Lesson Video: Equality of the Areas of Two Parallelograms | Nagwa Lesson Video: Equality of the Areas of Two Parallelograms | Nagwa

Lesson Video: Equality of the Areas of Two Parallelograms Mathematics • Second Year of Preparatory School

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.

14:49

Video Transcript

In this video, we’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. We’ll start by considering how we find the area of a parallelogram.

Let’s take the parallelogram 𝐴𝐵𝐶𝐷. 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 let’s say that we wanted to find the area of triangle 𝐴𝐵𝐶. We can define the base 𝑏 with the letter 𝑏, and the perpendicular height with the letter ℎ. So of course, the area of triangle 𝐴𝐵𝐶 is therefore 𝑏ℎ over two. However, since we’re interested in the area of the entire parallelogram, that means we’ll need to find the area of triangle 𝐴𝐷𝐶 as well. Its base is also 𝑏, since we know that a parallelogram has opposite sides congruent. The perpendicular height is the same as that of triangle 𝐴𝐵𝐶; it’s also ℎ. So the area of triangle 𝐴𝐷𝐶 is also 𝑏ℎ 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 𝐴𝐵𝐶𝐷. Adding the area of the two triangles, we would have 𝑏ℎ over two plus 𝑏ℎ over two, which simplifies to 𝑏ℎ. 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. Let’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 𝐴𝐵𝐶𝐷, which has all interior angles as right angles. We could define the base length 𝐷𝐶 as 𝑏. 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 𝑏 times ℎ.

Let’s consider then that we draw a parallelogram 𝐸𝐹𝐶𝐷. Notice that the base length 𝐶𝐷 is the same as that in the rectangle 𝐴𝐵𝐶𝐷. 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 𝐸𝐹𝐶𝐷 is also equal to 𝑏 times ℎ. Although 𝐴𝐵𝐶𝐷 is a rectangle, it’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. We’ll now see how we can use this property to find the areas of two parallelograms.

In the opposite figure, line 𝐴𝐵 is parallel to line 𝐶𝐹 and the distance between them is ℎ, where ℎ equals three centimeters and 𝐴𝐵 equals four centimeters. Find the areas of parallelograms 𝐴𝐵𝐶𝐷 and 𝐴𝐵𝐸𝐹, respectively.

It might be useful to begin by highlighting these two specific parallelograms. First, we have parallelogram 𝐴𝐵𝐶𝐷. And secondly, we have 𝐴𝐵𝐸𝐹. We can observe that both parallelograms share the same side of 𝐴𝐵. The heights of both of these parallelograms will be the distance between the parallel lines 𝐴𝐵 and 𝐶𝐹. This is also marked on the figure with the letter ℎ. Because these parallelogram share the same base 𝐴𝐵, we can use the property that parallelograms between a pair of parallel lines have the same area when they share a common base. So 𝐴𝐵𝐶𝐷 and 𝐴𝐵𝐸𝐹 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 𝐴𝐵𝐶𝐷, we take the base, which is given as four centimeters, and multiply it by the perpendicular height ℎ, which is given as three centimeters. This gives us an answer of 12 square centimeters. The area of parallelogram 𝐴𝐵𝐸𝐹 is also equal to 12 square centimeters. And so we have found the areas of both parallelograms.

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

In the following figure, 𝐴𝐵𝐶𝐷 and 𝐸𝐹𝐺𝐻 are two parallelograms, and line 𝐴𝐹 is parallel to line 𝐷𝐺. If 𝐴𝐵 equals 𝐸𝐹, 𝐶𝐷 equals five centimeters, and the sum of the areas of parallelogram 𝐸𝐹𝐺𝐻 and parallelogram 𝐴𝐵𝐶𝐷 is 40 square centimeters, find the shortest distance between line 𝐴𝐹 and line 𝐷𝐺.

The first thing we might note here is that the shortest distance between the two lines 𝐴𝐹 and 𝐷𝐺 is the perpendicular distance between these lines. Let’s define this perpendicular distance or perpendicular height with the letter ℎ. From the diagram and the information given in the question, we have that 𝐴𝐵 is congruent with 𝐸𝐹. 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 𝐴𝐵𝐶𝐷 and 𝐸𝐹𝐺𝐻 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 ℎ. 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 𝐶𝐷 is five centimeters, then we can work out the value of ℎ by using the formula to find the area of 𝐴𝐵𝐶𝐷. So we know that 20 is equal to five multiplied by ℎ. Dividing through by five, we have that ℎ is equal to four centimeters. We can therefore conclude that the shortest distance between the line 𝐴𝐹 and the line 𝐷𝐺 is equal to the perpendicular height, and that’s four centimeters.

In the next example, we’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 𝐴𝐵 is parallel to line 𝐶𝐸, line segment 𝐴𝐶 is parallel to line segment 𝐵𝐷, and 𝐴𝐵𝐸𝐹 is a rectangle. If 𝐵𝐸 equals four centimeters and 𝐴𝐵 equals three centimeters, find the area of parallelogram 𝐴𝐵𝐷𝐶.

From the information that we are given, we note that we have two pairs of parallel lines, which confirms that 𝐴𝐵𝐷𝐶 is indeed a parallelogram. We are further told that 𝐴𝐵𝐸𝐹 is a rectangle. Since a rectangle is simply a special case of parallelogram, it means we can also note that the line segments 𝐴𝐹 and 𝐵𝐸 are also parallel. We can use the information about 𝐴𝐵𝐸𝐹 to help us work out the area of the parallelogram 𝐴𝐵𝐷𝐶. We can use the information that 𝐵𝐸 is equal to four centimeters and 𝐴𝐵 is equal to three centimeters to help us work out the area of the rectangle 𝐴𝐵𝐸𝐹.

If you’re not sure why this is useful, let’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 𝐴𝐵𝐸𝐹 is a rectangle, that’s a special type of parallelogram, and the line segment 𝐴𝐵 is a common side to both 𝐴𝐵𝐸𝐹 and 𝐴𝐵𝐷𝐶. So if we work out the area of 𝐴𝐵𝐸𝐹, it’s going to be the same as the area of 𝐴𝐵𝐷𝐶.

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 𝐴𝐵𝐷𝐶 is going to be equal to this. So it’s also 12 square centimeters. We could also have worked out the area of 𝐴𝐵𝐷𝐶 directly. The area of a parallelogram is found by multiplying the base by the perpendicular height. The base of 𝐴𝐵𝐷𝐶 is three centimeters, and the perpendicular height is also the length of the line segment 𝐵𝐸, which is four centimeters. Either method would produce the result that the area of 𝐴𝐵𝐷𝐶 is 12 square centimeters.

In the next example, we’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 𝐴𝐵 is parallel to line 𝐸𝐷 and line segment 𝐴𝐶 is parallel to line segment 𝐵𝐷. If the area of triangle 𝐴𝐵𝐸 equals seven square centimeters, find the area of parallelogram 𝐴𝐵𝐷𝐶.

We can observe here that since the two lines 𝐴𝐵 and 𝐸𝐷 are parallel and the two line segments 𝐴𝐶 and 𝐵𝐷 are parallel, then we have indeed got a parallelogram in 𝐴𝐵𝐷𝐶. Given that the area of triangle 𝐴𝐵𝐸 is seven square centimeters, we need to work out the area of parallelogram 𝐴𝐵𝐷𝐶. To do this, we can use the fact that the triangle and the parallelogram share the same base of 𝐴𝐵. Importantly, they also share the same perpendicular height, which we can define with the letter ℎ.

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 𝐴𝐵𝐸 is equal to one-half multiplied by the length of the line segment 𝐴𝐵 multiplied by ℎ. 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 𝐴𝐵𝐷𝐶 can be calculated by the length of 𝐴𝐵 multiplied by ℎ.

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 𝐴𝐵𝐸 is seven square centimeters, we double it, giving us the answer that the area of parallelogram 𝐴𝐵𝐷𝐶 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.

We’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.

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