Pythagoras' Theorem
Pythagoras' theorem is one of the most important rules in GCSE and IGCSE Maths: in any right-angled
triangle, the square of the hypotenuse equals the sum of the squares of the other two sides
(a² + b² = c²). This page shows you how to find a missing side in a right-angled triangle,
how to check whether a triangle is right-angled, and how to apply the theorem to real problems involving
distances, heights and diagonals. Work through the worked examples, then use the free, randomly generated,
auto-marked practice questions below.
\(x\)
📚
Prior Knowledge
Before starting, be comfortable squaring numbers and finding
square roots,
and with
rearranging a simple equation
(you will rearrange \(a^2 + b^2 = c^2\) to make a shorter side the subject).
What Is Pythagoras' Theorem?
In any right-angled triangle, the square of the hypotenuse (the longest side, opposite the right angle) equals the sum of the squares of the other two sides:
\[ a^2 + b^2 = c^2 \]
Here \(c\) is the hypotenuse. Use the rule to find a missing side whenever you know the other two sides of a right-angled triangle, then round your answer to 3 significant figures.
Core Ideas
Right angle first
Only works in right-angled triangles. Identify the right angle before doing anything else.
Hypotenuse
Always the longest side and always opposite the right angle. In the formula, \(c\) is the hypotenuse.
Finding the hypotenuse
Square both shorter sides, add them, then square root: \(c = \sqrt{a^2 + b^2}\).
Finding a shorter side
Rearrange first: \(a = \sqrt{c^2 - b^2}\). Subtract, then square root.
Common Pythagorean Triples
A few right-angled triangles have whole-number sides. These are called Pythagorean triples, and learning to spot them lets you write the answer instantly and recognise a right angle without reaching for a calculator.
3, 4, 5\(3^2 + 4^2 = 5^2\)
5, 12, 13\(5^2 + 12^2 = 13^2\)
8, 15, 17\(8^2 + 15^2 = 17^2\)
7, 24, 25\(7^2 + 24^2 = 25^2\)
Tip: any multiple of a triple is also a triple. Scaling 3, 4, 5 by 2 gives 6, 8, 10, and by 3 gives 9, 12, 15. So whenever a triangle's sides are a multiple of a triple you know, it is right-angled.
Worked Examples
💡 Finding the Hypotenuse
Find side \(c\) in a right-angled triangle where \(a = 3\) and \(b = 4\).
1
Write the formula\[c^2 = a^2 + b^2\]
2
Substitute the values\[\begin{aligned} c^2 &= 3^2 + 4^2 \\ &= 9 + 16 \\ &= 25 \end{aligned}\]
3
Square root to find \(c\)\[\begin{aligned} c &= \sqrt{25} \\ &= 5 \end{aligned}\]
💡 Rounding to 3 Significant Figures
Find side \(c\) where \(a = 6\) and \(b = 9\). Give your answer to 3 significant figures.
1
Substitute the values\[\begin{aligned} c^2 &= 6^2 + 9^2 \\ &= 36 + 81 \\ &= 117 \end{aligned}\]
2
Square root (keep the surd)\[c = \sqrt{117}\]
3
Round only at the end\[c = 10.8 \text{ (3 s.f.)}\]
💡 Finding a Shorter Side
Find side \(a\) in a right-angled triangle where \(b = 5\) and \(c = 13\).
1
Rearrange for \(a^2\)\[a^2 = c^2 - b^2\]
2
Substitute the values\[\begin{aligned} a^2 &= 13^2 - 5^2 \\ &= 169 - 25 \\ &= 144 \end{aligned}\]
3
Square root to find \(a\)\[\begin{aligned} a &= \sqrt{144} \\ &= 12 \end{aligned}\]
💡 Is the Triangle Right-Angled?
A triangle has sides 6, 8 and 10. Is it right-angled?
1
Test: does \(a^2 + b^2 = c^2\)?Use the two shorter sides for \(a\) and \(b\), and the longest for \(c\).
2
Calculate both sides\[\begin{aligned} 6^2 + 8^2 &= 36 + 64 \\ &= 100 \\ 10^2 &= 100 \end{aligned}\]
3
ConcludeBoth sides equal 100, so yes, it is right-angled. If they were not equal, it would not be.
💡 Real Problem: Ladder Against a Wall
A 10 m ladder leans against a wall with its foot 6 m from the wall. How high up the wall does it reach?
1
Identify the sidesThe ladder is the hypotenuse (\(c = 10\)). The foot distance is a shorter side (\(b = 6\)). Find the height \(a\).
2
Rearrange and substitute\[\begin{aligned} a^2 &= 10^2 - 6^2 \\ &= 100 - 36 \\ &= 64 \end{aligned}\]
3
Square root for the answer\[\begin{aligned} a &= \sqrt{64} \\ &= 8 \text{ m} \end{aligned}\]
💡 Isosceles Triangle: Perpendicular Height
An isosceles triangle has a base of 10 cm and two equal sides of 13 cm. Find its perpendicular height.
1
Split it into two right trianglesDrop a perpendicular from the top vertex to the base. It bisects the base, giving two matching right-angled triangles, each with a short side of \(10 \div 2 = 5\) cm and a hypotenuse of 13 cm.
2
Find the height (a shorter side)\[\begin{aligned} h^2 &= 13^2 - 5^2 \\ &= 169 - 25 \\ &= 144 \end{aligned}\]
3
Square root for the answer\[\begin{aligned} h &= \sqrt{144} \\ &= 12 \text{ cm} \end{aligned}\]
💡 Distance Between Two Points
Find the distance between the points \((-2,\ 1)\) and \((3,\ 13)\).
1
The legs are the coordinate differencesThe horizontal leg is \(3 - (-2) = 5\) and the vertical leg is \(13 - 1 = 12\). A negative difference does not matter, because squaring makes it positive.
2
Apply the theorem\[\begin{aligned} d^2 &= 5^2 + 12^2 \\ &= 25 + 144 \\ &= 169 \end{aligned}\]
3
Square root for the answer\[\begin{aligned} d &= \sqrt{169} \\ &= 13 \end{aligned}\]
🔑 Key Points
- Only works in right-angled triangles.
- The hypotenuse \(c\) is always opposite the right angle (the longest side).
- Finding \(c\): square, add, square root.
- Finding \(a\) or \(b\): square, subtract, square root.
- Keep the surd, then round to 3 significant figures at the very end.
⚠️ Common Pitfalls
- Adding when you should subtract (finding a shorter side).
- Using the wrong side as the hypotenuse.
- Forgetting to square root at the end.
- Rounding too early, which throws off the final answer.
