Pythagoras’ theorem

Pythagoras’ theorem is probably the most famous theorem in all of maths. You can read about Pythagoras of Samos himself elsewhere. Now let’s turn to triangles.

The area of a square on the side of a triangle

Pythagoras’ theorem gives us a useful relationship between the areas of squares that we construct on the sides of a right-angled triangle. Imagine we have this right-angled triangle:

The square grid in the background gives us the dimensions of the triangle. The side AC, one of the two shorter sides (the legs), has length 4, and the side BC, the other shorter side, has length 3. But what is the length of the side AB, the hypotenuse of the triangle? At first glance we cannot tell. But we can try to work it out.

First we construct a square on the side BC. By that we mean a square whose sides all have the same length as the side BC, that is, three. In the diagram it looks like this:

We have the square BCDE, all of whose sides have length 3. What is the area of such a square? We work out the area of a square by multiplying the lengths of two of its sides, so the area of the square BCDE is 3 · 3 = 9. Now we draw in the square on the other shorter side, the side AC. This square has sides of length 4.

This square has area 4 · 4 = 16. Now we draw in the last square, on the hypotenuse AB:

We do not yet know the area of this new square, because we do not know the length of the side AB. But now we can make good use of Pythagoras’ theorem. It tells us that the area of this last square ABHI is equal to the sum of the areas of the previous two squares. The first two squares had areas 9 and 16, so the last square – according to Pythagoras’ theorem – has area 9 + 16 = 25. Let’s add it to the diagram:

Now there is just one question left: if the square ABHI has area 25, what is the length of its sides? It must be the square root of 25, so the length of the side AB is $\sqrt{25}=5$. If we check this by working out the area of a square with sides of length 5, we get the area 5 · 5 = 25. And now we can get on with the numbers and definitions.

The theorem

Pythagoras’ theorem goes something like this: “The area of the square on the hypotenuse of a right-angled triangle is equal to the sum of the areas of the squares on the other two sides.” In maths it is usually written like this:

$$c^2=a^2+b^2,$$

where a and b are the lengths of the two shorter sides of the triangle and c is the length of the hypotenuse.

So what does Pythagoras’ theorem say, and how does this notation relate to the previous diagram? The area of the square on a side of a triangle means that we take a square whose side length equals the length of that side and work out its area. We work out the area of a square by multiplying one side length by the other, so if the side has length c, the area S is S = c · c, which we can write using powers as S = c².

So we can read c² = a² + b² as “the area of a square with side length c is equal to the sum of the areas of squares with side lengths a and b”.

Let me stress once more that the theorem holds only in a right-angled triangle, not in a general one. Pythagoras’ theorem is classically used when you know the lengths of two sides and need to work out the length of the remaining side. If we know the lengths of both shorter sides a and b and want the length of the hypotenuse c, we work out the areas of the squares on those sides, that is, a² + b². That gives us the area of the square on the hypotenuse c, that is, c². To get the length of the side c, we just take the square root of the area we have worked out. This gives us the formula:

$$c=\sqrt{a^2+b^2},$$

where c is the length of the hypotenuse and a, b are the lengths of the two shorter sides. If we know the length of the hypotenuse and of one of the shorter sides and want the length of the other, we work it out in the same way, except that we first rearrange the equation to get one of the shorter sides on its own. So if we know c and b and want to work out a, then in the equation

$$c^2=a^2+b^2,$$

we get a² on its own by subtracting b²:

$$c^2-b^2=a^2,$$

swap the left-hand and right-hand sides:

$$a^2=c^2-b^2$$

and finally take the square root:

$$a=\sqrt{c^2-b^2}$$

First example

Take a triangle ABC in which we know the lengths of two sides: a = 3 and b = 4. It is the same triangle we had at the start. The question is: what is the length of the remaining side, the side c? Let’s work it out using the formula above. The following diagram shows the triangle:

We can see that we know the lengths of the two shorter sides and the hypotenuse is missing. So we use the first, unrearranged formula:

$$c=\sqrt{a^2+b^2}$$

We substitute the side lengths for the variables a and b:

$$c=\sqrt{3^2+4^2}$$

The formula contains a square – as a reminder, the following holds:

$$a^2=a\cdot a\to 4^2=4\cdot 4=16$$

So after working out the squares we get:

$$\begin{array}{rcl}c& =& \sqrt{3^2+4^2}\\ c& =& \sqrt{9+16}\\ c& =& \sqrt{25}\\ c& =& 5\end{array}$$

The result is that the length of the side c is five.

Second example

In this example we work out the length of one of the shorter sides when we know the other one and the length of the hypotenuse. So we have a triangle ABC with side lengths |AC| = 10 and |BC| = 6. The following diagram shows the triangle:

From the diagram we can see that the hypotenuse, that is, the longest side, is the side |AC|. We need to work out the length of the side |AB|. We substitute into the second formula, for working out the length of a shorter side:

$$a=\sqrt{c^2-b^2}$$

To use Pythagoras’ theorem correctly, we now need to be clear what each variable stands for. In this formula the variable a is the length of the side we want to work out, in our case the side |AB|. The variable c is the length of the hypotenuse, the longest side, in our case the side |AC|. And the variable b is the length of the shorter side whose length we know, in our case |BC|. So we substitute into the formula like this:

$$|AB|=\sqrt{|AC{|}^2-|BC{|}^2}$$

We substitute the specific side lengths that we know:

$$|AB|=\sqrt{{10}^2-6^2}$$

We work out the squares and subtract:

$$|AB|=\sqrt{100-36}=\sqrt{64}$$

And finally take the square root:

$$|AB|=8$$

The length of the side |AB| is eight.

Word problem

Imagine you are walking to a friend’s house along a straight road. The road is 250 metres long. After this quarter of a kilometre you turn left and walk another 100 metres, and you are at your friend’s house. The question is: how much shorter would the route be if you walked straight across the field?

As the first step, we work out the length of the route along the road. You first walk 250 metres straight ahead and then 100 metres to the left. That is 250 + 100 = 350 metres in total. Now it is time to work out the length of the route across the field. For that we will need a diagram.

One square on the diagram represents 50 metres. What would the straight route across the field look like? It would be the line segment from the point A to the point C. We draw it in the diagram in red.

Now we can see that to work out the length of the route across the field we just need to apply Pythagoras’ theorem in a suitable way. We are looking for the length of the hypotenuse and we know the lengths of both shorter sides, so we use the first formula and substitute like this:

$$|AC|=\sqrt{|AB{|}^2+|CB{|}^2}$$

We substitute the lengths:

$$|AC|=\sqrt{{250}^2+{100}^2}$$

We square the numbers:

$$|AC|=\sqrt{62500+10000}=\sqrt{72500}$$

Then take the square root and round:

$$|AC|=269$$

The length of the route across the field is 269 metres. The problem is not finished yet, because the question was how much shorter the route across the field is. So we subtract the lengths, and call the difference r:

$$r=350-269=81$$

The answer is that the route is shorter by approximately 81 metres.

The side length of a square

What is the side length of a square whose diagonal has length ten? How does Pythagoras’ theorem help us here? We need to find a right-angled triangle in the square that we can use to work out the side length. As a reminder: a square has four sides of equal length. Let’s draw the situation:

We can see that two triangles with a right angle have suddenly appeared in the square, and we can use them to work out the side length. Let’s take, for example, the triangle BCD. The side DB is the hypotenuse and the other two are the shorter sides. We do not know the lengths of the shorter sides, but we do know the length of the hypotenuse. These two sides have the same length, so |BC| = |CD|. We use the basic form of Pythagoras’ theorem:

$$|BD{|}^2=|BC{|}^2+|CD{|}^2$$

But since the two shorter sides are equal, we only need to work out the square of one of them and multiply it by two – we do not need to square both, because they are the same and we would get the same result. So we can write:

$$|BD{|}^2=2(|BC{|}^2)$$

We know the length of the side BD, which is the hypotenuse. It is equal to ten. We substitute into the equation:

$${10}^2=2(|BC{|}^2)$$

We square the ten:

$$100=2(|BC{|}^2)$$

Divide by two:

$$\frac{100}{2}=|BC{|}^2$$

We simplify the fraction:

$$50=|BC{|}^2$$

Now we are almost there. We know that the square of the side length of the square is equal to 50. To find the side length, we still need to take the square root of the equation, which we can do because we are working with positive numbers:

$$\sqrt{50}=\sqrt{|BC{|}^2}$$

We leave the square root of fifty in this form, but on the right-hand side of the equation we can cancel the square root and the square, because they cancel each other out.

$$\sqrt{50}=|BC|$$

If you like, you can also work out the square root of fifty. Rounded, it comes to about seven:

$$7=|BC|$$

So the side length of the square is approximately seven, or exactly the square root of fifty.

Calculator

You can work out the hypotenuse from both shorter sides, or one shorter side from the hypotenuse and the other shorter side; you can switch the units for each field.