### Video Transcript

In this video we’re going to look at similar figures and the relationship that exists between their areas. So first, let’s look at what is meant by the term similar figures. The definition’s on the screen here. Two figures are similar if first of all they have to be the same shape, so both squares or rectangles or triangles for example, and secondly corresponding lengths on these shapes have to be in the same ratio.

So if you think about a triangle for example, the bases of these two triangles must be in the same ratio as the heights of the two triangles. Let’s look at an example of this using rectangles. So the two rectangles on the screen here, they’re similar rectangles. And you can see this if you look at the ratio between corresponding pairs of sides. If I look at the width of the rectangle first of all, so the two and the three, this will give me what’s known as a length ratio of two to three.

If I look at the other dimension of the rectangle, so the four to six, then this would tell me that the length ratio is four to six. But of course this can be simplified by dividing both sides of this ratio by two. And if I do that, you’ll see that it simplifies to this same ratio of two to three. So because the corresponding pairs of sides are both in this ratio two to three when simplified, that means that these two rectangles are similar.

Now in this video we want to look specifically at the relationship between the areas of these similar figures. So I’ll work out both of the areas. Now as they’re both rectangles, that’s relatively straightforward, I just need to multiply their two dimensions together. So I have eight centimetres squared for the first rectangle and eighteen centimetres squared for the second.

Now let’s use those areas to write down an area ratio between the two rectangles. So the area ratio is going to be eight to eighteen, but that will simplify cause again I can divide both sides of the ratio by two. So that tells me that my simplified area ratio is four to nine.

Now the key point is how does this relate to the length ratio. Well the length ratio, remember, was two to three and the area ratio is four to nine. And perhaps you’ll spot that there’s a relationship between those sets of numbers, and it’s a square relationship; two squared gives us four and three squared gives us nine.

So another way of writing this area ratio would be two squared to three squared. Now it isn’t a coincidence that this relationship exists. It’s illustrative of a general rule when looking at the areas of similar figures and the general rule is this: if the length ratio between two similar figures is 𝑎 to 𝑏 then the area ratio between them will be 𝑎 squared to 𝑏 squared.

So this rule always holds true when working with similar figures. And now we’ll see how we can use it to answer some questions. So here’s the first question. We’re given a diagram of our polygon, in fact it’s a rectangle, on a coordinate grid and we’re asked to find the area of a similar polygon where 𝐴 dash 𝐷 dash is equal to six.

So let’s just look at the rectangle that we’re given first of all. Well the length of the base here from 𝐴 to 𝐵, that is two units. And the length of the vertical side here from 𝐴 to 𝐷, that is three units. So we can fill in those two measurements. We can also work out the area of this rectangle, so it’s gonna be two times three it’s gonna be six square units.

Now let’s think about this second polygon. So we’re told that it’s similar, but we’re told that the length of 𝐴 dash 𝐷 dash is six, whereas in our current polygon the length of 𝐴𝐷 is three units. This means we can write down the length ratio between these two polygons. So using that pair of corresponding sides, it’s going to be three to six.

But of course that ratio can be simplified; I can divide both sides of it by three. So my simplified length ratio is going to be one to two. What this means then in practical terms is that all of the lengths in the enlarged polygon are twice as long as they are in the existing polygon.

Now we want to know about the area of the larger polygon, so we need to recall that general rule that we saw before about the link between the length ratio and the area ratio. And the rule remember was this: if the length ratio is 𝑎 to 𝑏, then the area ratio is 𝑎 squared to 𝑏 squared. So that means I can use my known length ratio to work out the area ratio; I just need to square both sides of it.

So my area ratio is one squared to two squared, which of course is just one to four. What this means then is that the area of this enlarged polygon is four times the area of the smaller polygon.

So I have all the information I need in order to be able to calculate the area of this similar polygon. The area of the smaller polygon was six square units. And if this one is four times larger, than I just need to multiply six by four. So the area of this polygon six by four, which is of course twenty-four square units.

So a quick recap of what we did then. We used a pair of corresponding lengths to write down a length ratio. We then used our general rule to transform that to an area ratio by squaring both sides. Because we knew the area of the smaller polygon, we then combined it with the area ratio to work out the area of the larger polygon.

Okay the next question that we’re going to look at says corresponding sides of two similar polygons are eighteen centimetres and twenty-five centimetres. Given that the area of the smaller polygon is four hundred and eighty-six centimetres squared, we are asked to determine the area of the larger polygon. So let’s think about how to approach this then. We’re given this pair of corresponding sides, which means we can start off by writing down the length ratio between these two similar polygons.

And so here it is. The length ratio is eighteen to twenty-five, and that doesn’t simplify any further. Now we’re asked about areas, so we need to work out the area ratio. And remember that general rule, we need to square both sides of this in order to work out the area ratio.

So the area ratio will be eighteen squared to twenty-five squared. And so this is three hundred and twenty-four to six hundred and twenty-five, which again doesn’t simplify any further. So let’s use this area ratio to work out the area of the larger polygon. So we’re gonna give it a letter; I’m gonna call it A.

Now what this ratio means is if I take the area of the larger polygon, which is A, and divide it by the area of the smaller polygon, four hundred and eighty-six, then they’re in this ratio of three hundred and twenty-four to six hundred and twenty-five, which means I get the same result if I were to do six hundred and twenty-five divided by three hundred and twenty-four.

So what I’ve done here is set up an equation which I can then solve in order to work at the area of this larger polygon, using the area ratio that I know. The first step to solving this equation then is I need to multiply both sides of the equation by four hundred and eighty-six as that’s currently in the denominator on the left-hand side. So I’ll have six hundred and twenty-five over three hundred and twenty-four multiplied by four hundred and eighty-six.

And if I evaluate that and put in the units, it tells me that the area of the larger polygon is nine hundred and thirty-seven point five square centimetres. So exactly the same process as in the previous example: we wrote a length ratio first of all; squared both sides to form an area ratio; the bit that’s different from the last example is that previously we just had to multiply the area by four as it was just a one-to-four ratio; this time the ratio was a bit more complex, so we had to transform it into a fractional relationship, use that to form an equation, and then solve that equation to find the missing area.

Okay let’s look at a different type of question. We’re told that the two polygons below are similar, and we’re asked to calculate the value of 𝑥. Now 𝑥 is this missing length here in the smaller polygon. So let’s look at what we know. We haven’t been given a pair of corresponding lengths this time. We’ve been given a pair of corresponding areas because we can see that the two areas are thirty-five and three hundred and fifteen centimetres squared.

So we’re going to have to approach this question in a slightly different way. Rather than writing down a length ratio to start off with, we’re gonna instead right down an area ratio. So the ratio of these two areas is thirty-five to three hundred and fifteen. That can be simplified cause both sides can be divided by thirty-five.

So that gives me an area ratio of one to nine. Now we’d like to work backwards from knowing this area ratio to working out the length ratio, so remember the general rule that we saw previously. It was that whatever the length ratio is, to get the area ratio you have to square both sides. Now we’re gonna be working back the other way from the area ratio to the length ratio. So in order to go back the other way, rather than squaring both sides, we need to square root both sides.

So the length ratio is the square root of one to the square root of nine, which of course is just one to three. So what this tells us is that all the of lengths in the larger polygon are three times the lengths in the smaller polygon.

Now if I wanted to set it up as an equation just to demonstrate again what I did in the previous example, then I know that if I do 𝑥 divided by eighteen, so that corresponding pair of sides, then I’m gonna get the same result as if I do one divided by three.

So that’s using this length ratio of a third. So then I can solve this equation by multiplying both sides by eighteen, and so 𝑥 is equal to six. Another way of thinking about it without having to write down the equation is because it’s just a one-to-three ratio, then I could just divide eighteen by three to get this value of six.

So in this example, it was slightly different. We had to start with an area ratio and then work backwards by square rooting in order to calculate the length ratio that we needed.

Okay the final question says shapes X and Y are similar with sides in the ratio five to four. If each side length is tripled, what is the ratio of the areas of the enlarged shapes? Now you may have a gut feel for what the answer should be, or you may think perhaps there’s a detail you haven’t considered. Let’s go through some working out in order to see how we should approach this.

So we already have a length ratio for shapes X and Y; it’s five to four. That would mean that the area ratio for X and Y before we triple the sides and enlarge them, well remember we would square both sides of this ratio so the area ratio for X and Y would be twenty-five to sixteen.

Now let’s think about what happens when we triple the side lengths. Well we don’t know that these lengths are five and four, just that they’re in that ratio. But if we triple them, then they would be in the ratio fifteen to twelve. But here’s the key point. Because we’ve tripled both of them, that ratio of fifteen to twelve just simplifies straight back down to five to four again.

So tripling the side lengths actually has no effect whatsoever on the length ratio because we’ve done the same thing to both of the similar shapes. Therefore the area ratio of these enlarged shapes is gonna be the same as the area ratio of the original shapes.

So the area ratio is still twenty-five to sixteen. Now that may have been your gut feel. But if it wasn’t, then I hope that by going through the working out you’ll have been convinced of why this is the case.

So the key point from this example is if a question asks you about doubling or tripling the sides of both of the similar shapes, then it has no impact on the length ratio and consequently no impact on the area ratio. So you can just work with the original ratios that you’re given.

In summary then, we’ve seen the relationship between the length ratio and the area ratio of similar figures; we’ve seen how to apply this to calculating the area of a similar figure; and we’ve also seen how to apply it to calculating missing length by working backwards from knowing the area ratio to calculating the length ratio.