Maths SolverThe Pythagorean theorem is the statement that in any right-angled triangle, the square of the longest side equals the sum of the squares of the other two:
$$a^2 + b^2 = c^2.$$
That is the headline. What I want to do in this article is convince you that the headline is true — not because the textbook says so, but because it is genuinely obvious once you have seen one of the proofs done properly. I’ll walk through four of them: a visual rearrangement that needs no algebra at all, a short algebraic proof, a proof from similar triangles, and the converse you actually need on the SAT and at GCSE. There is also a calculator embedded later that handles the three common cases (find the hypotenuse, find a leg, check whether a triangle is right-angled) with the steps shown.
The vocabulary, briefly
A right-angled triangle has one $90^\circ$ angle. The two sides that touch that angle are called the legs — conventionally labelled $a$ and $b$. The third side, opposite the right angle, is called the hypotenuse and labelled $c$. The hypotenuse is always the longest side; this is not just convention, it is a consequence of the theorem.
Proof 1 — the rearrangement (no algebra)
Take a square of side $a + b$. Inside it, draw four copies of our right triangle, arranged so the four hypotenuses meet at the centre as a tilted square. The four triangles cover most of the big square; the white space left over in the middle is a square of side $c$, with area $c^2$.
Now rearrange the four triangles inside the same big square, this time pushing them into two of the corners. The four triangles still take up the same total area (we haven’t added or removed any), but now the white space is split into two squares: one of side $a$ and one of side $b$. The total white area is therefore $a^2 + b^2$.
The big square is the same in both pictures, and so is the area of the four triangles. Whatever is left over — the white space — must have the same total area in both. So
$$c^2 \;=\; a^2 + b^2.$$
This is the proof I show in class first, because it is the proof that makes the theorem feel true rather than just provable. There is no algebra; you are just rearranging cardboard.
Proof 2 — the short algebraic version
Take the same big square of side $a + b$ and just compute its area two ways. Direct expansion gives
$$(a + b)^2 = a^2 + 2ab + b^2.$$
The decomposition into four triangles plus the central tilted square gives
$$4 \cdot \tfrac{1}{2} ab + c^2 = 2ab + c^2.$$
These are two expressions for the same area, so they are equal:
$$a^2 + 2ab + b^2 = 2ab + c^2.$$
Subtract $2ab$ from both sides and you have $a^2 + b^2 = c^2$. Done.
Proof 3 — from similar triangles
Drop a perpendicular from the right angle to the hypotenuse. This splits the original triangle into two smaller right triangles, both of which turn out to be similar to the original (same angles, scaled side lengths).
Call the two pieces of the hypotenuse $p$ and $q$, with $p + q = c$. By the similarity of the smaller triangles to the original,
$$\frac{a}{c} = \frac{p}{a}, \qquad \frac{b}{c} = \frac{q}{b},$$
which rearrange to $a^2 = pc$ and $b^2 = qc$. Adding,
$$a^2 + b^2 = pc + qc = (p + q)\, c = c \cdot c = c^2.$$
This is essentially Euclid’s proof. It is the most efficient of the three for memorising, but the rearrangement proof is the one I use to explain why the theorem is true.
The converse, and why you need it
The converse of the theorem is the statement: if $a^2 + b^2 = c^2$ for the three sides of a triangle, then the triangle is right-angled, with the right angle opposite the side of length $c$. This is what you use when an exam question gives you three side lengths and asks “is this a right triangle?” The procedure is mechanical: identify the longest side as the candidate hypotenuse $c$, square the other two, add, and compare with $c^2$.
A worked example. Are sides $5$, $12$, $13$ the sides of a right triangle? The longest side is $13$, so $c = 13$. We need $5^2 + 12^2$ and $13^2$. $25 + 144 = 169$, and $13^2 = 169$. They match, so yes — this is a right triangle with the right angle opposite the side of length $13$. (The triple $5, 12, 13$ is one of the so-called Pythagorean triples; the other one worth memorising is $3, 4, 5$.)
The mistakes I see most often
1. Mis-identifying the hypotenuse
The hypotenuse is the side opposite the right angle, and is always the longest side. If a problem labels the sides $a, b, c$ but the right angle is not opposite $c$, then $c$ is not the hypotenuse and the formula in the form $a^2 + b^2 = c^2$ does not apply directly. Re-label or rewrite the equation; do not just plug in.
2. Forgetting to square-root at the end
If you find $a^2 + b^2 = 25$, that is $c^2$, not $c$. The hypotenuse is $5$, not $25$. This is by far the most common mistake on a timed test.
3. Using the formula on a non-right triangle
The Pythagorean theorem is only true for right-angled triangles. For any other triangle you need the law of cosines, $c^2 = a^2 + b^2 - 2ab\cos C$, which reduces to the Pythagorean theorem when $C = 90^\circ$ (because $\cos 90^\circ = 0$). If a question does not mention a right angle and there is not a right-angle marker on the diagram, do not use $a^2 + b^2 = c^2$.
4. Mixing units
$a$ and $b$ must be in the same units before you square them. If one side is given in metres and the other in centimetres, convert first; otherwise the answer is wrong by a factor of 100 or 10000, depending on which side you trust.
What the theorem is really doing
There is a subtlety worth saying out loud. The Pythagorean theorem is not just a fact about triangles; it is the definition of distance in flat two-dimensional space. The distance between points $(x_1, y_1)$ and $(x_2, y_2)$ in the plane is
$$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2},$$
which is just the Pythagorean theorem applied to the right triangle with legs $\Delta x$ and $\Delta y$. Every time you compute a distance in coordinate geometry — the radius of a circle, the magnitude of a vector, the closest approach of a line to a point — you are using the Pythagorean theorem. It generalises directly to three dimensions ($\sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2}$) and into linear algebra, where it becomes the foundational identity for inner products and norms. That is why this one statement, drilled in middle school, keeps reappearing for the next five years of maths.
References and further reading
- Euclid, Elements, Book I, Proposition 47 (the original similar-triangle proof, around 300 BCE).
- Wikipedia: Pythagorean theorem, Pythagorean triple, Law of cosines.
- Loomis, E. S. The Pythagorean Proposition (1940). Collects 370 different proofs — a curiosity, but the variety is striking.
- Khan Academy — Pythagorean theorem (free worked examples and exercises).