a^{2} + b^{2} = c^{2}.

That’s Pythagoras’ theorem. What it says is that in a right-angled triangle, the sum of the squares of the two smaller sides equal the square of the longest side (or hypotenuse).

From my experience, most GCSE students understand what Pythagoras’ theorem is, but when it comes to applying it to a question they struggle using a^{2} + b^{2} = c^{2}.

So, in this blog, I am going to show you a way to solve Pythagoras’ theorem problems without worrying about a, b and c. The method I am about to go through is simpler and doesn’t involve any algebra. I have had huge success with this method.

First of all, we need to understand right-angled triangles.

The right-angle will always be labelled by a square, as you see below. The longest side of the triangle (also known as the **hypotenuse**) is always the side opposite the right angle. Identifying the longest side is going to be pivotal in the method I am about to show you.

When answering a Pythagoras’ theorem question, the first question you should always ask yourself is “Am I finding the longest side?”. Let’s first work through an example where you are finding the longest side.**Example 1**

Find x.

Here, you can clearly see that we are being asked to find the longest side. As the side labelled “x” is the side opposite the right angle.

We need to follow four steps here to find x.

1. SQUARE

2. SQUARE

3. ADD

4. SQUARE ROOT

Let me explain what I mean by these four steps.

1. SQUARE: Square one of the given sides. 3^{2} = 9.

2. SQUARE: Square the other given side. 4^{2} = 16.

3. ADD: Add those two results together. 9 + 16 = 25.

4. SQUARE ROOT: Square root your result. The square root of 25 = 5.

So, x = 5 cm.

Now, let’s work through an example where we are not finding the longest side.**Example 2**

Find x.

Now, in this case, you will see that we are not finding the longest side. Because the side marked x is not opposite the right-angle.

These are the steps we need to follow to find x (they are almost exactly the same as in Example 1):

1. SQUARE

2. SQUARE**3. MINUS**

4. SQUARE ROOT

I have highlighted step 3 because you will notice that this is the only step that has changed from our previous example.

1. SQUARE: Square one of the given sides. 5^{2} = 25.

2. SQUARE: Square the other given side. 4^{2} = 16.

3. MINUS: *It is important that you minus the smaller number from the bigger number. 25 – 16 = 9.

4. SQUARE ROOT: Square root your result. root(9) = 3.

So x = 3 cm (you may notice that this triangle is identical to the triangle in Example 1, just with a different side missing).

*It is important we do this, because if we take away 25 from 16, we will get -9. If you then try to square root -9, you will be sorely disappointed as you cannot square root a negative number! (Plug it into your calculator and try it!). Your result from step 3 must always be positive.

So those are the steps. It does not involve any algebra. No a’s, b’s or c’s. This process and the two examples are summarised below (click here to view the PDF).

Now let’s do a couple of additional examples to reinforce the method.**Example 3**

Find x to 1 decimal place.

First of all, ask yourself – is x the longest side?

It is not opposite the right-angle, so the answer is NO. We will need to follow these steps:

1. SQUARE: 11^{2} = 121.

2. SQUARE: 7^{2} = 49.**3. MINUS: 121 – 49 = 72.**4. SQUARE ROOT: root(72) = 8.48528137424 = 8.5 to 1 decimal place.

So x = 8.5 cm.

Now, anyone who I have ever tutored will tell you that I regularly encourage students to check their answers! With Pythagoras’ theorem questions, there is always a very simple logical check you can do once you get your answer.

If I draw our right-angled triangle again but this time with the value of x we obtained…

Now, remember what I said at the start – the longest side in a right-angled triangle is the side opposite the right angle.

Which is the longest side in this triangle? It’s 11 cm, and that is the side opposite the right angle. So the answer of 8.5 cm makes sense.

If, however, I had got x = 18.5 cm, like below….

…this would definitely be wrong. Because if x were 18.5 cm, the 11 cm side that is opposite the right-angle would no longer be the longest side. So it doesn’t make sense.

This may seem obvious, but checks like this are vital. If you made a mistake and accidentally added instead of taken away in the 3^{rd} step, this could happen! And if you check it, you will be able to correct the mistake you made. This is incredibly important because as we all know, especially under exam pressure, mistakes can creep in very easily.**Example 4**

Find x to 1 decimal place.

First of all, ask yourself – is x the longest side?

It is opposite the right-angle, so the answer is YES on this occasion. So we will need to follow these steps:

1. SQUARE: 16^{2} = 256.

2. SQUARE: 9^{2} = 81.**3. ADD: 256 + 81 = 337.**

4. SQUARE ROOT: root(337) = 18.3575597507 = 18.4 to 1 decimal place.

So x = 18.4 cm.

Let’s do the check!

Here is the triangle with the value of x that we obtained.

Is the side opposite the right angle the longest side? Well, 18.4 is larger than both 16 and 9, so yes! Our answer makes sense.

So there it is – I believe it is a much simpler alternative to using a^{2} + b^{2} = c^{2}.

If you want to practice this method, try our Pythagoras worksheet.

If you need someone to explain this in person, then book in a free taster session here.