Lesson video

Hello, everyone, I am Mr. Gratton, and thank you for joining me in this similarity and enlargement lesson where we will look at how the perimeter of a shape changes as that shape is enlarged.

Pause here to have a look at some important keywords that we will be using today.

First up, we're going the long way around.

Let's look at the effect of enlargement on the lengths of sides of a shape.

So Jacob asks, what happens to the perimeter of an object when the object is enlarged? Well, Sam thinks that the best way to find out is to actually perform that enlargement.

So, given a centre of enlargement and some raise from the centre and through each vertex of the object, we can create this enlarged image.

Each side of the image has been enlarged by a scale factor of three.

This means that the length of each side of the image is three times the length of the corresponding side on the object.

As we can see with this width of three units being three times larger at nine units on the image, with the same thing happening from two units to six units.

But what about the perimeter? Well, let's complete all lengths around the shape, so three units and two units for a perimeter of the object of 10 units.

Whilst the image, we have nine units and six units for a perimeter of the image of 30 units in length.

By adding together all of the sides on the object, and again, for the image, we can conclude that the perimeter of the image is three times larger than the perimeter of the object.

For this check, we have an enlargement by a scale factor of five.

Pause here to find the lengths of the sides labelled a, b, and c.

A equals four times five equals 20 units, b is three times five equals 15 units, and c is five times five equals 25 units.

And pause here again to calculate the length of the perimeter of the object and the image by adding together the length of all of the sides on each shape.

The object has perimeter of 12 units, whilst the image has perimeter of 60 units.

It is also possible to find the perimeter of an enlargement without a coordinate grid as long as enough information is given about the original object.

For this object, we will enlarge it by a scale factor of six.

Therefore, each corresponding side of the object will be six times larger in the image, eight times six is 48, 11 times six is 66 and nine times six equals 54.

The perimeter of the image is therefore 210 units, the sum of all four sides of that shape.

Pause here to think about or discuss why the perimeter of this enlargement cannot be found from the given information.

We simply do not know the lengths of enough sides or the scale factor from object to image.

Right, for this object, we do know that the scale factor is two.

Pause here to find the perimeter of the image by first of all calculating the length of each of its four sides.

Here are all four side lengths and then when they're all summed together, the perimeter is 100 units in length.

If an object has hash marks, then all sides marked with the same number of hashes have equal length.

This also means that the corresponding sides on the image will also have its own hash marks like so.

The two horizontal sides on the object are equal in length whilst the side lengths of the image are likely different in size to the object, the corresponding two horizontal sides on the image are a pair of sides of equal length as well.

The object has been enlarged by a scale factor of four to become its image, meaning each side of the image is four times larger than on the object, such as 12 times four equals 48 and five times four equals 20.

But because of the hash marks, the opposite side is also 48 units, and this side of 20 units is also equal in length to the side opposite it, meaning that the perimeter of the image is 136 units in length.

For this check of an object that is enlarge by a scale factor of six, pause here to find the perimeter of the image.

Here are the side lengths of the enlarged image for a sum, a perimeter of 300 inches.

Similar again, but for a regular polygon and a scale factor of 20 this time, pause here to find the perimeter of the image.

80 millimetres for one side multiplied by the six sides for a perimeter of 480 millimetres.

Right, for this diagram, Jacob claims that the image on the right is an enlargement of the object on the left, simply because each side of the image is three units longer than that on the object.

However, Sam disagrees with this since they think that the length of each side has to be multiplied by the same scale factor, not added.

Pause here to think about or discuss whose statement is correct.

Sam's statement is correct.

This diagram is a non-example of enlargement.

If a shape is enlarged, the sides will not be a fixed length longer, such as plus three units longer, rather, each side will have its lengths multiplied by a constant value, such as with this example.

Here, each side length in the image is two times larger, a multiplication not addition when compared to its object.

Furthermore, all angles remain in variant after the enlargement.

What was a rectangle stays a rectangle.

Unlike this, enlarging a rectangle should not result in some other random parallelogram.

Here the angles have not remained invariant after the enlargement and therefore, the enlargement has been done incorrectly.

For this check, pause here to think about or discuss what else must Sam check to make sure that the image is definitely an enlargement of the object.

Whilst all the corresponding sides on the image are a constant three times larger, Sam should also measure the angles and check that all corresponding angles have not changed.

Great stuff, onto the practise.

Pause here for question one where you need to analyse the lengths and perimeters of each object-image pair.

For question two, sketch an image of this object after an enlargement by a scale factor of seven.

This sketch does not need to be to scale, rather, the only important thing is your labelling of each side length.

And for question three, find the perimeter of this pentagonal image.

Pause here for these two questions.

Good work and effort so far.

Let's look at some answers.

For object A in question one, the perimeter of the image is 110 centimetres, which is five times larger than the perimeter of the object.

And for object B, the perimeter of the image is 300 centimetres, which is 10 times larger than the perimeter of the object.

Pause here for the sketch of question two, whose perimeter is 154 units.

And for question three, the perimeter of that image is 242 units in length.

Right, we've calculated perimeters of enlargements the long way around, but is there a way to make our calculations more efficient? Well, let's have a look.

Right, here we have an object and its image after an enlargement.

Notice how the length of each side on the image is three times larger than the corresponding side of its object.

Let's check the perimeters.

What do you notice about them? The perimeter of the object is 20 centimetres whilst the image is 60 centimetres.

Wait, Jacobs spotted something.

The perimeter has increased by the exact same scale factor as each of its sides, and multiplied by three, also seen by the perimeter being three times larger as well.

Laura seems to get why this relationship happens.

Both side lengths and perimeters are both lengths so will be affected by the scale factor in the exact same way.

In fact, all sorts of lengths on an object will be multiplied by the scale factor when the object is enlarged.

These lengths aren't limited to only lengths of the sides of a shape as these lengths could also include the maximum height of the object, the perimeter of the object, and the distance between any two points on the object.

Using this information, Jacob thinks that we can find perimeters of images a lot quicker than before.

So assuming that the object and image are definitely similar to each other, we can find the perimeter of any enlarged image by, first of all, finding the perimeter of the object such as this object having a perimeter of 27 metres and then multiplying this object perimeter by the scale factor.

So with this scale factor being a multiply by three, the perimeter of the image will be 27 multiplied by three or 81 metres.

Jacob thinks that this is so much quicker than calculating the length of each and every side of the enlarged image first.

The following method will be a lot longer, don't you think? 11 times three is 33, four times three is 12 and 12 times three is 36, and then we have to add all of the side lengths together to get once again, 81 metres.

Those were a lot more steps and the number of steps would only increase with the number of sides on the original object, whereas finding the perimeter of the object first then multiplying by the scale factor is a relatively consistent method no matter the number of sides on the object and image.

Here we have a similar image after an enlargement.

First of all, pause here to find the perimeter of the original object.

That's 50 metres.

And now pause here again to find the perimeter of the image this time, knowing that the scale factor is six.

50 metres for the object multiplied by six for the scale factor equals 300 metres as the perimeter of the image.

And now we have only the description for a pentagon with these side lengths and a scale factor of 12.

Pause here to find the perimeter of the enlarged image.

The original pentagon had a perimeter of 40 centimetres.

We then times 40 centimetres by 12 to get 480 centimetres for the perimeter of the enlarged image.

This method is pretty great when finding the perimeter of images where the scale factor is non-integer, so a fraction or a decimal.

Here we have an image whose side lengths are a quarter of the lengths of the object and we have an object whose perimeter is 72 centimetres, so let's just apply the same scale factor of a quarter to that 72 centimetre perimeter.

The image has a perimeter of 72 divided by four, which equals 18 centimetres.

So using that knowledge, pause here to complete these sentences about the perimeter of the object and image.

The perimeter of the object is six plus 18 plus 21 equals 45, so 45 metres.

Therefore, the perimeter of the image is 1/3 of that at 15 metres.

One of the biggest benefits of using the perimeter of the object and multiplying by the scale factor is that we may avoid calculating fractional or decimal side lengths on the image even if the resultant perimeter of the image is integer in length.

For example, this object will be enlarged by a scale factor of 1/3 to make this image.

If we divided the length of each side by three, we'd get a fractional side length of 19 over three or 6.

33 as a decimal, and this fractional side length of 11 over three or 3.

66.

Decimals can be messy to deal with and prone to rounding errors.

The perimeter of the object is 60 centimetres, but the perimeter of the image involves adding all of these decimals together to get 20 centimetres.

Instead, we could have just taken the perimeter of the object and divided that by three straight away to get that same 20 centimetres, but without any fractions or decimals getting involved.

When there's a lot of information about side lengths, scale factors and perimeters, it can be helpful to bring everything together neatly in a table like this, showing the length of something on the object and image with the multiplying scale factor in between.

Right, for this check we have an object enlarged by a scale factor of 2.

5.

Pause now to calculate the perimeter of the enlarged image.

The perimeter of the object is 30 inches and so the perimeter of the image is 30 times by 2.

5, which equals 75 inches.

Brilliant, let's bring all of this efficient calculation to some practise questions.

Pause now to find the perimeters of the images of image A and hexagon for questions one and two.

And for question three, starting with the shortest pause here to put the perimeters of images B, C, and D in order of length.

Fantastic, here are the answers.

For question one, the perimeter of image A is 308 centimetres after adding all side lengths together and we can show this sum is correct by also multiplying the perimeter of the object at 44 centimetres by seven to get the same 308 centimetre perimeter for the image.

For question two, the perimeter of the object is 31 centimetres and therefore, the perimeter of the image is 93 centimetres.

For question three, the perimeter of image B is 55 centimetres and image C is 66 centimetres.

Image D has a perimeter of 60 centimetres and therefore the correct order is B, D, then C.

We've looked at how to calculate the perimeter of an image and in a pretty efficient way too, but can we use perimeters of objects and images to find out other information? Let's find out.

We can use the perimeter of an object and an image after an enlargement to find this scale factor between these two similar shapes.

For example, we have a trapezium and its enlargement, from the side lengths alone, we have no way of calculating the scale factor as no pair of corresponding sides have both lengths given, but from the perimeter, we can.

What is the scale factor from 52 to 312, 312 divided by 52, which simplifies to a multiply by six.

From the perimeters alone, we know that the object has been enlarged by a scale factor of six to make that image.

We can then use this scale factor to find the lengths of missing sides on either the object or the image.

The eight metre height on the object is multiplied by six to get a height of 48 metres on the image.

On the other hand, we have a 126 metres length on the image, this can be divided by six to find the corresponding length back on the original object to get 21 metres so onto you to find the remaining sides on the object and image.

Pause here to use that scale factor of six to find the lengths of the sides labelled A and B.

A is six metres whilst B is 102 metres.

Furthermore, comparing the sum of all the sides of a shape to the perimeter given to you is a good way of checking that all of your calculations are correct.

For example, six plus eight plus 21 plus 17 equals 52, which matches the 52 metres originally given as the perimeter of the object.

Perimeters and corresponding side lengths can be used to verify if an object and an image are similar or not.

For example, we want to verify if this object and that image are similar to each other, the perimeter of the object is four plus six plus eight or 18 metres.

Therefore, the scale factor from object to image is 2.

5.

Let's check if this pair of corresponding sides has this multiplicative relationship of 2.

5.

Well, six multiplied by 2.

5 is indeed 15, so this supports the image being similar to that object.

However, one side is not enough.

We need to check the same multiplicative relationship between all pairs of corresponding sides, so let's check here as well.

Ah, that's a problem, four multiplied by 2.

5 equals 10, not the 12 that is the true length of that side because this pair of corresponding size does not share the same multiplicative relationship of 2.

5 as the two perimeters and the other pair of corresponding sides, we know for absolute certain that that image is not similar to that object.

We can formalise these calculations on this scale factor table.

The perimeter of the object and image with a scale factor of 2.

5 between them.

We can also compare this pair of corresponding sides to see a scale factor, also of 2.

5.

However, with this pair of corresponding sides, the scale factor is not 2.

5.

All scale factors must be equal for these two shapes to be similar.

A table can be helpful to test for similarity when an image may have been rotated or reflected.

With object image pairs like this, we cannot rely on descriptions such as left, bottom, et cetera, as what is left on the object may be in a different orientation in the transformed image.

Therefore, labelling the vertices as A, B, C is helpful in labelling the lengths of the sides.

The object has a perimeter of 12 plus nine plus 15, which equals 36 centimetres and we know that the image has a perimeter of 162 centimetres, therefore the scale factor is 4.

5.

However, because the image has been rotated or reflected in some way, the side that is corresponding to that 60 centimetre length is not necessarily clear.

We test to see if 60 centimetres shares a scale factor of 4.

5 with any of the sides of the object.

If it does, then it is possible that they are a pair of corresponding sides.

Let's have a look at AB first.

The multiplicative relationship between them is five, not what we're looking for.

Let's try again with AC.

This time again, still not 4.

5, and finally, with BC we have four.

Now, it's too small because 60 centimetres is not 4.

5 times larger than any of the sides on the original object, the shapes are not similar to each other.

Imagine, however, that the side length is measured at 54 centimetres instead.

Even though we have a match, the side of 54 centimetres may be corresponding to AB because the scale factor between AB and 54 is also 4.

5, we cannot say for certain that this image is similar to the object.

This is because the other unknown corresponding side lengths may not share the same scale factor of 4.

5 as well.

However, we only need to know one more side length to confirm whether the shapes are similar or not.

For this final check, we have a table that shows lengths on both object and image.

Pause here to find the values in this table in alphabetical order.

The perimeter of the object is 44 centimetres, giving a scale factor between perimeters of five, the base of the object is eight centimetres, giving a scale factor between bases of seven.

The diagonal height of the object is 14 centimetres, giving a scale factor of heights of four.

The scale factor between each pair of corresponding lengths is different, therefore the object and image are definitely not similar to each other.

Right, onto the final practise task.

Pause here for question one.

And pause here again to represent given information in this table for question two.

Great effort, everyone.

Pause here to check your answers to question one, including the scale factor from object to image being a multiplied by seven.

And for question two, from the information given, we simply cannot tell.

The image may be similar to the object because both perimeter and a pair of corresponding sides shares a multiplicative relationship of six.

However, we still need to know the lengths of other sides to confirm their similarity or be told that both shapes are actually rectangles.

Amazing work everyone on a deep dive into perimeter and enlargement in a lesson where we have found the lengths of enlarged images by multiplying corresponding side lengths on the object by a constant scale factor.

Also, the perimeter of an enlargement can be found by calculating the lengths of every one of the sides of that image.

However, a more efficient way is to calculate the perimeter of the object first and then multiply that perimeter by the scale factor.

Furthermore, the perimeter of an object and its image can be used to find the scale factor between the two shapes or help check similarity between those two shapes.

That is all from me, Mr. Gratton, for this lesson.

Thank you all so much for your attention and effort today.

Until next time, everyone.

Take care and goodbye.