Lesson video

Hi, there.

My name is Mr. Tilstone.

I'm a teacher, and it's my great pleasure to be here with you today to guide you through this lesson, which is all about area and perimeter.

It's my favourite kind of maths today because it's going to involve some meaty problem solving and reasoning opportunities.

Sometimes in maths, the answer comes to you straight away.

Sometimes you know what to do straight away, but sometimes you have to do some thinking.

Sometimes you have to take some time to think exactly what's being asked of you and what your strategies might possibly be to solve the problems. And that's definitely going to be the case today.

So persist and persevere, and you'll get there in the end.

If you're ready, I'm ready.

Let's begin.

The outcome of today's lesson is I can reason about shapes using the relationship between the side lengths and perimeter.

And our keywords today.

My turn, area, your turn.

And my turn, perimeter, your turn.

I think you probably know what those words mean by now.

I hope so, but shall we just have a quick reminder? Area is a measurement of a flat surface and measures 2D space, and the distance around a 2D shape is its perimeter.

So do take some time to make sure that you've got those in the right order and you know which is which because they are all easily confused.

Our lesson today is split into two cycles.

The first will be reasoning about perimeter and the second reasoning about area, but they're going to be interlinked throughout.

And if you're ready then, let's start by thinking about perimeter and some problems to do with that.

And in today's lesson, you're going to meet Jun and Laura.

They're here today to give us a helping hand with our maths.

Have you met them before? Jun and Laura are exploring problems with rectangles in their maths lesson.

And Jun says, "To find the perimeter of a rectangle, I like to double the length, double the width and add them together." So that is one possible strategy.

Have you got a different strategy? He says, "I know I can multiply the length and width to calculate the area." That's true.

If the side lengths of the smaller rectangle were known, would it be possible to work out the perimeter of the larger rectangle? What do you think? Well, let's not even think about numbers at the moment.

Let's think about this as one side.

And you can see that side here again and again and again and again.

And what about this side, the shorter side? How many times can you see that? You can see it here and here and here and here.

Now let's start to think about some numbers.

So if the long side was five centimetres and the short side was two centimetres, could we work out the perimeter of that larger rectangle? There's lots of ways to do that.

Some are more efficient than others, but here's one.

Four lots of the long and short edge of the smaller rectangle make up the larger rectangle.

So it's most efficient to do seven centimetres multiplied by four.

So the larger rectangle has a perimeter of 28 centimetres.

This square has an area of 16 centimetres squared.

Remember that's how we read that small number two.

The square is folded in half and then folded in half again.

So folded in half.

And again, what's the perimeter of the final rectangle? Hmm.

Okay.

Well, this time, we need to think about area first and then perimeter.

So a square's got equal sides.

A number squared gives an area of 16 centimetres squared.

So what number is that? Let's think about a number.

Is it one squared? No, that's one.

Is it two squared? No, that's four.

Is it three squared? No, that's nine.

Is it four squared? Yes, that's 16.

So that means that the side length is four centimetres.

That's four centimetres by four centimetres.

Now, it's been folded.

That square's been folded.

One of those dimensions is going to change, but one's going to stay the same.

So what could the dimensions be this time? It's been folded in half, but it's been halved, that side length has been halved, so now it's four centimetres by two centimetres.

They're the dimensions.

Then it's folded in half again.

What's going to change? What's going to stay the same? There we go.

What are the dimensions? The four centimetres hasn't changed and one has been halved, so that's now four centimetres by one centimetre.

That still hasn't told us the perimeter though, but we have got enough information now to work out the perimeter.

What could you do? What would you do? Now you've got those dimensions.

Add them all together.

When the sides are added, it gives a perimeter of 10 centimetres.

And you might have a nice efficient way to do that.

For example, I knew double four was eight and I knew double one was two, so I just did eight add two.

Lots of ways to do it though.

You might have done four add one is five and double to make 10.

Either way, we've got the perimeter of 10 centimetres.

Let's have a check.

If the perimeter of a square is known, how can the length of one side be calculated? What do we need to do? Do we need to multiply the perimeter by four? Divide the perimeter by four? Or halve the perimeter? Pause the video.

What do you think? We know the perimeter.

So what do we do to calculate the length of one side? It's a square, so the sides are all equal in length.

We divide it, we divide it by four.

It's time for some practise.

So it's time to get those thinking caps on.

Remember to take your time with these.

Make sure you've understood the question before you dive in.

Have a good think.

If you can work with somebody else, that's always good.

So number one, a larger rectangle is made of three congruent rectangles.

You might have heard that word before.

Congruent.

They're all the same.

So what is the perimeter of the larger rectangle? Have a think about that.

Think about what you know and how you can use what you know.

Number two, the large square has an area of 100 centimetres squared.

So can you see a large square made up of four smaller squares? And the question is, what's the perimeter of one of the smaller squares? There's a few steps that you're going to need to go through to find the answer to that.

It won't come to you straight away.

And then B, what would the perimeter of each small square be if the large square had an area of 16 centimetres squared, 36 centimetres squared, and 64 centimetres squared? And number three, this square has an area of 36 centimetres squared and it's been cut into three equal rectangles.

So it's a little bit like that folding problem, isn't it, that we looked at before? What is the perimeter of the smaller rectangle? Have fun investigating that.

Work together if you can.

And I'll see you soon for some feedback.

Welcome back.

How did you get on? Have you been doing lots of thinking and lots of problem solving? Well, number one, did you find the perimeter of the larger rectangle? Now, you might have noticed something.

The way the shape is composed, it's possible to see that the short side of a smaller rectangle is half the length of the long side.

Did you notice that? That's what good mathematicians do.

They notice things.

That means that side length is eight centimetres and we can use that, because now we can say what all of the parts of the perimeter are, so we've got lots of 16 centimetre bits and lots of eight centimetre bits.

And in whatever efficient way you chose to do it, you can add those together.

You might have used a bit of multiplication there rather than adding, but the perimeter is 80 centimetres.

Very well done if you've got that.

And number two, the large square has an area of 100 centimetres square.

What's the perimeter of one of the smaller squares? Well, each small square must have an area of 25 centimetres squared because it's one-quarter of the area of the large square.

So that's step one.

The side length of the small square must therefore be five centimetres because five squared or five times five equals 25.

So therefore, each small square side length is five centimetres.

And then five centimetres times four is 20 centimetres.

So the perimeter of the small square, each small square is 20 centimetres.

Well done if you got that.

That took a few different steps, didn't it? But if you break it down and treat it as some different small steps, you get there in the end.

And B, the same strategy can be applied for different areas.

So if it's 16 centimetres squared, that makes the perimeter of each small square eight centimetres.

And you might start to notice a little pattern here.

If it's 36 centimetres squared, it's 12 centimetres for each small square perimeter.

And if it's 64 centimetres squared, it's 16 centimetres for each small square perimeter.

Did you notice that the areas are squares of even numbers, four, six, and eight, and that the resulting perimeter of the smaller square increased by four centimetres each time.

Maybe you carried on, maybe you investigated the next even square number.

Number three, the square has an area of 36 centimetres squared.

It's been cut into three equal rectangles.

What's the perimeter of the smaller rectangle? So a little bit like the folding problem from before, but this time into three parts.

Well, the side lengths of the square are six centimetres because six squared equals 36.

The short side of the small rectangles is one-third the length or two centimetres.

So the longer side remains six centimetres and then six centimetres plus two centimetres plus six centimetres plus two centimetres equals 16 centimetres.

And you might have had a more efficient way of getting to the answer.

And the perimeter is therefore 16 centimetres.

Very, very well done if you got that.

You're doing it ever so well and it's time now to move on to the second cycle, which is reasoning about area.

Jun and Laura are playing a game.

Laura has a rectangle with one of the dimensions and area already given.

Can you see that? So the area's given, that's 120 centimetres squared, and just one of the dimensions is given, which is eight centimetres.

She's covered up one of the dimensions with a sticky note and challenged Jun to work out the missing dimension.

He says, "I know that if you multiply one dimension by the other, you get the area." So eight times something equals 120.

Eight times 15 equals 120.

You might have used division to get there as well, but it's eight times 15.

So therefore, the missing side length is 15 centimetres.

That gives the area eight times 15 is 120.

Well done, Jun.

This can be thought of as a multiplication question as Jun has or as a division question.

You could have thought of it as 120 divided by 80 equals 15.

Let's do a quick check.

Let's play the game with you.

So we've got the area, it's 70 centimetres squared.

We've got one of the dimensions, which is 14 centimetres.

What must the other one be? You might use multiplication here and you might use division.

Let's have a go.

Pause the video.

Five centimetres, well done if you got five centimetres.

Five times 14 is 70 or 70 divided by 14 equals five.

Two possible ways to arrive at that answer.

If the side lengths of the smallest rectangle were known, would it be possible to work out the area of the larger rectangle? So this might look familiar to you.

You might remember this from the first cycle.

This time, we're thinking about area.

So if we knew those side lengths, what could you do to work out the total area? What would you do? The larger rectangle is four times the size of the smaller rectangle.

It's got four times the area if you like.

So multiplying the two known side lengths gives the area of the smaller rectangle.

And then multiply it by four to get the area of the larger rectangle.

So yeah, it's possible if you know those dimensions.

The larger rectangle has a size twice as long as a smaller rectangle.

Did you notice this? So the side lengths could be doubled and then multiplied together.

So that's an alternative strategy.

Let's have a check.

Let's put some values in there.

What's the area of the larger rectangle? If the side lengths are five centimetres and two centimetres for the smaller one, what's the area of the larger rectangle? Pause the video and have a go.

Well, the arithmetic here was fairly gentle, so if you knew what you were doing, I don't think it would've been too complicated.

Five times two is 10, so the area of the small rectangle is 10 centimetres squared and we just multiply that by four.

So the area of the larger rectangle is 40 centimetres squared.

Well done if you got that, you're on track.

This picture, including the frame, of a picture of Jun, has an area of 72 centimetres squared.

So the picture plus the frame equals 72 centimetres squared.

The frame is one centimetres in width all the way around.

Let me show you what I mean by that.

So each part of the frame has a width of one centimetres.

What possible dimensions could the outer edge of the frame have? Hmm.

Well, the area is 72 centimetres squared.

So let's start first of all by thinking about the factors of 72.

The numbers that multiply together to make 72.

Factors come in pairs.

What are the factor pairs for 72? Lots of them.

You can be systematic about how you approach this as well.

I always like to start with the lowest possible number, which is one, so one times 72 is 72.

Two times 36.

What could the next one be? Three times 24 equals 72.

Is there another one? What's the next one? It's four times 18.

Is there another one? Could we use five? No, that's not a factor of 72, but six is.

Six times 12.

So times tables factor as well.

Is there another one? Yes, eight centimetres times nine.

Another times tables factor.

They're factors of 72 as well.

So we've got some possible lengths and widths.

But the question is, do all of those seem reasonable for the dimensions of a photo frame? Hmm? Which ones could it be? Could you, for example, have a photo frame that was one centimetre, one centimetre in width and 72 centimetres in length? Oh, that'd be a really odd photograph frame, wouldn't it? It'd be really long and thin.

So some of them are just not plausible, but some of them are.

The rest are, so it could be four times 18, could be six times 12, could be eight times nine.

They could be the outer dimensions.

Let's use these dimensions for the outer edge of the frame.

Let's pick one.

It could have been any of the three, but let's go for six centimetres by 12 centimetres.

That sounds plausible.

So six for the short side, 12 for the long side.

Just like that.

So the whole picture has an area of 72 centimetres squared including the frame.

That means the frame has dimensions of six centimetres by 12 centimetres in this case.

But remember the frame is one centimetre in width all around the photo.

So that means a photo has got different dimensions.

Smaller dimensions.

What would they be? It's one centimetre less.

Can we calculate the dimensions in area of just the photo if we've got enough information? Yes, we have.

There's one centimetre of frame on each side of the photo.

So the dimensions are two centimetres shorter than the frame.

Its dimensions in this case then must be four centimetres and 10 centimetres.

Do we now have enough information to work out the area of the photo? Yes, we do.

What do we do with those two numbers? We multiply them together, four centimetres times 10 centimetres equals 40 centimetres squared.

And that's the area of the photo part.

The photo inside has dimensions of four centimetres and 10 centimetres and an area of 40 centimetres squared.

So we know the photo is 40 centimetres squared in area.

We know the photo plus the frame is 72 centimetres squared.

So what about the area just of the frame? What could we do? Hmm.

Could use subtraction.

72 centimetres squared subtract 40 centimetres squared equals 32 centimetres squared.

And that would be the area of just the frame.

Let's have a check.

A picture frame has an area of 72 centimetres squared.

Just like before.

Let's pick a different pair of dimensions.

If the dimensions of the frame were eight by nine, eight centimetres times nine centimetres, and the frame was still one centimetre in width all the way around, what would the area of the photo be and what would the area of the frame be? So use the same steps as we've just done.

You might not remember exactly what they were, and if not, that's fine.

Just have a little think about what you could do.

Pause the video, work with somebody else if you can, and compare strategies and answers and I'll see you soon for some feedback.

Let's have a look.

Did you come up with an answer? Well, the dimensions of the photo would be six centimetres by seven centimetres because it's two centimetres less for each of the dimensions because of the photo frame.

And the area of the photo will therefore be 42 centimetres squared.

And that's a times tables fact.

So hopefully you got that straight away.

Hopefully that was automatic for you.

Now, we know the whole area was 72 centimetres squared.

So if we subtract the 42 centimetres squared, which is the photo part, it gives us the area of the frame part, which is 30 centimetres squared.

Very well done if you got that.

You're definitely on track and you're definitely ready for some independent practise.

And here are the practise questions.

So number one, calculate the missing dimensions.

So you've got the area and you've got one dimension.

What must the other dimension be? And remember, you might want to think of it as multiplication or division, whichever suits you.

Number two, the shape is composed of three identical rectangles.

What's the area of the shape? It doesn't look like you've got much information there, but you have got enough information.

You've just got through a little bit of thinking and a little bit of noticing.

And number three, a picture including the frame has an area of 84 centimetres squared.

The frame has a width of one centimetre all the way around.

So it's a bit like before, but the picture in the frame got a slightly bigger area this time.

So what could the dimensions of the frame be? And B, choose some reasonable possible dimensions.

So not too narrow.

For each of them, what must the area of the photo be and what must the area of the frame be? So there's lots of possible answers you can give here.

Don't stop at just one, keep going.

And number four, if a window plus the frame around it have an area of 960 centimetres squared.

We're looking at bigger numbers here.

It's quite similar though to the previous question, isn't it? What could the length and width of the frame reasonably be? Think about factors.

And then B, the frame is two centimetres in width this time all the way around.

Not one centimetre this time, two centimetres.

What could the area of the glass be and what could the area of the frame be? Give as many answers as possible.

Remember, don't stop at just one.

There's lots of answers here.

Good luck with all of that.

If you can collaborate, I can't recommend that enough.

Share your ideas and strategies.

Okay, good luck and I'll see you soon for some feedback.

Welcome back.

How do you find that? Hopefully your brain's hurting a little bit.

That's good if it is, that means you've been doing some thinking.

So number one, calculate the missing dimensions, so you could use division or multiplication here.

But in the first case, eight times 14 is 112.

So 14 is a missing side length, 14 centimetres.

And for B, how many times does 15 go into 165 or 15 times what is 165? Well, I know that ten 15s is 150, so therefore eleven 15s is 165.

So the missing dimension is 11 centimetres.

Well done if you got that.

You might have done it differently to me, that's fine.

And the shape is composed of three identical rectangles.

What's the area of the shape? Doesn't look like we've got much information, but we have because you might have noticed that each short side length of the rectangle is seven centimetres.

I've noticed that from that.

So the way it's composed, you can see that, so that's seven centimetres.

And now we've got enough information to work out the area of one rectangle.

It's 98 centimetres squared and we've got three of those and three times 98, you might have found an efficient way to do that.

For example, I started up by thinking about three times 100, but whatever strategy you chose, the area is 294 centimetres squared.

Very well done if you got that.

And for three, the picture and the frame have got an area of 84 centimetres squared and the frame's got a width of one centimetre all the way around.

So we've done a similar question to this before.

We'd start off by finding the factor pairs, and some of these would be reasonable side lengths for the photo frame and some of them would not.

So we've got one times 84, two times 42, three times 28, four times 21, six times 14, and seven times 12.

And being systematic helped me to find all of them.

Now, these dimensions are particularly reasonable or plausible.

I could imagine photo frames with those dimensions.

Okay, now what could the area of the photo be and what must the area of the frame be? So let's pick one of those.

So let's say six centimetres times 14 centimetres.

The dimensions of the photo are four centimetres by 12 centimetres and the area of the photo is therefore 48 centimetres squared.

That's not quite answered all of it.

That's the photo.

What about the frame? We're gonna use our subtraction here.

84 centimetres squared subtract 48 centimetres squared equals 36 centimetres squared.

And this is the area of the frame.

Now let's take a different one.

If it was seven by 12, that would make the dimensions of the photo five by 10, so 50.

50 centimetres squared for the photo area.

84 centimetres squared subtract 50 centimetres squared equals 34 centimetres squared.

And that is the area of the frame.

But number four, factor pairs to 96 can be used as a starting point to give factor pairs to 960.

So for example, if you know 24 times four is a factor pair to 96, that means 24 by 40 is a factor pair of 960.

And that's one of many possibilities.

So the dimensions of the glass will be four centimetres less than the dimension of the frame this time because it was a two centimetre width, so in this case 20 times 36.

Well, I did 10 times 36 and doubled it.

That's 720.

Now to finish off that question, 960 centimetres squared, that's the entire area, the total area, subtract 720 centimetres squared equals 240 centimetres squared, and that will be the area of the window frame.

That's one of many possible answers.

Well done if you got that and well, and if you got different ones as well.

We've come to the end of the lesson.

Today, we've been reasoning about shapes using the relationship between side length and area and perimeter.

In rectangles, there's a direct link between side length, perimeter, and area.

They're all interconnected.

When one of these pieces of information is missing, it's often possible to work out the other pieces of information.

You just gotta do a bit of thinking.

These skills can be applied in problem-solving situations, which you've just done there.

Very well done on your amazing accomplishments and achievements today.

It's been a tricky lesson, but I hope you've enjoyed that challenge.

Give yourself a little pat on the back.

It's very well deserved.

I hope I get the chance to spend another maths lesson with you in the near future.

But until then, enjoy the rest of your day whatever you've got in store.

Take care and goodbye.