The Pythagorean Theorem

Right triangles and their sides

A right triangle is a triangle with a right angle, a square corner of exactly $90^\circ$ like the ones you met in the lesson on angles. Its three sides do different jobs, so they get different names. The two sides that form the right angle are the legs. The third side, the one across from the right angle, is the hypotenuse. It is the only side that does not touch the right angle, which is the quickest way to find it whichever way the triangle is turned.

It is usual to call the legs $a$ and $b$ and the hypotenuse $c$. The letters are only a habit. What matters is the position: $c$ always stands for the side across from the right angle. Which leg you call $a$ and which you call $b$ makes no difference to anything below.

One more fact about the hypotenuse will turn out to be true: it is always the longest of the three sides. That is not obvious from a sketch, so it waits until the rule below can prove it.

Squares on the three sides

Take the right triangle with legs $3$ and $4$ units and a hypotenuse of $5$ units. Build a square outward on each side, and rule each square into unit squares so you can count its area.

The square on the leg of length $3$ holds $3^2 = 9$ unit squares, and the square on the leg of length $4$ holds $4^2 = 16$. The tilted square on the hypotenuse holds $5^2 = 25$. Now notice how those three numbers fit together:

The two squares on the legs, put together, have exactly the area of the square on the hypotenuse. The next section shows that this happens in every right triangle, not only in this one.

The right angle is not optional. Try the same sum on a triangle with no right angle, such as one whose three sides are all $2$ units long. It has no legs at all, but suppose you squared two of its sides and added anyway. You would get $2^2 + 2^2 = 8$, while the third side squared is only $2^2 = 4$. Without a right angle the sum can fail, so everything below is about right triangles.

Why the squares always balance

Counting unit squares for one triangle proves nothing about the next. The argument below works for every right triangle at once.

The figure shows a right triangle with a short leg $a$ and a long leg $b$, drawn in the ratio $5$ to $12$, and the same triangle completed to an $a$ by $b$ rectangle. Its other two angles are marked $x$ (at the far end of leg $a$) and $y$ (at the far end of leg $b$).

The second figure shows the big square that the proof fills in two different ways. Keep both figures in view.

That result is the Pythagorean theorem: in a right triangle with legs $a$ and $b$ and hypotenuse $c$,

In words, the square of the hypotenuse equals the sum of the squares of the two legs. Use it once you know the triangle has a right angle.

It also settles the claim from the first section. For legs $5$ and $12$, the hypotenuse’s square is $169 = 25 + 144$. So $169$ is more than $144$ and more than $25$, and a bigger number has a bigger square root, so $13$ is longer than $12$ and longer than $5$. In general, $c^2$ is $a^2$ with $b^2$ added on, and $b^2$ is more than zero because $b$ is a length greater than zero. So $c^2$ is bigger than $a^2$, and $c$ is longer than $a$. The same reasoning shows $c$ is longer than $b$. The hypotenuse is always the longest side of a right triangle.

Finding the hypotenuse

When you know both legs, the theorem hands you $c^2$, and one square root finishes the job. Square each leg, add the two squares, and take the square root of the total. Only the non-negative root counts, because a length cannot be negative, which is exactly what the radical sign gives.

Most pairs of whole-number legs do not give a perfect square. When the sum of their squares is a whole number that is not a perfect square, the hypotenuse is irrational. You leave it as an exact square root, simplified the way you learned with square roots.

Finding a missing leg

The same equation works backward. Take a right triangle with a hypotenuse of $10$ and one leg of $6$. The theorem says

Take $36$ away from both sides, just as you would undo an addition in any equation. That leaves $b^2 = 100 - 36 = 64$, so $b = 8$. Nothing depended on $10$ and $6$. For any hypotenuse $c$ and known leg $a$, removing $a^2$ from both sides gives

So for a leg you subtract. This only works when the hypotenuse is longer than the leg you know. If it is not, the subtraction gives zero or less, and no right triangle has those sides.

Measuring with the theorem

Any side of a right triangle can be found this way once you know the other two, as long as all the lengths are in the same unit. A ladder against a wall is the classic case, provided the wall stands at a right angle to level ground.

A rectangle’s diagonal is the same kind of length. Its corners are right angles, so the diagonal cuts it into two right triangles with the rectangle’s sides as legs. A park $20$ m by $21$ m has a diagonal path of $\sqrt{20^2 + 21^2} = \sqrt{841} = 29$ m. Cutting across saves $12$ m on the $41$ m walk round two edges.

The builders’ rope from the start of the lesson works the same way. If a corner is square, marks $3$ and $4$ steps along its two sides must be $5$ steps apart. So if the gap comes out at anything other than $5$, the corner is certainly not square. Does a gap of exactly $5$ prove the corner is square? That is the theorem run backward, starting from the sides and concluding the angle. It is a different statement, and this lesson does not prove it. Geometry takes it up.

Common mistakes