Lesson video

Hello, everyone.

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

Today, we will look at the relationship between the area scale factor, linear scale factor, areas and lengths of a pair of similar shapes.

Pause here to have a look at some of the keywords that we'll be using today.

First up, let's find missing lengths when given the areas of two similar shapes.

Right, to start, here we have an object and its image, and they are both similar to each other.

The linear scale factor between them can be found by identifying the multiplicative relationship between one pair of corresponding lengths.

And these lengths can be anything, a side length, a height, or even a perimeter.

For this pair of similar shapes, the linear scale factor between the base lengths of 36 and 4 is 36 divided by 4, which simplifies to a multiply by 9.

The linear scale factor between any two corresponding lengths on A and B is nine.

Furthermore, the multiplicative relationship between the areas of these two shapes, which is also known as the area scale factor, is always the square of the linear scale factor.

Therefore, the area scale factor is 9 squared or 81.

Therefore, we can take the area of A, which is 30 centimetre squared and multiply it by 81 to get the area of the enlarged shape B at 2,430 centimetres squared.

Okay, for this check, pause here to identify which of these statements are correct.

The linear scale factor is a multiply by 6 and therefore, the area scale factor is 6 squared at 36.

We can represent corresponding lengths and areas in this, a scale factor table, in order to find the linear scale factor, the area scale factor, and the areas of two similar shapes.

The height, a pair of corresponding lengths between E and F are 64 and 20, meaning that the linear scale factor is 64 over 20, which simplifies to 16 over 5.

Not all scale factors are integers.

Non-integer scale factors can be kept as simplified fractions.

We know that the area of E is 1,050 and the area scale factor is the square of 16 over 5, which is 256 over 25.

Therefore, the area of F is 1,050 multiplied by 256 over 25, that area scale factor, which gives a total of 10,752 millimetres squared.

Notice how the area increases a lot in comparison to the increase in the lengths of the corresponding sides.

This is because any linear scale factor greater than one results in an area scale factor that is larger than the linear scale factor.

For this check, pause here to place the correct information into this scale factor table and then calculate scale factors in order to find the values of A to E.

The linear scale factor is three, and so the area scale factor is 3 squared equals 9.

9 times the area of 105 gives an area of 945.

Next up, we have object J that has been enlarged by a linear scale factor of 15.

Pause here to complete this scale factor table and answer questions A and B.

If the linear scale factor from J to K is a multiply by 15, then we can divide a length on K by 15 to get the corresponding length on J.

Therefore, 30 divided by 15 equals 2.

Furthermore, if the linear scale factor is 15, then the area scale factor is 15 squared equals 225.

Therefore, 1,125 divided by 225 equals 5.

Therefore, the area of J is 5 centimetres squared.

And lastly, all these scale factor relationships apply to all shapes, not just polygons.

The linear scale factor from L to M is 3 over 2.

Pause here to find the area of shape M.

The area scale factor is 3 over 2 squared, which is 9 over 4.

And in terms of pi, the area of M is 117 pi centimetre squared.

It is also possible to calculate the linear scale factor between two similar shapes if we are given the areas of both of those shapes.

If the area scale factor is the square of the linear scale factor, then the linear scale factor is the square root of the area scale factor.

For these two similar shapes, their areas are 1,152 and 18, giving an area scale factor of 64.

If the area scale factor from P to Q is 64, then the linear scale factor from P to Q is the square root of 64, which we place here.

The square root of 64 is 8.

Therefore, the linear scale factor from P to Q is eight.

Using that knowledge of the relationship between the area scale factor and the linear scale factor, pause here to complete both of these sentences.

If the area scale factor is 121, then the linear scale factor is the square root of 121, which is 11.

And once again, pause here to use the relationship between the area scale factor and the linear scale factor.

This is the completed scale factor table, and the linear scale factor is nine.

Okay, we can find linear scale factors from areas, but so what? Well, we can use those linear scale factors to calculate lengths between two similar shapes.

For example, here we have shapes V and W.

Their areas are 126 and 14, giving an area scale factor of nine.

Therefore, the linear scale factor is the square root of nine, which is three.

The linear scale factor applies to all types of lengths, so the height, another length, also has that same linear scale factor of three.

If the height of V is six, then the height of W is 6 times 3, which is 18 centimetres squared.

Conversely, if the perimeter of W is 60, then we can divide by the scale factor to find the perimeter of V at 20 centimetres.

We have used this linear scale factor to find lengths on W and on V.

Okay, for this check, we have two similar shapes with lengths c centimetres and d centimetres that are unknown.

Pause here to find the values of c and d using this scale factor table.

This is the fully filled-in table, therefore, c centimetres is seven centimetres and d centimetres is 28 centimetres.

Well done, everyone.

Good effort so far.

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

Find the linear and area scale factor between them and use this information to find the area of A.

And for question two, find the area scale factor and linear scale factor from C to D as simplified fractions.

Pause now for these two questions.

Okay, for question three, using all of the information on similar shapes, E and F, fill in that scale factor table to find the radius of F and the perimeter of E.

Pause now for question three.

And finally, question four, pause here to find the lengths of x, y, and z and then by imagining or sketching a third similar shape, find the area of that third shape.

Brilliant work.

Onto the answers.

For question one, the linear scale factor is 11 and the area scale factor is 121.

The area of shape A is 5 centimetres squared.

For question 2, the area scale factor is 49 over 4 when fully simplified.

Therefore, the linear scale factor is 7 over 2 or 3.

5.

For question three, pause here to compare your scale factor table to the one on screen.

And the radius of F is 48 centimetres whilst the perimeter of E is 71 centimetres.

For question 4a, pause here to compare your calculations to the ones on screen in this scale factor table and x equals 96, y equals 6, and z equals 108.

And finally, question 4b, the linear scale factor from the larger shape to this new shape is 1 over 20.

Therefore, the area scale factor is 1 over 400.

This is because the length x is 96 centimetres long.

We can see from these scale factors that this new shape is really small in comparison to the two original shapes, so we expect the area of this new shape to be much smaller than both of them.

The area of the large shape is 1,728.

Multiplied by this scale factor of 1 over 400 gives an area of this new shape of just 4.

32 centimetres squared.

Right, now that we know how to find lengths when given the areas of two similar shapes, let's take all of the information that we know and more to find the perimeters of shapes from areas and other information.

So, if you are given two similar shapes, then sometimes, we can calculate the perimeter when given enough information about their areas and enough other information about the rest of the shape.

These are the two areas and they have a scale factor of 16 from A to B, meaning that every single length has a linear scale factor of four.

We can see that triangle A has a length on the right-hand side of three centimetres, meaning that the corresponding side on B has a length of 12 centimetres.

Similarly, B has a length on the left-hand side of 28 centimetres, meaning that the corresponding side on A has a length of seven centimetres.

Ah, but for the base lengths, we already have both lengths for A and B.

We can see that 8 times 4 does actually equal 32.

This pair of corresponding side lengths agrees with the linear scale factor of four.

Four was calculated in two different ways, showing that the area scale factor that we calculated was likely correct.

Now that we know the lengths of all three sides of both triangle A and B, we can calculate each of their perimeters.

The sum of all of the sides on A gives a perimeter of 18 centimetres whilst the sum of all of the side lengths on B gives a perimeter of 72 centimetres.

Notice how all of the linear scale factors for each of the individual lengths are a multiply by four.

This should absolutely be the same for the perimeter scale factor since the perimeter is just another length.

For this check, pause here to complete this scale factor table in order to find the perimeters of shapes C and D.

The area scale factor is 25 and therefore, the linear scale factor is five.

The perimeter of C is 46 centimetres whilst the perimeter of D is 230 centimetres.

Jacob sensibly asks whether it is always possible to find the perimeter if you know the areas of both shapes like with this example where we know the area of E is 160 centimetres squared and the area of F is 1,440 centimetres squared.

But Sam thinks not as sometimes, you might not get given the right extra information about these shapes, such as not enough unique side lengths across both object and image.

If a pair of corresponding side lengths on both object and image are not known, then it is likely not possible to calculate the perimeters of either shape.

Once again, using a scale factor table to place all of the information in a neat and tidy way can help identify whether enough information is known or can be figured out using scale factors.

For this scale factor table, we have two areas related by an area scale factor of nine, meaning that the linear scale factor is three across all different lengths.

We know a pair of corresponding side lengths at the top of each shape.

These two side lengths agree with our multiplied by 3 scale factor because 12 times 3 does equal 36.

We know this 33-centimeter length on F, so the corresponding length on E is 11 centimetres.

Similarly, this small diagonal on the left of E is one centimetre, so the corresponding length on F is three centimetres.

But we know neither the long right-hand diagonal sides on either E or F.

Because we can't find out the length of either shape, neither the perimeters of E nor F can ever be found.

Here are a few checks to do with these two similar triangles.

Jacob says that because three side lengths are known, we have enough information to find either perimeter.

Pause here to think about or discuss why Jacob's comment is incorrect.

It's incorrect because we have a pair of corresponding sides, the right-hand side of each triangle, whose lengths are unknown.

So pause here to figure out what information we can find out, given the current information on G and H.

We can find out the linear scale factor, the area scale factor, the length of side x, and the vertical height of H.

We can find out, so let's do so.

Pause here to find all four of these pieces of information about these two triangles.

The linear scale factor is two and therefore, the area scale factor is four.

The length of x is 8 divided by the linear scale factor of 2, which equals four.

And the vertical height of H is 48, divided by that base of 8, then multiplied by 2 to get 12 centimetres.

Jacob uses what we've just discussed to conclude that the perimeters of these two shapes cannot be found since we only know two lengths and that's just not enough information.

Usually, I would agree with what Jacob has said.

However, pause here to think about or discuss what is particularly special about these two triangles and what can we do with that extra information to find the lengths of more sides that we couldn't find with other less special triangles.

Sam believes that we can use Pythagoras' theorem to find the length of a third side of one of the two triangles.

This is because the special property of these triangles is that they are right-angled triangles.

However, if we take one triangle as it currently is, we still cannot find the hypotenuse.

Since only one side length is known, therefore, we cannot use Pythagoras' theorem as we need two separate side lengths to find the length of the third side.

What we can do, however, is find the area scale factor in order to find the lengths of the sides corresponding to 16 and 21 centimetres.

This gives us two sides of each triangle, meaning that we can use Pythagoras' theorem to find the length of the third side.

So here is our scale factor table with one row for each of the three sides of the triangle and one row for the perimeter and one row for the area.

The two areas give us an area scale factor of nine, meaning that the linear scale factors are three.

The base on J is 16 centimetres, so the corresponding base on K is 48 centimetres.

Similarly, the height on K is 21 centimetres, and so the height on J is seven centimetres.

We currently know neither hypotenuse, but we do now have enough information to use Pythagoras' theorem to find the lengths of each hypotenuse.

In fact, we only need to use Pythagoras' theorem once to find the length of one hypotenuse.

The other hypotenuse can be found using the linear scale factor.

However, using Pythagoras' theorem a second time on the second triangle should verify the length that you calculated from scale factors is correct in a second different way.

So let's use Pythagoras theorem on that smaller triangle J.

We have 7 squared plus 16 squared equals the hypotenuse squared of c squared, which simplifies to 305 equals c squared.

C is therefore the square root of 305, which is 17.

46 centimetres after rounding.

The linear scale factor is three, and so the hypotenuse on K is 3 times the length of the hypotenuse on J.

3c equals 3 lots of the square root of 305, which is 52.

39 centimetres after some rounding.

It is possible that you might have calculated this hypotenuse to be 52.

38 centimetres instead by multiplying 17.

46 by 3.

This small difference is due to rounding at different stages, and 52.

39 is the slightly more accurate value.

So now we do know the hypotenuse of each triangle from Pythagoras' theorem, meaning that each perimeter can be found.

We can check that our calculations make sense from the same linear scale factor of three as with all of the other lengths in our triangles.

It is even possible to find the perimeters of two similar right-angle triangles if only one side length is known across both shapes as long as both areas are given.

Pause here to calculate the length of y centimetres.

Y centimetres is five centimetres.

So, using this given information, pause here to find the area scale factor and linear scale factor from L to M.

The area scale factor is 2,240 divided by 35, which equals 64.

And the linear scale factor is the square root of 64 at 8.

Now we know the lengths of two sides, pause here to use Pythagoras' theorem to find the hypotenuse of L.

When rounded to two decimal places, the hypotenuse is 14.

87 centimetres.

And now, using all the known information that we have, find the perimeter of M.

The perimeter of L is 33.

86 centimetres, and so the perimeter of M is eight times longer as the linear scale factor is eight, giving a perimeter of 270.

93 ish centimetres.

Your answer to the perimeter of M might be marginally different, depending on your rounding at different stages.

Great effort, everyone.

Onto the practise task.

For question one, complete the scale factor table to find the perimeters of A and B.

And for question two, explain why it is impossible to find the perimeters of these two shapes.

Pause now for questions one and two.

And for question three, we have three similar shapes.

By drawing a modified scale factor table or two different scale factor tables, pause here to find the perimeter of each shape.

And finally, question four, Lucas designs a metal plate cover using some computer software, as you can see on this diagram.

A machine then cuts out a real-life version of this design to real-life scale.

The real metal plate has an area of 6,318 millimetres squared.

Pause here to calculate the distance that Lucas traced around on the real metal plate cover.

Amazing work on those questions.

Here are the answers.

Pause here to check your scale factor table for question one and compare it to the one on screen.

The perimeter of A is 35 centimetres and the perimeter of B is 140 centimetres.

For question two, the perimeter is impossible to find because the length of that right-hand diagonal side is not known on either shape, meaning its length cannot be calculated using scale factors.

We also do not know for certain that either shape is a right-angled trapezium and so we cannot rely on Pythagoras' theorem due to a lack of right angles.

However, the lengths of 6 centimetres, 22 centimetres, and 21 centimetres can be found.

For question three, pause here to compare your calculations to the ones on screen and note that the perimeters of A, B, and C are A is 30 centimetres, B is 90 centimetres, and C is 180 centimetres.

And finally, question four, the computer design had an area of 19.

5 millimetres squared and a perimeter of approximately 26.

22 millimetres.

The perimeter was found by first using Pythagoras' theorem on the two diagonal sides of the design.

Using the areas of 19.

5 and 6,318, the area scale factor is 324.

Therefore, the linear scale factor is 18.

The perimeter of the real metal plate cover is 472 millimetres after some rounding.

Amazing work, everyone, on this lesson where we have used the linear scale factor k and the area scale factor of k squared.

If given the areas of two similar shapes, its area scale factor can be found and therefore, the linear scale factor can also be found by square rooting the area scale factor.

It is important to be careful in first, identifying corresponding values between two similar shapes and whether those values are lengths or areas.

And lastly, we can use other mathematical skills such as Pythagoras' theorem to help find missing sides of certain triangles.

Thank you all so much for your effort today.

I've been Mr. Gratton, and you've been a great mathematician.

Until next time, everyone.

Take care and goodbye.