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Lesson Explainer: Equality of the Areas of Two Parallelograms Mathematics

In this explainer, 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.β€Œ

Parallelograms are one of the most fundamental shapes in geometry and mathematics, appearing throughout fields such as physics, mechanics, and calculus. Finding the area of parallelograms is an important skill that is used in these fields for reasons such as optimization and problem-solving.

Another reason parallelograms are so important is that squares, rectangles, and rhombuses are all special cases of parallelograms, so any result we prove for parallelograms also applies to these shapes. Finally, all parallelograms are the combination of two triangles; this means that results involving triangles and parallelograms often apply to the other.

For example, we recall that the area of a triangle is half the length of its base multiplied by its height. We can use this to find the area of a parallelogram 𝐴𝐡𝐢𝐷.

We note that triangles 𝐴𝐡𝐢 and 𝐢𝐷𝐴 have all sides congruent, so they are congruent by the SSS criterion. Thus, these triangles have the same area. In particular, the area of the parallelogram is twice the area of one of the triangles.

Hence, areaarea(𝐴𝐡𝐢𝐷)=2Γ—(𝐴𝐡𝐢)=2Γ—12(π‘β„Ž)=π‘β„Ž.

We see that the area of a parallelogram is the length of its base times its perpendicular height.

This highlights the relationship between properties of triangles and properties of parallelograms. Results about one shape can help us show results for the other.

Let’s use this motivation to find some useful results and properties of parallelograms. We can start by noting that the perpendicular distance between parallel lines remains constant.

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, 𝐴𝐡𝐢𝐷 has all interior angles as right angles.

This means that 𝐴𝐷=𝐡𝐢=β„Ž.

This is a useful result since a parallelogram has opposite sides that are parallel, and its area is calculated by using its perpendicular height. In particular, since the distance between parallel lines remains constant, we can note that parallelograms between the same pair of parallel lines will have the same height.

Now, since the area of a parallelogram is the length of its base times its perpendicular height, we can see that these two parallelograms will only have the same area if their bases are the same length. We have shown that, in general, the following result holds.

Theorem: Equality of Areas of Parallelograms between Parallel Lines

Parallelograms between a pair of parallel lines have the same area when their bases are the same length or when they share a common base.

We can see why this is true if the parallelograms between a pair of parallel lines share a common same base by considering an example.

We see that parallelograms 𝐴𝐡𝐢𝐷 and 𝐴𝐡𝐸𝐹 are between the same pair of parallel lines, so their heights are both equal to β„Ž. They also share the same base of length 𝐴𝐡, this confirms that they have the same area.

Let’s see an example of using this property to find the areas of two parallelograms.

Example 1: Finding the Areas of a Pair of Parallelograms That Share a Common Base

In the opposite figure, ⃖⃗𝐴𝐡⫽⃖⃗𝐢𝐹 and the distance between them is β„Ž, where β„Ž=3cm and 𝐴𝐡=4cm. Find the areas of parallelograms 𝐴𝐡𝐢𝐷 and 𝐴𝐡𝐸𝐹 respectively.

Answer

Let’s start by marking the two parallelograms 𝐴𝐡𝐢𝐷 and 𝐴𝐡𝐸𝐹 on the given diagram.

We note that both parallelograms share the side 𝐴𝐡, and their heights are given by the distance between the parallel lines ⃖⃗𝐴𝐡 and ⃖⃗𝐢𝐹. Since the distance between parallel lines is constant and the area of a parallelogram is base times perpendicular height, we can conclude that the parallelograms have equal areas.

We are told that the distance between the parallel lines is 3 cm and 𝐴𝐡=4cm. These are the values of the perpendicular heights of both parallelograms and the lengths of their bases respectively.

Hence, both parallelograms have areas of 3Γ—4=12cm.

In our next example, we will use the property of the equality of the areas of parallelograms to determine the distance between parallel lines using the areas of two parallelograms between the parallel lines with congruent bases.

Example 2: Finding the Distance between Two Parallel Lines given the Sum of the Areas of Parallelograms Drawn

In the following figure, 𝐴𝐡𝐢𝐷 and 𝐸𝐹𝐺𝐻 are two parallelograms, and ⃖⃗𝐴𝐹⫽⃖⃗𝐷𝐺. If 𝐴𝐡=𝐸𝐹, 𝐢𝐷=5cm, and the sum of the areas of parallelogram 𝐸𝐹𝐺𝐻 and parallelogram 𝐴𝐡𝐢𝐷 is 40 cm2, find the shortest distance between ⃖⃗𝐴𝐹 and ⃖⃗𝐷𝐺.

Answer

We want to find the shortest distance between ⃖⃗𝐴𝐹 and ⃖⃗𝐷𝐺; this is the perpendicular distance between the lines.

We notice that parallelograms 𝐴𝐡𝐢𝐷 and 𝐸𝐹𝐺𝐻 are between two parallel lines and they have congruent bases since 𝐴𝐡=𝐸𝐹. Since the distance between parallel lines remains constant, we can conclude that the areas of the parallelograms are equal since they have bases of the same length and equal heights.

Therefore, since the sum of their areas is 40 cm2, we can halve this value to see that each parallelogram has an area of 40Γ—12=20cm.

Since the area of a parallelogram is the length of its base times its perpendicular height, we can use the fact that the bases have lengths 5 cm and their areas are 20 cm2 to get 20=5Γ—β„Žβ„Ž=205=4.cm

It is worth noting that the heights of the parallelograms are the perpendicular distances between the parallel lines, so they are also the shortest distances between the lines.

Hence, the shortest distances β„Ž between ⃖⃗𝐴𝐹 and ⃖⃗𝐷𝐺 is 4 cm.

In our next example, we will once again use the same property, but this time one of the parallelograms is a rectangle, to determine the area of the parallelogram.

Example 3: Finding the Area of a Parallelogram Using the Area of a Rectangle

In the figure below, ⃖⃗𝐴𝐡⫽⃖⃗𝐢𝐸, 𝐴𝐢⫽𝐡𝐷, and 𝐴𝐡𝐸𝐹 is a rectangle. If 𝐡𝐸=4cm and 𝐴𝐡=3cm, find the area of parallelogram 𝐴𝐡𝐷𝐢.

Answer

To determine the area of parallelogram 𝐴𝐡𝐷𝐢, we can recall that its area is the length of its base times its perpendicular height. We see that the base 𝐴𝐡 has a length of 3 cm. We can find the perpendicular height by noting that 𝐴𝐡𝐸𝐹 is a rectangle and ⃖⃗𝐴𝐡⫽⃖⃗𝐢𝐸. We recall that, since the lines are parallel, the perpendicular distance between the lines stays constant. We note that π΅πΈβŸ‚βƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΄π΅ and 𝐡𝐸=4cm. Thus, the perpendicular height of parallelogram 𝐴𝐡𝐷𝐢 is 4 cm.

Hence, areacm(𝐴𝐡𝐷𝐢)=3Γ—4=12.

It is worth noting this is not the only method to answer this question. We can also use the fact that parallelograms between a pair of parallel lines with congruent bases have the same area. We note that rectangles are special cases of parallelograms and that parallelogram 𝐴𝐡𝐷𝐢 and rectangle 𝐴𝐡𝐸𝐹 are between the same parallel lines and they share the base 𝐴𝐡. Therefore, they have same area.

We then note that areaareacm(𝐴𝐡𝐷𝐢)=(𝐴𝐡𝐸𝐹)=3Γ—4=12.

In the above example, we saw that this property involving parallelograms can also be used with rectangles. In a similar manner, since parallelograms are the combination of two congruent triangles, we can also apply this result using the area of a triangle to determine the area of a parallelogram, as we will see in our next example.

Example 4: Finding the Area of a Parallelogram given the Area of a Triangle That Shares the Same Base

In the opposite figure, ⃖⃗𝐴𝐡⫽⃖⃗𝐸𝐷 and 𝐴𝐢⫽𝐡𝐷. If the area of △𝐴𝐡𝐸=7cm, find the area of parallelogram 𝐴𝐡𝐷𝐢.

Answer

We recall that the area of a triangle is half the length of its base multiplied by its perpendicular height. We are given the area of △𝐴𝐡𝐸, and we can note that this triangle shares a common base with parallelogram 𝐴𝐡𝐷𝐢.

Both the parallelogram and triangle have bases of the same length (since 𝐴𝐡 is common to both) and they also have the same perpendicular height β„Ž since they lie between the same pair of parallel lines. The area of △𝐴𝐡𝐸 is given by area△𝐴𝐡𝐸=12Γ—(𝐴𝐡)Γ—β„Ž.

We now note that the area of parallelogram 𝐴𝐡𝐷𝐢 is twice that of the area of △𝐴𝐡𝐸: areaarea(𝐴𝐡𝐷𝐢)=(𝐴𝐡)Γ—β„Ž=2Γ—ο€Ό12Γ—(𝐴𝐡)Γ—β„Žοˆ=2×△𝐴𝐡𝐸.

We are told that areacm△𝐴𝐡𝐸=7, so we can substitute this into the equation to get areacm(𝐴𝐡𝐷𝐢)=2Γ—7=14.

In the previous 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.

Theorem: Areas of Parallelograms and Triangles between Parallel Lines

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.

In our final example, we will consider the areas of two triangles between parallel lines.

Example 5: Using the Area of a Triangle to Solve Parallelogram Problems

If 𝐴𝐡𝐢𝐷 is a parallelogram where the perimeter = 79.5 cm, 𝐡𝐢=2𝐴𝐡, the area of △𝐴𝐡𝐢=106cm, 𝐸 is the midpoint of 𝐡𝐢, and 𝐹𝐺 is perpendicular to 𝐡𝐢.

  1. Find 𝐹𝐺.
  2. Find the area of △𝐴𝐸𝐢.

Answer

Part 1

We want to determine the length of 𝐹𝐺. We can note that this is the height of the parallelogram and the heights of triangles △𝐴𝐡𝐢 and △𝐴𝐸𝐢. We know the area of triangle △𝐴𝐡𝐢, and we know that this area is equal to half the length of its base multiplied by its perpendicular height.

We can, therefore, determine the length of 𝐹𝐺 if we can find the length of the base of this triangle: 𝐡𝐢.

We note that we are given the perimeter of the parallelogram and that 𝐡𝐢 is twice the length of 𝐴𝐡. Since opposite sides in a parallelogram are equal in length and the perimeter is the sum of the side lengths, we have 79.5=𝐴𝐡+𝐴𝐡+𝐡𝐢+𝐡𝐢=𝐴𝐡+𝐴𝐡+2𝐴𝐡+2𝐴𝐡=6𝐴𝐡.

We divide both sides of the equation by 6 to get 𝐴𝐡=79.56=13.25.cm

Since 𝐡𝐢=2𝐴𝐡, we have 𝐡𝐢=26.5cm.

We have area△𝐴𝐡𝐢=12×𝐡𝐢×𝐹𝐺.

We know that 𝐡𝐢=26.5cm and areacm△𝐴𝐡𝐢=106, so 106=12Γ—26.5×𝐹𝐺𝐹𝐺=10626.5Γ—2=8.cm

Part 2

We want to determine the area of △𝐴𝐸𝐢, to do this we need to find the length of its base and perpendicular height.

Now, we can note that 𝐸 is the midpoint of 𝐡𝐢, so 𝐸𝐢 is half the length of 𝐡𝐢. So, 𝐸𝐢=12Γ—26.5=13.25.cm

We know that its height is 8 cm, so areacm△𝐴𝐸𝐢=12×𝐸𝐢×𝐹𝐺=12Γ—13.25Γ—8=53.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • Parallelograms between a pair of parallel lines have the same area when their bases are congruent or share a common base.
  • 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.
  • 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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