The Pythagorean theorem: a² + b² = c²
Why the squares on the two short sides of a right triangle add up to the square on the long one — with the picture that makes it obvious.
What this shows
In a right triangle — one with a 90° angle — let the two legs have lengths a and b and the hypotenuse (the side opposite the right angle) have length c. The Pythagorean theorem says:
a² + b² = c²
The figure builds a literal square on each of the three sides and labels its area. For the classic 3-4-5 triangle, the squares have areas 9, 16, and 25 — and 9 + 16 = 25. The theorem holds for every right triangle, not just the integer-sided ones.
There are over 400 distinct proofs of the theorem on record. The easiest to believe is the "rearrangement" proof: take four copies of the same right triangle, arrange them inside a bigger square in two different ways, and observe that the leftover space in one arrangement is a² + b² and in the other is c². Both arrangements have the same total area, so the leftover spaces match.
Where it shows up
The theorem is the working tool behind almost every distance calculation in two or three dimensions: the straight-line distance between two points in the plane is √((x₂ − x₁)² + (y₂ − y₁)²), which is the Pythagorean theorem on the horizontal and vertical legs. GPS, computer graphics, physics — they all lean on it.
Pythagorean triples (whole-number solutions: 3-4-5, 5-12-13, 8-15-17, …) appear in surveying, in computer-graphics ray intersections, and in cryptography (where they motivate the much harder problem of finding integer solutions to a^n + b^n = c^n, which Fermat famously claimed could never be done for n > 2).
Frequently asked questions
Does it only work for the 3-4-5 triangle?
No — it works for every right triangle. The 3-4-5 is just a convenient example because all three side lengths are whole numbers. For arbitrary triangles you can have any positive a and b, and c is whatever √(a² + b²) comes out to.
What if the triangle isn't right-angled?
Then a² + b² ≠ c² in general. The generalisation is the Law of Cosines: c² = a² + b² − 2ab·cos(C), where C is the angle opposite the side c. When C is 90°, cos(C) = 0 and the −2ab·cos(C) term vanishes, recovering Pythagoras.
Why is it true?
Many proofs exist. The cleanest builds two squares of the same outer side length (a + b) and tiles each with four copies of the right triangle. One arrangement leaves a c-by-c square in the middle; the other leaves an a-by-a square plus a b-by-b square. Both tilings have the same total area, so c² = a² + b².
Does it work in 3D?
Yes — the diagonal of a rectangular box with sides a, b, c has length √(a² + b² + c²). This is just Pythagoras applied twice: once in the base (giving √(a² + b²)) and once vertically with that diagonal as one leg and c as the other.