### Video Transcript

In this video, weβll see how we can
identify that two parallelograms will have the same area when their bases are equal
in length and the vertices opposite these bases are on a line which is parallel to
them. Weβll start by considering how we
find the area of a parallelogram.

Letβs take the parallelogram
π΄π΅πΆπ·. And of course, remember that a
parallelogram is defined as a quadrilateral that has two pairs of opposite sides
parallel. In order to derive a formula to
find the area of a parallelogram, we can consider this parallelogram divided into
two triangles. We can recall that the area of a
triangle is found by multiplying the base by the perpendicular height and dividing
by two.

So letβs say that we wanted to find
the area of triangle π΄π΅πΆ. We can define the base π with the
letter π, and the perpendicular height with the letter β. So of course, the area of triangle
π΄π΅πΆ is therefore πβ over two. However, since weβre interested in
the area of the entire parallelogram, that means weβll need to find the area of
triangle π΄π·πΆ as well. Its base is also π, since we know
that a parallelogram has opposite sides congruent. The perpendicular height is the
same as that of triangle π΄π΅πΆ; itβs also β. So the area of triangle π΄π·πΆ is
also πβ over two.

As an aside, we could in fact prove
that these two triangles are congruent, and we could do that by using the SSS
congruence criterion. However, we can use the results of
the two areas of these triangles to work out the area of π΄π΅πΆπ·. Adding the area of the two
triangles, we would have πβ over two plus πβ over two, which simplifies to
πβ. We have therefore demonstrated that
the area of this parallelogram, and indeed any parallelogram, can be found by
multiplying the base and the perpendicular height.

We can use this fact to give us
some useful results and properties of parallelograms. Letβs start by considering a pair
of parallel lines. As the lines are parallel, then the
perpendicular distance between the parallel lines is always the same. The reason for this is that any two
lines perpendicular to the same line are parallel to each other. So these lines give us a
rectangle. For example, we have rectangle
π΄π΅πΆπ·, which has all interior angles as right angles. We could define the base length
π·πΆ as π. And to find the area of this
rectangle, we know that we multiply the length by the width. So in this case, we would have π
times β.

Letβs consider then that we draw a
parallelogram πΈπΉπΆπ·. Notice that the base length πΆπ· is
the same as that in the rectangle π΄π΅πΆπ·. As we have already demonstrated,
the area of this parallelogram is equal to the base multiplied by the perpendicular
height. So the area of parallelogram
πΈπΉπΆπ· is also equal to π times β. Although π΄π΅πΆπ· is a rectangle,
itβs also just a special type of parallelogram. So these two parallelograms have
the same area but only because the base is the same length.

We can generalize this in the
following theorem. Parallelograms between a pair of
parallel lines have the same area when their bases are the same or when they share a
common base. Weβll now see how we can use this
property to find the areas of two parallelograms.

In the opposite figure, line π΄π΅
is parallel to line πΆπΉ and the distance between them is β, where β equals three
centimeters and π΄π΅ equals four centimeters. Find the areas of parallelograms
π΄π΅πΆπ· and π΄π΅πΈπΉ, respectively.

It might be useful to begin by
highlighting these two specific parallelograms. First, we have parallelogram
π΄π΅πΆπ·. And secondly, we have π΄π΅πΈπΉ. We can observe that both
parallelograms share the same side of π΄π΅. The heights of both of these
parallelograms will be the distance between the parallel lines π΄π΅ and πΆπΉ. This is also marked on the figure
with the letter β. Because these parallelogram share
the same base π΄π΅, we can use the property that parallelograms between a pair of
parallel lines have the same area when they share a common base. So π΄π΅πΆπ· and π΄π΅πΈπΉ will
actually have the same area.

To find the area of either of these
parallelograms, we can use the formula that the area of a parallelogram is equal to
the base multiplied by the perpendicular height. To find the area of π΄π΅πΆπ·, we
take the base, which is given as four centimeters, and multiply it by the
perpendicular height β, which is given as three centimeters. This gives us an answer of 12
square centimeters. The area of parallelogram π΄π΅πΈπΉ
is also equal to 12 square centimeters. And so we have found the areas of
both parallelograms.

In the next example, weβll use the
property of the equality of the areas of parallelograms to determine the distance
between two parallel lines.

In the following figure, π΄π΅πΆπ·
and πΈπΉπΊπ» are two parallelograms, and line π΄πΉ is parallel to line π·πΊ. If π΄π΅ equals πΈπΉ, πΆπ· equals
five centimeters, and the sum of the areas of parallelogram πΈπΉπΊπ» and
parallelogram π΄π΅πΆπ· is 40 square centimeters, find the shortest distance between
line π΄πΉ and line π·πΊ.

The first thing we might note here
is that the shortest distance between the two lines π΄πΉ and π·πΊ is the
perpendicular distance between these lines. Letβs define this perpendicular
distance or perpendicular height with the letter β. From the diagram and the
information given in the question, we have that π΄π΅ is congruent with πΈπΉ. So in fact, these two
parallelograms have congruent bases. And importantly, since the distance
between the parallel lines remains constant, this means that the areas of the two
parallelograms π΄π΅πΆπ· and πΈπΉπΊπ» must be equal. And as we are given that the sum of
their areas is 40 square centimeters, then we can halve this value to see that the
area of each parallelogram must be 20 square centimeters.

Of course, we still need to find
this shortest distance, or the value of β. And to do this, we need to recall
the formula to find the area of a parallelogram. This is equal to the base
multiplied by the perpendicular height. Given the information that the
length of the line segment πΆπ· is five centimeters, then we can work out the value
of β by using the formula to find the area of π΄π΅πΆπ·. So we know that 20 is equal to five
multiplied by β. Dividing through by five, we have
that β is equal to four centimeters. We can therefore conclude that the
shortest distance between the line π΄πΉ and the line π·πΊ is equal to the
perpendicular height, and thatβs four centimeters.

In the next example, weβll use the
same property to determine the area of a parallelogram, but this time one of the
parallelograms is a rectangle.

In the figure below, line π΄π΅ is
parallel to line πΆπΈ, line segment π΄πΆ is parallel to line segment π΅π·, and
π΄π΅πΈπΉ is a rectangle. If π΅πΈ equals four centimeters and
π΄π΅ equals three centimeters, find the area of parallelogram π΄π΅π·πΆ.

From the information that we are
given, we note that we have two pairs of parallel lines, which confirms that
π΄π΅π·πΆ is indeed a parallelogram. We are further told that π΄π΅πΈπΉ
is a rectangle. Since a rectangle is simply a
special case of parallelogram, it means we can also note that the line segments π΄πΉ
and π΅πΈ are also parallel. We can use the information about
π΄π΅πΈπΉ to help us work out the area of the parallelogram π΄π΅π·πΆ. We can use the information that
π΅πΈ is equal to four centimeters and π΄π΅ is equal to three centimeters to help us
work out the area of the rectangle π΄π΅πΈπΉ.

If youβre not sure why this is
useful, letβs recall an important property about parallelograms created between two
parallel lines. Parallelograms between a pair of
parallel lines with congruent bases have the same area, so even though π΄π΅πΈπΉ is a
rectangle, thatβs a special type of parallelogram, and the line segment π΄π΅ is a
common side to both π΄π΅πΈπΉ and π΄π΅π·πΆ. So if we work out the area of
π΄π΅πΈπΉ, itβs going to be the same as the area of π΄π΅π·πΆ.

And of course, to work out the area
of a rectangle, we multiply the length by the width. Three times four is 12, and the
area units will be square centimeters. The area of π΄π΅π·πΆ is going to be
equal to this. So itβs also 12 square
centimeters. We could also have worked out the
area of π΄π΅π·πΆ directly. The area of a parallelogram is
found by multiplying the base by the perpendicular height. The base of π΄π΅π·πΆ is three
centimeters, and the perpendicular height is also the length of the line segment
π΅πΈ, which is four centimeters. Either method would produce the
result that the area of π΄π΅π·πΆ is 12 square centimeters.

In the next example, weβll see how
we can find the area of a parallelogram given the area of a triangle which shares
the same base.

In the opposite figure, line π΄π΅
is parallel to line πΈπ· and line segment π΄πΆ is parallel to line segment π΅π·. If the area of triangle π΄π΅πΈ
equals seven square centimeters, find the area of parallelogram π΄π΅π·πΆ.

We can observe here that since the
two lines π΄π΅ and πΈπ· are parallel and the two line segments π΄πΆ and π΅π· are
parallel, then we have indeed got a parallelogram in π΄π΅π·πΆ. Given that the area of triangle
π΄π΅πΈ is seven square centimeters, we need to work out the area of parallelogram
π΄π΅π·πΆ. To do this, we can use the fact
that the triangle and the parallelogram share the same base of π΄π΅. Importantly, they also share the
same perpendicular height, which we can define with the letter β.

Now to find the area of a triangle,
we recall that this is equal to one-half multiplied by the base multiplied by the
perpendicular height. So the area of triangle π΄π΅πΈ is
equal to one-half multiplied by the length of the line segment π΄π΅ multiplied by
β. Equally, we can recall that to find
the area of a parallelogram, we multiply the base by the perpendicular height. Therefore, the area of
parallelogram π΄π΅π·πΆ can be calculated by the length of π΄π΅ multiplied by β.

If we then observe the right-hand
side of both of these equations, we can observe that the parallelogram is simply
double the area of the triangle. Therefore, given that the area of
triangle π΄π΅πΈ is seven square centimeters, we double it, giving us the answer that
the area of parallelogram π΄π΅π·πΆ is 14 square centimeters.

In this example, we showed that if
a triangle and a parallelogram share a common base and lie between the same pair of
parallel lines, then the parallelogram has twice the area of the triangle. This result also holds true if they
have congruent bases.

This result can be formalized in
the following theorem. If a triangle and a parallelogram
share a common base or have congruent bases and lie between the same pair of
parallel lines, then the parallelogram has twice the area of the triangle.

Weβll now recap some of the key
points of this video. Parallelograms between a pair of
parallel lines have the same area when their bases are congruent, or they share a
common base. Likewise, if a parallelogram and
rectangle are between the same pair of parallel lines and have congruent bases or
share a common base, then they have the same area. Finally, as we saw demonstrated in
the previous example, if a triangle and a parallelogram share a common base or have
congruent bases and lie between the same pair of parallel lines, then the
parallelogram has twice the area of the triangle.