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In this video, we will learn how to use the properties of similar polygons to solve algebraic expressions and equations.
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We will begin by recalling some definitions and facts about similar polygons.
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Similar polygons are two or more polygons with the same shape but not the same size.
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They have corresponding angles that are congruent and corresponding sides that are proportional.
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One way to consider this proportionality is to think of scale factors.
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We can think about similar polygons as enlarging or shrinking the same shape.
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For example, our smaller rectangle has dimensions two centimeters and three centimeters.
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The larger rectangle has a four-centimeter length and an unknown length.
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As two multiplied by two is equal to four, the scale factor here is two.
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We can then calculate the missing length in the larger rectangle by multiplying three centimeters by two.
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This is equal to six centimeters.
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If two polygons are similar, we know the lengths of the corresponding sides are proportional.
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Finding the scale factor is straightforward when we are given corresponding lengths in the polygons.
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However, it is also important to be able to solve problems where we are asked if the two polygons are similar.
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Letβs consider two rectangles with dimensions six centimeters and eight centimeters, and 15 centimeters and 20 centimeters.
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If the two rectangles are similar, the ratio of their corresponding sides must be equal.
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In this case, six-eighths must be equal to fifteen twentieths.
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We can simplify the first fraction by dividing the numerator and denominator by two.
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This means that six-eighths in its simplest form is three-quarters.
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15 and 20 have a highest common factor of five.
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Therefore, we can divide the numerator and denominator by five.
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This also gives us three-quarters.
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As the ratio of the corresponding sides simplifies to the same fraction, we know that the two rectangles are similar.
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We will now look at some questions using the properties of similar polygons to solve expressions and equations.
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Given that the two polygons are similar, find the value of π₯.
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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.
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One pair of corresponding sides are π½π and π
π.
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A second pair of corresponding sides are πΆπ½ and ππ
.
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As the corresponding sides are proportional, we know that their ratios are the same.
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Two π₯ plus six over seven π₯ minus seven must be equal to 24 over 28.
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In order to find the value of π₯, we could cross multiply at this stage.
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However, it is easier to simplify our fractions first.
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We can factor the numerator and denominator of the left-hand fraction.
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Two π₯ plus six becomes two multiplied by π₯ plus three.
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And seven π₯ minus seven becomes seven multiplied by π₯ minus one.
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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.
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The denominators on both sides are divisible by seven.
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And the numerators are divisible by two.
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This leaves us with a simplified equation of π₯ plus three over π₯ minus one is equal to three.
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We can multiply both sides of this equation by π₯ minus one.
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Distributing the parentheses gives us π₯ plus three is equal to three π₯ minus three.
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We can then subtract π₯ and add three to both sides of the equation.
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This gives us six is equal to two π₯.
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Finally, dividing both sides by two gives us π₯ is equal to three.
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If the two polygons are similar, the value of π₯ is three.
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We can check this by substituting three back into the expressions for the smaller shape.
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Two multiplied by three is equal to six, and adding six to this gives us 12.
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Seven multiplied by three is equal to 21 and subtracting seven gives us 14.
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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.
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We will now look at a second similar example.
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Given that ππ
ππ is similar to ππππ, find the value of π₯.
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Note that the approximation symbol here means that the two rectangles are similar.
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When two polygons are similar, we know that their corresponding sides are proportional.
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In this question, we have corresponding sides ππ and ππ along with ππ and ππ.
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This means that the ratio of six π₯ minus 16 to 13 will be the same as nine π₯ minus 33 to 15.
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Writing this in fractional form gives us six π₯ minus 16 over nine π₯ minus 33 is equal to 13 over 15.
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In order to calculate π₯, we could cross multiply immediately.
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However, it is often useful to try and simplify the fractions first.
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The numerator and denominator on the left-hand side can be factored.
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Six π₯ minus 16 is equal to two multiplied by three π₯ minus eight.
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And nine π₯ minus 33 is equal to three multiplied by three π₯ minus 11.
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The denominators have a common factor of three, so we can divide both of these by three.
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Cross multiplying at this stage gives us 10 multiplied by three π₯ minus eight is equal to 13 multiplied by three π₯ minus 11.
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We get the 10 by multiplying five by two.
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Redistributing our parentheses gives us 30π₯ minus 80 is equal to 39π₯ minus 143.
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Adding 143 to both sides of this equation gives us 30π₯ plus 63 is equal to 39π₯.
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Subtracting 30π₯ from both sides gives us 63 is equal to nine π₯.
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Finally, dividing both sides of this equation by nine gives us a value of π₯ equal to seven.
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We can then check this answer by substituting π₯ equals seven into our expressions on the first rectangle.
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Six multiplied by seven is equal to 42, and subtracting 16 gives us 26.
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Nine multiplied by seven is equal to 63.
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Subtracting 33 from this gives us 30.
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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.
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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.
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We will now look at a question where it is not so clear which sides are corresponding.
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Given that the two polygons are similar, find the value of π₯.
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We know that any similar polygons have corresponding angles that are congruent and corresponding sides that are proportional.
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Due to the orientation of these shapes, it may not immediately be obvious which sides are corresponding.
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In order to work this out, it is useful to identify the corresponding angles first.
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One pair of corresponding sides are π΅πΆ and ππ».
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A second pair of corresponding sides are therefore πΆπ· and π»π½.
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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.
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Writing this in fractional form, we have two over four π₯ minus 37 is equal to six over two π₯ minus 11.
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Both of the numerators here are divisible by two.
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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.
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Distributing our parentheses gives us two π₯ minus 11 is equal to 12π₯ minus 111.
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Adding 111 to both sides of this equation gives us two π₯ plus 100 is equal to 12π₯.
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We can then subtract two π₯ from both sides of this equation, which gives us 100 is equal to 10π₯.
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Finally, dividing both sides of this equation by 10 gives us a value of π₯ equal to 10.
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We can then substitute this value back into the expressions for the lengths of π΅πΆ and ππ» to check our answer.
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Four multiplied by 10 is equal to 40.
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Subtracting 37 from this gives us three.
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Two multiplied by 10 is equal to 20, and subtracting 11 from this gives us nine.
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The ratios two to six and three to nine are identical as they can both be simplified to one to three.
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An alternative method in this question would be to initially recognize that the scale factor was three.
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This is because the length π½π» is three times the length of πΆπ·.
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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.
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Following this method would also have got us a value of π₯ equal to 10.
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We will now consider a final question where there are two unknowns.
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Given that π΄π΅πΆπ· is similar to πΈπΉπΊπ», find the values of π₯ and π¦.
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We know that in any similar polygon, the corresponding sides are proportional.
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This means that our first step is to identify the corresponding sides.
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The side πΆπ· is corresponding to πΊπ», π΄π· is corresponding to πΈπ», and π΅πΆ is corresponding to πΉπΊ.
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This means that the ratio of these three sides must be equal.
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The ratios 10 to five, eight to two π¦ minus 14, and π₯ to eight must all be equal.
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We could set these up in fractional form to calculate the value of π₯ and π¦.
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However, it is clear from the first ratio that this simplifies to two to one.
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This means that all the lengths in the second trapezium will be half the lengths of the first trapezium.
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The scale factor to go from trapezium π΄π΅πΆπ· to πΈπΉπΊπ» is one-half.
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We can therefore say that two π¦ minus 14 is equal to a half of eight, and a half of eight is four.
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Adding 14 to both sides of this equation gives us two π¦ is equal to 18.
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Dividing by two gives us π¦ is equal to nine.
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We can use the same method to calculate the value of π₯.
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The length πΊπ», which is equal to eight, is half the value of πΆπ·, which is π₯.
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Multiplying both sides of this equation by two gives us 16 is equal to π₯.
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The missing values are π₯ equals 16 and π¦ equals nine.
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We will now summarize the key points from this video.
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Firstly, we know that the corresponding angles in similar polygons are congruent.
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The corresponding sides on the other hand are proportional.
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We can think of similar polygons as being enlarged or shrunk.
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And the ratio of one corresponding side to another is the scale factor.
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This scale factor will be the same for all corresponding sides in the polygon.
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We can use our knowledge of ratios and fractions to set up equations that can be solved.
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This allows us to calculate any variables on the lengths of the sides of our polygons.
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In order to prove that two or more polygons are similar, we need to show that the ratios of the corresponding sides are equal.