### 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.

Letβs 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 π₯.

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
π½π and π
π. A second pair of corresponding
sides are πΆπ½ and ππ
. As the corresponding sides are
proportional, we know that their ratios are the same. Two π₯ plus six over seven π₯ minus
seven must be equal to 24 over 28.

In order to find the value of π₯,
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 π₯ plus six becomes two
multiplied by π₯ plus three. And seven π₯ minus seven becomes
seven multiplied by π₯ 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 π₯ plus three over π₯ minus one is equal to three. We can multiply both sides of this
equation by π₯ minus one. Distributing the parentheses gives
us π₯ plus three is equal to three π₯ minus three. We can then subtract π₯ and add
three to both sides of the equation. This gives us six is equal to two
π₯. Finally, dividing both sides by two
gives us π₯ is equal to three.

If the two polygons are similar,
the value of π₯ 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 ππ
ππ is similar to
ππππ, find the value of π₯.

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 ππ and ππ along with ππ and ππ. This means that the ratio of six π₯
minus 16 to 13 will be the same as nine π₯ minus 33 to 15. Writing this in fractional form
gives us six π₯ minus 16 over nine π₯ minus 33 is equal to 13 over 15.

In order to calculate π₯, 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 π₯ minus 16 is equal to two
multiplied by three π₯ minus eight. And nine π₯ minus 33 is equal to
three multiplied by three π₯ 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 π₯ minus eight is equal to 13 multiplied by three π₯
minus 11. We get the 10 by multiplying five
by two. Redistributing our parentheses
gives us 30π₯ minus 80 is equal to 39π₯ minus 143. Adding 143 to both sides of this
equation gives us 30π₯ plus 63 is equal to 39π₯. Subtracting 30π₯ from both sides
gives us 63 is equal to nine π₯. Finally, dividing both sides of
this equation by nine gives us a value of π₯ equal to seven.

We can then check this answer by
substituting π₯ 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 ππ
ππ to ππππ 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 π₯.

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
π΅πΆ and ππ». A second pair of corresponding
sides are therefore πΆπ· and π»π½.

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. Writing this in fractional form, we
have two over four π₯ minus 37 is equal to six over two π₯ minus 11. Both of the numerators here are
divisible by two. 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.

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

We can then substitute this value
back into the expressions for the lengths of π΅πΆ and ππ» 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 π½π» is
three times the length of πΆπ·. 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. Following this method would also
have got us a value of π₯ equal to 10.

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

Given that π΄π΅πΆπ· is similar to
πΈπΉπΊπ», find the values of π₯ and π¦.

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 πΆπ· is corresponding to
πΊπ», π΄π· is corresponding to πΈπ», and π΅πΆ is corresponding to πΉπΊ. This means that the ratio of these
three sides must be equal. The ratios 10 to five, eight to two
π¦ minus 14, and π₯ to eight must all be equal. We could set these up in fractional
form to calculate the value of π₯ and π¦. 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 π΄π΅πΆπ· to πΈπΉπΊπ» is one-half.

We can therefore say that two π¦
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 π¦ is equal to 18. Dividing by two gives us π¦ is
equal to nine. We can use the same method to
calculate the value of π₯. The length πΊπ», which is equal to
eight, is half the value of πΆπ·, which is π₯. Multiplying both sides of this
equation by two gives us 16 is equal to π₯. The missing values are π₯ equals 16
and π¦ 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.