# Lesson Video: Areas of Similar Polygons Mathematics

In this video, we will learn how to calculate areas of similar polygons given two corresponding side lengths or the scale factor between them and the area of one of the polygons.

17:16

### Video Transcript

In this video, we will learn how to calculate areas of similar polygons given two corresponding side lengths or the scale factor between them and the area of one of the polygons. Weβll begin by recapping what it means for two polygons to be similar.

Two polygons with the same number of sides are similar if two conditions are satisfied. Firstly, corresponding angles are congruent and secondly, corresponding sides are proportional. For example, consider these two rectangles. Both shapes have four sides. And as all the interior angles in a rectangle are right angles, corresponding angles are congruent. Considering the side length, if we divide the length of the longest side of the first rectangle by the length of the longest side of the second, we have π΄π΅ over ππ, which is equivalent to πΆπ· over ππ. And this is equal to eight over four, which is two.

If we consider the shorter side lengths, we have π΅πΆ over ππ, which is the same as π·π΄ over ππ. This is three over 1.5, which is also equal to two. So, the ratio is the same for each pair of corresponding sides, and hence corresponding sides are indeed proportional. Notice that the two rectangles in this example were drawn in different orientations. The corresponding sides are vertical on one rectangle, but horizontal on the other. So this is something we do need to be aware of when working with similar polygons. We can state that π΄π΅πΆπ· is similar to ππππ. And the ordering of the letters is important here because it reflects which vertices of the two polygons are corresponding to one another.

To find the scale factor from one polygon to another, so letβs first consider the scale factor from π΄π΅πΆπ· to ππππ, we divide any of the side lengths of ππππ by the corresponding side length of π΄π΅πΆπ·. So, for example, the scale factor of π΄π΅πΆπ· to ππππ is ππ over π΄π΅. Thatβs four over eight, which is equal to one-half. Scale factors are always multiplicative. So this means that to get from a length on π΄π΅πΆπ· to a length on ππππ, we multiply by one-half. To go the other way, we multiply by two.

We should always check that any scale factors weβve calculated make sense. If weβre going from the larger polygon to the smaller one, then the scale factor should be less than one. Whereas if weβre going in the other direction, the scale factor should be greater than one. In general, if the scale factor in one direction is π, then in the opposite direction it is the reciprocal, one over π.

An alternative way of expressing the relationship between the side lengths of two similar polygons is using a ratio. For example, the ratio of π΄π΅πΆπ· to ππππ is the ratio of the length of side π΄π΅ to the length of side ππ. Thatβs eight to four, which simplifies to two to one. Now, all of what weβve discussed so far should be a recap of our knowledge of similar polygons. Weβre now going to extend this to understand how the areas of similar polygons are related to one another.

So suppose we have two similar polygons in which the scale factor of similarity is π. Our first thought may be that the scale factor between the areas is also π. Letβs test this for the two similar triangles shown here. The area of the first triangle is its base multiplied by its perpendicular height over two. The area of the second triangle is its base multiplied by its perpendicular height over two. Thatβs ππ multiplied by πβ over two. We can rewrite this as π squared multiplied by πβ over two. And as the area of the first triangle is πβ over two, we find that the area of the second triangle is π squared multiplied by the area of the first. So in fact, the scale factor for the areas isnβt π; itβs π squared.

If we think about it, this makes sense because area is two dimensional and both dimensions have been enlarged by this factor of π. So, the effect on the area is to enlarge it by a factor of π squared. This result is true for all polygons. We therefore need to be a little more specific when referring to scale factors for similar polygons. And use the term length scale factor to define the scale factor between the lengths and area scale factor to describe the scale factor between the areas. We can state this result generally as if the length scale factor between two similar polygons is π, then the area scale factor is π squared.

In terms of the ratios, we can also state that if the length ratio of two similar polygons is π to π, then the ratio of their areas is π squared to π squared. So now that weβve determined the relationship between the areas of similar polygons, letβs consider some examples in which we apply these results.

Given the following figure, find the area of a similar polygon π΄ prime π΅ prime πΆ prime π· prime, in which π΄ prime π΅ prime equals six.

So weβre told that there exists a similar polygon to the one in the figure in which the length of the side π΄ prime π΅ prime is six units. Itβs reasonable to assume that the same letters have been used to represent corresponding vertices on the two polygons. So letβs compare the length of π΄ prime π΅ prime to the length of π΄π΅ on the polygon in the figure. Side π΄π΅ is horizontal and goes from two to five, so it has a length of three units. We can therefore calculate the length scale factor of π΄π΅πΆπ· to π΄ prime π΅ prime πΆ prime π· prime by dividing the length of π΄ prime π΅ prime by the length of π΄π΅. Thatβs six over three, which of course is equal to two. This means then that the lengths on polygon π΄ prime π΅ prime πΆ prime π· prime are each twice as long as the corresponding lengths on polygon π΄π΅πΆπ·.

Now it is in fact the area of the second polygon that weβre interested in. We recall then that if the length scale factor for two similar polygons is π, the area scale factor is π squared. So the area scale factor of π΄π΅πΆπ· to π΄ prime π΅ prime πΆ prime π· prime is two squared, which is four. Another way of saying this is that the area of π΄ prime π΅ prime πΆ prime π· prime is four times the area of π΄π΅πΆπ·. We can find the area of polygon π΄π΅πΆπ· from the figure. Itβs a rectangle with one dimension of length three units and the other dimension of length five units. Its area is therefore three multiplied by five, which is 15 square units.

The area of π΄ prime π΅ prime πΆ prime π· prime then is four multiplied by 15, which is equal to 60. So by recalling that if the length scale factor between two similar polygons is π, then the area scale factor is π squared, we found that the area of the similar polygon π΄ prime π΅ prime πΆ prime π· prime is 60 square units.

Letβs consider another example.

Rectangle ππππ is similar to rectangle π½πΎπΏπ with their sides having a ratio of eight to nine. If the dimensions of each rectangle are doubled, find the ratio of the areas of the larger rectangles.

So weβre told that the length ratio between the sides of the original rectangles is eight to nine. We can recall that if the length ratio of two similar polygons is π to π, then the ratio of their areas is π squared to π squared. Now weβre told that dimensions of each rectangle are doubled, but this doesnβt actually affect the length ratio because the lengths in each polygon have been multiplied by the same factor of two. This would just be equivalent to making the length ratio two π to two π. But of course, we could then simplify this ratio by dividing both sides by two so we get back to π to π. The length ratio of the enlarged rectangles is still eight to nine. Using the result we wrote down then, the ratio of the areas of the larger rectangles is eight squared to nine squared, which is 64 to 81.

This illustrates a general point. If we have a pair of similar polygons with a given length ratio, and if the same scale factor is applied to both polygons, then the length ratio remains the same, and furthermore the area ratio will also remain the same. Weβll now consider some examples in which we use the perimeter of similar shapes to solve problems involving their areas.

Square A is an enlargement of square B by a scale factor of two-thirds. If the perimeter of square A equals 56 centimeters, what is the area of square B? Give your answer to the nearest hundredth.

Weβre given that square A is an enlargement of square B by a scale factor of two-thirds. And as this scale factor is less than one, this means that square A is actually smaller. All squares are similar to one another. So we can use the properties of similarity here. Weβre told that the perimeter of square A is 56 centimeters. And we know that for a square the perimeter is four times the side length. So four times π sub A, which is representing the side length of A, is equal to 56. Dividing both sides of this equation by four, we find that the side length of square A is 14 centimeters.

We can then proceed in two different ways. In our first method, weβll calculate the side length of square B using the length scale factor and then the area. As the length scale factor from square B to square A is two-thirds, the length scale factor in the opposite direction is the reciprocal of this; itβs three over two. So, the side length of square B is three over two multiplied by the side length of A. Thatβs three over two multiplied by 14, which is 21. To find the area of a square, we square its side length. So, the area of square B is 21 squared, which is 441.

The second method we could use is to find the area of square A and then consider the relationship between the areas of these two similar shapes. The area of A is its side length squared. Thatβs 14 squared, which is 196 centimeters squared. We then recall that if the length scale factor between two similar polygons is π, the area scale factor is π squared. So, the area scale factor from A to B is three over two squared, which is nine over four. The area of square B then is nine over four multiplied by the area of square A. Thatβs nine over four multiplied by 196, which is again 441. The question asked though that we give our answer to the nearest hundredth. So, we find that the area of square B is 441.00 square centimeters.

Weβll now look at another example in which we consider the relationship between the areas and perimeters of similar polygons.

The areas of two similar polygons are 361 centimeters squared and 81 centimeters squared. Given that the perimeter of the first is 38 centimeters, find the perimeter of the second.

Weβve been given the areas of the two similar polygons. And so, we can write down the area ratio; itβs 361 to 81. We want to find the perimeter of the second polygon given the perimeter of the first. And to do this, we need to know the length ratio. We can recall that for two similar polygons with corresponding sides in a length ratio of π to π, the area ratio is π squared to π squared, which means if you want to work backwards from the area ratio to calculating the length ratio, we need to square root both parts. So the length ratio is the square root of 361 to the square root of 81, which is 19 to nine.

Now the perimeters of similar shapes are in the same ratio as their lengths because the perimeter is just the sum of the individual lengths. So, the perimeters which are 38 centimeters and a currently unknown value are also in the ratio 19 to nine. To get from 19 to 38, we have to multiply by two. So doing the same to both parts of the ratio, 19 to nine is equivalent to 38 to 18. So by working backwards from knowing the area ratio between these two similar polygons to calculating the length ratio and hence the ratio of the perimeters, we found that the perimeter of the second polygon is 18 centimeters.

Letβs now consider one final example in which weβll apply the theory of the areas of similar polygons to a real-world problem.

It costs 3799 pounds to fit wooden flooring in a class with dimensions 28 meters and 10 meters. How much would it cost to fit wooden flooring in a similar room with dimensions 84 meters and 30 meters?

We can assume that as weβre not told the shape of the rooms, then they are rectangles. Now weβre told that these two rooms are similar, which we can assume means mathematically similar. But letβs check this. We need to establish that corresponding angles are congruent and corresponding sides are proportional. We know that corresponding angles are congruent because all the interior angles in a rectangle are 90 degrees. To establish whether corresponding sides are proportional. Letβs check the ratio between the dimensions of the rooms.

Using the longer side of each, we have 84 over 28, which is equal to three. And using the shorter sides, we have 30 over 10, which is also equal to three. The ratio is the same. So corresponding sides are indeed proportional, and the two rectangles are mathematically similar. We need to work out the cost of fitting wooden flooring in the larger room, which will be dependent on its area. Weβre told the cost for the smaller classroom is 3799 pounds. And we can calculate the area of this room using the formula for the area of a rectangle. Its length multiplied by width, which is 28 multiplied by 10, 280 square meters.

We can also find the area of the larger room. Itβs 84 multiplied by 30, which is 2520 square meters. The area ratio for the two rooms is 2520 to 280. And in fact, this simplifies to nine to one. We now know that the area of the bigger room is nine times the area of the smaller room. And assuming the cost is directly proportional to the area, the cost of the flooring for the larger room will be nine times the cost for the smaller room. Thatβs nine times 3799, which is 34191.

We could also have determined that the area scale factor was nine. By recalling that if the length scale factor between two similar polygons is π, then the area scale factor is π squared. We found the length scale factor to be three. So the area scale factor is three squared, which is nine. We found that the cost of fitting wooden flooring in the larger room is 34,191 pounds.

Letβs now summarize the key points from this video. Two polygons with the same number of sides are similar if corresponding angles are congruent and corresponding sides are proportional. If the length scale factor between two similar polygons is π, then the area scale factor is π squared. If the length ratio of two similar polygons is π to π, then the ratio of their areas is π squared to π squared. And we saw that we can work in the opposite direction to calculate the length ratio given the area ratio. Finally, as perimeter is a length, we can also say that the ratio of the areas of two similar polygons is equal to the square of the ratio of their perimeters.