Lesson video

Hello, my name is Dr.

Rowlandson and I'm thrilled that you're joining me in today's lesson.

Let's get started.

Welcome to today's lesson from the unit of geometrical properties and Pythagoras' theorem.

This lesson is called checking understanding of similarity.

And by the end of today's lesson, we'll be able to enlarge a shape and state what changes and what is invariant.

Here are some previous keywords that you may have come across before, and we'll be using these words in today's lesson.

So you may wanna pause the video at this point while you remind yourselves the meanings of these words before pressing play to continue.

This lesson contains two learn cycles.

In the first learn cycle, we're gonna focus on enlargement.

And then the second learn cycle, we'll focus on similarity.

But let's start off with checking understanding of enlargement.

Let's start off by considering this question here.

Enlarge the object by a scale factor of three from the centre of enlargement.

And we're gonna consider a few different ways of doing this.

For example, how could this question be answered using the grid? Pause the video while you think about how you might do this and then press play when you're ready to continue.

Well, Lucas says, "We could count squares between the centre of enlargement and each vertex on the object and we can then multiply by three to find the vertices on the image." For example, if we take the bottom right-hand vertex of this triangle, we can see that it's four squares away from the centre of enlargement.

So the bottom right-hand vertex of our image will be four times three, which is 12 squares to the right of our centre of enlargement.

And then the top left vertex is three squares above the centre of enlargement.

Multiply that by three and we get nine squares above the centre of enlargement.

And then we can see to get to the last vertex, we can go four squares to the right and three squares up from the centre of enlargement.

If we multiply both of those by three, we get 12 squares to the right and nine squares up.

So we have all three of our points here.

Join them up and we have our image, which is three times the size of the object.

How about if we didn't have a grid? If this question was on a plain background, how could this be done using a ruler? Pause the video while you think about this.

You could even draw a triangle and a centre of enlargement and have a go at it yourself, but once you've thought about it, press play when you're ready to continue.

Well, Aisha says, "We could measure the distance from the centre of enlargement to each vertex on the object and then we could multiply by three to find the vertices on the image," the same as we did with the grid, but we're using our ruler to measure instead of the grid lines.

For example, the bottom right-hand vertex of this triangle is four centimetres away from the centre of enlargement.

If we keep our ruler still and don't adjust it from this point, we could measure 12 centimetres away from the centre of enlargement and plot the point of our new vertex.

We could do the same this way, it's three centimetres away.

Keeping our rulers still, we could measure nine centimetres away and then this third point might be a bit different to what we did in the grid because rather than having to go right and then up, we can go directly.

We can measure it to be five centimetres away, multiply it by three, then plot our new vertex at 15 centimetres away.

Once we've done that, we've got our three vertices, we can join them up to create our image.

So what about a slightly different challenge? How about if we didn't have a ruler but what we did have was a straight edge that didn't have a scale on it? Something like a plain piece of wood or a plain piece of plastic or the side of something that is straight and sturdy.

If we had that, a straight edge, and a pair of encompasses, how could we enlarge this shape by a scale factor of three using that equipment? Pause the video while you think about this and again, if you have it to hand, you might wanna have a go at it yourself, but when you're ready to continue, press play and we'll do it together.

Well, Sam's gonna talk us through this one.

Sam says, "If the straight edge doesn't have a scale on it, then we can't use it to measure distances." That's a problem.

We don't know how far each vertex is away from the centre of enlargement.

However, we could use it to draw ray lines through each vertex, starting from the centre of enlargement, like so.

We could open the pair of compasses to the distance between the centre of enlargement and a vertex.

Now, that bottom right-hand vertex there, we don't know how far that is away from the centre of enlargement, but what we do know is that it's the same as the distance between the pencil tip and the needle of the compass.

So this could be used then to mark three lots of this distance along the ray line.

So here we have one lot of that distance.

Now we have two lots of that distance and now we have three lots of that distance.

So that third marker on that ray line there, that is three times the distance from the centre of enlargement as the vertexes.

Let's do it again on another vertex.

Here we have opened the pair of compasses, so the distance between the needle and the pencil tip is the same as the distance between a centre of enlargement and the top left vertex.

And now we've got two lots of that and three lots of that.

And if we do it again on the remaining vertex, we've got one lot of that distance, two lots of it, three lots of it.

And then we can plot our vertices at those points, which are all three times the distance from the centre of enlargement and join 'em up to create our image.

So let's check what we've learned.

Jacob is enlarging a shape by a scale factor of five.

He measures the distance from the centre of enlargement to the first vertex.

How far will its corresponding vertex be from the centre of enlargement? Pause the video while you write down an answer and press play when you're ready to continue.

Well, the vertex on the object is four centimetres away from the centre of enlargement.

If we multiply it by five, then the corresponding vertex on the image will be 20 centimetres away from the centre of enlargement.

Sofia is enlarging a shape by a scale factor of two.

She has plotted the positions of the vertices on the image, but one of them is incorrect.

Which vertex has been plotted incorrectly? Is it A, B, C or D? Pause the video while you write down answer and press play when you're ready to continue.

It's point C which is in the wrong position.

And there's a few ways we can check this.

One way, if you had a paper copy to hand, you would be able to draw ray lines from the centre of enlargement through each vertex on the object and see if there are any points that don't lie on the ray lines.

If the ray lines have been drawn accurately and there's a point that is missing the ray line, then you know that's in the wrong place.

If they were all on the ray lines, then it could be that one is plotted incorrectly but just happens to be on a ray line.

For example, if you imagine point A there, if that was one square to the right, it would still be on the ray line but it wouldn't be in the right place.

So another way we could check is by looking at the distances.

We look at a distance from the centre of enlargement to the point C on the object, then we go one down and four across.

If you multiply it by two, it should be two down and eight across.

So we can see point C should really be one square to the left.

So now that we've looked at how to enlarge a shape, let's consider what happens after a shape has been enlarged.

Here we have an object.

The object was enlarged by a scale factor of three to create the image.

Alex measures the length on the object as three centimetres, four centimetres and five centimetres.

He's about to measure the lengths on the image.

What should he expect? What would you expect those lengths on the image to be? Pause the video while you think about it and press play when you're ready to continue.

Alex says, "Each length on the image will be three times its corresponding length on the object." So that length which is four centimetres on the object.

If we multiply it by three, we get 12 centimetres on the image.

The length that is three centimetres on the object times three is nine centimetres on the image and the five centimetre length on the object times by three gives 15 centimetres on the image.

So here we can see all the lengths on the image are all three times the lengths on the object.

Alex then measures the angles on the object.

He's about to measure the angles on the image.

What should he expect this time? Pause the video while you think about this and press play when you're ready to continue.

Alex says, "Each angle on the image will be three times its corresponding angle on the object." What do you think about that? Laura says, "But that would make the angles in the image sum to more than 180 degrees.

That can't be right." And Laura's got a point here.

If we think about how angles in the object sum to 180 degrees, if we multiply each of those three numbers by three, then those numbers would add up to three lots of 180 degrees and that's not what triangles sum to.

So what could we take away from that? Well, when a shape is enlarged, the lengths may change but the angles do not change.

And if we have an object and an image, we can check that the angles are the same by cutting out the object and placing it at the different vertices on the image.

For example, if we put it in the top left vertex, we can see that those angles are the same.

In the top right vertex, those angles are the same.

On the bottom right vertex, we can see that those angles are the same.

So all the angles on the image must be the same as the angles on the object.

And there's a few visual checks we can do as well.

For example, we can see that both triangles are right angled.

If we had doubled the right angle on the object, if we double 90 degrees, we would've got 180 degrees and we can see that actually no, it's not that on the image, it's still a right angle.

We can also see that both triangles contain two acute angles.

In the object, if we look at that 53 degrees, if we had times that by two, we would've got 106 degrees.

That is obtuse.

But we can see visually that the angle is definitely acute in the image.

And also, we can check that all the angles in each triangle sum to 180 degrees because they should do.

Let's now think about scale factors.

Here we have two shapes and one of the shapes is an enlargement of the other.

How could this scale factor be calculated between these shapes? Well, Andeep says, "We could calculate a scale factor by dividing a pair of corresponding lengths." But Izzy says, "The order of the division will depend on which shape is the object and which is the image." Are we going from the small shape to the big shape or are we going from the big shape to the small shape? And that will affect our division when it comes to calculating the scale factor.

If A is the object and B is the image, so we're taking the small rectangle and we're enlarging it to create the big rectangle, we could calculate the scale factor by doing 10 divided by five to get two or we could do six divided by three to get two.

So the scale factor is two.

If B was the object and A was the image, so we're going from the big rectangle and we're enlarging it to make the small rectangle, then the scale factor's going need to be less than one to make it so that it gets smaller.

Our division here would be five divided by 10 is 1/2, or we could do three divided by six is 1/2, which means a scale factor to get from B to A is 1/2.

In other words, all the lengths of B, if you halve them, you get all the lengths of A.

So let's check what we've learned.

The object has been enlarged by a scale factor of five to create the image.

What is the value of X? Pause the video while you write down an answer and press play when you're ready to continue.

Well, the scale factor is five and the length we have is six.

Do six times five, we get 30 for our value of X.

The object has been enlarged by a scale factor of 1/2 to create the image.

Now, what is the value of X this time? Pause the video while you write down an answer and press play when you're ready to continue.

X would be 70 because angle sizes are preserved during enlargement.

The object has been enlarged to create the image.

What is the scale factor? Pause the video while you choose either A, B, C or D and press play when you're ready for an answer.

Well, we can see that the object is bigger than the image and that the length on the image is 1/3 the length of the object, so the scale factor is 1/3.

Over to you now for task A.

This task contains two questions and here is question one.

In this question, it says first enlarge the triangle by a scale factor of three from the point marked.

Then once you've done that, we'll go check our answer in a couple of ways.

First, use a ruler to measure all the lengths on each shape and check that each length on your image is three times its corresponding length on the object.

And then use a protractor to measure all the angles in each shape.

And check that each angle on the image is equal to its corresponding angle on the object.

Now, you'll have to measure as carefully as possible if you wanna check these correctly.

Pause the video while you have a go and then press play when you're ready for question two.

And here is question two.

Jun wants to enlarge the shape by a scale factor of two, but the problem is that grid is pretty small so he needs to choose a centre of enlargement carefully so that the image stays on the grid.

Find all the points which Jun could use for his centre of enlargement and mark them on the grid.

Pause the video while you have a go at this and then press play when you're ready to work through some answers.

Right, great effort with that, let's now check how we've done.

We need to start by enlarging the triangle by a scale factor of three from the point marked and it should look something a bit like this.

Now checking this is gonna be a bit tricky 'cause depending on how your worksheet was printed, the length and distances may vary slightly, but however it's been printed, your bottom left vertex on your image should be directly below the bottom right vertex of the object.

And then when we measure the lengths, once again, these lengths may vary depending on the printing, but an example of a combination lengths are what's on the screen here.

You could have three centimetres, four centimetres, and five centimetres on your object, which means you'd have nine centimetres, 12 centimetres and 15 centimetres on your image.

But for whatever lengths you got, if you divide each length on the image by its corresponding length on the object, you should get three or something very close to three, depending on how accurately you measured.

And then if the worksheet has been printed in proportion, then the angles should match the ones below.

That is 37 degrees, 53 degrees and a right angle, 90 degrees, on each your object and your image.

Now, the angles may vary if the proportions were not maintained in the printing.

For example, if it's stretched vertically more than it stretched horizontally.

But either way, the angles in the image should match your angles in the object.

An additional check could be you could add up your angles and check that they sum to 180 degrees.

And then question two, Jun wants to enlarge a shape by a scale factor of two.

He needs to choose the centre of enlargement carefully so the image stays on the grid.

What points could Jun use? Well, Jun can use any points in this region here, including on the lines you can see.

It doesn't just have to be on the grid lines, it can also be in the middle of the squares as well.

It just makes it a bit tricky when you're trying to actually do the enlargement, but still, anywhere in that region is absolutely fine and it'll keep the image on the grid.

If you wanna explore this a bit more, the slide deck contains a link to a GeoGebra file where you can play with this interactively and see where the image ends up as you move the centre of enlargement around.

Great work so far.

Let's now move on to the second learn cycle, which is checking understanding of similarity.

Here is a photograph of Liverpool, and Sofia and Jun both copy the photograph and make a bigger version of it.

But hang on a minute, Jun's copy doesn't look right.

The buildings in that copy look really thin and tall compared to the original photograph.

Why does Jun's photograph look all distorted? Pause the video while you think about this and press play when you're ready to continue.

Well, the reason behind this is all to do with similarity.

When an object is enlarged to create an image, the object and the image are similar and two shapes are similar if the lengths are in the same proportions.

If we look at these two photographs here, we could say that these shapes are similar for a few different reasons.

We could say they are similar because the multiplicative relationships between the corresponding sides on each shape are equal.

So here we've got four times two is eight and six times two is 12.

The fact that both of those multipliers are the same means that they're similar.

Or we can say the multiplicative relationships within each shape are equal.

On the original copy of the photograph, the smaller one, we can see that the width of that photograph, four, if we times it by 1.

5, we get the length, which is six and that's the same as well in the bigger copy of the photograph.

The width of the eight and we times that by 1.

5, we get the length which is 12.

The fact that both those multipliers are the same means they're similar and we can also say they're similar by looking at the ratios between the lengths on each shape and seeing that they are equivalent.

The ratio four to six is equivalent to the ratio eight to 12.

They both simplify to two to three.

And when we think about those multipliers that we just saw, we can also see those on our ratio as well.

So what's going on with these two photographs then? Why does it look so distorted? Well, these shapes are not similar and we can say that they're not similar because for a few reasons again.

The multiplicative relationships between the corresponding sides of each shape are not equal.

We can see that we've multiplied one measurement by two, but the other measurement by one.

Different multipliers, they're not similar.

Or we can say the multiplicative relationships within the shapes are not equal.

We can see that in one case, we multiply one length by 1.

5 to get the other, but in the other case, we multiply by 0.

75.

Those multipliers are different, they're not similar.

All we can say that the ratios between the lengths in each shape are not equivalent.

We can see that the ratio four to six is not equivalent to the ratio eight to six.

And we can see that also because the multipliers differ within this pair of ratios.

So here's a question to consider.

Are all circles similar? And for whatever you think, how could you justify your answer? Pause the video while you think about this and press play when you're ready to continue.

I wonder what we thought.

Let's look at a few different ways we can reason about this.

Jacob says the radii of a circle is the same length in every direction.

So for example, a1 is equal to a2 and b1 is equal to b2.

That means the multipliers between corresponding radii will be equal because if we wanted to get these multipliers, we would do b1 divided by a1 or we could do b2 divided by a2, but because b1 and b2 are the same as each other and also, a1 and a2 are equal to each other, then these divisions will give the same answers.

Aisha says something similar.

"The radii of a circle is the same length in every direction.

So the multipliers between radii within each circle are equal.

They're always equal to one." The multiplier between a1 and a2 would be one 'cause they're the same length and also the same as well between b1 and b2.

So that means they're similar.

And Jun says the ratio between the diameter and the circumference in every circle is always the same.

What's the multiplier between the diameter and the circle? It's always pi.

So three different ways of thinking about why all circles are similar.

So let's put all this together now and summarise what we know about similar shapes.

Two shapes are similar if the lengths are in the same proportions, but the orientations of the shapes may differ.

So here we have two triangles.

They are in different orientations, but the shapes are similar, which means that the multipliers between corresponding lengths are equal.

If we take the lengths on the larger triangle, nine, 12, and 15, and divide 'em by the corresponding lengths of the smaller triangle, three, four, and five, we should get the same an every time, three, three, and three.

We also know that the ratios between the lengths in each shape are equivalent.

The ratio of the lengths in the smaller triangle is three to four to five, and the ratio of the lengths in the larger triangle is nine to 12 to 15.

Those two ratios are equivalent.

We also know that corresponding angles are equal to each other as well because angles are invariants during enlargement.

So let's check what we've learned.

Which shape is similar to triangle T? Is it A, B, or C? Pause the video while you make a choice and press play when you're ready for an answer.

The answer is B.

What makes this question tricky is that the triangles are all in different orientations to the one which is labelled T.

However, we can match up the longest side in each triangle, we can match up the shorter side in each triangle and the one in between to C, where the proportions are all the same.

With triangle A, each of those lengths is two centimetres more than the lengths in T.

Now remember, it's a multiplicative relationship for similar shapes, not an additive one.

So adding two to every single length will not create a similar shape.

And for triangle C, we can see that for two of the lengths, they have been halved, 12 divided by two is six and six divided by two is three, but hang on a minute, nine divided by two is not four.

So that's not right there.

That one's not similar either.

So the answer is B, and we can see that because if we go from the smallest length to the biggest one, 6, 9, and 12 on shape T, if we divide those by three, we get two, three, and four, which is on the triangle labelled B.

True or false? A rectangle that is six metres by 12 metres is similar to a rectangle that is 10 feet by 20 feet.

Pause the video while you choose true or false and also choose a justification of the answer as well and then press play when you're ready to continue.

The answer is true.

It doesn't matter that they're in different units.

The key thing is that the proportions between the measurements in each rectangle are the same.

We can see that in the rectangle measured in metres, 12 is double six and also in the rectangle that is measured in feet, 20 is double 10.

Right, let's now put all those bits and pieces that we've learned together to solve this multi-step question.

The two shapes below are similar and we need to find the missing lengths.

Now, we can see that the shapes are in different orientations, so we'll have to deal with that.

We'll need to get a scale factor as well.

And then we'll have to use that scale factor to get a length on the larger triangle and also a length on the smaller triangle.

So maybe pause the video at this point while you think about how you might do all those things and then press play when you're ready to work through it together.

Let's start by trying to get a scale factor.

Now, if we look at the 13 on the small triangle, that's the longest edge of that triangle, but we can see that the longest edge of the other triangle isn't labelled.

So we can't use the 13 to get a scale factor.

Let's look at the 12.

That's the second longest edge of the smallest triangle, and we can see that the second longest edge on the larger triangle is labelled 18.

So we can get a scale factor by using the 12 and 18.

Let's do 18 divided by 12 to get three over two.

Now, that scale factor, because we did 18 divided by 12, that tells us that 12 times three over two makes 18.

So that scale factor is what we multiply the lengths of the smaller triangle by to get the lengths of the bigger triangle.

So let's use it to get the longer side.

We can do 13 multiplied by three over two to get 19.

5.

But then if I want to get the length on the smaller triangle on the shorter side, because I'm going the other way, I'm gonna have to use my scale factor differently.

I'll have to use 7.

5 divided by three over two to get five.

Here's another question to consider.

Alex and Sam are estimating the height of the traffic lights on a sunny day.

And the reason why it's important that it's sunny is they've got some shadows.

How could they do this by using similar shapes? Pause the video while you think about this and press play when you're ready to continue.

Well, Sam says, "If we know Alex's height, then we could estimate the height of the traffic lights by measuring the shadows." They know that Alex's height is 156 centimetres.

They measure his shadow as 117 centimetres and they measure the shadow of the traffic light as three metres, and then they could get the height of the traffic lights by using similar shapes.

You can even imagine this as a triangle if you join a line up from the tip of the traffic light to the tip of its shadow and do the same with Alex's head if you wanted to.

It doesn't really matter either way because we can do 156 divided by 117 to get the multiplier within Alex and his shadow to get four over three.

And that multiplier should be the same on the traffic lights as well.

So we can do three times four over three to get four.

So the estimated height of the traffic lights is four metres.

Let's check what we've learned there.

These two shapes are similar.

Find the value of X.

Pause the video while you work this out and press play when you're ready for an answer.

Well, we could do this in a couple of ways.

One way could be to get the multiplier between the shapes by doing 20 divided by eight to get five over two, and then do four multiplied by five over two to get 10.

Or we could look at the multipliers between sides in the same shape.

Four times two is eight, so we could do X times two to get 20, which means X is 10.

Over to you now for task B.

This task contains two questions and here is question one.

Here we have two scenarios and I'd like you to use similarity to find the missing lengths in each scenario.

Pause the video while you have a go at this and then press play when you're ready for question two.

And here is question two.

Match each triangle with its similar triangle.

And then once you've done that, find a scale factor between each pair of similar triangles.

Pause the video while you work these out and then press play when you're ready for some answers.

Well done with that.

Let's now work through some answers.

In question a, you should have got 6.

1 metres for the height of that flagpole.

And in question 1b, you should have got 35 inches for the width of the small snooker table.

And in question two, let's match each triangle with its similar triangle and then find the scale factor.

Well, A is similar to E and the scale factor is either three or 1/3, depending on whether you're going from A to E or E to A.

B is similar to D and the scale factor is either two or 1/2, depending on which way you are going.

C is similar to G and the scale factor is either three over two or two over three.

And F is similar to H with a scale factor of either two or 1/2.

Great work today.

Let's now summarise what we've learned in this lesson.

An enlargement means a change of size.

Now, it could be getting bigger or it could be getting smaller.

It's a change of size.

The length of the lines may change when a shape is enlarged, but the angles inside the object do not change when it's enlarged.

The angles remain invariant.

When an object is enlarged, the image is similar to the object.

These two shapes are similar, which means that the orientations may differ, but the lengths are in the same proportions.

Weld done today.

Thank you very much.