三平方の定理

In a right-angled triangle, the square of the hypotenuse (the longest side, opposite the right angle) equals the sum of the squares of the other two sides. Writing the hypotenuse as $c$ and the two shorter sides as $a$ and $b$, we have $a^2+b^2=c^2$. Remember: the hypotenuse is always the longest side and is the one opposite the right angle.

When the two shorter sides $a$ and $b$ are known, compute $c^2=a^2+b^2$ and then take the square root: $c=\sqrt{a^2+b^2}$. Since the hypotenuse is the longest side, your answer should be larger than each of the two given sides — a quick sanity check.

When the hypotenuse $c$ and one shorter side are known, rearrange the formula into a subtraction. To find side $a$, use $a^2=c^2-b^2$, so $a=\sqrt{c^2-b^2}$. Be careful to subtract from the square of the hypotenuse (the larger value); getting the order wrong gives a negative number.

Given three side lengths, label the longest side $c$ and check whether $a^2+b^2=c^2$. If it holds, the triangle is right-angled, with the right angle opposite the longest side (the converse of Pythagoras' Theorem). If $a^2+b^2>c^2$ the triangle is acute; if $a^2+b^2<c^2$ it is obtuse.

A set of three whole-number sides that form a right-angled triangle is called a Pythagorean triple. The ones that come up most often are $3:4:5$, $5:12:13$, $8:15:17$ and $7:24:25$. Any multiple of these — for example $6:8:10$, which is $3:4:5$ doubled — is also right-angled. In an exam, spotting that the given sides match one of these ratios often lets you write the answer without a square-root calculation.

Pythagoras' Theorem appears whenever a right angle creates a straight-line distance: a ladder leaning against a wall (ground, wall and ladder form a right triangle), the distance between two points on a coordinate grid, or the diagonal of a rectangle. The trick is to sketch the situation, mark the right angle, and identify which side is the hypotenuse before writing the equation. The distance between points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$, which also comes from Pythagoras.

Pythagoras' Theorem also works in three dimensions (this is Extended content in IGCSE 0580). For a cuboid with length $a$, width $b$ and height $c$, the space diagonal $d$ joining the two farthest corners is $d=\sqrt{a^2+b^2+c^2}$. The idea is to use Pythagoras twice: first find the base diagonal $\sqrt{a^2+b^2}$, then apply the theorem again to a right triangle whose sides are that base diagonal and the height $c$.