Pythagorean Theorem

In a right triangle, $a^2 + b^2 = c^2$, where $c$ is the hypotenuse — the side facing the right angle. Written out, it says that the square built on the long side has exactly the same area as the two squares built on the short sides put together, and that is worth pausing on, because it is a statement about areas that lets you calculate a length. That is the whole trick, and it is why the squares and the square root are there rather than something simpler.

There are hundreds of proofs; the one worth carrying around takes two copies of a square of side $a + b$. In the first, arrange four copies of the triangle in the corners so the leftover space forms two squares, of areas $a^2$ and $b^2$. In the second, slide the same four triangles into a pinwheel so the leftover space is a single tilted square of area $c^2$. Same big square, same four triangles removed, so what remains must match. No algebra, and you can do it with paper. The converse is true as well, and it is the half that earns its living on site: if three measured lengths satisfy the relation, the angle between the two short ones is square. That is why a layout crew measures 3, 4 and 5 to set a corner rather than trusting a framing square.

Use the biggest triangle the space allows — a 3–4–5 in feet leaves a corner good to perhaps half a degree, while a 12–16–20 divides that error by four. A rafter run of 5.4 m with a rise of 2.2 m needs a length of $\sqrt{5.4^2 + 2.2^2} = 5.83$ m before the tail cut. Two triples are worth memorising because they come out whole: 3–4–5 and 5–12–13.

Three ways it goes wrong. The first is putting a leg where the hypotenuse belongs: $c$ is always the longest side, so if the answer comes back shorter than something you typed, the sides are in the wrong slots. The second is dropping the squares — the legs 3 and 4 do not make a 7, they make a 5, and the shortcut across a rectangular lot saves far less than people expect. The third is applying it to a triangle that has no right angle at all; for those, the law of cosines carries a correction term and reduces to this the moment the angle reaches 90°. Solving for a leg has a built-in honesty check: $a = \sqrt{c^2 - b^2}$ needs $c > b$, and if it is not, the square root turns imaginary because the triangle you described cannot be drawn.