Video Transcript
In this video, we will learn how to
calculate areas of similar polygons given two corresponding side lengths or the
scale factor between them and the area of one of the polygons. We’ll begin by recapping what it
means for two polygons to be similar.
Two polygons with the same number
of sides are similar if two conditions are satisfied. Firstly, corresponding angles are
congruent and secondly, corresponding sides are proportional. For example, consider these two
rectangles. Both shapes have four sides. And as all the interior angles in a
rectangle are right angles, corresponding angles are congruent. Considering the side length, if we
divide the length of the longest side of the first rectangle by the length of the
longest side of the second, we have 𝐴𝐵 over 𝑃𝑄, which is equivalent to 𝐶𝐷 over
𝑅𝑆. And this is equal to eight over
four, which is two.
If we consider the shorter side
lengths, we have 𝐵𝐶 over 𝑄𝑅, which is the same as 𝐷𝐴 over 𝑆𝑃. This is three over 1.5, which is
also equal to two. So, the ratio is the same for each
pair of corresponding sides, and hence corresponding sides are indeed
proportional. Notice that the two rectangles in
this example were drawn in different orientations. The corresponding sides are
vertical on one rectangle, but horizontal on the other. So this is something we do need to
be aware of when working with similar polygons. We can state that 𝐴𝐵𝐶𝐷 is
similar to 𝑃𝑄𝑅𝑆. And the ordering of the letters is
important here because it reflects which vertices of the two polygons are
corresponding to one another.
To find the scale factor from one
polygon to another, so let’s first consider the scale factor from 𝐴𝐵𝐶𝐷 to
𝑃𝑄𝑅𝑆, we divide any of the side lengths of 𝑃𝑄𝑅𝑆 by the corresponding side
length of 𝐴𝐵𝐶𝐷. So, for example, the scale factor
of 𝐴𝐵𝐶𝐷 to 𝑃𝑄𝑅𝑆 is 𝑃𝑄 over 𝐴𝐵. That’s four over eight, which is
equal to one-half. Scale factors are always
multiplicative. So this means that to get from a
length on 𝐴𝐵𝐶𝐷 to a length on 𝑃𝑄𝑅𝑆, we multiply by one-half. To go the other way, we multiply by
two.
We should always check that any
scale factors we’ve calculated make sense. If we’re going from the larger
polygon to the smaller one, then the scale factor should be less than one. Whereas if we’re going in the other
direction, the scale factor should be greater than one. In general, if the scale factor in
one direction is 𝑘, then in the opposite direction it is the reciprocal, one over
𝑘.
An alternative way of expressing
the relationship between the side lengths of two similar polygons is using a
ratio. For example, the ratio of 𝐴𝐵𝐶𝐷
to 𝑃𝑄𝑅𝑆 is the ratio of the length of side 𝐴𝐵 to the length of side 𝑃𝑄. That’s eight to four, which
simplifies to two to one. Now, all of what we’ve discussed so
far should be a recap of our knowledge of similar polygons. We’re now going to extend this to
understand how the areas of similar polygons are related to one another.
So suppose we have two similar
polygons in which the scale factor of similarity is 𝑘. Our first thought may be that the
scale factor between the areas is also 𝑘. Let’s test this for the two similar
triangles shown here. The area of the first triangle is
its base multiplied by its perpendicular height over two. The area of the second triangle is
its base multiplied by its perpendicular height over two. That’s 𝑘𝑏 multiplied by 𝑘ℎ over
two. We can rewrite this as 𝑘 squared
multiplied by 𝑏ℎ over two. And as the area of the first
triangle is 𝑏ℎ over two, we find that the area of the second triangle is 𝑘 squared
multiplied by the area of the first. So in fact, the scale factor for
the areas isn’t 𝑘; it’s 𝑘 squared.
If we think about it, this makes
sense because area is two dimensional and both dimensions have been enlarged by this
factor of 𝑘. So, the effect on the area is to
enlarge it by a factor of 𝑘 squared. This result is true for all
polygons. We therefore need to be a little
more specific when referring to scale factors for similar polygons. And use the term length scale
factor to define the scale factor between the lengths and area scale factor to
describe the scale factor between the areas. We can state this result generally
as if the length scale factor between two similar polygons is 𝑘, then the area
scale factor is 𝑘 squared.
In terms of the ratios, we can also
state that if the length ratio of two similar polygons is 𝑎 to 𝑏, then the ratio
of their areas is 𝑎 squared to 𝑏 squared. So now that we’ve determined the
relationship between the areas of similar polygons, let’s consider some examples in
which we apply these results.
Given the following figure,
find the area of a similar polygon 𝐴 prime 𝐵 prime 𝐶 prime 𝐷 prime, in which
𝐴 prime 𝐵 prime equals six.
So we’re told that there exists
a similar polygon to the one in the figure in which the length of the side 𝐴
prime 𝐵 prime is six units. It’s reasonable to assume that
the same letters have been used to represent corresponding vertices on the two
polygons. So let’s compare the length of
𝐴 prime 𝐵 prime to the length of 𝐴𝐵 on the polygon in the figure. Side 𝐴𝐵 is horizontal and
goes from two to five, so it has a length of three units. We can therefore calculate the
length scale factor of 𝐴𝐵𝐶𝐷 to 𝐴 prime 𝐵 prime 𝐶 prime 𝐷 prime by
dividing the length of 𝐴 prime 𝐵 prime by the length of 𝐴𝐵. That’s six over three, which of
course is equal to two. This means then that the
lengths on polygon 𝐴 prime 𝐵 prime 𝐶 prime 𝐷 prime are each twice as long as
the corresponding lengths on polygon 𝐴𝐵𝐶𝐷.
Now it is in fact the area of
the second polygon that we’re interested in. We recall then that if the
length scale factor for two similar polygons is 𝑘, the area scale factor is 𝑘
squared. So the area scale factor of
𝐴𝐵𝐶𝐷 to 𝐴 prime 𝐵 prime 𝐶 prime 𝐷 prime is two squared, which is
four. Another way of saying this is
that the area of 𝐴 prime 𝐵 prime 𝐶 prime 𝐷 prime is four times the area of
𝐴𝐵𝐶𝐷. We can find the area of polygon
𝐴𝐵𝐶𝐷 from the figure. It’s a rectangle with one
dimension of length three units and the other dimension of length five
units. Its area is therefore three
multiplied by five, which is 15 square units.
The area of 𝐴 prime 𝐵 prime
𝐶 prime 𝐷 prime then is four multiplied by 15, which is equal to 60. So by recalling that if the
length scale factor between two similar polygons is 𝑘, then the area scale
factor is 𝑘 squared, we found that the area of the similar polygon 𝐴 prime 𝐵
prime 𝐶 prime 𝐷 prime is 60 square units.
Let’s consider another example.
Rectangle 𝑄𝑅𝑆𝑇 is similar
to rectangle 𝐽𝐾𝐿𝑀 with their sides having a ratio of eight to nine. If the dimensions of each
rectangle are doubled, find the ratio of the areas of the larger rectangles.
So we’re told that the length
ratio between the sides of the original rectangles is eight to nine. We can recall that if the
length ratio of two similar polygons is 𝑎 to 𝑏, then the ratio of their areas
is 𝑎 squared to 𝑏 squared. Now we’re told that dimensions
of each rectangle are doubled, but this doesn’t actually affect the length ratio
because the lengths in each polygon have been multiplied by the same factor of
two. This would just be equivalent
to making the length ratio two 𝑎 to two 𝑏. But of course, we could then
simplify this ratio by dividing both sides by two so we get back to 𝑎 to
𝑏. The length ratio of the
enlarged rectangles is still eight to nine. Using the result we wrote down
then, the ratio of the areas of the larger rectangles is eight squared to nine
squared, which is 64 to 81.
This illustrates a general
point. If we have a pair of similar
polygons with a given length ratio, and if the same scale factor is applied to both
polygons, then the length ratio remains the same, and furthermore the area ratio
will also remain the same. We’ll now consider some examples in
which we use the perimeter of similar shapes to solve problems involving their
areas.
Square A is an enlargement of
square B by a scale factor of two-thirds. If the perimeter of square A
equals 56 centimeters, what is the area of square B? Give your answer to the nearest
hundredth.
We’re given that square A is an
enlargement of square B by a scale factor of two-thirds. And as this scale factor is
less than one, this means that square A is actually smaller. All squares are similar to one
another. So we can use the properties of
similarity here. We’re told that the perimeter
of square A is 56 centimeters. And we know that for a square
the perimeter is four times the side length. So four times 𝑆 sub A, which
is representing the side length of A, is equal to 56. Dividing both sides of this
equation by four, we find that the side length of square A is 14
centimeters.
We can then proceed in two
different ways. In our first method, we’ll
calculate the side length of square B using the length scale factor and then the
area. As the length scale factor from
square B to square A is two-thirds, the length scale factor in the opposite
direction is the reciprocal of this; it’s three over two. So, the side length of square B
is three over two multiplied by the side length of A. That’s three over two
multiplied by 14, which is 21. To find the area of a square,
we square its side length. So, the area of square B is 21
squared, which is 441.
The second method we could use
is to find the area of square A and then consider the relationship between the
areas of these two similar shapes. The area of A is its side
length squared. That’s 14 squared, which is 196
centimeters squared. We then recall that if the
length scale factor between two similar polygons is 𝑘, the area scale factor is
𝑘 squared. So, the area scale factor from
A to B is three over two squared, which is nine over four. The area of square B then is
nine over four multiplied by the area of square A. That’s nine over four
multiplied by 196, which is again 441. The question asked though that
we give our answer to the nearest hundredth. So, we find that the area of
square B is 441.00 square centimeters.
We’ll now look at another example
in which we consider the relationship between the areas and perimeters of similar
polygons.
The areas of two similar
polygons are 361 centimeters squared and 81 centimeters squared. Given that the perimeter of the
first is 38 centimeters, find the perimeter of the second.
We’ve been given the areas of
the two similar polygons. And so, we can write down the
area ratio; it’s 361 to 81. We want to find the perimeter
of the second polygon given the perimeter of the first. And to do this, we need to know
the length ratio. We can recall that for two
similar polygons with corresponding sides in a length ratio of 𝑎 to 𝑏, the
area ratio is 𝑎 squared to 𝑏 squared, which means if you want to work
backwards from the area ratio to calculating the length ratio, we need to square
root both parts. So the length ratio is the
square root of 361 to the square root of 81, which is 19 to nine.
Now the perimeters of similar
shapes are in the same ratio as their lengths because the perimeter is just the
sum of the individual lengths. So, the perimeters which are 38
centimeters and a currently unknown value are also in the ratio 19 to nine. To get from 19 to 38, we have
to multiply by two. So doing the same to both parts
of the ratio, 19 to nine is equivalent to 38 to 18. So by working backwards from
knowing the area ratio between these two similar polygons to calculating the
length ratio and hence the ratio of the perimeters, we found that the perimeter
of the second polygon is 18 centimeters.
Let’s now consider one final
example in which we’ll apply the theory of the areas of similar polygons to a
real-world problem.
It costs 3799 pounds to fit
wooden flooring in a class with dimensions 28 meters and 10 meters. How much would it cost to fit
wooden flooring in a similar room with dimensions 84 meters and 30 meters?
We can assume that as we’re not
told the shape of the rooms, then they are rectangles. Now we’re told that these two
rooms are similar, which we can assume means mathematically similar. But let’s check this. We need to establish that
corresponding angles are congruent and corresponding sides are proportional. We know that corresponding
angles are congruent because all the interior angles in a rectangle are 90
degrees. To establish whether
corresponding sides are proportional. Let’s check the ratio between
the dimensions of the rooms.
Using the longer side of each,
we have 84 over 28, which is equal to three. And using the shorter sides, we
have 30 over 10, which is also equal to three. The ratio is the same. So corresponding sides are
indeed proportional, and the two rectangles are mathematically similar. We need to work out the cost of
fitting wooden flooring in the larger room, which will be dependent on its
area. We’re told the cost for the
smaller classroom is 3799 pounds. And we can calculate the area
of this room using the formula for the area of a rectangle. Its length multiplied by width,
which is 28 multiplied by 10, 280 square meters.
We can also find the area of
the larger room. It’s 84 multiplied by 30, which
is 2520 square meters. The area ratio for the two
rooms is 2520 to 280. And in fact, this simplifies to
nine to one. We now know that the area of
the bigger room is nine times the area of the smaller room. And assuming the cost is
directly proportional to the area, the cost of the flooring for the larger room
will be nine times the cost for the smaller room. That’s nine times 3799, which
is 34191.
We could also have determined
that the area scale factor was nine. By recalling that if the length
scale factor between two similar polygons is 𝑘, then the area scale factor is
𝑘 squared. We found the length scale
factor to be three. So the area scale factor is
three squared, which is nine. We found that the cost of
fitting wooden flooring in the larger room is 34,191 pounds.
Let’s now summarize the key points
from this video. Two polygons with the same number
of sides are similar if corresponding angles are congruent and corresponding sides
are proportional. If the length scale factor between
two similar polygons is 𝑘, then the area scale factor is 𝑘 squared. If the length ratio of two similar
polygons is 𝑎 to 𝑏, then the ratio of their areas is 𝑎 squared to 𝑏 squared. And we saw that we can work in the
opposite direction to calculate the length ratio given the area ratio. Finally, as perimeter is a length,
we can also say that the ratio of the areas of two similar polygons is equal to the
square of the ratio of their perimeters.
