# The area of a parallelogram: what is it and how is it calculated?

🏆Practice area of a parallelogram

## How is the area of a parallelogram calculated?

We can calculate the area of a parallelogram by multiplying one of its sides by its relative height.

To understand it better, we can use the following figure and the accompanying formula:

$A=DC\times H1=BC\times H2$

It can be seen that: $H1$ and $H2$ are the two heights corresponding to the bases $DC$ and $BC$ respectively.

Area of a Parallelogram

## Test yourself on area of a parallelogram!

A parallelogram has a length equal to 6 cm and a height equal to 4.5 cm.

Calculate the area of the parallelogram.

## What is a parallelogram?

#### Property of parallelograms

• The opposite angles of the parallelogram have the same size.
• The opposite sides of the parallelogram have the same length.
• Parallelograms have two intersecting diagonals that create two pairs of triangles. In addition, the four triangles that are formed have the same area.
• The angles of the parallelogram complement each other until they reach $180^o$ degrees.
• The sum of the squares of its diagonals is equal to the sum of the squares of the four sides of the parallelogram.

In other words:

$KM^{2}+LN^{2}=KL^{2}+LM^{2}+MN^{2}+NK^{2}$

Or, in other words:

$KM^{2}+LN^{2}=2KL^{2}+2LM^{2}$

## Examples of parallelograms

• Rectangle: is a parallelogram in which all its angles are right angles, that is, they measure $90^o$ degrees and its two diagonals have the same length.

Rectangle

• Rhombus: a parallelogram whose four sides are of equal length (and its two diagonals intersect at right angles, that is, they are perpendicular).

Rhombus

• Square: is a parallelogram that meets the definition of rectangle and rhombus (but also its two diagonals are perpendicular and have the same length).

### Square

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## Practice exercises for finding the area of a parallelogram

### Exercise 1

Find the area of the parallelogram $KLMN$ illustrated in the figure below using the data provided:

• $MN=10cm$
• $KP=5cm$

Area of a Parallelogram

Exercise 1

Solution:

This is a fairly simple exercise in which we must substitute the given data in the formula corresponding to the area of a parallelogram:

$A=MN\cdot KP=10\cdot5=50cm²$

Answer: The area of the parallelogram $KLMN$ is $50cm²$.

### Exercise 2

Analyze the illustration below and indicate if there are any errors in the data given. Explain your answer.

Solution:

This exercise deals with the area of a parallelogram. As we have already said, the area of this geometric shape can be calculated in two ways. With the first one, we must use as base the side $DC$ and consider as its relative height $AS$; the other way, is to consider the adjacent side $BC$ as the base and its relative height $AF$. The answer we obtain by applying both methods must be the same.

We substitute the data in the formula and we obtain the following:

$A=DC\cdot AS=9\cdot3=27$

$A=BC\cdot AF=6\cdot5=30$

As we can see, we have obtained a different result by applying one or the other method and, therefore, the given data are wrong.

Do you know what the answer is?

### Exercise 3

Find the area of the parallelogram $DEFG$ according to the illustration and the data below:

• $DE=12\operatorname{cm}$
• $KG=5\operatorname{cm}$
• $DK=9\operatorname{cm}$

Solution:

If we look at the illustration, we see that $DK$ refers to the external height of the parallelogram $DEFG$.

According to the characteristics of the parallelogram that we have just learned, the opposite sides of a parallelogram are identical and parallel to each other, that is: $DE= GF=12$ and $DE$ parallel to $GF$.

To calculate the area of this parallelogram we do not need the data about the length of $KG$ since this information is not useful for such a calculation, but was given to us only to confuse us. To calculate the area of a parallelogram, we only need the length of a side and its relative height.

That said, we substitute the data into the formula and we will get the following:

$A=GF\cdot DK=12\cdot9=108cm²$

Answer: The area of the parallelogram $DEFG$ is $108 cm²$.

### Exercise 4

Inside the parallelogram $ABCD$ is the rectangle $AECF$ with a perimeter of $24$.

$AE = 8$

What is the area of the parallelogram?

Solution:

In the first step we must find the length $EC$, which we will identify as $X$.

We know that the perimeter of the rectangle is equal to the sum of its sides $(AE+EC+CF+FA)$.

Because in the rectangle the opposite sides are equal, we can write the formula like this: $2AE+2EC=24$

We substitute the known data:

$2\times8+2X=24$

$16+2X=24$

We clear the $X$

$2X=8$

And divide by $2$

$X=4$

Now, we can use the Pythagorean formula to calculate $EB$.

Pythagoras -$A^2+B^2=C^2$

$EB^2+4^2=5^2$

$EB^2+16=25$

We clear $EB$

$EB^2=9$

Calculate the root

$EB=3$

The area of the parallelogram is the product of the side $AB$ by its relative height EC $AB\times EC$

$AB=\text{ AE}+EB$

On the other hand,

$AB=8+3=11$

Substitute the data into the area formula:

$11\times4=44$

Answer: $44$

### Exercise 5

Given that:

The perimeter of the parallelogram $ABCD$ is equal to $22 cm$. $DL = 3cm$

$AC = 4cm$

The height $=?$

And the side of the parallelogram

$DL = 3cm$

Calculate the area of the parallelogram. $ABCD$

Solution:

Parallel opposite sides are equal $AC=BD=4cm$

Parallel opposite sides are equal $AB=CD=Xcm$

$AB+BD+CD+AC\text{ }=$ Perimeter of the parallelogram

$X+4+X+4=22$

$2X+8=22$ /-8

$2X=14$ /:2

$X=7$

$CD=7$

Area $ABCD=CD\cdot LD$ (height)

Area $ABCD=7\cdot3$

Area $ABCD = 21$

Area of the parallelogram: $ABCD=21 cm²$

### Exercise 6

Consignment

Given the parallelogram $ABCD$

The area of the parallelogram is $98\operatorname{cm}²$

$\frac{AE}{DC}=\frac{1}{2}$

Objective:

Find a $DC$

Solution

According to the existing data we can calculate a $AE$

$AE=\frac{1}{2}DC$

$\text{ABCD}=DC\cdot AE=$

We replace the data accordingly

$98=\frac{1}{2}DC\cdot DC$

Multiply by $2$

$196=DC^2$

Take the root

$DC=14$

$14$

Do you think you will be able to solve it?

### Exercise 7

Reference

The area of the parallelogram $ABCD$ is $72\operatorname{cm}²$

Find a $DC$

Solution

$AE$ is the external height $DC$

$\text{ABCD}=DC\cdot AE=$

Replace the data accordingly

$72y=DC\cdot9$

Divide by $9$

$\frac{72y}{9}=DC$

$8y=DC$

$8y$

### Exercise 8

Consignment

Given the parallelogram $ABCD$

The relationship between $AE$ and $DC$ is $4:7$

Find the area of the parallelogram $ABCD$

Solution

According to the existing data we first calculate a $DC$

$\frac{AE}{DC}=\frac{4}{7}$

Replace a $AE$

$\frac{8}{DC}=\frac{4}{7}$

Multiply by cross

$8\cdot7=4\cdot DC$

Divide by $4$

$DC=\frac{8\cdot7}{4}=7\cdot2=14$

$\text{ABCD}=DC\cdot AE=$

Replace accordingly

$8\cdot14=112$

$112$

### Exercise 9

Assignment

Given the parallelogram of the figure

Its area is equal to $40\operatorname{cm}²$

Find a $AE$

Solution

$\text{ABCD}=DC\cdot AE=$

$DC=AB=8$

In the parallelogram the opposite sides are equal to each other.

Replace the data accordingly

$40=AE\cdot8$

Divide by $8$

$AE=5$

$5$

### Exercise 10

Consignment

Given the parallelogram $ABCD$

Its area is equal to $100\operatorname{cm}²$

Find a $AD$

Solution

$\text{ABCD}=DC\cdot AE=$

Replace the data accordingly

$100=6\cdot AD$

Divide by $6$

$AD=16.67$

$16.67$

Do you know what the answer is?

## examples with solutions for area of a parallelogram

### Exercise #1

Look at the parallelogram in the figure.

Its area is equal to 70 cm².

Calculate DC.

### Step-by-Step Solution

The formula for the area of a parallelogram:

Height * The side to which the height descends.

We replace in the formula all the known data, including the area:

5*DC = 70

We divide by 5:

DC = 70/5 = 14

And that's how we reveal the unknown!

$14$ cm

### Exercise #2

Look at the parallelogram in the figure below.

Its area is equal to 40 cm².

Calculate AE.

### Step-by-Step Solution

Given that ABCD is a parallelogram,$AB=CD=8$According to the properties of the parallelogram, each pair of opposite sides are equal and parallel.

To find AE we will use the area given to us in the formula to find the area of the parallelogram:

$S=DC\times AE$

$40=8\times AE$

We divide both sides of the equation by 8:

$8AE:8=40:8$

$AE=5$

$5$ cm

### Exercise #3

ABCD parallelogram, it is known that:

BE is perpendicular to DE

BF is perpendicular to DF

Calculate the area of the parallelogram in 2 different ways

### Step-by-Step Solution

In this exercise, we are given two heights and two sides.

It is important to keep in mind: The external height can also be used to calculate the area

Therefore, we can perform the operation of the following exercise:

The height BF * the side AD

8*6

The height BE the side DC
4
*12

The solution of these two exercises is 48, which is the area of the parallelogram.

48 cm²

### Exercise #4

AE is the height of the parallelogram ABCD.

AB is 3 cm longer than AE.

The area of ABCD is 32 cm².

Calculate the length of side AB.

### Step-by-Step Solution

Keep in mind that AB is 3 cm greater than AE, so we must pay attention to the data when we put the formula to calculate the parallelogram:

Height multiplied by the side of the height:

$AB\times AE=S$

We will mark AE with the letter a and therefore AB will be a+3:

$a\times(a+3)=32$

We open the parentheses:

$a^2+3a=32$

We use the trinomial/roots formula:

$a^2+3a-32=0$$(a+8)(a-5)=0$

That means we have two options:

$a=-8,a=5$

Since it is not possible to place a negative side in the formula to calculate the area$a=5$

Now we can calculate the sides:

$AE=5$

$AB=5+3=8$

8 cm

### Exercise #5

ABCD is a parallelogram.

CE is its height.

CB = 5
AE = 7
EB = 2

What is the area of the parallelogram?

### Step-by-Step Solution

To find the area,

first, the height of the parallelogram must be found.

To conclude, let's take a look at triangle EBC.

Since we know it is a right triangle (since it is the height of the parallelogram)

the Pythagorean theorem can be used:

$a^2+b^2=c^2$

In this case: $EB^2+EC^2=BC^2$

We place the given information: $2^2+EC^2=5^2$

We isolate the variable:$EC^2=5^2+2^2$

We solve:$EC^2=25-4=21$

$EC=\sqrt{21}$

Now all that remains is to calculate the area.

It is important to remember that for this, the length of each side must be used.
That is, AE+EB=2+7=9

$\sqrt{21}\times9=41.24$