# Lesson Video: Equality of the Areas of Two Parallelograms Mathematics

In this video, we will learn how to identify Parallelograms that have the same area when their bases are equal in length and the vertices opposite to these bases are on a parallel line to them.

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### Video Transcript

In this video, weβll 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. Weβll start by considering how we find the area of a parallelogram.

Letβs take the parallelogram π΄π΅πΆπ·. 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 letβs say that we wanted to find the area of triangle π΄π΅πΆ. 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΅πΆ is therefore πβ over two. However, since weβre interested in the area of the entire parallelogram, that means weβll need to find the area of triangle π΄π·πΆ 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΅πΆ; itβs also β. So the area of triangle π΄π·πΆ is also πβ 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΅πΆπ·. Adding the area of the two triangles, we would have πβ over two plus πβ over two, which simplifies to πβ. 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. Letβs 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΅πΆπ·, which has all interior angles as right angles. We could define the base length π·πΆ 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.

Letβs consider then that we draw a parallelogram πΈπΉπΆπ·. Notice that the base length πΆπ· is the same as that in the rectangle π΄π΅πΆπ·. 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 πΈπΉπΆπ· is also equal to π times β. Although π΄π΅πΆπ· is a rectangle, itβs 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. Weβll 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 πΆπΉ 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΅πΆπ· and π΄π΅πΈπΉ, respectively.

It might be useful to begin by highlighting these two specific parallelograms. First, we have parallelogram π΄π΅πΆπ·. And secondly, we have π΄π΅πΈπΉ. 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 πΆπΉ. 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΅πΆπ· and π΄π΅πΈπΉ 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΅πΆπ·, 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΅πΈπΉ is also equal to 12 square centimeters. And so we have found the areas of both parallelograms.

In the next example, weβll 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΅πΆπ· and πΈπΉπΊπ» are two parallelograms, and line π΄πΉ is parallel to line π·πΊ. If π΄π΅ equals πΈπΉ, πΆπ· equals five centimeters, and the sum of the areas of parallelogram πΈπΉπΊπ» and parallelogram π΄π΅πΆπ· is 40 square centimeters, find the shortest distance between line π΄πΉ and line π·πΊ.

The first thing we might note here is that the shortest distance between the two lines π΄πΉ and π·πΊ is the perpendicular distance between these lines. Letβs 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 πΈπΉ. 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΅πΆπ· and πΈπΉπΊπ» 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 πΆπ· is five centimeters, then we can work out the value of β by using the formula to find the area of π΄π΅πΆπ·. 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΄πΉ and the line π·πΊ is equal to the perpendicular height, and thatβs four centimeters.

In the next example, weβll 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 πΆπΈ, line segment π΄πΆ is parallel to line segment π΅π·, and π΄π΅πΈπΉ is a rectangle. If π΅πΈ equals four centimeters and π΄π΅ equals three centimeters, find the area of parallelogram π΄π΅π·πΆ.

From the information that we are given, we note that we have two pairs of parallel lines, which confirms that π΄π΅π·πΆ is indeed a parallelogram. We are further told that π΄π΅πΈπΉ is a rectangle. Since a rectangle is simply a special case of parallelogram, it means we can also note that the line segments π΄πΉ and π΅πΈ are also parallel. We can use the information about π΄π΅πΈπΉ to help us work out the area of the parallelogram π΄π΅π·πΆ. We can use the information that π΅πΈ 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΅πΈπΉ.

If youβre not sure why this is useful, letβs 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΅πΈπΉ is a rectangle, thatβs a special type of parallelogram, and the line segment π΄π΅ is a common side to both π΄π΅πΈπΉ and π΄π΅π·πΆ. So if we work out the area of π΄π΅πΈπΉ, itβs going to be the same as the area of π΄π΅π·πΆ.

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·πΆ is going to be equal to this. So itβs also 12 square centimeters. We could also have worked out the area of π΄π΅π·πΆ directly. The area of a parallelogram is found by multiplying the base by the perpendicular height. The base of π΄π΅π·πΆ is three centimeters, and the perpendicular height is also the length of the line segment π΅πΈ, which is four centimeters. Either method would produce the result that the area of π΄π΅π·πΆ is 12 square centimeters.

In the next example, weβll 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 πΈπ· and line segment π΄πΆ is parallel to line segment π΅π·. If the area of triangle π΄π΅πΈ equals seven square centimeters, find the area of parallelogram π΄π΅π·πΆ.

We can observe here that since the two lines π΄π΅ and πΈπ· are parallel and the two line segments π΄πΆ and π΅π· are parallel, then we have indeed got a parallelogram in π΄π΅π·πΆ. Given that the area of triangle π΄π΅πΈ is seven square centimeters, we need to work out the area of parallelogram π΄π΅π·πΆ. 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΅πΈ 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·πΆ 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΅πΈ is seven square centimeters, we double it, giving us the answer that the area of parallelogram π΄π΅π·πΆ 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.

Weβll 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.