# Area of a Parallelogram - Examples, Exercises and Solutions

## 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

## 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

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

$\sqrt{21}\times9=41.24$

41.24

### Exercise #5

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 #6

ABCD is a parallelogram with a perimeter of 38 cm.

AB is twice as long as CE.

AD is three times shorter than CE.

CE is the height of the parallelogram.

Calculate the area of the parallelogram.

### Step-by-Step Solution

Let's call CE as X

According to the data

$AB=x+2,AD=x-3$

The perimeter of the parallelogram:

$2(AB+AD)$

$38=2(x+2+x-3)$

$38=2(2x-1)$

$38=4x-2$

$38+2=4x$

$40=4x$

$x=10$

Now it can be argued:

$AD=10-3=7,CE=10$

The area of the parallelogram:

$CE\times AD=10\times7=70$

70 cm²

### Exercise #7

ABCD is a parallelogram whose perimeter is equal to 24 cm.

The side of the parallelogram is two times greater than the adjacent side (AB>AD).

CE is the height of the side AB

The area of the parallelogram is 24 cm².

Find the height of CE

### Step-by-Step Solution

The perimeter of the parallelogram is calculated as follows:

$S_{ABCD}=AB+BC+CD+DA$ Since ABCD is a parallelogram, each pair of opposite sides is equal, and therefore, AB=DC and AD=BC

According to the figure that the side of the parallelogram is 2 times larger than the side adjacent to it, it can be argued that$AB=DC=2BC$

We inut the data we know in the formula to calculate the perimeter:

$P_{ABCD}=2BC+BC+2BC+BC$

We replace the given perimeter in the formula and add up all the BC coefficients accordingly:

$24=6BC$

We divide the two sections by 6

$24:6=6BC:6$

$BC=4$

We know that$AB=DC=2BC$We replace the data we obtained (BC=4)

$AB=DC=2\times4=8$

As ABCD is a parallelogram, then all pairs of opposite sides are equal, therefore BC=AD=4

To find EC we use the formula:$A_{ABCD}=AB\times EC$

We replace the existing data:

$24=8\times EC$

We divide the two sections by 8$24:8=8EC:8$

$3=EC$

3 cm

### Exercise #8

The parallelogram ABCD contains the rectangle AEFC inside it, which has a perimeter of 24.

AE = 8

BC = 5

What is the area of the parallelogram?

### Step-by-Step Solution

In the first step, we must find the length of EC, which we will identify with an X.

We know that the perimeter of a rectangle is the sum of all its sides (AE+EC+CF+FA),

Since in a rectangle the opposite sides are equal, the formula can also be written like this: 2AE=2EC.

We replace the known data:

$2\times8+2X=24$

$16+2X=24$

We isolate X:

$2X=8$

and divide by 2:

$X=4$

Now we can use the Pythagorean theorem to find EB.

(Pythagoras: $A^2+B^2=C^2$)

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

$EB^2+16=25$

We isolate the variable

$EB^2=9$

We take the square root of the equation.

$EB=3$

The area of a parallelogram is the height multiplied by the side to which the height descends, that is$AB\times EC$.

$AB=\text{ AE}+EB$

$AB=8+3=11$

And therefore we will apply the area formula:

$11\times4=44$

44

### Exercise #9

The area of trapezoid ABCD is X cm².

The line AE creates triangle AED and parallelogram ABCE.

The ratio between the area of triangle AED and the area of parallelogram ABCE is 1:3.

Calculate the ratio between sides DE and EC.

### Step-by-Step Solution

To calculate the ratio between the sides we will use the existing figure:

$\frac{A_{AED}}{A_{ABCE}}=\frac{1}{3}$

We calculate the ratio between the sides according to the formula to find the area and then replace the data.

We know that the area of triangle ADE is equal to:

$A_{ADE}=\frac{h\times DE}{2}$

We know that the area of the parallelogram is equal to:

$A_{ABCD}=h\times EC$

We replace the data in the formula given by the ratio between the areas:

$\frac{\frac{1}{2}h\times DE}{h\times EC}=\frac{1}{3}$

We solve by cross multiplying and obtain the formula:

$h\times EC=3(\frac{1}{2}h\times DE)$

We open the parentheses accordingly:

$h\times EC=1.5h\times DE$

We divide both sides by h:

$EC=\frac{1.5h\times DE}{h}$

We simplify to h:

$EC=1.5DE$

Therefore, the ratio between is: $\frac{EC}{DE}=\frac{1}{1.5}$

$1:1.5$

### Exercise #10

ABCD is a parallelogram.

Angle ACB is equal to angle EBC.

BF = 6

CE = 9

BF is perpendicular to DE.

Calculate the area of the parallelogram.

### Step-by-Step Solution

Given that angle ACB is equal to angle CBE, it follows that AC is parallel to BE

since alternate angles between parallel lines are equal.

As we know that ABCD is a parallelogram, AB is parallel to DC and therefore AB is also parallel to CE since it is a line that continues DC.

Given that AC is parallel to BE and, in addition, AB is parallel to CE, it can be argued that ABCE is a parallelogram and, therefore, each pair of opposite sides in a parallelogram are parallel and equal.

From this it is concluded that AB=CE=9

Now we calculate the area of the parallelogram ABCD according to the data.

$S_{ABCD}=AB\times BF$

We replace the data accordingly:

$S_{ABCD}=9\times6=54$

54 cm²

### Exercise #11

Below is a circle bounded by a parallelogram:

All meeting points are tangential to the circle.
The circumference is 25.13.

What is the area of the parallelogram?

### Step-by-Step Solution

First, we add letters as reference points:

Let's observe points A and B.

We know that two tangent lines to a circle that start from the same point are parallel to each other.

Therefore:

$AE=AF=3$
$BG=BF=6$

And from here we can calculate:

$AB=AF+FB=3+6=9$

Now we need the height of the parallelogram.

We know that F is tangent to the circle, so the diameter that comes out of point F will also be the height of the parallelogram.

It is also known that the diameter is equal to two radii.

Since the circumference is 25.13.

Circumference formula:$2\pi R$
We replace and solve:

$2\pi R=25.13$
$\pi R=12.565$
$R\approx4$

The height of the parallelogram is equal to two radii, that is, 8.

And from here you can calculate with a parallelogram area formula:

$AlturaXLado$

$9\times8\approx72$

$\approx72$

### Exercise #12

The following is a circle enclosed in a parallelogram:

All meeting points are tangent to the circle.
The circumference is 25.13.

What is the area of the zones marked in blue?

### Step-by-Step Solution

First, we add letters as reference points:

Let's observe points A and B.

We know that two tangent lines to a circle that start from the same point are parallel to each other.

Therefore:

$AE=AF=3$
$BG=BF=6$

From here we can calculate:

$AB=AF+FB=3+6=9$

Now we need the height of the parallelogram.

We know that F is tangent to the circle, so the diameter that comes out of point F will also be the height of the parallelogram.

It is also known that the diameter is equal to two radii.

It is known that the circumference of the circle is 25.13.

Formula of the circumference:$2\pi R$
We replace and solve:

$2\pi R=25.13$
$\pi R=12.565$
$R\approx4$

The height of the parallelogram is equal to two radii, that is, 8.

And from here it is possible to calculate the area of the parallelogram:

$\text{Lado }x\text{ Altura}$$9\times8\approx72$

Now, we calculate the area of the circle according to the formula:$\pi R^2$

$\pi4^2=50.26$

Now, subtract the area of the circle from the surface of the trapezoid to get the answer:

$72-56.24\approx21.73$

$\approx21.73$

### Exercise #13

ABCD is a parallelogram
BFCE is a deltoid

What is the area of the parallelogram ABCD?

### Step-by-Step Solution

First, we must remember the formula for the area of a parallelogram:$\text{Lado }x\text{ Altura}$.

In this case, we will try to find the height CH and the side BC.

We start from the side

First, let's observe the small triangle EBG,

As it is a right triangle, we can use the Pythagorean theorem (

$A^2+B^2=C^2$)

$BG^2+4^2=5^2$

$BG^2+16=25$

$BG^2=9$

$BG=3$

Now, let's start looking for GC.

First, remember that the deltoid has two pairs of equal adjacent sides, therefore:$FC=EC=9$

Now we can also do Pythagoras in the triangle GCE.

$GC^2+4^2=9^2$

$GC^2+16=81$

$GC^2=65$

$GC=\sqrt{65}$

Now we can calculate the side BC:

$BC=BG+GT=3+\sqrt{65}\approx11$

Now, let's observe the triangle BGE and DHC

Angle BGE = 90°
Angle CHD = 90°
Angle CDH=EBG because these are opposite parallel angles.

Therefore, there is a ratio of similarity between the two triangles, so:

$\frac{HD}{BG}=\frac{HC}{GE}$

$\frac{HD}{BG}=\frac{7.5}{3}=2.5$

$\frac{HC}{EG}=\frac{HC}{4}=2.5$

$HC=10$

Now that there is a height and a side, all that remains is to calculate.

$10\times11\approx110$

$\approx110$

### Exercise #14

The parallelogram ABCD is shown below.

BC is the diameter of the circle whose circumference is equal to $10\pi$ cm.

ECFD is a rhombus whose area is 24 cm².

What is the area of ABCD?

### Step-by-Step Solution

Let's try to calculate the area in two ways.

In the first method, we will try to use the rhombus ECFD:

Let's try to calculate according to the formula $area=DC\times h_{DC}$

We will lower a height to DC and see that we do not have enough data to calculate, so we will not be able to calculate the area of the parallelogram using the rhombus.

In the second method , we will try to use the circle:

$area=BC\times h_{BC}$We will lower a height to BC and see that we do not have enough data to calculate, so we will not be able to calculate the area of the parallelogram using the circle.

From this it follows that we do not have enough data to calculate the area of parallelogram ABCD and therefore the exercise cannot be solved.

It is not possible to calculate.

### Exercise #15

Calculate the area of the parallelogram based on the data in the figure: