WEBVTT
00:00:01.320 --> 00:00:09.280
In this video, we will learn how to use the properties of similar polygons to solve algebraic expressions and equations.
00:00:09.800 --> 00:00:14.160
We will begin by recalling some definitions and facts about similar polygons.
00:00:14.720 --> 00:00:20.600
Similar polygons are two or more polygons with the same shape but not the same size.
00:00:20.960 --> 00:00:26.840
They have corresponding angles that are congruent and corresponding sides that are proportional.
00:00:27.360 --> 00:00:31.800
One way to consider this proportionality is to think of scale factors.
00:00:32.240 --> 00:00:37.840
We can think about similar polygons as enlarging or shrinking the same shape.
00:00:38.280 --> 00:00:43.520
For example, our smaller rectangle has dimensions two centimeters and three centimeters.
00:00:44.000 --> 00:00:48.200
The larger rectangle has a four-centimeter length and an unknown length.
00:00:48.720 --> 00:00:53.520
As two multiplied by two is equal to four, the scale factor here is two.
00:00:54.040 --> 00:00:59.840
We can then calculate the missing length in the larger rectangle by multiplying three centimeters by two.
00:01:00.320 --> 00:01:02.280
This is equal to six centimeters.
00:01:02.800 --> 00:01:08.240
If two polygons are similar, we know the lengths of the corresponding sides are proportional.
00:01:08.760 --> 00:01:13.600
Finding the scale factor is straightforward when we are given corresponding lengths in the polygons.
00:01:14.120 --> 00:01:20.800
However, it is also important to be able to solve problems where we are asked if the two polygons are similar.
00:01:21.320 --> 00:01:27.440
Letβs consider two rectangles with dimensions six centimeters and eight centimeters, and 15 centimeters and 20 centimeters.
00:01:27.880 --> 00:01:32.720
If the two rectangles are similar, the ratio of their corresponding sides must be equal.
00:01:33.120 --> 00:01:37.360
In this case, six-eighths must be equal to fifteen twentieths.
00:01:37.880 --> 00:01:43.280
We can simplify the first fraction by dividing the numerator and denominator by two.
00:01:43.760 --> 00:01:48.400
This means that six-eighths in its simplest form is three-quarters.
00:01:48.840 --> 00:01:52.440
15 and 20 have a highest common factor of five.
00:01:52.920 --> 00:01:57.560
Therefore, we can divide the numerator and denominator by five.
00:01:57.960 --> 00:01:59.920
This also gives us three-quarters.
00:02:00.360 --> 00:02:06.760
As the ratio of the corresponding sides simplifies to the same fraction, we know that the two rectangles are similar.
00:02:07.240 --> 00:02:14.840
We will now look at some questions using the properties of similar polygons to solve expressions and equations.
00:02:15.560 --> 00:02:19.400
Given that the two polygons are similar, find the value of π₯.
00:02:20.160 --> 00:02:30.000
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.
00:02:30.520 --> 00:02:34.680
One pair of corresponding sides are π½π and π
π.
00:02:35.200 --> 00:02:39.160
A second pair of corresponding sides are πΆπ½ and ππ
.
00:02:39.680 --> 00:02:43.920
As the corresponding sides are proportional, we know that their ratios are the same.
00:02:44.480 --> 00:02:49.600
Two π₯ plus six over seven π₯ minus seven must be equal to 24 over 28.
00:02:50.080 --> 00:02:54.280
In order to find the value of π₯, we could cross multiply at this stage.
00:02:54.720 --> 00:02:57.880
However, it is easier to simplify our fractions first.
00:02:58.400 --> 00:03:02.480
We can factor the numerator and denominator of the left-hand fraction.
00:03:03.040 --> 00:03:06.920
Two π₯ plus six becomes two multiplied by π₯ plus three.
00:03:07.200 --> 00:03:11.120
And seven π₯ minus seven becomes seven multiplied by π₯ minus one.
00:03:11.760 --> 00:03:24.640
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.
00:03:25.120 --> 00:03:28.320
The denominators on both sides are divisible by seven.
00:03:28.680 --> 00:03:30.680
And the numerators are divisible by two.
00:03:31.280 --> 00:03:37.360
This leaves us with a simplified equation of π₯ plus three over π₯ minus one is equal to three.
00:03:37.760 --> 00:03:41.520
We can multiply both sides of this equation by π₯ minus one.
00:03:42.160 --> 00:03:47.240
Distributing the parentheses gives us π₯ plus three is equal to three π₯ minus three.
00:03:47.640 --> 00:03:51.840
We can then subtract π₯ and add three to both sides of the equation.
00:03:52.360 --> 00:03:54.880
This gives us six is equal to two π₯.
00:03:55.480 --> 00:04:00.040
Finally, dividing both sides by two gives us π₯ is equal to three.
00:04:00.600 --> 00:04:04.480
If the two polygons are similar, the value of π₯ is three.
00:04:05.080 --> 00:04:09.640
We can check this by substituting three back into the expressions for the smaller shape.
00:04:10.200 --> 00:04:15.120
Two multiplied by three is equal to six, and adding six to this gives us 12.
00:04:15.680 --> 00:04:20.640
Seven multiplied by three is equal to 21 and subtracting seven gives us 14.
00:04:21.200 --> 00:04:31.440
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.
00:04:32.080 --> 00:04:35.080
We will now look at a second similar example.
00:04:35.920 --> 00:04:40.880
Given that ππ
ππ is similar to ππππ, find the value of π₯.
00:04:41.160 --> 00:04:47.280
Note that the approximation symbol here means that the two rectangles are similar.
00:04:47.800 --> 00:04:52.600
When two polygons are similar, we know that their corresponding sides are proportional.
00:04:53.040 --> 00:05:00.480
In this question, we have corresponding sides ππ and ππ along with ππ and ππ.
00:05:01.120 --> 00:05:08.160
This means that the ratio of six π₯ minus 16 to 13 will be the same as nine π₯ minus 33 to 15.
00:05:08.640 --> 00:05:16.360
Writing this in fractional form gives us six π₯ minus 16 over nine π₯ minus 33 is equal to 13 over 15.
00:05:16.800 --> 00:05:20.800
In order to calculate π₯, we could cross multiply immediately.
00:05:21.200 --> 00:05:24.880
However, it is often useful to try and simplify the fractions first.
00:05:25.440 --> 00:05:29.400
The numerator and denominator on the left-hand side can be factored.
00:05:29.920 --> 00:05:34.320
Six π₯ minus 16 is equal to two multiplied by three π₯ minus eight.
00:05:34.720 --> 00:05:39.360
And nine π₯ minus 33 is equal to three multiplied by three π₯ minus 11.
00:05:39.880 --> 00:05:44.160
The denominators have a common factor of three, so we can divide both of these by three.
00:05:44.840 --> 00:05:53.800
Cross multiplying at this stage gives us 10 multiplied by three π₯ minus eight is equal to 13 multiplied by three π₯ minus 11.
00:05:54.360 --> 00:05:56.920
We get the 10 by multiplying five by two.
00:05:57.480 --> 00:06:04.440
Redistributing our parentheses gives us 30π₯ minus 80 is equal to 39π₯ minus 143.
00:06:04.960 --> 00:06:12.080
Adding 143 to both sides of this equation gives us 30π₯ plus 63 is equal to 39π₯.
00:06:12.480 --> 00:06:17.560
Subtracting 30π₯ from both sides gives us 63 is equal to nine π₯.
00:06:18.000 --> 00:06:23.760
Finally, dividing both sides of this equation by nine gives us a value of π₯ equal to seven.
00:06:24.200 --> 00:06:30.840
We can then check this answer by substituting π₯ equals seven into our expressions on the first rectangle.
00:06:31.320 --> 00:06:36.920
Six multiplied by seven is equal to 42, and subtracting 16 gives us 26.
00:06:37.360 --> 00:06:40.400
Nine multiplied by seven is equal to 63.
00:06:40.800 --> 00:06:43.200
Subtracting 33 from this gives us 30.
00:06:43.720 --> 00:06:52.240
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.
00:06:52.800 --> 00:07:03.840
The scale factor to get from rectangle ππ
ππ to ππππ is a half as the corresponding sides of the second rectangle are half the size of the first one.
00:07:04.560 --> 00:07:08.520
We will now look at a question where it is not so clear which sides are corresponding.
00:07:09.200 --> 00:07:13.520
Given that the two polygons are similar, find the value of π₯.
00:07:14.080 --> 00:07:21.480
We know that any similar polygons have corresponding angles that are congruent and corresponding sides that are proportional.
00:07:21.920 --> 00:07:28.200
Due to the orientation of these shapes, it may not immediately be obvious which sides are corresponding.
00:07:28.680 --> 00:07:33.400
In order to work this out, it is useful to identify the corresponding angles first.
00:07:33.960 --> 00:07:35.920
One pair of corresponding sides are π΅πΆ and ππ».
00:07:35.920 --> 00:07:42.120
A second pair of corresponding sides are therefore πΆπ· and π»π½.
00:07:42.120 --> 00:07:52.120
As the corresponding sides are proportional, we know that the ratios two to six and four π₯ minus 37 to two π₯ minus 11 must be equal.
00:07:52.720 --> 00:08:00.360
Writing this in fractional form, we have two over four π₯ minus 37 is equal to six over two π₯ minus 11.
00:08:00.880 --> 00:08:03.400
Both of the numerators here are divisible by two.
00:08:03.920 --> 00:08:11.800
We can then cross multiply to give us one multiplied by two π₯ minus 11 is equal to three multiplied by four π₯ minus 37.
00:08:12.320 --> 00:08:18.040
Distributing our parentheses gives us two π₯ minus 11 is equal to 12π₯ minus 111.
00:08:18.520 --> 00:08:24.120
Adding 111 to both sides of this equation gives us two π₯ plus 100 is equal to 12π₯.
00:08:24.720 --> 00:08:31.040
We can then subtract two π₯ from both sides of this equation, which gives us 100 is equal to 10π₯.
00:08:31.600 --> 00:08:37.120
Finally, dividing both sides of this equation by 10 gives us a value of π₯ equal to 10.
00:08:37.600 --> 00:08:44.560
We can then substitute this value back into the expressions for the lengths of π΅πΆ and ππ» to check our answer.
00:08:45.120 --> 00:08:47.560
Four multiplied by 10 is equal to 40.
00:08:48.000 --> 00:08:50.640
Subtracting 37 from this gives us three.
00:08:51.200 --> 00:08:56.280
Two multiplied by 10 is equal to 20, and subtracting 11 from this gives us nine.
00:08:56.880 --> 00:09:03.960
The ratios two to six and three to nine are identical as they can both be simplified to one to three.
00:09:04.760 --> 00:09:10.400
An alternative method in this question would be to initially recognize that the scale factor was three.
00:09:11.000 --> 00:09:15.040
This is because the length π½π» is three times the length of πΆπ·.
00:09:15.560 --> 00:09:25.360
We could then have set up the equation two π₯ minus 11 is equal to three multiplied by four π₯ minus 37 as the length ππ» is three times the length of π΅πΆ.
00:09:25.800 --> 00:09:30.840
Following this method would also have got us a value of π₯ equal to 10.
00:09:31.640 --> 00:09:35.520
We will now consider a final question where there are two unknowns.
00:09:36.120 --> 00:09:43.240
Given that π΄π΅πΆπ· is similar to πΈπΉπΊπ», find the values of π₯ and π¦.
00:09:43.920 --> 00:09:48.080
We know that in any similar polygon, the corresponding sides are proportional.
00:09:48.440 --> 00:09:52.760
This means that our first step is to identify the corresponding sides.
00:09:53.280 --> 00:10:00.400
The side πΆπ· is corresponding to πΊπ», π΄π· is corresponding to πΈπ», and π΅πΆ is corresponding to πΉπΊ.
00:10:00.920 --> 00:10:04.200
This means that the ratio of these three sides must be equal.
00:10:04.680 --> 00:10:11.200
The ratios 10 to five, eight to two π¦ minus 14, and π₯ to eight must all be equal.
00:10:11.640 --> 00:10:16.720
We could set these up in fractional form to calculate the value of π₯ and π¦.
00:10:17.240 --> 00:10:21.880
However, it is clear from the first ratio that this simplifies to two to one.
00:10:22.480 --> 00:10:28.120
This means that all the lengths in the second trapezium will be half the lengths of the first trapezium.
00:10:28.600 --> 00:10:32.240
The scale factor to go from trapezium π΄π΅πΆπ· to πΈπΉπΊπ» is one-half.
00:10:32.240 --> 00:10:41.120
We can therefore say that two π¦ minus 14 is equal to a half of eight, and a half of eight is four.
00:10:41.520 --> 00:10:46.280
Adding 14 to both sides of this equation gives us two π¦ is equal to 18.
00:10:46.720 --> 00:10:49.880
Dividing by two gives us π¦ is equal to nine.
00:10:50.400 --> 00:10:53.640
We can use the same method to calculate the value of π₯.
00:10:54.200 --> 00:10:59.240
The length πΊπ», which is equal to eight, is half the value of πΆπ·, which is π₯.
00:10:59.720 --> 00:11:04.800
Multiplying both sides of this equation by two gives us 16 is equal to π₯.
00:11:05.320 --> 00:11:09.800
The missing values are π₯ equals 16 and π¦ equals nine.
00:11:10.760 --> 00:11:13.280
We will now summarize the key points from this video.
00:11:13.880 --> 00:11:19.160
Firstly, we know that the corresponding angles in similar polygons are congruent.
00:11:19.600 --> 00:11:22.240
The corresponding sides on the other hand are proportional.
00:11:22.720 --> 00:11:26.600
We can think of similar polygons as being enlarged or shrunk.
00:11:27.080 --> 00:11:30.560
And the ratio of one corresponding side to another is the scale factor.
00:11:31.040 --> 00:11:34.920
This scale factor will be the same for all corresponding sides in the polygon.
00:11:35.400 --> 00:11:41.640
We can use our knowledge of ratios and fractions to set up equations that can be solved.
00:11:42.160 --> 00:11:46.560
This allows us to calculate any variables on the lengths of the sides of our polygons.
00:11:47.120 --> 00:11:54.320
In order to prove that two or more polygons are similar, we need to show that the ratios of the corresponding sides are equal.