then put around the entire perimeter of the rectangle and along the diagonal path. The fencing costs $12.50 per metre. Find the total cost of the fencing.
Step-by-step solution
1. Step 1
  Find the diagonal using Pythagoras.
  $d^{2} = 24^{2} + 7^{2} = 576 + 49 = 625$
2. Step 2
  Square root.
  $d = 25\,\text{m}$
3. Step 3
  Total length of fencing = perimeter + diagonal $= 2(24 + 7) + 25$ .
  $L = 62 + 25 = 87\,\text{m}$
4. Step 4
  Multiply by the unit cost.
  $\text{Cost} = 87 \times 12.50 = 1087.50$
Answer
$1{,}087.50
Examiner tip
A* candidates often lose the final mark by forgetting to add the perimeter and the diagonal together before multiplying. Lay the calculation out in clear stages.
10Diagonal of a rectangle
Core• Adapted from 0580/12 Feb/Mar 2023 Q9• rectangle, diagonal
Question
A rectangle has length $15$ cm and width $8$ cm. Find the length of its diagonal.
Step-by-step solution
1. Step 1
  The diagonal is the hypotenuse of a right-angled triangle whose legs are the length and width.
  $d^{2} = 15^{2} + 8^{2}$
2. Step 2
  Compute.
  $d^{2} = 225 + 64 = 289$
3. Step 3
  Square root.
  $d = \sqrt{289} = 17\,\text{cm}$
Answer
$d = 17$ cm
Examiner tip
Treat a rectangle as two right-angled triangles split along the diagonal — a useful pattern that recurs in many 0580 paper 2 questions.

Key Formulae — Pythagoras Theorem

The formulae you need to memorise for pythagoras theorem on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Pythagoras' theorem
$a^{2} + b^{2} = c^{2}$
$a, b$
legs of the right-angled triangle
$c$
hypotenuse (side opposite the right angle)
When to use
Whenever you have a right-angled triangle and need to find a side.
Cuboid body diagonal
$D = \sqrt{a^{2} + b^{2} + c^{2}}$
$a, b, c$
edge lengths
When to use
Direct shortcut for the longest internal diagonal of a cuboid.

Key Definitions and Keywords — Pythagoras Theorem

Definitions to memorise and the exact keywords mark schemes credit for pythagoras theorem answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Hypotenuse
Examiner keyword
The longest side of a right-angled triangle — the one opposite the right angle.
Legs (of a right-angled triangle)
The two shorter sides forming the right angle.
Converse of Pythagoras
If $a^{2} + b^{2} = c^{2}$ holds for the three sides, the triangle is right-angled.
Example
$3, 4, 5$ satisfy $9 + 16 = 25$ , so this is a right-angled triangle.

Common Mistakes and Misconceptions — Pythagoras Theorem

The traps other students keep falling into on pythagoras theorem questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Identifying the wrong side as the hypotenuse
0580/22 — recurring
Why it happens
Students label the longest given side as $c$ even when they're trying to find it.
How to avoid it
The hypotenuse is opposite the right angle — find that vertex first.
✕Adding when finding a leg (should subtract)
Why it happens
Memorising $a^{2} + b^{2} = c^{2}$ as a single line of arithmetic.
How to avoid it
If finding a LEG, rearrange first: $a^{2} = c^{2} - b^{2}$ .
✕Forgetting the final square root
Why it happens
Stopping at $c^{2} = 169$ and writing $c = 169$ .
How to avoid it
Always: square the legs, sum, take square root.
✕Doing only one Pythagoras step in a 3D problem
Why it happens
Students compute the base diagonal but forget to combine with the height.
How to avoid it
3D body diagonal needs TWO Pythagoras applications (or the cuboid shortcut).

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Pythagoras Theorem

Short Study Notes in the form of Slides

Short Study Notes — Pythagoras Theorem

Detailed Notes

What you’ll learn

How it’s examined

1Find the hypotenuse

2Find a missing leg

3Pythagoras with decimals

4Distance between two points

5Pythagoras in 3D

6Ladder against a wall (real-world)

7Find the height of an isosceles triangle

8Decide whether a triangle is right-angled (converse)

9Multi-step: perimeter and cost

10Diagonal of a rectangle

Pythagoras' theorem

Cuboid body diagonal

Hypotenuse

Legs (of a right-angled triangle)

Converse of Pythagoras

✕Identifying the wrong side as the hypotenuse

✕Adding when finding a leg (should subtract)

✕Forgetting the final square root

✕Doing only one Pythagoras step in a 3D problem

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Pythagoras Theorem — frequently asked questions