The Pythagorean Theorem: a² + b² = c²

The theorem

In a right triangle, the two shorter sides are the legs and the longest — opposite the right angle — is the hypotenuse.

$$a^2 + b^2 = c^2$$

With legs $3$ and $4$:

$$3^2 + 4^2 = 9 + 16 = 25 \quad\Rightarrow\quad c = \sqrt{25} = 5$$

Why it is a statement about areas, not lengths

The name “squared” is not incidental here. Draw an actual square on each side of the triangle.

The square on the hypotenuse has area $c^2$. The squares on the two legs have areas $a^2$ and $b^2$. The theorem says the two smaller squares, put together, hold exactly as much area as the big one.

$$\underbrace{9}_{3 \times 3} + \underbrace{16}_{4 \times 4} = \underbrace{25}_{5 \times 5}$$

That is a claim about area, not about lengths — $3 + 4 \ne 5$, and it was never supposed to. Only after squaring do the sides fit together.

Finding a leg instead

The formula rearranges like any other:

$$a^2 = c^2 - b^2$$

The hypotenuse’s square is the largest, so the legs’ squares come out of it. Adding when you should subtract is the error to watch for, and it always produces an impossible answer: a leg longer than the hypotenuse.

You know	You want	Do
both legs	the hypotenuse	add the squares
the hypotenuse and a leg	the other leg	subtract the squares

Triples worth recognising

Some right triangles have whole-number sides. Spotting one saves the arithmetic.

Triple	Check
$3, 4, 5$	$9 + 16 = 25$
$5, 12, 13$	$25 + 144 = 169$
$8, 15, 17$	$64 + 225 = 289$
$7, 24, 25$	$49 + 576 = 625$

Any multiple of a triple is also a triple: $6, 8, 10$ and $9, 12, 15$ are both the $3, 4, 5$ triangle scaled up. Scaling multiplies every side by the same number, so the relationship survives.

It only holds for right triangles

The theorem needs the right angle. On a triangle without one it is false.

The converse is true as well, and useful: if the three sides of a triangle satisfy $a^2 + b^2 = c^2$, then the triangle must be right-angled. That is how builders check a corner is square — measure $3$ along one wall, $4$ along the other, and confirm the diagonal is exactly $5$.

Worked examples

Common mistakes

Practice problems