# Pythagoras' theorem proof

Categories: gcse trigonometry pythagoras

In this article, we will see how to prove Pythagoras' theorem.

There are many proofs of the theorem, we will just look at two of them:

- A simple visual proof.
- A proof using algebra.

Here is a video on the topic:

## Pythagoras' theorem

Pythagoras' theorem applies to right-angled triangles like this one:

This triangle has sides *a*, *b* and *c*. Side *c* is the *hypotenuse*. The hypotenuse is the side that is opposite the right angle. It is also the longest of the three sides.

Pythagoras' theorem is a formula that relates the lengths of the three sides:

We will prove that this is true.

## Visual proof

The visual proof is quite simple. We form a square using four identical triangles, like this:

The outer square has sides of length *a + b*.

The inner square (the yellow area) is a square whose sides have length *c* because each side of the square is made from the hypotenuse of one of the triangles, which we previously said has length *c*:

So the yellow square has area:

Now we rearrange the four triangles to form two rectangles:

The important thing to notice here is that the outer square still has length *a + b*, so it is exactly the same size as the previous step.

The rearranged triangles leave a yellow space that can be divided into two squares, one of side *a* (with area *a²*) and the other with side *b* (with area *b²*). So the total yellow area is:

Since both outer squares are the same size, and both contain four triangles of the same size, it follows that the two yellow areas must also be equal. This proves that:

## Be careful with visual proofs!

A problem with visual proofs is that it is easy to make assumptions - if it looks right, you might assume it must be right.

We still need to check that it is valid.

In this case, we are assuming that the yellow area in step 1 is a square. Just because it looks square doesn't mean it is square! It could be a rhombus with sides of length *c* but the angles might not be right angles. That would change its area.

It isn't difficult to verify that the shape is a square. One way is to use rotational symmetry. The original shape has rotational symmetry of order four because the outer boundary is a square and the corners of the yellow area are on the equivalent points on the sides of the boundary.

This means that if we rotate the shape through 90° the whole figure (including the yellow area) will map onto itself:

This proves that the two sides marked with a tick must be equal, and the two marked angles must also be equal.

If we rotate the shape through 90° twice more, it proves that all four sides are equal, and all four angles are equal.

Since the angles in a quadrilateral add up to 360°, if the angles are all equal then each angle must be 90°.

A quadrilateral with four equal sides and four right angles must be a square, so the visual proof is valid.

## An algebraic proof

If you prefer to use algebra rather than shuffling shapes around, there is a different proof based on the same idea.

We will start from the same diagram as before:

We will calculate the total area of the full square in two different ways. First, we will use the outer dimensions of the square:

This square has side length *a + b* so its area is:

Expanding the brackets gives:

Simplifying this gives:

Now we will find the area in a different way. These shapes completely cover the full square:

There are four identical blue triangles. Each triangle has area:

So the total area of the four blue triangles and the yellow square is

This is equal to the total area we calculated previously:

So:

Subtracting *2ab* from both sides gives us Pythagoras' formula:

## See also

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