Lesson Video: Similarity of Polygons | Nagwa Lesson Video: Similarity of Polygons | Nagwa

# Lesson Video: Similarity of Polygons Mathematics

In this video, we will learn how to identify and prove the similarity of polygons, write the order of the corresponding vertices, and use the similarity to solve problems.

17:55

### Video Transcript

In this video, we’ll learn how to identify and prove the similarity of polygons, write the order of the corresponding vertices, and use similarity to solve problems. Let’s start by recalling that polygons are two-dimensional shapes with straight sides. For example, squares, rectangles, triangles, and even hexagons are all polygons. It’s important to remember that polygons that have exactly the same shape and size are called congruent, whereas similar polygons have the same shape but maybe a different size. So let’s define exactly what we mean by similar polygons. We say that two polygons are similar if their corresponding angles are congruent, that means they have the same measure, and their corresponding sides are in proportion.

Let’s look at an example of what we mean by this. Here we have quadrilaterals 𝐴𝐵𝐶𝐷 and 𝑃𝑄𝑅𝑆. If we want to say that these two quadrilaterals are similar, we can use this wavy line, which means that they are similar. So 𝐴𝐵𝐶𝐷 is similar to 𝑃𝑄𝑅𝑆. And because they are similar, we know that corresponding angles are congruent. So for example, the measure of angle 𝐴 is equal to the measure of angle 𝑃. The corresponding angle to the vertex 𝐵 is that at the vertex 𝑄. So the measure of angle 𝐵 is equal to the measure of angle 𝑄. The measure of angle 𝐶 is congruent to the measure of angle 𝑅. And the measure of angle 𝐷 is congruent to the measure of angle 𝑆.

We can also observe the corresponding sides. These are the line segments 𝐴𝐵 and 𝑃𝑄, the line segments 𝐵𝐶 and 𝑄𝑅, the line segments 𝐶𝐷 and 𝑅𝑆, and the line segments 𝐷𝐴 and 𝑆𝑃. Now, since we know that corresponding sides are in the same proportion, we can write that 𝐴𝐵 over 𝑃𝑄 is equal to 𝐵𝐶 over 𝑄𝑅 is equal to 𝐶𝐷 over 𝑅𝑆 is equal to 𝐷𝐴 over 𝑆𝑃. But it wouldn’t matter if we wrote all of these numerators and denominators swapped in the entire statement like this because we’re really just writing the proportionality relationship between the corresponding sides.

Before we look at some examples, there’s one final point to note about the similarity statement. Before, we wrote that 𝐴𝐵𝐶𝐷 is similar to 𝑃𝑄𝑅𝑆, but there are other ways that we could have written this. For example, we could’ve said that 𝐵𝐶𝐷𝐴 is similar to 𝑄𝑅𝑆𝑃 or even that 𝐶𝐵𝐴𝐷 is similar to 𝑅𝑄𝑃𝑆. The important thing is that the corresponding vertices are given in the same position in the similarity relationship. So now we have seen the notation involved in similarity. We’ll see how we can use this definition of similar polygons to identify if two polygons are similar.

Are the two polygons similar?

We can recall that two polygons are similar if their corresponding angles are congruent and their corresponding sides are in proportion. So let’s have a look at the two quadrilaterals that we are given. We notice that we have two pairs of congruent angles. The measure of angle 𝐴𝐵𝐶 is equal to the measure of angle 𝐹𝐺𝐻 because they’re both 103 degrees. And the measure of angle 𝐵𝐶𝐷 is equal to the measure of angle 𝐺𝐻𝐸 as they are both 84 degrees. We can determine the value of this angle at 𝐷𝐴𝐵 by recalling that the angles in a quadrilateral sum to 360 degrees. So if we add the three angles of 95 degrees, 103 degrees, and 84 degrees and subtract that from 360 degrees, we would find the measure of angle 𝐷𝐴𝐵. And when we do that, we get an angle measure of 78 degrees.

We could use the same method and the fact that the angles in a quadrilateral sum to 360 degrees to calculate the measure of angle 𝐹𝐸𝐻. But if we think about this logically, we already know that if we have three angles of 78, 103 degrees, and 84 degrees, then we must add 95 degrees in order to get an answer of 360 degrees. And so the measure of angle 𝐹𝐸𝐻 must be 95 degrees.

We can now observe that we have fulfilled the first part of this similarity criteria, that is, that the corresponding angles are congruent. But we also need to check that the corresponding sides are in proportion. The side which corresponds to 𝐴𝐵 is 𝐹𝐺. The side corresponding to 𝐵𝐶 is 𝐺𝐻. And the two remaining corresponding pairs are 𝐶𝐷 and 𝐻𝐸 and 𝐷𝐴 and 𝐹𝐸. If we could say that all of these proportions are equal, then the polygons would be similar. We can check this by filling in the length information that we’re given. Beginning with 𝐴𝐵 and 𝐹𝐺, 𝐴𝐵 is 20 centimeters and 𝐹𝐺 is 14 centimeters. The proportion 20 over 14 can be simplified to 10 over seven. Next, we have 𝐵𝐶 over 𝐺𝐻, and that’s equal to 16 over 11.

Now we could at this point continue to write the other proportions in this following way. However, we know that 16 over 11 is not the same as 10 over seven. That’s enough to demonstrate that these two polygons do not have corresponding sides in proportion. Although we may have another pair of sides in the same proportion as 𝐴𝐵 over 𝐹𝐺, they are not all in the same proportion. We can therefore give the answer no, the two polygons are not similar.

We’ll now think about how similar polygons may be considered as a dilation of each other. If the scale factor or ratio of enlargement is one, then the polygons would be congruent. Let’s now think about how we can determine the ratio of enlargement, which will be useful when we’re finding the length of an unknown side. Let’s say that we’re told that triangle 𝐽𝐾𝐿 is similar to triangle 𝑀𝑁𝑂. Because we know that they are similar, then the corresponding sides are in proportion. As the sides 𝐽𝐾 and 𝑀𝑁 are corresponding, we can write the proportion as 𝑀𝑁 over 𝐽𝐾, which is equal to eight over four, and that simplifies to two. The ratio of enlargement or scale factor from triangle 𝐽𝐾𝐿 to triangle 𝑀𝑁𝑂 is two.

If we wanted to find the scale factor in the opposite direction, we put the side of the triangle that we’re going to as the numerator and the side on the triangle that we’re coming from as the denominator. This time, the ratio is four over eight, or one-half. Alternatively, if we want to find the scale factor in the reverse direction and we know that in this direction it’s multiplying by two, then we find the reciprocal or we divide by two. But scale factors must be given in terms of a multiplier, and dividing by two is equivalent to multiplying by half.

So let’s say for example that we wanted to find the length of the line segment 𝐽𝐿. Well we know that the corresponding side 𝑀𝑂 is 11 centimeters and the ratio of enlargement in the correct direction is one-half. 11 multiplied by one-half is 5.5 centimeters. It’s worth noting that this is equivalent to finding the length of the line segment 𝐽𝐿 by writing a single equation. For example, we could say that 𝐽𝐿 over 𝑀𝑂 is equal to 𝐽𝐾 over 𝑀𝑁. Filling in the side lengths, we would have that 𝐽𝐿 over 11 is equal to four-eighths. By rearranging and simplifying, we would determine that 𝐽𝐿 is equal to 5.5 centimeters. So if we wanted to find the length of an unknown side, either method would work. We’ll now see another example.

Given that 𝐴𝐵𝐶𝐷 is similar to 𝑍𝑌𝑋𝐿, find the measure of angle 𝑋𝐿𝑍 and the length of the line segment 𝐶𝐷.

We are told here that 𝐴𝐵𝐶𝐷 is similar to 𝑍𝑌𝑋𝐿. And that means that their corresponding angles are congruent and their corresponding sides are in proportion. We are first asked to find the measure of angle 𝑋𝐿𝑍. But we noticed that we don’t have enough information about the angles in polygon 𝑍𝑌𝑋𝐿 to work out this angle measure. However, as these polygons are similar, we know that this angle measure of 𝐵𝐶𝐷 is corresponding to angle 𝑌𝑋𝐿, and so the angle 𝑌𝑋𝐿 must have a measure of 85 degrees. We can then use the fact that the internal angle measures in a quadrilateral add to 360 degrees to work out that the measure of angle 𝑋𝐿𝑍 must be equal to 360 degrees subtract 85 degrees plus 109 degrees plus 105 degrees. This gives us 61 degrees.

Next, we need to find the length of the line segment 𝐶𝐷. We note that the length of 𝐶𝐷 corresponds with the length 𝑋𝐿 in polygon 𝑍𝑌𝑋𝐿. The proportion of these sides will be the same proportion as that between all other pairs of corresponding sides in the polygon. When we are answering problems involving similar polygons, we need to look for or be able to calculate another pair of corresponding sides. Here we are given that 𝐴𝐵 is 75 centimeters and 𝑍𝑌 is 150 centimeters. We could therefore write that 𝐶𝐷 over 𝑋𝐿 is equal to 𝐴𝐵 over 𝑍𝑌.

Filling in the length measurements, we have that 𝐶𝐷 over 246.2 is equal to 75 over 150. We can take out our common factor of 75 from the numerator and denominator on the right-hand side. And then we can rearrange by multiplying through by 246.2. The length of 𝐶𝐷 is therefore 123.1 centimeters. Alternatively, we could’ve found the scale factor or ratio of enlargement between the two polygons. Using the side length of 𝑥𝑦 and 𝐴𝐵, we could’ve found that to go to the polygon 𝐴𝐵𝐶𝐷 from the polygon 𝑍𝑌𝑋𝐿, we multiply by one-half. Multiplying the length 246.2 centimeters by one-half would have given us the same value of 123.1 centimeters.

As we have now answered both parts of this question, we can give the answer that the measure of angle 𝑋𝐿𝑍 is 61 degrees and the length of 𝐶𝐷 is 123.1 centimeters.

We will now consider similarity in regular polygons. Let’s recall that a regular polygon is a polygon where all the angles are congruent and all the sides are congruent. Some examples of regular polygons are equilateral triangles, squares, regular pentagons, regular hexagons, and so on. Let’s consider two squares of different side lengths. Since we know that in each square all the angles are congruent, then each of these angles are also congruent to their corresponding angles in the other square. We also know that the proportion of corresponding side lengths will be the same for each side length. We can therefore say that these two squares are similar.

In fact, we can say that any regular 𝑛-sided polygon is similar to another regular 𝑛-sided polygon, where the values of 𝑛 are the same. In other words, that means that all equilateral triangles are similar to one another. All squares are similar to one another. All regular pentagons are similar to one another, and so on. But of course this does only apply to regular polygons. Since similar polygons must have corresponding angles congruent and corresponding sides in the same proportion, we could not say, for example, that this square and rectangle are similar simply because the angles are all 90 degrees in each. We will now see one final example.

A polygon has sides of length two centimeters, four centimeters, three centimeters, eight centimeters, and four centimeters. A second similar polygon has a perimeter of 31.5 centimeters. What are the lengths of its sides?

Let’s begin by recalling that similar polygons have corresponding angles congruent and corresponding sides in proportion. Here we are given the five side lengths of this polygon, which must be a pentagon, and we need to work out the side lengths of the second polygon using only the information about its perimeter. Because the side lengths of similar polygons are in proportion, then the perimeter, which is also a measure of length, will be in the same proportion. We can calculate the scale factor from the first polygon to the second polygon by working out the perimeter of the second polygon divided by the perimeter of the first polygon. We are given that the perimeter of the second polygon is 31.5 centimeters.

And we can calculate the perimeter of the first polygon by adding the side lengths. Two plus four plus three plus eight plus four centimeters will give us a perimeter of 21 centimeters. The scale factor is therefore 31.5 over 21, which can be simplified to three over two. So now to find the side length in the second polygon, we multiply each corresponding side length in the first polygon by the scale factor of three over two. The first side length will therefore be calculated as two times three over two. This will be equal to three centimeters. Next, four centimeters times three over two is six centimeters. The third side is three centimeters times three over two, which is 4.5 centimeters. The fourth and fifth sides can be calculated as 12 centimeters and six centimeters.

As a useful check of our answer, we know that the perimeter of the second polygon should be 31.5. And when we add three centimeters, six centimeters, 4.5 centimeters, 12 centimeters, and six centimeters, we do indeed get a perimeter of 31.5 centimeters, which confirms our answer for the five side lengths.

We can now finish this video by summarizing the key points. Two polygons are similar if their corresponding angles are congruent and their corresponding sides are in proportion. We must ensure that the letter ordering is correct when writing a similarity relationship between polygons. We can calculate the length of an unknown side by writing the proportional relationship between two pairs of corresponding sides or by first calculating the dilation scale factor. We also saw that all regular polygons are similar to other regular polygons with the same number of sides. And finally, as we saw in the previous example, the scale factor between the perimeters of two similar polygons is the same as that between corresponding side lengths.