Pythagoras' theorem

Why a² + b² = c² holds, spotting the hypotenuse, the add-or-subtract decision, testing whether a triangle is right-angled, distance between two coordinates, and Pythagoras in 3D.

Geometry & measures · Pythagoras & right-angled trigonometry

Pythagoras' theorem

Why a² + b² = c² holds, spotting the hypotenuse, the add-or-subtract decision, testing whether a triangle is right-angled, distance between two coordinates, and Pythagoras in 3D.

Why it works

In a right-angled triangle, the square on the hypotenuse equals the sum of the squares on the other two sides:

a^2 + b^2 = c^2 \qquad (c \text{ is the hypotenuse}).

Here's one way to see it. Take four copies of the right-angled triangle and arrange them inside a square of side

a + b

, leaving a tilted square of side

c

in the middle. The big square's area is

(a+b)^2

; it is also four triangles (

4 \times \frac{1}{2}ab = 2ab

) plus the tilted square (

c^2

). So

(a+b)^2 = 2ab + c^2

, and since

(a+b)^2 = a^2 + 2ab + b^2

, the

2ab

cancels to leave

a^2 + b^2 = c^2

.

Find the hypotenuse first — always. It is the side opposite the right angle, and it is the longest side. It is not always the one drawn sloping, and not always the one labelled

c

.

Add or subtract? Let the hypotenuse decide.

Looking for the hypotenuse: add the squares, then square-root.
Looking for a shorter side: subtract the smaller square from the

hypotenuse's square, then square-root.

A sanity check catches every version of this slip: the hypotenuse must come out longest. If a "shorter side" comes out bigger than the hypotenuse, the operation was the wrong way round.

Testing for a right angle. If

a^2 + b^2 = c^2

for the three sides (with

c

the longest), the triangle is right-angled; if the two sides don't balance, it isn't. Show both sides of the comparison as numbers, then state the conclusion.

Distance between two points. The gap between

(x_1, y_1)

and

(x_2, y_2)

is the hypotenuse of a triangle with horizontal side

x_2 - x_1

and vertical side

y_2 - y_1

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

Squaring kills any minus signs, so the order of subtraction doesn't matter.

In 3D, use Pythagoras twice. For the long diagonal of a cuboid, first find the diagonal across the base (

\sqrt{l^2 + w^2}

), then use that as one side of a new right-angled triangle with the height. Keep the base diagonal squared rather than rounding it — rounding halfway through is what makes 3D answers drift off the mark scheme.