Lesson video

Hello, everyone and welcome to another maths lesson.

It is great to see you joining me, Mr. Gratton, here for a lesson where we will be exploring relationships between squares that are joined together at their vertices to create different types of triangles, including right-angled ones.

We won't be looking explicitly at Pythagoras's theorem until much later on in the lesson, but pause here if you want a sneak peek at what we'll be looking at today.

Let's refine our focus onto just right-angled triangles and see if we can consistently generate them from different squares.

Let's go.

Laura says if she picks any two squares and knows the areas of those two squares, it is possible to instantly know the size of this third square needed to create a right-angled triangle.

This is because the areas of the two smaller squares must add up to make the area of the larger square as we can see here.

16 + 24 = 40.

The area of the largest square must have an area of exactly 40 square units.

But Alex raises a very good point.

What if you do not know the area of the two smaller squares? Well, it's possible to calculate the area by measuring one side length of a square, in this case, five units in length.

Then squaring that length as the area of a square is its base multiplied by its height, which is of equal length to each other.

So the area of this square is five times five equals 25 square units.

Therefore, the area of this third square needed to create a right-angled triangle is 30 + 25 = 55 square units.

So for this check, what is the area of square B in units squared? Pause now to have a think.

I take the length, which is four units, and square it.

Four squared is 16 units squared.

For the next question, what is the area of square C now that I know the area of square B? Pause now to do this next question.

And as usual, I take the areas of the two smaller squares, 18 and 16 in this case, and add them together.

Therefore, 18 plus 16 is 34, and so the area of square C is 34 squared units.

And for this next check, find the area of each of these three squares.

Pause now to do this.

The area of square A is four squared, which is 16.

The area of square B is three squared, which is nine.

The area of square C is those two areas added together.

16 + 9 is 25 units squared.

Next up, what is the value of X? The side length of square C? Pause now to do this.

The side length of square C is going to be 25 square rooted, which is five.

Therefore, the side length of square C is five units.

And for this last check, once again, find the areas of squares A, B, and C.

Pause now to do this.

For A, three squared is nine.

For C, seven squared is 49.

And for B, I've got 49 take away 9, which is 40 units squared.

For B, I had to subtract rather than add because I know for certain that square B is not the biggest square in this diagram because it is not opposite the right angle, which is the largest angle in that triangle.

And onto the practise.

For question one, find the area of each unlabeled square.

Pause now to do this.

And similarly for question two, pause here to find the area of every square in each diagram.

And lastly, for question three, pause here to find the area of each square and find the missing length of each square that is labelled with a letter.

Onto the answers for question one.

A had a missing area of 22 units squared.

B had a missing area of 30 units squared.

And C had a missing area of 15 units squared.

And for question two, pause here to see if your answers match those that are on screen.

And once again, for question three, pause here to match your answers to those that are on screen.

We've looked so much at triangles and squares, but where does Pythagoras' theorem come into all of this? Let's see if it's anything new.

Sam seems to think not.

They believe that we've already done Pythagoras' theorem.

They say that the area of square A plus the area of square B equals the area of square C, which seems familiar when we think about the work that we've done in the previous cycles in this lesson.

And Alex agrees and identifies that we've all already done this, but only when dealing with right-angled triangles.

This is because we're looking at an equality.

The area of square A plus the area of square B equals the area of square C.

We've seen triangles with, for example, obtuse angles where the area of the largest square is massive compared to the combined area of the two other smaller squares, not simply equal to them.

Lucas raises a very good point.

What if we don't know the area of the squares? Oh, but haven't we asked this question before? What was the solution last time? Alex remembers, find the length of a side of the square, then square it to find its area.

And so there are many equivalent but different ways of looking at Pythagoras' theorem.

Area of square A plus area of square B equals the area of square C.

Or if we do not know the areas of the squares, we can consider the length of side A, then square it plus the length of side B, then square that, equals the length of side C also squared.

And this way of describing Pythagoras' theorem is a lot closer to one that you might have heard of before.

This rule can be further condensed down into a form like this.

Rather than saying the length of a side squared, simply say A squared plus B squared equals C squared where A squared, B squared and C squared are just the areas of each square found by squaring the length of each of those three squares.

But it is important to note that C squared must be the area of the largest square since the largest square is equal in area to the sum or addition of the other two squares.

However, Lucas is a bit sceptical.

We've seen it's true for some right-angled triangles, but the following proof can be adapted to any right-angled triangle no matter the size of the squares that make it.

And here's the proof.

We can take the side of the largest square that is shared by the triangle and translate it down onto one of the smaller squares.

This is usually the middle sized square, not the largest or smallest one.

We can then repeat this exact same process, but for one of the perpendicular sides of the largest square, like so, again translating it down onto the middle sized square.

What we've done now is split that middle square into four congruent quadrilaterals.

We can then take each of these four quadrilaterals and translate them into the largest square, fitting them within each corner of that square.

Doing this will leave a square sized gap in the middle.

And what do you notice? Oh, the area of that gap is exactly the area of the smallest square.

So what we've shown is a method of manipulating the two smaller squares from a right-angled triangle in such a way that they fit perfectly into the largest square from that same triangle.

This visual proof is not something that you need to learn or replicate, but it is important in visually showing why Pythagoras' theorem actually works.

And onto the check.

Select the statement that completes this sentence.

For Pythagoras' theorem to be used, three squares must join together to make what? Pause now to choose the correct option.

We use Pythagoras' theorem when we are dealing with a right-angled triangle.

And for this next check, which of these are suitable ways of describing Pythagoras' theorem for three squares which join at their vertices to create a right-angled triangle? Pause now to look through all four options.

And actually, all four of them are correct.

Each of these four options are things that we've looked at throughout this lesson.

And onto the practise questions.

On which of these three diagrams is Pythagoras' theorem true? And explain your reasoning and what does that tell you about those triangles? Pause now to do this question.

Which of these four options are valid representations of Pythagoras' theorem for these three squares which enclose that right-angled triangle? Pause now to do this question.

And lastly, pause here to answer question number four.

And here are the answers.

Only on diagram B is Pythagoras' theorem true.

This is because Pythagoras' theorem suggests that the sum of the areas of the two smaller squares equals the area of the largest square, and 14 plus 14 is 28.

This means that only B creates a right-angled triangle.

And for question three, A, B, and D are all valid representations of Pythagoras' theorem for this diagram.

C is not valid because it suggests that square F is the largest square.

We can see here that square H is the largest instead.

And pause here to have a look at the answers for question number four.

And thank you all so much for your effort today.

We have explored so many triangles from a range of different squares in a lesson where we have joined different combinations of three squares, sometimes where some of the squares are congruent and sometimes where none of them are congruent.

Depending on the combinations chosen, we can create equilateral, isosceles and scaling triangles.

We've also observed that the largest angle in the triangle can either be acute, right or obtuse.

And its size depends on how large the area of the largest square is when compared to the sum of the areas of the two smaller squares.

And importantly, if the areas of the two smaller squares are equal in size to the area of the largest square, a right-angled triangle is created and that is Pythagoras's theorem.

Thank you so much for making the decision to join me today.

I hope you can use this understanding for the rest of the Pythagoras' theorem topic.

I've been Mr. Gratton and until our next math lesson together, take care and enjoy the rest of your day.

Goodbye.