Lesson video

Hello and welcome.

Thank you for joining me, Mr. Gratton in this lesson on similarity and enlargement.

Today, we will look at the relationship between the areas of an object and its enlarged image.

Pause here, have a look at some of the important keywords for this lesson.

Before we look rigorously at the maths linking scale factors and area, let's visualise how objects may fit inside their enlargements.

When an object is enlarged by a scale factor of k, this means each of its side lengths are k times larger on the image in comparison to the original object.

For example, shape A is enlarged by a scale factor of two to make B.

What was two units in length on the object is four units on the image.

What was five units on the object is 10 units on the image and so on.

The perimeter of shape A is 16 units, whilst the perimeter of shape B is 32 units.

There is an enlargement by a scale factor of two and therefore the length that is the perimeter is also multiplied by this scale factor of two.

The perimeter of an image is also k times larger than its object.

For this check, we have an object that is enlarged by a scale factor of 12.

By first finding the perimeter of the object, find the perimeter of the image.

Pause now to do this.

The perimeter of the object is 40 units.

Multiply this by the scale factor of 12 gives the perimeter of the image of 480 units.

Perimeter multiplied by scale factor equals the enlarged perimeter.

Here, we have rectangle C that is enlarged by a scale factor of four to create rectangle D.

The enlarged height is four times four equals 16 centimetres.

Whilst the enlarged width is seven times four equals 28 centimetres.

The perimeter of object C is 22 centimetres, whilst the perimeter of image D is 88 centimetres.

Jacob's spots that the perimeter is four times larger because the scale factor is four.

Jacob concludes that the area of D must also be four times larger.

However, Sam is not so sure.

Sam's advice is pretty sensible.

Let's all check by calculating the area of both shapes.

The area of object C is 28 centimetres squared.

However, the area of D is 448 centimetres squared.

That multiplier does not seem to be a multiplier of four and it's not.

This multiplier between the areas is 16 instead.

Jacob understandably does not quite understand why the multiplier for the area is different to the scale factor between the lengths of each side or the perimeter from object two image.

It is possible to visualise why the area multiplier from C to D is 16 by placing copies of C into D.

Along the top of D, we have one, two, three, four.

The area multiplier from C to D is not four as if it were the entire area of D would be filled by these four congruent copies of C.

However, we can still fit more copies of C into D, therefore this area multiplier is going to be greater than four.

In total we have 8, 12, 16, we can fit exactly 16 congruent copies of C into its enlargement, D.

Therefore, the area of D is equal to exactly 16 lots of the area of C.

Here, we have object E with an area of 91 millimetres squared.

We also have its enlargement, F.

We can place congruent copies of rectangle E that fit perfectly inside rectangle F.

Pause here to find the area of rectangle F.

There are exactly nine congruent copies of E each with an area of 91 millimetres squared inside F.

Therefore, the total area of F is 91 times by nine millimetres squared, which is 819 millimetres squared.

Remember, perimeter and area follow different rules when dealing with a scale factor.

The perimeter is a length and so it follows the scale factor directly.

Pause here to find the perimeter of F if the perimeter of E is 40 millimetres and the scale factor from E to F is three.

Each length is three times as long and, therefore, the perimeter of F will be three lots of the perimeter of E, so 40 times by three millimetres is 120 millimetres.

Right, Jacob's inquiry is very sensible.

We've spotted that this strategy works with rectangles, but does it work with other shapes like triangles? Well, let's have a look.

Sam believes it is possible, but maybe we have to do some rotation or reflection of some of the congruent copies of object G first.

Here we have triangle H.

That is an enlargement of G by a scale factor of two.

Its base and perpendicular height are 18 centimetres and 24 centimetres, respectively.

The area of G is 54 centimetres squared whilst the area of the enlargement H is 216 centimetres squared.

The multiplier between these two areas is a multiply by four, so let's see if we can fit exactly four lots of G into that enlarged triangle, H.

That's one, two, oh, that's only three copies of G.

Where's the fourth? Well, Sam has a keen eye and has spotted that the empty space in the middle of triangle H is that fourth upside down copy of G.

But whilst the upside down triangle does look like the exact same shape as G, Jacob's advice is absolutely brilliant.

Let's use tracing paper to check that the empty space is actually congruent to G and not a random triangle that looks close to G.

I take the tracing paper and draw G as accurately as possible with a ruler and check that all four triangles are congruent to G.

Well, these three are and so is the fourth one after rotating that tracing paper 180 degrees.

We have demonstrated that all four triangles inside H are congruent to G.

Therefore, we have shown visually why the multiplier between the areas of G and H is a multiply by four.

Here we have triangular object J and its enlargement, K.

Jacob uses tracing paper to draw triangles that are congruent to J along the base of K.

Pause here to calculate the scale factor from J to K.

Five triangle Js fit along the base of K, so the scale factor is five.

This will also have been true if we try to fit triangles along the vertical height of K exactly five will fit like this.

We can see that five triangle Js fit along the base, five along the vertical height and five along the diagonal side of K.

Using this information and the fact that the perimeter of J is 12 centimetres, pause here to calculate the perimeter of K.

As the scale factor is five, five, lots of the perimeter J will give the perimeter of K at 60 centimetres.

And lastly, by first counting the number of triangles congruent to J inside of K, pause here to calculate the area of K.

There are 25 congruent copies of J inside K.

Therefore, 25 copies multiplied by the six for one area gives 150 centimetre squared as the total area of K.

Great stuff everyone.

Let's have a look at some independent practise.

For question one, we have a triangular object and its image.

Explain how you know that the scale factor is four, and then, think about how you would find the area multiplier from A to B and then using that area multiplier, find the area of B.

For question two, some congruent copies of rectangle C have been placed inside D.

Enough copies of C have been placed inside D in order for you to answer parts A and B, but feel free to draw the rest in if it helps you to visualise the enlargement.

Using what information you have, find the area of D.

Pause now for questions one and two.

And for question three, grab some tracing paper to figure out how many congruent copies of E you can fit inside F.

Pause now to answer all five parts to question three.

Wow, great effort on all of those questions everyone.

The answers to question one.

Four triangle A fit along the base or on the right hand side or on the left hand diagonal side of triangle B.

Therefore, the scale factor from A two B is four.

For part B, there are a total of 16 triangle A inside triangle B, and so the area multiplier from A to B is 16, therefore the area of B is 160 centimetres squared.

For question two, the length of the base of D is a length that is five times as long as that of C because five copies of C can fit along the base of D.

The same can be said for the height.

For part B, because you can fit a total of 25 rectangle Cs in D, the area of D is 150 centimetres squared.

And for question three, here's what your shape F would have looked like if you used the tracing paper correctly.

The scale factor is three as three Es fit along the base of F.

However, the area multiplier is nine because there are a total of nine copies of E inside F.

For question 3D, the perimeter of F is 51 units and the area of F is 108 units squared.

Now that we've visualised how congruent copies of an object may fit into its enlarged image, let's formalise what we've seen for all sorts of different shapes and their enlargements.

Okay, here we have an object, a trapezium that we want to enlarge by a scale factor of four and Jacob asks whether we can fit congruent copies of the trapezium into an enlargement of it.

However, some wonders if it's even possible to do this with every shape.

Well, let's have a look.

Here's an enlargement and here's an attempt at fitting congruent copies of the original object into it.

No matter how you try to arrange these trapeziums in whatever orientation or configuration, it is absolutely impossible to fully cover the entire area of the enlargement with only congruent copies of that object.

There will always be some space left over that is not in the shape of a trapezium, and here's a more obvious example of this.

A circle's enlargement can never be fully filled by only congruent copies of the original circle.

Sometimes it is impossible for a shape to tessellate perfectly inside an enlargement of itself even if that shape does tessellate on a plane.

And sometimes it is absolutely impossible because the shapes cannot tessellate at all.

Jacob loses hope that the area of an enlarged shape can even be found if you can't even fit congruent copies of the original object into the enlargement.

However, it is definitely still possible to calculate the area of an enlarged image.

There is a relationship between the skill factor used to enlarge an object and the multiplier between the area of the object and its enlarged image.

The area multiplier is also known as the area scale factor, a scale factor between the areas of the object and the image.

So, we have area scale factor as the scale factor between the areas of an object and its image.

So, what's this? What's the linear scale factor? Well, linear scale factor is just a more detailed description of the scale factor that we are already familiar with.

The scale factor between the lengths of an object and its image.

L for length also means L for linear scale factor.

You might see the phrase scale factor being used.

Most of the time, scale factor actually means linear scale factor.

However, it is best to try and understand whether it actually means linear scale factor or area scale factor by looking at the context that scale factor is being used in.

Does the context reference side lengths and perimeters or does it reference areas? For clarity, during the rest of this video, we'll always use either linear or area scale factor explicitly.

So, how do we even go about finding the area scale factor? Well, in this example, each side of C has been enlarged by a linear scale factor of four to make D.

The area of C is five times nine equals 45 centimetres squared, whilst the area of the enlargement D is five times four and nine times four multiplied together.

The five and the nine can be rearranged because five times nine is the area of C at 45.

Whilst we have this, four times four left over, which can be simplified to four squared centimetres squared.

By rearranging our calculation for the area of D, we can see that the area of D is four squared times larger than C.

Therefore, the area scale factor from C to D is four squared equals 16.

But Jacob inquires, what's so significant about four squared to which Sam spots that four squared is the square of the linear scale factor? If the scale factor is four, then the area scale factor is four squared.

So, if the area of C is 45 centimetres squared, then the area of D is 45 times by four squared, giving 720 centimetres squared.

This relationship is true for all linear and area scale factors.

If the linear scale factor is k, where k is just any number, then the enlargement with side lengths that are k times larger than the side lengths of the object give us an area scale factor that is always k squared.

If an object has an area of ab, then the enlarged image has an area of ab times by k squared.

Only one pair of corresponding lengths are needed on any two similar shapes in order to find both the linear scale factor and the square of the linear scale factor, that is the area scale factor.

We can find the linear scale factor by identifying the multiplicative relationship between these two corresponding lengths and then square this linear scale factor to find the area scale factor.

For this check, let's bring together our knowledge of lengths for the linear scale factor and areas for the area scale factor.

Pause here to complete these sentences for object E and its enlargement, F.

The height of F is five times larger than the height of E because eight times five equals 40.

The linear scale factor from E to F is therefore five and the area scale factor from E to F is 25, the square of the linear scale factor.

The area of F is 400 inches squared because the area of F is just the area of E multiplied by that area scale factor of 25.

We can calculate the area of any shape using this area scale factor relationship between two similar shapes.

These two trapeziums are similar to each other.

We first need to find the linear scale factor by identifying two corresponding lengths.

The vertical heights of 24 and four are corresponding, giving a linear scale factor of six.

Therefore, the area scale factor is the square of this at six squared equals 36.

First, we need to find the area of trapezium, G, which is 30 centimetres squared.

The area of the similar shape, H, is 30 times by the area scale factor of 36, giving 1,080 centimetres squared.

We can represent all relevant lengths and areas in this, a scale factor table.

One row showing lengths and the linear scale factor with the other row showing the areas and the area scale factor.

We have two corresponding lengths, the heights of 24 and four, giving a scale factor of 24 over four or six.

The area of G is 30 and the area scale factor is just the linear scale factor squared, six squared equals 36.

And the area of H is the area of G multiplied by this area scale factor giving 1,080.

Okay, for this check, J has been enlarged by a scale factor to make K.

Pause here to fill in all five parts of this scale factor table.

The scale factor is three, giving a length of K at 33 centimetres.

The area of J is 22 centimetre squared multiplied by this area scale factor of nine, giving an area for K at 198 centimetres squared.

It is also possible to divide using the area scale factor to find the area of an original object if you are given the area of the enlarged image.

Pause here to fill in all six parts of this scale factor table.

The diagonal lengths of 42 and six give a linear scale factor of seven, meaning that the area scale factor is 49.

If the area of shape M is 392, then the area of the object shape L is 392 divided by 49, which is eight centimetres squared.

And lastly, linear and area scale factors may be non-integer.

Where possible, it can be helpful to leave the scale factors as simplified fractions.

Pause here to complete this scale factor table to find the area of shape, N.

If the linear scale factor is seven over two, then the area scale factor is seven squared over two squared at 49 over four.

Therefore, the area of N is 60 centimetres squared.

Brilliant effort on this cycle so far.

For question one of this practise task, we have two similar shapes.

Identify the scale factors between them and, therefore, find the area of B.

And for question two, calculate the lengths, areas, and area scale factor from C to D.

Pause now for questions one and two.

For question three, pause here to complete this scale factor table and, therefore, find the area of F.

For question four, let's find the area of triangle H in two different ways.

Way one by using two lengths found from the linear scale factor and way two, using an area and the area scale factor.

And for question five, construct your own scale factor table from scratch in order to find the area of J.

Pause now for these final two questions.

Amazing effort everyone.

Here are the answers.

For question one, we have a linear scale factor of two and, therefore, the area scale factor is four.

Four lots of 60 gives an area of 240 centimetres squared.

For question two, the length of r millimetres is 20 millimetres.

The area scale factor is 10 squared equals 100, and, therefore, 100 lots of 3.

5 gives an area of 350 millimetres squared.

For question three, pause here to check your scale factor table with the one on screen and the area of F is 490 centimetres squared.

And for question four, the calculation that uses two lengths is 100, that is the base, multiplied by 45, that is the height, divided by two.

Whilst the area calculation is 90 times 25, with that area scale factor being 25, both methods give exactly the same answer of 2,250 centimetres squared.

And finally, question five.

Pause here to check your calculations and compare them to the ones on screen.

The area of J is 104 millimetres squared.

Great effort, everyone, in all of your applications of scale factors and a lesson where we have found the areas of enlargements by fully filling and enlargement with congruent copies of the original object.

We've also honed our definition of scale factor two, a linear scale factor that is a multiplier between lengths of an object and its enlarged image, whilst an area scale factor is a multiplier between areas.

If the linear scale factor is k, then the area scale factor is always k squared.

Once again, I thank you all so much for joining me, Mr. Gratton, in this maths lesson.

So, until our next maths lesson together, stay safe, take care, and have an amazing rest of your day.