In any right-angled triangle, the square on the longest side equals the two squares on the other sides added together. That single fact lets you find any one side from the other two, and it is the foundation of almost everything that comes later in geometry and trigonometry.
What each letter means
| Symbol | Meaning |
|---|---|
| c | The hypotenuse — the side opposite the right angle, and always the longest of the three. |
| a and b | The two shorter sides, called the legs. It does not matter which you call which; swapping them changes nothing. |
The one thing you must get right is which side is the hypotenuse. It is not “the one at the bottom” or “the sloping one” — it is the side that does not touch the right angle. Find the right angle first, then look straight across from it.
When it applies, and when it does not
The theorem holds for right-angled triangles and nothing else. If the triangle has no right angle, a² + b² = c² is simply false, and a question that hands you three sides with no right angle is asking for the cosine rule instead.
Read backwards, it becomes a test: if a triangle’s three sides satisfy a² + b² = c², the triangle must be right-angled. That is how you check whether a corner is square without a protractor, and it is why builders measure 3 metres along one wall, 4 along the other, and check the diagonal reads exactly 5.
A few side triples come out as whole numbers, and they are worth knowing on sight because exam questions lean on them: 3-4-5, 5-12-13, 8-15-17, and any multiple of those, so 6-8-10 and 9-12-15 as well.
A worked example
A ladder leans against a wall. Its foot is 1.5 m from the wall and it reaches 3.6 m up. How long is the ladder?
The wall and the ground meet at a right angle, so the ladder is the hypotenuse:
c² = a² + b² = 1.5² + 3.6² = 2.25 + 12.96 = 15.21
c = √15.21 = 3.9 m
Now the same question backwards. A 3.9 m ladder must reach 3.6 m up a wall — how far out can its foot be? This time the unknown is a leg, so subtract rather than add:
a² = c² − b² = 3.9² − 3.6² = 15.21 − 12.96 = 2.25
a = √2.25 = 1.5 m
That is the whole difference between the two cases. Looking for the hypotenuse, you add. Looking for a leg, you subtract — and you subtract from the hypotenuse, never the other way round.
Where marks get lost
Adding when you should subtract. By far the most common error. If the side you want is the longest one, add; otherwise subtract. A quick check: your answer for a leg must come out smaller than the hypotenuse, and your answer for a hypotenuse must come out larger than both legs.
Forgetting the square root. The formula gives you c², not c. Stopping at 15.21 and writing it as the answer is a very easy mark to drop when you are working quickly.
Mixed units. Centimetres in one side and metres in another will produce a confident, completely wrong number. Convert everything to the same unit before you start, not afterwards.
Rounding too early. Round only at the final line. Rounding √15.21 to 3.9 partway through a longer question and then squaring it again drags the error along with you.
Questions students actually ask
How do I know which side is the hypotenuse?
Find the right angle, then look directly opposite it. That side is the hypotenuse. It is always the longest side of the triangle, so if your answer says otherwise, you have labelled it wrongly.
Can I use Pythagoras’ theorem on a triangle that has no right angle?
No. For a triangle with no right angle you need the cosine rule, c² = a² + b² − 2ab·cos C, which reduces to Pythagoras when C is 90° because cos 90° is zero.
What if I get a negative number under the square root?
It means the value you called the hypotenuse is not the longest side, so no such triangle exists. Check that you have not swapped the hypotenuse with one of the legs.
Does the theorem work in three dimensions?
Yes, applied twice. For a box with edges a, b and c, the long diagonal is √(a² + b² + c²) — you use the theorem once across the base, then again from that diagonal up to the far corner.
Why is it named after Pythagoras?
The relationship was in use in Babylon and Egypt well over a thousand years before him. His school is credited with proving it holds for every right-angled triangle rather than just noticing that it worked for particular ones, and proof is the part that mattered.
Related formulas
The quadratic formula · Projectile motion · Arithmetic progression
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