## An introduction to tackling Pythagoras questions

- 262Views
- September 15, 2018
- 0 Comments
- GCSE Maths

# An introduction to tackling Pythagoras questions

Pythagoras is one of those maths things that sounds a lot more complicated than it actually is. As this is an intro to it, we’ll take a little look at where it all came from first. The Pythagorean theorem was developed by Pythagoras, a mathematician from Ancient Greece, or at least it’s credited to him no-one really knows if it was him for definite, but let’s be honest, it doesn’t matter who developed and proved it, all that matters is that we need to do it on our GCSE papers.

## So what is it useful for?

**Great Question!** All too often students will use the argument about topics in maths, when will I ever use this? Pythagoras is like most things in maths and it has a definite use in the real world. To put it simply, Pythagoras is used to finding the length of a missing side in a right-angled triangle. As long as you have 2 out of the 3 sides you can find the missing one by using this handy theorem.

You still don’t know what it’s useful for in the real world though? It’s not like you’re going to have to find the missing length on a right-angled triangle in order to catch the bus, are you? Well, don’t worry about the real-world use for it, we’ll get to that later and it will blow your mind.

## How Does it Work?

To put it simply it’s just a case of putting numbers into a formula and working out the answer. That’s all it is. What’s the formula? It’s an easy one to remember, a2 + b2 = c2 . That is all it is. So, we can end the article, there right? Awesome, see you next time…

## That doesn’t help that much

^{2}+ b

^{2}= c

^{2}mean? Well, the easiest way to describe it is, the c stands for the hypotenuse. The a and b stand for the other two sides. All we have to do is substitute the values we have into the formula and we can work our answer out from there. So how does that work then? Let’s try looking at a question to see how we would go about doing it. Find the value of x.

Here we can see that the hypotenuse is labelled x. So, if we were to put it into our formula the x would substitute in place of the c. This means that the numbers 20 and 22 would substitute in place of the a and the b. It doesn’t matter which way round we put the a and b, just as long as we put 20 and 22 where they should be. To help us do this, on our exam paper we can simply label the triangle. That instantly allows us to see what goes where and it will make it easier for us to solve the question.

Once we’ve labelled our triangle it will look a little something like this, just with better handwriting. Here we can see that by substituting the values from our triangle into the formula, a^{2} + b^{2} = c^{2} , we can begin to work out what the value of x is. Once we have substituted the values into the formula it should look like this.

22^{2} + 20^{2} = x^{2}

For the next bit, you’re not going to be happy, we need to use a little bit of algebra, kind of anyway. So, work out 22 squared and 20 squared for your next step.

For the next bit, you’re not going to be happy, we need to use a little bit of algebra, kind of anyway. So, work out 22 squared and 20 squared for your next step.

484 + 400 = x^{2}

Which gives us a total of

884 = x^{2}

We’re finished now, right? Not quite. Our next step is our final step, find the value of x. You see, if you look at the question closely you will see that we haven’t actually found what x is worth. We have found out what x squared is worth. That’s not our answer. If we leave it like this, we will still pick up some marks, but we won’t get full marks. We’ve come this far, we may as well get the full marks.

So here is the algebra, kind of. We know that x squared equals 884. When we want to rearrange something in algebra, we need to do the opposite to something to move it from one side to the other. Knowing that useful piece of information is how we’re going to find out what the value of x is.

If x squared is 884, we need to know what x is. So, what’s the opposite of squaring something? To square root it! So, the value of x is the square root of 884, or √884.

Depending on the question you could get the final mark for leaving it as √884, but it’s better to be safe and work it out as a decimal number, so just pop it into your calculator which will give you 29.7321374946 or to 1 decimal place 29.7. This makes our final answer to this question 29.7. Easy right? Let’s go back over how we did it just to make sure we’ve taken it all in.

So, in this case, we turned the c into the x. We turned the an into 22 and we turned the b into 20. To help us with this we labelled the triangle as well, just to make sure we could visualise exactly what we needed to do. This left us with 222 + 202 = x^{2}

For this we had to find out the answer to 22 squared and 20 squared. This gave us 484 for 22 squared and it gave us 400 for 20 squared. At this stage we were left with 484 + 400 = x^{2}

Now we need to add together the values we had after we had squared the a and the b. So, this meant that we were adding together 484 and 400. Here we were left with 884 = x^{2}

Our final step is to square root whatever answer we got from adding the squared values of a and b together. In this case we needed to find out the value of √884. This means that after rounding our answer to 1 decimal place our final answer is x = 29.7

That complete working out would look like this on your exam paper.

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

22^{2} + 20^{2} = x^{2}

484 + 400 = x^{2}

884 = x^{2}

√884 = x

x = 29.7

^{2}+ b

^{2}= c

^{2}to help us find the answer. So, let’s have a look at a question where we need to find the a or the b to give ourselves an idea how to solve them. Find the value of x.

Ahhh, look at it. It looks similar but the tactics we used before won’t work on this one. What are we going to do? Don’t panic, just like you shouldn’t panic during your exam, we don’t need to panic now either. We’re going to use a little bit of logic, the Pythagorean formula and good old-fashioned maths to work out the value of x. We might kind of use algebra again too, but don’t worry about it, you won’t even notice.

Straight away we can see that there is one significant difference with our problem, we already have the value for the hypotenuse, so we will be finding out the value of either a or b. We’ll call the x value a this time. Remember, on your exam paper you can label the triangle with a, b and c if you like. If you struggle to see things in your head then this might help you to visualise it, in fact even if you don’t struggle to see things in your head it still will help you remember where you are with your working.

This time, just like before we are going to substitute the values we have into the formula. So, put your values into a^{2} + b^{2} = c^{2} and we can get to work. Remember that we are going to put x into our formula as the a value. This will give us x^{2} + 14^{2} = 17^{2} and from here we can work out the value of x.

We need to work out the squared values next, just like if we were finding out the value of the hypotenuse, this will give us x^{2} + 196 = 289 it’s after this step that we will do things a little differently to how we did it the first time.

As you can see, you can’t add the two values together for a and b because a is the value that we are missing. So, we will need to do some almost algebra in order to work out the value of x. It’s pretty easy though so don’t panic. All we need to do is move the b value over to the same side of the equals side as the c value. So as the b value is positive 196, we need to do the inverse (opposite). That all sounds a little bit complicated, so let’s go through it in a little bit more detail.

x^{2} + 196 = 289

To move the 196 over to the equals side we will need to make it into the inverse. That will look like this.

x^{2} = 289 – 196

See how the + 196 from the left-hand side moved over to the right-hand side and became – 196? That’s what we mean by making it the inverse. So, after we have carried out this operation we are left with.

x^{2} = 93

Remember, we’re still not finished yet. We have the value for x^{2} but we don’t have the value for x on its own. In order to find that we will need to do the inverse of squaring to our value, this means we will need to find the square root of 93 or √93 as we would write it onto our calculator.

This gives us the answer 9.643650761 or if we’re doing it to 1 decimal place 9.6. Our final answer to this problem is x = 9.6. It’s as simple as that. Let’s just go back over it all to make sure we got every step right.

So, in this case, we turned the an into the x. We turned the b into 14 and we turned the c into 17. To help us with this we can also label the triangle as well, just to make sure we could visualise exactly what we needed to do. This left us with x^{2} + 14^{2} = 17^{2}

For this we had to find out the answer to 14 squared and 17 squared. This gave us 196 for 14 squared and it gave us 289 for 17 squared. At this stage we were left with x^{2} + 196 = 289

This time we will need to move the b value over to the c side. This requires us to do the inverse to it. Once we had completed it, we were left with x^{2} = 289 – 196 which we can clean up a little bit to be x^{2} = 93

Our final step is to square root whatever answer we got after we moved the b over to the same side as the c and then subtracted b from c. In this case we needed to find out the value of √93. This means that after rounding our answer to 1 decimal place our final answer is x = 9.6.

These are the two types of question you will encounter when you first begin to learn about Pythagoras’ theorem. There is everything you need inside here to be able to solve them without any trouble whatsoever. We’re not quite finished yet though, remember near the start when I said that you could use Pythagoras in a real-life situation? Neither of those examples are really useful in the real world, are they?

There are a number of different useful real-life examples, but we’ll just share two for you to help you understand why it is so useful. The first example would be placing a ladder up against a wall. When you put a ladder up against a wall it needs to be a certain distance away otherwise it won’t be safe and stable. If you need to get 5 metres up the wall and the ladder needs to be 3 metres away from the wall to stay safe, you will be able to use Pythagoras in order to work out how long your ladder will need to be. This isn’t just a really useful real-world application, but it could also save your life! You don’t get much more useful than that.

What else can we use Pythagoras for? What about finding out which size wall bracket you need for your TV? If you don’t know the size of your TV, you can just measure the width and height and from there you can work out the diagonal length of your TV. This will give you the size of your TV for the purposes of purchasing the correct bracket and you can avoid any TV bracket-based accidents. Some people would say that ensuring your TV is okay is more important than saving your life, but that’s a debate for another day.

Of course, these are just real-world applications that anyone can use, there are many of other uses for Pythagoras’ theorem that would interest architects and builders. It is one of the most practical and useful areas of maths that there is.

With this introduction you will be able to access it and in the future move forwards onto more complex and difficult problems, as well as solving real world examples. Just remember the differences between the two basic problems, when you need to find the hypotenuse you add the a and b, when you already have the hypotenuse you move the b over to the same side as the c and subtract. If you remember that, you won’t go wrong.