Pythagoras' theorem states that, in any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares on the other two sides.

The longest side of a right-angled triangle is the hypotenuse. The hypotenuse is always opposite the right angle.

Pythagoras’ theorem can be used to calculate the length of any side in a right-angled triangle. Pythagoras’ theorem can be applied to solve 3-dimensional problems.

In this lesson we are learning about Pythagoras's theorem. We're going to find the length of a side on a right angle triangle using this theorem. As you can see, at the top of the screen, there is is a lilac box called that says key terms. And I will be referring to these key terms as we go through the lesson. So the main aims of today's lesson are to learn Pythagoras theorem and understand how it relates to the areas of squares on a right angle triangle.

So we're initially going to actually derive the theorem, and then we're going to use the theorem to answer geometrical questions that are related to exam style questions as well. So here we have a right angle triangle on the left side of the screen. So Pythagoras was a mathematician who wanted to see if he could find a link between the sides of a rectangle triangle. So he decided to draw squares on each of the sides of the triangle. So he looked at the areas of the squares that he drew on to see if there was a link between those three.

So if I show you a diagram of this, so you can actually see the areas. So if we look at this pink square here, the area of the small square is nine. The next smaller square, which is this orange one here, the area of that is 16, and the area of the largest square is 25. Now you may have found the link between 916 and 25. So Pythagoras basically summarised that if you find the area of the two smaller lengths and you add them together, it will give you the area of the largest square.

And this is now known as Pythagoras's Theorem. So remember, square, all the sides are the same. So if I square this length, it will give me the area of the square at the bottom. And if I square this height, it will give me the area of this square. And we can put that together into a theorem.

A squared plus B squared equals C squared. So A, B and C all relate or refer to the sides of your triangle. So what's really important is some of these key terms. The hypotenuse. The hypotenuse is the longest side on your triangle.

And what's really important is that this side is labelled as C. The other two shorter sides can be A or B. So I could have had the bases B and the height of A. What's really important is that your Hypotenuse is C. Okay?

So once we've labelled our three sides, we can use the theorem to find the length of C. Here is an example and we'll work through this together. So we've got a base of six, a height of eight, and a hypotenuse length of X. So we need to label this triangle A, B and C. So don't forget that the hypotenuse, the slope, is all they see.

Sometimes they can rotate these diagrams around and so the slope ends up being on the base. Just remember, it's always the side opposite, directly opposite the 90 degree symbol. Okay? So that's always the Hypotenuse, the slant. The other two will be A or B.

So those are my sides labelled. And to find my value of X or C, I need to substitute the given values into this theorem. So the value of A is six, the value of B is eight, and the value of C is X. So, as you can see, I've replaced A with six, B with eight, and C with X. Now, a lot of students at this point will get a calculator out and plug this straight into their calculator.

But I always say, even if you use a calculator, it's fine, but always have another line of working here and that is just squaring these two values. It's actually very good for when you're going back to cheque your work, because you'd be able to note any numerical errors. You've made it's clear to the examiner what you've done as well, and it's just mathematically quite neat for a solution as well. So you should lay your working out like this. Okay, so once you've squared your two values, add them together and we've got a square with an area of 100.

But remember, I only want this one side length. So if X squared is 100, how could we find X? We would square root, that is the opposite of square on something, or that's how you undo a square. So the square root of 100 is ten. And don't forget the units which were given in the diagram as centimetres.

So my answer is ten centimetres. So to summarise, label your triangle, substitute your values into your theorem and then work accordingly. And at the end, once you've found C squared square, read it to find justc so that's how we use the theorem for finding the hypotenuse. But sometimes you can be asked to find a shorter side. So all you have to do is to actually take the two given sides away.

So it's the longest side, take away one of the shorter sides. Now, the reason why I've got two theories there is because it depends on what you label as your missing site. So remember, A and B were interchangeable. So if you're looking for A, you will use this theorem, and if you're looking for B, you will use this one. So let's have a look at another example.

And we'll be using the finding a shorter length. This time, label your three sides. Again, always label your hypotenuse first as C and then the other two as A or B. And as you can see, I'm looking for B. So I'm going to use the formula that asked me to find B squared.

And just like in the previous example, I'm going to plug in or substitute in my values. So C was 13 and a was five and B was X. Again, the next line of solution should be just working out what 13 squared is and what five squared is. Remember, that's 13 times 13 and five times five. Take the two values away and you've got X squared.

We are looking for just X. So I'm going to square root that gives you a length of twelve centimetres. So we've just used the theorem to find not only the hypotenuse, but also one of the shorter sides as well.

So what I've got to finish this lesson off is two exam style questions for you to try. You may use a calculator for them both. I would like you to lay it out out in the same way as we've laid out the solutions to the other two examples. And feel free to leave your solutions on my profile so that I can have a look and cheque it for you. Thank you very much.


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Sanaa A

Sanaa is currently Deputy Head of Maths at a secondary school and has been tutoring GCSE Maths students for over 10 years. She tailors her approach so that no student is left behind.

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