Lesson video

Hello, I'm Mrs. Lashley and I'm gonna be working with you as we go through the lesson today.

I hope you're ready to make a start.

So in today's lesson we're gonna look at calculating the perimeter of compound shapes, which are made from rectangles, triangles, parallelograms, and trapeziums. And we're gonna look at trying to do that in the most efficient way.

So some keywords that we are familiar with from our prior learning, but you might wish to pause the video, just read them again before we make a start.

So the first learning cycle is going to be focusing primarily on finding the perimeter of compound shapes.

And the second one is going to be still finding the perimeter of compound shapes, but using other math skills that we have as well.

So let's make a start as just reminding ourselves and checking our understanding of finding the perimeter of compound shapes.

So the perimeter is the distance around a 2D shape and perimeter is a length.

And so it can be measured with units of length such as centimetres, which is metric, or inches, which are imperial.

What other units could be used to measure perimeter? Pause the video whilst you think about that.

So you're thinking about units of measure, which measure length and press play when you wanna check.

So you may have come up with millimetres where we use mm, or metres, which we just use the m or kilometres where we use km.

So they are metric units of length or on the imperial sort of measurements, feet, ft.

yards, yd or miles.

Did you think of any others? So this shape is a polygon and we could classify it as an octagon as it has eight edges and eight interior angles.

But we would also call this a compound or a composite shape.

So we are looking at finding the perimeter of compound shapes, but I might use the word composite and that means the same thing.

But here we have a shape that we would classify as compound or a composite shape, although we would call it an octagon as well.

So what basic shapes can this octagon be created from? So again, just pause the video whilst you look at that octagon and decide what basic shapes, because a composite shape is made from basic two or more basic shapes would create this.

So triangles and a rectangle.

You may not have split it in the same places as I did, but you would still be using triangles a rectangle to create it, so two or more basic shapes.

And our basic shapes we tend to say are rectangles, squares, triangles, parallelograms and trapeziums because they we're working with perimeter today, but if you wanted the area, they have a formula to calculate the area of those basic shapes where these more irregular shapes do not have a formula to find the area.

So we need to think about what they've been built from instead.

But that's a future lesson thinking about area.

But for now, for today we're thinking about perimeter, but it will be important later on to wonder and think about these basic shapes.

So the perimeter in its most basic form can be calculated by just summing the edges together.

If you stood at one vertex and imagined walking around the boundary or the edges and how far have you walked, it's a length.

So imagine laying this all out into a straight line and being able to measure it with a ruler.

So for this octagon, what is the perimeter? Well, it's three.

So I'm starting on this very top vertex and I am going around clockwise, let's say so three at six, at five, at four, at four, at five, at six at three.

And that equals to 36 metres.

So the perimeter of this octagon or this compound shape is 36 metres.

Maybe you are already thinking about being a bit more efficient than that.

So here's a check for you.

So what is the perimeter of this compound shape? So pause the video and calculate the perimeter.

And when you're ready to check, press play.

I'm hoping you got to 41 centimetres.

The unit on all of them was centimetres, so there wasn't any conversion necessary.

And then we can just sum all of those given edges.

We had the edge length for all of the edges, so there was no missing parts to our perimeter.

So to find the perimeter of any shape, we need to sum the edges, but it's not always as straightforward in terms of just get given them.

So Jun says, "Not all edges are labelled with a length." And look at that composite shape there.

His true only some of them have got those arrows to indicate their length.

"But some of the edges are equal to each other." How does Jun know that some of the edges are equal to each other? Have a look at the diagram.

What is it that's telling Jun that there are some equal edges? Well, it's the hash marks.

So to be more efficient, we might look at using this equality between some of these sites.

So the perimeter could be two lots of 6.

5 'cause we've got two edges with the one hash mark.

Two lots of eight because there are two edges with the double hash mark and then all the remaining edges, six, four and two.

So rather than doing 6.

5 plus 6.

5, we could just double it because there are two of them.

So the perimeter of this compound shape is 41 metres.

For a check, use the hash marks to calculate the perimeter of this shape.

Pause the video whilst you're doing that, when you're ready to check the answer, press play.

So the answer was 33.

7 centimetres.

That is the perimeter of this shape.

I'm hoping you looked at using the equality of some of the edges.

So there were two four centimetre edges because we had two with the one hash.

There were two seven centimetre edges because there was two that had the three hash marks and there were two 3.

8 edges because they're double hash marks.

And then now 4.

1 centimetre edge was the only one.

The only thing to be mindful with this calculation is making sure you've included all of the edges in your perimeter.

So maybe by doing a quick count of what type of shape this is, is it octagon? Is it a heptagon? Is it a hexagon? And then in your calculation, have you included that many edges? So this one was a heptagon, a seven sided polygon.

So are there seven lengths within your perimeter? Well you've got two lots of four, so that's two, two lots of seven so that's makes four edges.

Two lots of 3.

8 that makes six edges.

And the 4.

1, which is seven.

So you have included all seven.

So although it's more efficient than doing maybe lots of adding, you need to be careful because you're not necessarily walking around the shape or imagining that you're going around the boundary of the shape.

So this compound shape has a maximum width of 20 metres and a maximum height of 12 metres.

What is its perimeter? So this type of compound shape is also known as a rectilinear compound shape because of all those right angles that show that it's only made of rectangles.

Alex says, "We haven't got enough information to calculate the perimeter of this compound shape!" What do you think? We've got this two edge lengths marked.

Is that enough to get the perimeter? Laura says, "We do because we know that the vertical edges total 24 metres and the horizontal edges total 40 metres." And so Alex, ah, okay, "So the perimeter is equal to a rectangle of 12 metres by 20 metres.

And so the efficient way of finding the perimeter of a rectangle is to sum the width and the length and then double it because you have two widths and two lengths within your perimeter." The reason that this is equal is because we can imagine, well hopefully you can imagine that the vertical edge could be sort of pulled out to the right hand side and that would have the same total height of 12 metres.

So when Laura said that the vertical height summed 24, that was because all of the vertical edges would sort of line up.

So a quick check on that and thinking about this idea of moving some of the edges out.

Which shape has the same perimeter as the rectangle here? So this rectangle has the dimensions A by B.

So which of these composite shapes has got the same perimeter? Pause the video and when you're ready to check press play.

So it was C because A is the maximum vertical height.

And if you look at the other two vertical edges, they would sum to A, and then if you do the same for the horizontal, the maximum horizontal is B.

And the two other horizontal edges they sum to B.

So the the perimeter would be 2a add 2b, which is the same as the rectangle.

If you go to the compound shape on A, there are more vertical edges that would need to be added up.

So you would have A, on the right hand side that is labelled it's opposite edge would also be A, so that's already 2a.

And then there are two more vertical edges within the middle of that shape.

So that would be too much for the perimeter of the rectangle.

And then the one in the middle, the part B composite shape, we've got a slant there on the edge.

It's sort of got a corner chopped off if you like.

And so it wouldn't have the same distance because the diagonal would be longer than the vertical and would also be longer than the horizontal.

So we're up to a task for this first part.

So question one, there's two parts.

I want you to find the perimeter of these compound shapes.

So pause the video and when you're ready to move to the next question, press play.

So here is the final question of this task.

So by calculating the perimeter of each of the compound shapes, then place them in ascending order.

So ascending means from the least up to the most.

So the smallest perimeter up to the greatest perimeter.

So pause the video whilst you're calculating the perimeters and then ordering them, and then when you're ready to go through the answers to task A press play.

So 1a, we were given all of the edges lengths, so this was just to sum them.

You may have used the idea that this was equivalent to a rectangle of 9 by 10 to work it out.

That was another way to do it.

But just by going back to the basic definition of perimeter, which is the sum of all of the edges, we could do 9 plus 3 plus 5 plus 7 plus 4 plus 10, which equals 38, the unit was metres.

For part B, we couldn't use that this was equivalent to a rectangle because of those 4.

5 on one side and the 4.

5 on the other.

But then we've got two vertical edges sort of within the central part of the shape.

So that wouldn't work.

But what we could use was the fact that there were edges that were equal to each other.

So we had three, three centimetre edges.

They were all of the ones with the one hash mark.

We had two 4.

5 centimetre edges.

They were ones with the double hash mark and there were two 1.

5 centimetre edges, they had the triple.

And then lastly there was that base edge of six, the sum was 27 centimetres.

Question two, you needed to calculate the perimeter and then order them.

So A had a perimeter of 28 metres.

B had a perimeter of 29.

4 metres, and C had a perimeter of 26 metres.

So in ascend order it was C, then A and then B.

So we're now up to the second learning cycle, which is still focusing on finding the perimeter of a compound shape, but we're gonna use other math skills as well.

So this Pentagon, it's got five edges.

This Pentagon can be split into a square and an equilateral triangle.

What does this tell you about the perimeter of the pentagon? So just have a think about that.

Remember we're trying to think about other math skills to find the perimeter.

So if it's split into a square and an equilateral triangle, what does that tell you about its perimeter? Well, it would be five lots of x, where x is the length of one edge.

The square will have equal edges because it is a regular quadrilateral.

And the equilateral triangle would have equal edges because it is a regular triangle and they're joined along one of their edges, which means that the edge lengths would be the same in both the square and the equilateral triangle.

So for this perimeter it would be five lots of x.

It doesn't have to be x, obviously, but where x is the length of one.

So if one of the edges was 12 centimetres, then what is the perimeter? Well, it would be five lots of 12 and that would be 60 centimetres.

So here is a check for you.

This hexagon, it's got six edges, can be split into two equilateral triangles and a square.

The edge of the square is eight centimetres.

So what is the perimeter of the hexagon? So pause the video and then when you're ready to check your answer, just press play.

So the perimeter would be 48 centimetres because once again the square and the equilateral triangles are all gonna have the same edge lengths.

So six lots of eight is 48 centimetres.

To find the perimeter, we need to know all the individual edge lengths in order to sum them or the sum of groups of edges.

So here we have a compound shape and Jacob says, "Using the hash marks, I know that the other edge is eight centimetres because it was marked with one single hash mark and that was labelled as eight centimetres.

So any other edge is marked with one hash mark would also be eight centimetres." And so he said, "which means this pentagon can be split into a square and a right-angled triangle." So we notice it's a square because it's eight and eight.

So the opposite side would also need to be eight.

And as would the other side to make it a square.

It says, I know that the sum of the height of the triangle and the four centimetre edge of the pentagon needs to be eight centimetres.

So look at the diagram, the height of the right angle triangle and the four centimetre edge, which was given on the Pentagon has to total eight centimetres.

So they are the vertical heights, which means that the height of the triangle would be four centimetres.

Four at four gives him the eight, which is an edge of the square.

Also, the base of the triangle would be the difference between the 11 centimetre edge and the eight centimetre of the square, which means that that part would be three centimetres.

But he thinks he's at a problem still.

"I still do not have all the edges of the pentagon though? So I cannot find the perimeter of this shape!" So Jacob has done a really good thing in maths that he's started by adding some information to the diagram that he was given information that he knew he wasn't assuming, but he knew he could add onto the diagram.

But he feels like he's found himself at a sort of dead end because he knows the perimeter is the edges and there is one edge that he doesn't have and therefore the perimeter cannot be found.

But is Jacob correct? So I want you just to think about everything you know, everything you've learned up to this point.

Is Jacob correct that he cannot find the perimeter of this shape? There is just nothing he can do.

So no, he can calculate the length of that edge using Pythagoras' Theorem.

So he's right that currently he cannot find the perimeter because there is one edge length that is unknown, but it's not unknown and impossible to find because there is that right-angled triangle.

So he can use Pythagoras' Theorem.

It's the hypotenuse that is trying to calculate 'cause it's the longest edge opposite the right angle within the triangle.

And so using Pythagoras' Theorem, if we call that edge x, then x squared is equal to four squared plus three squared.

We can evaluate those squares and that means that x squared is equal to 25 and we can square root it to get 5.

We're taking the positive square root because this is a distance and so therefore it cannot be negative.

So now we can get the perimeter because we've got all of the external edges of this pentagon.

So I'd like you to pause the video and calculate the length of y.

You can use a calculator, think about what you know.

You might wanna sketch this out.

And when you're ready to check, press play.

So the answer was 13, so the length of y is 13.

And there is the working out of the way to get there on the screen.

So 10 minus 5, that's to get the sort of vertical height of the right angle triangle at the top of the shape.

You knew that the total height of this shape was 10.

There an edge on the opposite side, which was five.

So you knew that that had to be five.

And then you needed to work out the sort of base edge of the right angle triangle, which was the total of all horizontal edges.

None of them overlapped each other.

So it was the six, then the five, and then the one.

Then we could use Pythagoras' Theorem because it is a right angle triangle.

It's the hypotonus that you're trying to calculate.

So 5 squared plus 12 squared is 169 and the square root would be 13.

So you're into your last task for today's lesson.

And so for question one, I want you to find the perimeter of these compound shapes.

There are three parts to the question and you are told how they have been created and you're given a edge length.

So pause the video whilst you're calculating the perimeter of those compound shapes.

And then when you're ready to move on, press play.

Okay, so here is question two.

There's two parts to this.

So find the perimeter of these compound shapes.

So remember you are going to have to use other skills within this.

It's not just given to you straight away in terms of being able to find the perimeter.

So pause the video and work through question part A and part B.

Think about adding to the diagram what you can straight away by using the information given within the diagram.

And then think about the other skills you could use.

Press play when you're ready to go to question three.

Question three, you need to find the printer of this compound shape.

So pause the video whilst you're working through question three.

And when you press play, we will be going through the answers to task B.

So question one was all about using properties of shape.

You were told that they were created by squares and run by in some sort of fashion.

So for part A, it was two squares and a rhombus, which means that all of the edges were equal.

So you needed to count how many there were.

It was an octagon.

So eight lots of nine is 72 centimetres.

On part B, it was created by two squares and a rhombus.

But again, different arrangement, there were still eight equal edges.

So this one would have a perimeter of 28.

8 centimetres.

And lastly it was two rhombi and a square.

Again, if you count there are eight, it's an octagon.

So eight lots of 8.

3 is 66.

4 centimetres.

So the other skill here was knowledge of shapes, properties of shape that you then could use that in order to be efficient when finding the perimeter.

Question two, this is only part A and then I'll move on to part B.

So you needed to find the perimeter of these compound shapes.

You were given some hash marks, which meant that you knew things were equal and there was some markings to indicate that there was right angle triangles.

There were two edges that you would need to use Pythagoras' Theorem to calculate.

And I've labelled those x and y and I've redrawn just the right angle triangles.

The working out for x and the working out for y is also on the screen.

So you can see they were not integer values.

So I've left the dot, dot, dot to indicate that there is more digits to that.

I've used the most accurate on my calculator and then rounded the final answer.

So my perimeter of this compound shape is 115.

7 centimetres to one decimal place.

You may have rounded it to a different degree of accuracy.

One decimal place seems sensible in the unit of centimetres because then that would be indicating how many millimetres as well.

Part B, we had again, some hash marks where you could know that that one edge was four.

And also we needed to find this length that I've marked x and it would be Pythagoras' Theorem.

Again, this one wasn't integer value, so our answer is not rounded.

It is the exact answer of 56 centimetres.

And then question three, it was a similar question to one we looked at in the lesson.

So find the perimeter of this compound shape? Again, the hash mark indicates that this is a square.

By knowing that it's a square we can sort of chop or dissect the bottom edge into the 24 part.

And then this base of the right angle triangle.

We weren't finding the hypotonus this time, we were finding a short edge using Pythagoras' Theorem and it was five.

And then we could calculate the perimeter.

And the perimeter was 102 centimetres or 1.

02 metres.

Perhaps you converted it into metres as well.

So to summarise, the perimeter of a compound shape is found the same way as finding a perimeter of any shape, which is to find the sum of all the edges.

It's not always necessary though to know all the individual evidence to find the perimeter as the sum of groups of sides can be deduced.

Other mathematical knowledge may be necessary to calculate the perimeter, for example, using properties of shape and Pythagoras' Theorem.

But it's not only those two parts of mathematics that could be used.

Really well done today and I look forward to working with you again in the future.