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×H1=BC×H2 A=DC\times H1=BC\times H2

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

Area of a Parallelogram

1 - Area of a Parallelogram

Practice 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.

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Video Solution

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!

Answer

14 14 cm

Exercise #2

Look at the parallelogram in the figure below.

Its area is equal to 40 cm².

Calculate AE.

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Video Solution

Step-by-Step Solution

Given that ABCD is a parallelogram,AB=CD=8 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×AE S=DC\times AE

40=8×AE 40=8\times AE

We divide both sides of the equation by 8:

8AE:8=40:8 8AE:8=40:8

AE=5 AE=5

Answer

5 5 cm

Exercise #3

ABCD parallelogram, it is known that:

BE is perpendicular to DE

BF is perpendicular to DF

BF=8 BE=4 AD=6 DC=12

Calculate the area of the parallelogram in 2 different ways

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Video Solution

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.

 

Answer

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.

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Video Solution

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×AE=S AB\times AE=S

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

a×(a+3)=32 a\times(a+3)=32

We open the parentheses:

a2+3a=32 a^2+3a=32

We use the trinomial/roots formula:

a2+3a32=0 a^2+3a-32=0 (a+8)(a5)=0 (a+8)(a-5)=0

That means we have two options:

a=8,a=5 a=-8,a=5

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

Now we can calculate the sides:

AE=5 AE=5

AB=5+3=8 AB=5+3=8

Answer

8 cm

Exercise #5

ABCD is a parallelogram.

CE is its height.

CB = 5
AE = 7
EB = 2

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What is the area of the parallelogram?

Video Solution

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:

a2+b2=c2 a^2+b^2=c^2

In this case: EB2+EC2=BC2 EB^2+EC^2=BC^2

We place the given information: 22+EC2=52 2^2+EC^2=5^2

We isolate the variable:EC2=52+22 EC^2=5^2+2^2

We solve:EC2=254=21 EC^2=25-4=21

EC=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

21×9=41.24 \sqrt{21}\times9=41.24

Answer

41.24

examples with solutions for area of a parallelogram

Exercise #1

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

AE = 8

BC = 5

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What is the area of the parallelogram?

Video Solution

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:

2×8+2X=24 2\times8+2X=24

16+2X=24 16+2X=24

We isolate X:

2X=8 2X=8

and divide by 2:

X=4 X=4

Now we can use the Pythagorean theorem to find EB.

(Pythagoras: A2+B2=C2 A^2+B^2=C^2 )

EB2+42=52 EB^2+4^2=5^2

EB2+16=25 EB^2+16=25

We isolate the variable

EB2=9 EB^2=9

We take the square root of the equation.

EB=3 EB=3

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

AB= AE+EB AB=\text{ AE}+EB

AB=8+3=11 AB=8+3=11

And therefore we will apply the area formula:

11×4=44 11\times4=44

Answer

44

Exercise #2

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.

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Video Solution

Step-by-Step Solution

Let's call CE as X

According to the data

AB=x+2,AD=x3 AB=x+2,AD=x-3

The perimeter of the parallelogram:

2(AB+AD) 2(AB+AD)

38=2(x+2+x3) 38=2(x+2+x-3)

38=2(2x1) 38=2(2x-1)

38=4x2 38=4x-2

38+2=4x 38+2=4x

40=4x 40=4x

x=10 x=10

Now it can be argued:

AD=103=7,CE=10 AD=10-3=7,CE=10

The area of the parallelogram:

CE×AD=10×7=70 CE\times AD=10\times7=70

Answer

70 cm²

Exercise #3

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

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Video Solution

Step-by-Step Solution

The perimeter of the parallelogram is calculated as follows:

SABCD=AB+BC+CD+DA 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 thatAB=DC=2BC AB=DC=2BC

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

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

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

24=6BC 24=6BC

We divide the two sections by 6

24:6=6BC:6 24:6=6BC:6

BC=4 BC=4

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

AB=DC=2×4=8 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:AABCD=AB×EC A_{ABCD}=AB\times EC

We replace the existing data:

24=8×EC 24=8\times EC

We divide the two sections by 824:8=8EC:8 24:8=8EC:8

3=EC 3=EC

Answer

3 cm

Exercise #4

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.

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Video Solution

Step-by-Step Solution

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

AAEDAABCE=13 \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:

AADE=h×DE2 A_{ADE}=\frac{h\times DE}{2}

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

AABCD=h×EC A_{ABCD}=h\times EC

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

12h×DEh×EC=13 \frac{\frac{1}{2}h\times DE}{h\times EC}=\frac{1}{3}

We solve by cross multiplying and obtain the formula:

h×EC=3(12h×DE) h\times EC=3(\frac{1}{2}h\times DE)

We open the parentheses accordingly:

h×EC=1.5h×DE h\times EC=1.5h\times DE

We divide both sides by h:

EC=1.5h×DEh EC=\frac{1.5h\times DE}{h}

We simplify to h:

EC=1.5DE EC=1.5DE

Therefore, the ratio between is: ECDE=11.5 \frac{EC}{DE}=\frac{1}{1.5}

Answer

1:1.5 1:1.5

Exercise #5

Below is a circle bounded by a parallelogram:

36

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

What is the area of the parallelogram?

Video Solution

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

And from here we can calculate:

AB=AF+FB=3+6=9 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:2πR 2\pi R
We replace and solve:

2πR=25.13 2\pi R=25.13
πR=12.565 \pi R=12.565
R4 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 AlturaXLado

9×872 9\times8\approx72

Answer

72 \approx72

examples with solutions for area of a parallelogram

Exercise #1

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.

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Video Solution

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.

SABCD=AB×BF S_{ABCD}=AB\times BF

We replace the data accordingly:

SABCD=9×6=54 S_{ABCD}=9\times6=54

Answer

54 cm²

Exercise #2

The following is a circle enclosed in a parallelogram:

36

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

What is the area of the zones marked in blue?

Video Solution

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

From here we can calculate:

AB=AF+FB=3+6=9 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:2πR 2\pi R
We replace and solve:

2πR=25.13 2\pi R=25.13
πR=12.565 \pi R=12.565
R4 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:

Lado x Altura \text{Lado }x\text{ Altura} 9×872 9\times8\approx72

Now, we calculate the area of the circle according to the formula:πR2 \pi R^2

π42=50.26 \pi4^2=50.26

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

7256.2421.73 72-56.24\approx21.73

Answer

21.73 \approx21.73

Exercise #3

ABCD is a parallelogram
BFCE is a deltoid

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What is the area of the parallelogram ABCD?

Video Solution

Step-by-Step Solution

First, we must remember the formula for the area of a parallelogram:Lado x Altura \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 (

A2+B2=C2 A^2+B^2=C^2 )

BG2+42=52 BG^2+4^2=5^2

BG2+16=25 BG^2+16=25

BG2=9 BG^2=9

BG=3 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 FC=EC=9

Now we can also do Pythagoras in the triangle GCE.

GC2+42=92 GC^2+4^2=9^2

GC2+16=81 GC^2+16=81

GC2=65 GC^2=65

GC=65 GC=\sqrt{65}

Now we can calculate the side BC:

BC=BG+GT=3+6511 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:

HDBG=HCGE \frac{HD}{BG}=\frac{HC}{GE}

HDBG=7.53=2.5 \frac{HD}{BG}=\frac{7.5}{3}=2.5

HCEG=HC4=2.5 \frac{HC}{EG}=\frac{HC}{4}=2.5

HC=10 HC=10

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

10×11110 10\times11\approx110

Answer

110 \approx110

Exercise #4

The parallelogram ABCD is shown below.

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

ECFD is a rhombus whose area is 24 cm².

What is the area of ABCD?

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Video Solution

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×hDC 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×hBC 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.

Answer

It is not possible to calculate.

Exercise #5

Calculate the area of the parallelogram according to the data in the diagram.

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Video Solution

Answer

70

Topics learned in later sections

  1. Area
  2. Parallelogram
  3. Perimeter of a Parallelogram
  4. Perimeter