Lesson video

Hello there.

I'm Mr. Tilstone.

I'm a teacher and I just love maths.

So today it's a real pleasure to be working with you on this lesson, which is all about perimeter.

You might have had lots of recent experience learning about perimeter and you may be becoming something of a perimeter expert.

Let's see if we can tie all of that knowledge together to solve some perimeter-based problems. Are you ready? Let's go.

The outcome of today's lesson is, "I can solve problems involving the perimeter and side lengths of polygons." And we've got two keywords.

My turn, perimeter.

Your turn.

And my turn, regular.

Your turn.

You might have been using those words quite a lot recently.

Hopefully you have and you've got a bit of knowledge, but why not have a little reminder.

Perimeter, the distance around a two-dimensional shape is called the perimeter.

Here's a rectangle and there's its perimeter.

And a regular polygon has all sides equal and all angles equal.

So some polygons are irregular, but we're gonna focus today on regular polygons.

And here are some examples.

They're regular polygons.

Our lesson is split into two cycles.

The first will be different polygons with the same perimeter, and the second, different polygons with the same side length.

Let's start by focusing on different polygons with the same perimeter.

In this lesson you're going to meet Laura and Lucas.

You might have met them before.

You might have met them recently.

They're going to help us out with our math today.

A regular triangle.

So picture a regular triangle and a square, which is a regular shape, have the same perimeter, and it's a whole number.

What could the perimeter be and what could the side lengths be? Give as many answers as possible.

You might want to have a bit of time thinking about that before we explore it together.

Lucas says, "I'm going to think about the triangle." Okay.

"It's got three sides." Yes, he's right.

Three sides all the same length.

"If I start with sides of one centimetre, I can use this to find different perimeters." That's a good idea, Lucas.

He's being systematic about it.

So yeah, just like this, that would give a perimeter of three centimetres.

Because one centimetre times three equals three centimetres.

Yes, I see what he's doing.

Now he's trying two centimetres for the side lengths.

Two centimetres times three is six centimetres.

What do you think he's going to try next? Well done if you said three centimetres for the side lengths.

Three centimetres multiplied by three equals nine centimetres.

The perimeter increases by three each time.

Each time the side length goes up by one centimetre, the perimeter increases by three.

"I can use this to find other perimeters without even drawing." That's a good idea.

That will save time.

Let's create a table to show that.

Lucas loves tables, by the way, and I can't fault him for that.

It's a good way to present your findings in maths.

Here we go.

So he's making a table.

This is the first column of a table and it's thinking about the perimeter of a triangle.

So remember that first one was three centimetres and then we're going up in threes.

So six, nine.

Can you see what's going on here? 12.

Going up in threes.

15, 18.

What's coming next? 21.

And what's next? 24.

They're multiples of three.

Laura says, "I'm going to add another column to this table." Good teamwork, good collaboration.

I like this.

"For the perimeter of the square." Okay, so let's think about the perimeter of the square.

If we're being systematic, like Lucas was.

Got four sides, so its perimeter goes up in fours.

So yes, if each side length was one centimetre, that would be four centimetres, two centimetres, it would be eight centimetres.

It's going up in fours this time.

Four, eight.

Can you do this with me? 12, 16, 20, 24, 28.

And finally 32.

What do you think we can do now? We haven't answered the question yet.

We've done lots of work, but we've not answered the question.

Remember the question was, "A regular triangle and a square have the same perimeter." Same perimeter, what could it be? Well, Laura says, "Let's see which perimeters appear in both columns." Good idea, Laura.

Hmm? Can you spot any? Can you spot a number that comes up in both columns? Yes.

12.

So the triangle is possible for the triangle to have a perimeter of 12 centimetres.

And it's also possible for the square to have a perimeter of 12 centimetres.

So they could have that perimeter, they could both have a 12 centimetre perimeter.

Was there another one? There was.

24 centimetres.

24 is a multiple of three.

And 24 is a multiple of four.

That number appears in both lists.

So the triangle could have a perimeter of 24 centimetres and the square the same.

Now that's not fully answered the question yet.

That's what could the perimeter be, what about what could the side length be? Well let's start with the 12 centimetre possibility.

That was the first one that we found out.

So there's a sketch of both shapes, regular triangle and a square with a perimeter of 12 centimetres.

What can we now do to find out the side length of that triangle? Hmm.

And what about the square? It's got three sides.

So three times something equals 12.

Three times four equals 12.

So we're using our times table's knowledge.

Lucas says, "The triangle has side lengths of four centimetres If the perimeter is 12 centimetres." What about the square there? Well, four times something equals 12.

The square's got four sides.

So four times three equals 12.

So the side lengths of the square would be three centimetres.

Let's try a different possibility.

We also said that 24 appeared in both lists.

So that's a possible perimeter.

Let's see if we can work out what the side lengths will be.

Use the same principle as before.

Same calculations.

What are we gonna do this time with that 24? Three times something equals 24.

Three times eight equals 24.

You can do that with division as well.

24 divided by three equals eight.

So the side lengths this time are going to be eight centimetres.

What about the square? Well, the square's got four sides.

So what can we do with that? Four times something equals 24 or 24 divided by four equals something.

And that is six.

Again using our times table skills.

So the side lengths will be six centimetres in this case.

The square has side lengths of six centimetres.

Now we've answered the question.

Laura and Lucas record their results in a new table.

I told you they love tables.

Tables are great.

So that perimeter, when it was 12, the triangle side length was four, because three times four is 12.

And the square equals three, because four times three equals 12.

And when it was 24 the triangle side length was eight because three times eight equals 24.

And the square side length was six because four times six equals 24.

Now can you start to see something? Do you notice something? Lucas notices that the perimeters have a difference of 12.

Did you notice that? 12, 24.

Hmm.

He wonders if adding 12 to 24 will give another equal perimeter.

What do you think? Hmm? Is that the next possibility? It is.

So if the perimeter is 36, what would the triangle side length be? Three times what equals 36? Three groups of 12 is 36.

So that's 12.

What about the square side length? Four times something equals 36.

It's still a times tables fact.

Four groups of nine is 36.

A square side length is nine centimetres.

And we could go on and on.

It works though.

He's right.

Let's have a check.

A square and a regular pentagon this time.

"A square and a regular pentagon have the same perimeter, which is a whole number." Just like before.

"What could the perimeter be and what could the side lengths be? Give as many answers as possible." Now you're not going to do that, but I want you to explain to a partner the steps that you could take to find some solutions.

So if you were to do it, what would you do? How would you do it? Pause the video and discuss that.

What did you say? Well you might have said something like this.

Make a table, like Laura and Lucas did, counting in fours for the square 'cause it's got four sides and five for the pentagon because it's got five sides.

Find which perimeter numbers appear in both columns.

And straight away I can think of 20.

That would be in both columns.

That's a multiple of four and a multiple of five.

When you have found a solution, divide it by four for the square and divided by five for the pentagon.

And this will give the side lengths.

Repeat for other possibilities.

Well done if you said something along those lines.

Time for some practise.

Number one, "A square and a regular pentagon have the same perimeter, which is a whole number." So just like we've just explored, What could the perimeter be and what could the side lengths be? Give as many answers as possible.

And then number two.

You may wish to explore different pairs of regular shapes.

And number three, for an extra challenge, if you're up for that, explore three regular shapes with the same perimeter.

Hmm, three regular shapes.

Alright, good luck with that.

If you can work with somebody, do that.

Look what happened when Laura and Lucas worked together.

Brilliant things happen.

So if you can, you do it too.

Alright, pause the video.

Good luck.

Enjoy and I will see you soon.

Welcome back.

How was your exploring? Were you systematic? Did you find some different possibilities? Let's have a look.

So square and a regular pentagon.

Square's got four sides.

Pentagon's got five.

What could the perimeter be? What could the side lengths be? Well, I'm gonna count in fours for the square.

Going to count in fives for the perimeter.

I'm gonna look for a number that appears in both.

20.

So you may have used your times tables to work systematically.

For the first six multiples of four and five, a common perimeter is 20.

And you may have extended your table all the way up to 12 times four and 12 times five and found a common perimeter of 40.

So 40 also appears in both lists.

So we've got 20 and 40.

I think I know what the next one would be as well.

Do you? So here we go.

So if the perimeter's 20 for each of the shapes, we need four times something equals 20 and five times something equals 20.

Times tables Facts again.

Four times five is 20, five times four is 20.

So they will be the side lengths.

You may have explored some further possibilities.

So you might have looked at 40 for the perimeter.

The square side length will be 10, the pentagon eight.

You may have gone even further.

60 is a possible perimeter.

And that would give a square a side length of 15.

Now four times 15 is not a times tables fact, but we could find it by adding together the previous two times tables facts.

That would work.

And this is a times tables fact though.

Five times 12 equals 60.

So 12 will be the pentagon side length.

What about something like that? You might have gone really, really far and got something like 200.

Now there's a relationship here between the 20 and the 200.

I wonder if we could use that.

Four lots of five are 20, then four lots of five tens are 20 tens or 200.

So that would mean the square side length will be 50 and the pentagon side length 40.

You're doing really, really well.

Let's kick it up a notch.

Let's look at different polygons with the same side length.

Shape A is a square with a perimeter of 20 centimetres.

A new shape has been composed from copies of shape A.

What is its perimeter? So let's have a look at that.

You can see shape A and then you can see shape A four times to make a new square.

Laura says, "Well that's easy! If the little square has a perimeter of 20 centimetres and there are four of them in the big square," What do you think she's gonna say? Four times 20 centimetres equals 80 centimetres.

Hmm? Is Laura right? Is it that straightforward? Hmm.

No I'm afraid not.

Some of the sides would now be inside the big square.

So it's not just a case of multiplying by four.

Have a look at this.

So here's two of the sides of the square, the little square.

And look, they'd be inside so you're not counting all of the sides.

"The square is regular and has four sides," says Lucas.

20 centimetres divided by four equals five centimetres.

That's giving a side length.

So the side length is five centimetres.

"I can use this information on the new shape." I think Lucas is onto something here.

"Now I can count in fives." Okay, so if you know that's five, we've got five lots more times as well though.

So here and here and here and here and here and here and here.

"Or I can make it even quicker by using my times tables." Yes.

What times tables fact can you see there? Five times something? How many times can you see five? Five centimetres times eight equals 40 centimetres.

Is Lucas right? Yes.

But what could be more efficient? Have you got a good idea? Instead of five centimetres times eight, I can see 10 centimetres times four.

Yes, those two five centimetres go together to make a 10 centimetre side length.

And you can see that four times.

That's another times table's fact.

10 centimetres times four equals 40 centimetres.

So like a good mathematician, Laura and Lucas kept on finding even more efficient ways to solve that.

Great things can happen when you solve problems together in maths.

Absolutely.

Shape A is a square with a perimeter of 20 centimetres.

A new shape has been composed from copies of shape A.

So can you see that? You can see shape A three times? What is its perimeter? Hmm.

What do you think? Laura says, "I think this shape will have a smaller perimeter than the big square because it's made of three small squares, not four." Hmm.

Have a think about that.

Chew that one over.

Hmm.

And Lucas says, "I think it will have a longer perimeter than the big square because it's now a hexagon." Oh right.

So with more sides.

Hmm.

What do you think? Do you agree with Laura? Do you agree with Lucas or do you disagree with them both? Hmm.

Well, let's investigate.

It's five centimetres for each side length.

This time the shape's not regular so we can't do what we did before.

Now if it was, can you remember that previous shape that was composed of all fives like that? It's not that dissimilar because all we've gotta do is think about those two extra five centimetres and if you fold them inwards, there we go.

That's a perimeter.

So it's actually got the same perimeter as before.

So they were both incorrect.

So very well done to you if you knew that.

Five centimetres times eight equals 40 centimetres, it was the same perimeter.

Let's have a check.

Shape A is a square with a perimeter of 16 centimetres.

It's been duplicated.

So copies have been made to make a new shape.

What is the perimeter of the new shape? Pause the video, work with somebody if you can and compare answers and strategies and I'll give you the answer very shortly.

Did you come to an agreement about the answer here? Did you use the same strategy to work it out? Well let's first of all find out one side length.

So if it's got a perimeter of 16 centimetres, 16 centimetres divided by four equals four centimetres.

So each of those side lengths is four centimetres.

And you can see that four centimetres eight times or you might have said eight centimetres four times.

But either way, four centimetres times eight equals 32 centimetres.

Well done if you got that.

Here we go.

This is looking at the same shape has side lengths of eight centimetres.

Still gives it 32 centimetres for the perimeter.

It's time for some final practise.

Shape A has a perimeter of 20 centimetres.

Three larger squares have been composed from copies of shape A.

What is the perimeter of each new square? Number two, shape C is a regular triangle with a perimeter of 12 centimetres.

Each side is four centimetres.

The hexagon has been composed of copies of shape C.

What is the perimeter of the hexagon? Three, shape A is a regular hexagon with a perimeter of 12 centimetres.

Each side is two centimetres.

Copies of shape A have been arranged to make a new shape.

What's the perimeter of the new shape? Number four, shape D is a square with a perimeter of 16 centimetres and new shapes being composed of copies of shape D.

What is it's perimeter? Good luck tackling those problems. Have a good think.

Remember the answer won't come to you straight away.

Sometimes you need to do a little bit of digging, a little bit of thinking, a little bit of changing plans if things aren't working.

But keep persisting and keep persevering and get there in the end.

Alright, pause the video and I'll see you soon.

How successful were you there? Shall we find out? So here we go, shape A's got a perimeter of 20 centimetres.

What's the perimeter of each new square? Well we can use that 20 centimetres to work out that the side length of A must be five centimetres and that's the side length of the small square.

So that's got eight of those.

So eight times five is 40 centimetres.

That's got 12 of those, 12 times five is 60 centimetres.

This one's got 16 of those.

That's not a times tables fact though, is it? Hmm? Have you got a good strategy to work that one out? I think there's something you can do.

However, each side of the large square is 20 centimetres long, giving an easier calculation.

So if you think of one side as 20 centimetres, you can do 20 times four because you know two times four.

So four times 20 is 80 centimetres.

And two, shape C is a regular triangle with a perimeter of 12 centimetres, that makes each side four centimetres.

The hexagon's been composed of copies of shape C.

What's the perimeter? So that means each side length is four and we can see that side length six times because it's a hexagon and that's 24 centimetres.

The perimeter is 24 centimetres.

And shape A is a regular hexagon with a perimeter of 12 centimetres.

Each side is two centimetres.

Copies of that have been arranged to make a new shape.

What's the perimeter of the new shape? So how many times can we see that two centimetre side length in the new shape? Here we go.

There's one.

But we can see that 12 times.

So two times 12, it's a times tables fact, that's 24 centimetres.

The perimeter is 24 centimetres.

And shape D's a square with a perimeter of 16 centimetres.

A new shape's being composed of copies of it.

What's its perimeter? So we can work out the side length because it's a square.

16 divided by four or four times something equals 16.

That's four centimetres.

How many times can we see that four centimetre side length there.

? 10 times.

So that's four.

Four times 10 is 40.

There's 10 lots of the sides of the square.

So the perimeter is 40 centimetres.

We've come to the end of the lesson.

Today's lesson has been solving problems involving the perimeter and side lengths of polygons.

It's been quite a tough lesson, you've been having to do a lot of thinking, lots of reasoning.

So well done if you were successful today.

Different regular polygons can have the same perimeter.

When it is known, the side lengths of each can be determined by dividing the perimeter by the side length or using our times tables facts.

Different polygons can have the same side lengths.

And when the side length is known, new shapes can be created, both regular and irregular, with different perimeters.

You've been absolutely amazing.

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

It might be about perimeter or it might be about something else entirely, but a very well done on your accomplishments and your achievements today.

Take care.

Goodbye.