Lesson video

Hi, welcome to today's lesson on the perimeter of polygons.

By the end of today's lesson, you'll be able to use the properties of a range of polygons to deduce their perimeters, including generalising a perimeter formula where it's appropriate.

There are two sections to today's lesson, and we're gonna begin with section one on efficiently calculating perimeter.

Perimeter is the distance around a 2D shape.

Now, we can calculate perimeter perfectly fine by summing all of the edges of our shape together.

Repeated addition can be calculated more efficiently, though, through either the use of number bonds or through multiplication.

Now, the perimeter of this square can be found in a variety of ways.

Pause the video and write down as many ways as you can think to calculate the perimeter of this square.

Welcome back.

How many ways do you have? Do you have all the same as me? Well, our first one is, I could sum the four sides.

I could of course just multiply one of the sides by four.

That'll get me to the same because I've got four sides at the same length.

That's four lots of, in this case five, so 20.

Now I could have been a bit less efficient than five lots of four, or four lots of five, but been more efficient than just writing five add five add five add five.

I could have done two lots of five plus two lots of five.

I could have done three lots of five plus one lot of five but these are sort of in between.

They're not quite as efficient as four lots of five but they are more efficient than just repeated addition.

On the screen you can see Lucas's and Andeep's methods for calculating the perimeter of this irregular pentagon.

Which method do you prefer? Is there a scenario where you might use Lucas's method and a different situation where you might use Andeep's? What would determine whether one was better than the other? In other words, when might you use Lucas's method and when might you use Andeep's? Pause the video while you read through their methods and have a go at answering my questions.

Welcome back.

Let's have a look at what you've said.

Well, you could have said that you prefer Andeep's method because it's a lot shorter.

You could have said, for example, that Lucas's method, although you prefer Andeep's, it does still have a place.

Lucas's method of adding all the sides together is going to be useful if all the values are different.

Because remember, with Andeep's method, he spotted that he can see four, or rather length four, three times.

So he is going to multiply rather than add four repeatedly.

But if it wasn't the same length, then we wouldn't be able to do Andeep's method.

So actually, both methods have a place.

It just depends on what values the different length in our shape have.

In this task, I'd like you to calculate the perimeter of each polygon.

When you've done that, evaluate your approach and determine if you can be more efficient.

So what we mean is, for a, work out the perimeter of that polygon.

And when you've done it, look at your working and think carefully.

Could any of your steps been done more efficiently? Could you have skipped some steps? Remember, it is important to show your working.

Do you need every single step? Could you have replaced two steps with one different step, maybe? It's just as important to evaluate your working as it is to actually calculate the answer because by doing so, you can decide if you are being as efficient as you can be and look for ways to improve.

Pause the video now while you have a go at this task.

Welcome back.

Now it's time for d and e.

These polygons are a little more complicated and that we have quite a lot of sides here.

But we're asking you to do exactly the same thing.

Calculate the perimeter for each polygon and then evaluate your approach and determine if you could be more efficient.

It could be quite hard to spot how to be more efficient right at the start so don't worry if you don't spot it straight away.

Sometimes, it can take us looking back on our work to spot ways that we could be more efficient, but that's okay.

We can then learn that lesson and apply it to our future working.

Pause the video now while you have a go at d and e.

Welcome back.

Let's go through our solutions.

In a, the most efficient way to calculate this polygon is to realise that all of the sides are the same length.

We can tell that because of the notation.

Since I have here a hexagon, and therefore six sides, all of length three, I can calculate the perimeter by simply doing six lots of three.

In other words, 18 millimetres.

This is the most efficient method.

It's far more efficient than repeated addition because there are less steps overall in my calculation.

But again, if you've got the right answer by doing repeated addition, that's great.

It's just spotting if you can be more efficient.

We like to do that because as mathematicians, we prefer efficiency.

Because who wants to do more work than we have to? Let's look at b.

Did you spot that there are two sides of the same length? In fact, we have two pairs of sides of the same length.

Now, you could have written five add five add four add four.

You could have done two lots of five plus two lots of four.

What I've done here though is that I've used my understanding of the distributive law and said, well, if I add the five and the four together I can then just double it to get to the answer.

It's okay if you didn't remember this one.

After all, the distributive law is something we did look at a couple of units ago, but it may have slipped your mind.

It's nice to be reminded of it though.

And here, you can see it written out using brackets.

Two lots of the result of adding four and five together.

In other words, two lots of nine which is 18 centimetres.

Now this is more efficient than repeated addition because there are less steps in this calculation.

It is more efficient than saying two lots of four or two lots of five because the way we've got here is only two steps rather than three.

Let's look at c.

Now in c, I've had to use addition.

I couldn't spot a way to combine these numbers to make them into more efficient numbers to calculate with and I couldn't see any numbers that were the same so I couldn't use multiplication in place of repeated addition.

Now in c, I've just got to sum the sides together, which is what I've done to give a total perimeter of 28.

2 inches.

Let's move on and look at d.

Now you might be surprised at how short, efficient my working is considering the shape that we are greeted with.

Now I could see lots of ways to work this perimeter out.

But what I did realise is that there were two lots of six in my diagram.

I have a six on the left of my diagram and if I look at the two vertical sides on the right, I have four foot add two foot, that makes six.

So I actually have two lots of six if I consider all the vertical sides.

If I look at the horizontal sides, so that's the one at the bottom, the one at the very top, and the one in the middle.

Across the bottom, it's four foot.

And the two other horizontal sides are each two foot.

Well hang on, two plus two is four.

So I've actually got two lots of four there.

So I realised I have two lots of six and two lots of four.

But just like in question b, I can write that more efficiently.

It's two lots of six add four, making a total of 20.

Again, it's okay if you didn't come up with this efficient method.

You may have come up with one that is a bit less efficient but still better than adding them all together.

Again, if you didn't, and all you did was sum the sides that you can see, did you get to 20? If you did, then brilliant, because you calculated the perimeter.

The awareness you could have been more efficient is good and it's something to work towards but you've got the right answer and you should be pleased with yourself because of that.

Now, let's look at e.

e initially looks quite tricky, lots of decimal values there.

I'm not initially seeing anything the same at all.

But when I glance at the working I've done, it says 15 lots of three, 45.

How on earth did I get that? What did I do and what did I notice? I noticed I had some number bonds here.

8.

6 and 6.

4 sum to make 15.

I wonder if there are any other pairs of 15.

Well, I looked at 5.

2, and I'd want a number just under 10, so that would be, hmmm, 9.

8 looks possible.

And indeed, when I add 5.

2 and 9.

8 together, I get 15 again.

What about the 11.

3 and 3.

7? Summing them also makes 15.

Actually, what I've got here is three pairs of sides that all sum to 15.

Well, that's just three lots of 15 then, because three pairs of 15, three lots of 15, 45 metres.

Again, don't worry if you didn't spot this one.

I really had to know my numbers really well to be able to clock this.

And this is something that comes with practise.

So if you didn't spot it, all this means is a little bit more practise is needed and you'll be able to do this as well.

Let's now move on and have a look at section two, where this time we're gonna use formulas to calculate perimeter.

Perimeter can always be calculated through summing all of our edge lengths, and that's the thing to remember.

That will always work.

There are just times we can be more efficient.

Now we can use the properties of certain polygons in order to write a formula to calculate the perimeter in an efficient manner.

Our square has four sides of equal length.

Now we know that because it's a square.

And the properties of a square state that it is a quadrilateral, where all four sides are the same length.

If I know that one of the side lengths is a, can I write a formula to tell me how to calculate perimeter? Well, in order to find the perimeter, I need four lots or four multiplied by the length of one side.

In other words, to find the perimeter, or P, I need to do four times the length of one side or four times a.

So the perimeter of my square, or P, is equal to 4a.

You are presented with a regular octagon.

An octagon, remember, has eight sides.

And because it's regular, each side has the same length.

The perimeter of this octagon can be written, how? Pause the video and write a formula for the perimeter of your octagon.

Did you write P equals 8b? If you did, well done.

Remember, because we have a regular shape here, we know that the side lengths are the same.

There are eight lengths, all of length b, so that's eight lots of b.

In other words, P, perimeter, equals 8b.

Now let's consider the rectangle.

We know that our rectangle has a length and a width, or a base and a height, or a long side and a short side.

How could I write the perimeter of this rectangle? Well, I know that two of the sides are the same length and the other two sides of the same length.

So the perimeter of my rectangle can be written as two lots of a plus two lots of b.

In other words, the sum of the two short sides or two lots of the short side, so two lots of a, plus two lots of the long side, or two lots of b.

It's your turn.

And this time you have a kite.

The perimeter of the kite can be written as, what? Pause the video and have a go.

Did you write the perimeter was 2d plus 2e? Well done.

We know that kites have two sides of the same length and another pair of sides of the same length.

It's now time for your final task.

In this one, you have multiple steps.

The first is to match the shape to a formula that could be used to find its perimeter.

Now in this case, you'll need to work out what the a, b, c, t, r, w, or k, refer to in a particular shape.

So you might find it handy to try and label some of the sides using these letters to see which one you think goes with which shape.

Pause the video now while you have a go.

In part b, we'd now like you to calculate the perimeter of each shape using your chosen formula.

To help you out, I've now put some lengths on the shapes.

You should just be able to substitute these into your formula.

Substitution, remember, was a topic we covered when we looked at our algebra.

If you need a bit of a reminder, don't hesitate.

Go back and look at that lesson.

But if you feel confident, then use your formula to calculate the perimeter for each shape.

Pause the video now while you do this.

And finally, consider each answer.

Was the formula actually useful? Pause the video and write down your thoughts.

Welcome back.

In part a, we asked you to match each shape to a formula that could be used to find its perimeter.

The star that we can see has all of its sides the same length, and there are 10 of them.

Which formula shows 10 lots of one side? And that's right, it's the formula on the end where the perimeter is 10 lots of a side.

For the trapezium, I know I've got two sides that are different and two sides that happen to be the same.

So I'm looking for probably three different lengths, one of which is multiplied by two.

It must be the first formula then.

And then the last one, with my triangle.

It's not clear that any of these sides are the same length, so I need three different unique lengths.

No multiplying here at all.

Must be the middle formula.

In b, we asked you to calculate the perimeter of each shape using the formula.

Well, for our star then, the perimeter is 10 lots of one length, so that's 10 lots of five or 50 centimetres.

In b, the formula was a plus two lots of b plus Cc.

It doesn't actually matter which of the two sides is a or c, whether you use 3.

4 or eight.

The important bit is the side that's doubled is the six centimetre length because that's our slanted length and it's the one that's marked as being equal to the other slanted side.

So 3.

4 plus two lots of six plus eight, giving us a total of 23.

4 centimetres.

Now let's look at our final shape.

In our triangle, remember it just said to add the three lengths together.

So five add four add three gives us a total of 12 centimetres.

It's now time to reflect on whether or not the formula was useful to calculate the perimeter of the shape.

Now there are lots of different things you might have said.

For example, the formula for the star was helpful because I didn't have to do lots of addition.

It told you it's 10 lots of one side.

The formula for the isosceles trapezium was sort of helpful in that, well, I didn't have to add the six again, I could just double it.

But I was only doubling one value.

So perhaps not as useful as it might have been.

And the formula of triangle was not helpful at all because it was just adding everything together.

In fact, you might even have said that the formula for the triangle wasn't necessary because you know to find the perimeter, you're summing the sides.

So all it did was tell you what you already knew.

It's worth remembering that there isn't always an easier method or a more efficient method than one we already know, which is why we evaluate.

So we can be critical about what we are doing and whether what we are doing can be improved or not.

Let's summarise everything we've done in our lesson today.

Perimeter is the distance around a 2D shape.

You can always find the perimeter by summing the length of all the edges.

For certain polygons, it may be more efficient to consider multiplication and/or number bonds.

Well done in today's lesson.

You've worked really hard.

I look forward to seeing you in our next lesson.