Lesson Video: Applications of Similar Polygons Mathematics

Video Transcript

In this video, we will learn how to use the properties of similar polygons to solve algebraic expressions and equations. We will begin by recalling some definitions and facts about similar polygons.

Similar polygons are two or more polygons with the same shape but not the same size. They have corresponding angles that are congruent and corresponding sides that are proportional. One way to consider this proportionality is to think of scale factors. We can think about similar polygons as enlarging or shrinking the same shape. For example, our smaller rectangle has dimensions two centimeters and three centimeters. The larger rectangle has a four-centimeter length and an unknown length.

As two multiplied by two is equal to four, the scale factor here is two. We can then calculate the missing length in the larger rectangle by multiplying three centimeters by two. This is equal to six centimeters. If two polygons are similar, we know the lengths of the corresponding sides are proportional. Finding the scale factor is straightforward when we are given corresponding lengths in the polygons. However, it is also important to be able to solve problems where we are asked if the two polygons are similar.

Let’s consider two rectangles with dimensions six centimeters and eight centimeters, and 15 centimeters and 20 centimeters. If the two rectangles are similar, the ratio of their corresponding sides must be equal. In this case, six-eighths must be equal to fifteen twentieths. We can simplify the first fraction by dividing the numerator and denominator by two. This means that six-eighths in its simplest form is three-quarters. 15 and 20 have a highest common factor of five. Therefore, we can divide the numerator and denominator by five. This also gives us three-quarters. As the ratio of the corresponding sides simplifies to the same fraction, we know that the two rectangles are similar. We will now look at some questions using the properties of similar polygons to solve expressions and equations.

Given that the two polygons are similar, find the value of 𝑥.

When dealing with any question involving similar polygons, we know that the corresponding angles are congruent or the same and the corresponding sides are proportional. One pair of corresponding sides are 𝐽𝑊 and 𝑅𝑆. A second pair of corresponding sides are 𝐶𝐽 and 𝑃𝑅. As the corresponding sides are proportional, we know that their ratios are the same. Two 𝑥 plus six over seven 𝑥 minus seven must be equal to 24 over 28.

In order to find the value of 𝑥, we could cross multiply at this stage. However, it is easier to simplify our fractions first. We can factor the numerator and denominator of the left-hand fraction. Two 𝑥 plus six becomes two multiplied by 𝑥 plus three. And seven 𝑥 minus seven becomes seven multiplied by 𝑥 minus one. Dividing the numerator and denominator of our right-hand fraction by four gives us six over seven as 24 divided by four is six and 28 divided by four is equal to seven.

The denominators on both sides are divisible by seven. And the numerators are divisible by two. This leaves us with a simplified equation of 𝑥 plus three over 𝑥 minus one is equal to three. We can multiply both sides of this equation by 𝑥 minus one. Distributing the parentheses gives us 𝑥 plus three is equal to three 𝑥 minus three. We can then subtract 𝑥 and add three to both sides of the equation. This gives us six is equal to two 𝑥. Finally, dividing both sides by two gives us 𝑥 is equal to three.

If the two polygons are similar, the value of 𝑥 is three. We can check this by substituting three back into the expressions for the smaller shape. Two multiplied by three is equal to six, and adding six to this gives us 12. Seven multiplied by three is equal to 21 and subtracting seven gives us 14. It is therefore clear that our two polygons are similar with a scale factor of two as 12 multiplied by two is 24 and 14 multiplied by two is 28.

We will now look at a second similar example.

Given that 𝑄𝑅𝑃𝑀 is similar to 𝑆𝑉𝑍𝑊, find the value of 𝑥.

Note that the approximation symbol here means that the two rectangles are similar. When two polygons are similar, we know that their corresponding sides are proportional. In this question, we have corresponding sides 𝑄𝑀 and 𝑆𝑊 along with 𝑀𝑃 and 𝑊𝑍. This means that the ratio of six 𝑥 minus 16 to 13 will be the same as nine 𝑥 minus 33 to 15. Writing this in fractional form gives us six 𝑥 minus 16 over nine 𝑥 minus 33 is equal to 13 over 15.

In order to calculate 𝑥, we could cross multiply immediately. However, it is often useful to try and simplify the fractions first. The numerator and denominator on the left-hand side can be factored. Six 𝑥 minus 16 is equal to two multiplied by three 𝑥 minus eight. And nine 𝑥 minus 33 is equal to three multiplied by three 𝑥 minus 11. The denominators have a common factor of three, so we can divide both of these by three.

Cross multiplying at this stage gives us 10 multiplied by three 𝑥 minus eight is equal to 13 multiplied by three 𝑥 minus 11. We get the 10 by multiplying five by two. Redistributing our parentheses gives us 30𝑥 minus 80 is equal to 39𝑥 minus 143. Adding 143 to both sides of this equation gives us 30𝑥 plus 63 is equal to 39𝑥. Subtracting 30𝑥 from both sides gives us 63 is equal to nine 𝑥. Finally, dividing both sides of this equation by nine gives us a value of 𝑥 equal to seven.

We can then check this answer by substituting 𝑥 equals seven into our expressions on the first rectangle. Six multiplied by seven is equal to 42, and subtracting 16 gives us 26. Nine multiplied by seven is equal to 63. Subtracting 33 from this gives us 30. We can therefore see that our first rectangle is twice the size of our second rectangle as 26 is double 13 and 30 is double 15. The scale factor to get from rectangle 𝑄𝑅𝑃𝑀 to 𝑆𝑉𝑍𝑊 is a half as the corresponding sides of the second rectangle are half the size of the first one.

We will now look at a question where it is not so clear which sides are corresponding.

Given that the two polygons are similar, find the value of 𝑥.

We know that any similar polygons have corresponding angles that are congruent and corresponding sides that are proportional. Due to the orientation of these shapes, it may not immediately be obvious which sides are corresponding. In order to work this out, it is useful to identify the corresponding angles first. One pair of corresponding sides are 𝐵𝐶 and 𝑇𝐻. A second pair of corresponding sides are therefore 𝐶𝐷 and 𝐻𝐽.

As the corresponding sides are proportional, we know that the ratios two to six and four 𝑥 minus 37 to two 𝑥 minus 11 must be equal. Writing this in fractional form, we have two over four 𝑥 minus 37 is equal to six over two 𝑥 minus 11. Both of the numerators here are divisible by two. We can then cross multiply to give us one multiplied by two 𝑥 minus 11 is equal to three multiplied by four 𝑥 minus 37.

Distributing our parentheses gives us two 𝑥 minus 11 is equal to 12𝑥 minus 111. Adding 111 to both sides of this equation gives us two 𝑥 plus 100 is equal to 12𝑥. We can then subtract two 𝑥 from both sides of this equation, which gives us 100 is equal to 10𝑥. Finally, dividing both sides of this equation by 10 gives us a value of 𝑥 equal to 10.

We can then substitute this value back into the expressions for the lengths of 𝐵𝐶 and 𝑇𝐻 to check our answer. Four multiplied by 10 is equal to 40. Subtracting 37 from this gives us three. Two multiplied by 10 is equal to 20, and subtracting 11 from this gives us nine. The ratios two to six and three to nine are identical as they can both be simplified to one to three.

An alternative method in this question would be to initially recognize that the scale factor was three. This is because the length 𝐽𝐻 is three times the length of 𝐶𝐷. We could then have set up the equation two 𝑥 minus 11 is equal to three multiplied by four 𝑥 minus 37 as the length 𝑇𝐻 is three times the length of 𝐵𝐶. Following this method would also have got us a value of 𝑥 equal to 10.

We will now consider a final question where there are two unknowns.

Given that 𝐴𝐵𝐶𝐷 is similar to 𝐸𝐹𝐺𝐻, find the values of 𝑥 and 𝑦.

We know that in any similar polygon, the corresponding sides are proportional. This means that our first step is to identify the corresponding sides. The side 𝐶𝐷 is corresponding to 𝐺𝐻, 𝐴𝐷 is corresponding to 𝐸𝐻, and 𝐵𝐶 is corresponding to 𝐹𝐺. This means that the ratio of these three sides must be equal. The ratios 10 to five, eight to two 𝑦 minus 14, and 𝑥 to eight must all be equal. We could set these up in fractional form to calculate the value of 𝑥 and 𝑦. However, it is clear from the first ratio that this simplifies to two to one. This means that all the lengths in the second trapezium will be half the lengths of the first trapezium. The scale factor to go from trapezium 𝐴𝐵𝐶𝐷 to 𝐸𝐹𝐺𝐻 is one-half.

We can therefore say that two 𝑦 minus 14 is equal to a half of eight, and a half of eight is four. Adding 14 to both sides of this equation gives us two 𝑦 is equal to 18. Dividing by two gives us 𝑦 is equal to nine. We can use the same method to calculate the value of 𝑥. The length 𝐺𝐻, which is equal to eight, is half the value of 𝐶𝐷, which is 𝑥. Multiplying both sides of this equation by two gives us 16 is equal to 𝑥. The missing values are 𝑥 equals 16 and 𝑦 equals nine.

We will now summarize the key points from this video. Firstly, we know that the corresponding angles in similar polygons are congruent. The corresponding sides on the other hand are proportional. We can think of similar polygons as being enlarged or shrunk. And the ratio of one corresponding side to another is the scale factor. This scale factor will be the same for all corresponding sides in the polygon. We can use our knowledge of ratios and fractions to set up equations that can be solved. This allows us to calculate any variables on the lengths of the sides of our polygons. In order to prove that two or more polygons are similar, we need to show that the ratios of the corresponding sides are equal.