Pythagoras

The area of the two smaller squares always adds up to the area of the largest one.

Every lesson in this chapter has been building toward one result. This is it: a right triangle, O, P and Q, with a genuine square built outward on each of its three sides. O, P and Q are all draggable, and the angle at O always stays a right angle.

A right triangle, O, P and Q, with a genuine square built outward on each of its three sides. O, P and Q are all draggable, and the angle at O always stays a right angle. A live reading gives the area of each of the three squares, and shows that the two legs' squares always add up to the hypotenuse's. This needs JavaScript switched on.

Drag O, P or Q. The two smaller squares always add up to the largest one.

There is no counterexample to find here, no matter how hard you drag — try anyway before reading on. Every arrangement gives the same relationship between the three square areas, which is worth a name. The longest side of a right triangle, the one opposite the right angle, is the hypotenuse — a word this course has used once before, in passing, without needing it. It is needed now: Pythagoras’s theorem is a statement about the hypotenuse specifically, not about either of the other two sides.

Proving it with area alone

Picture four copies of this exact right triangle, legs $a$ and $b$ , hypotenuse $c$ , arranged around a square of side $a + b$ so their right angles fill the four corners and their hypotenuses form a smaller square in the middle. The big square’s own area is

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

The same big square is also exactly the four triangles plus the smaller square in the middle. Each triangle has area $\frac{ab}{2}$ , so the four together are $2ab$ :

(a+b)^2 = 2ab + (\text{middle square's area})

Set the two expressions for $(a+b)^2$ equal and cancel the $2ab$ both sides already share:

a^2 + 2ab + b^2 = 2ab + (\text{middle square's area}) \quad\Longrightarrow\quad a^2 + b^2 = \text{middle square's area}

The middle square’s sides are the four triangles’ hypotenuses — each one length $c$ — and G5.3 already settled that four equal sides meeting at right angles is enough to call it a genuine square, so its area is $c^2$ . That gives $a^2 + b^2 = c^2$ .

This is one proof out of hundreds that exist for this theorem, and the reason to know this one specifically is that it never uses anything beyond area and algebra — no trigonometry, no coordinates, nothing this course hasn’t already justified.

One to watch out for.

The middle square being a genuine square, not just a quadrilateral that happens to look square, is doing real work in this proof — it is why the last chapter’s lesson on rectangles, rhombuses and squares as special cases is a dependency here, not a coincidence.

Two triangles to recognise on sight

A right triangle with legs 6 and 8.

6^2 + 8^2 = 36 + 64 = 100 = 10^2

so the hypotenuse is 10.

A right triangle with legs 5 and 12.

5^2 + 12^2 = 25 + 144 = 169 = 13^2

so the hypotenuse is 13.

3-4-5 and 5-12-13 are the two smallest whole-number right triangles, and they turn up often enough that spotting them on sight saves working through the arithmetic each time.

Do the exercise

This lesson counts as done once you get its exercise right. Nothing to tick off by hand.