Education

How to find the area of a Parallelogram?

Math

Math

Have you ever come across a rectangle-shaped geometrical figure which also looks like a quadrilateral and then may even fit a bunch of squares inside it? Well, let us tell you that this figure is a special case of a Quadrilateral which is designed in such a way that it can not only fit a square inside it but also a rectangle and a rhombus! Interesting, isn’t it? Now, let us explore more about this mysterious figure!

Parallelogram

A Parallelogram is a 4-sided closed figure formed by 2 pairs of equal and parallel opposite sides. It is a 2-dimensional special type of Quadrilateral. The sum of the interior angles of a Quadrilateral is 360 degrees. Since the opposite sides of a Parallelogram are parallel, its opposite angles and sides are congruent. Let us understand this by studying the properties of a parallelogram.

Properties of a Parallelogram

  • Opposite sides are congruent.
  • Opposite angles are congruent.
  • The opposite sides of a parallelogram are equal in length.
  • Adjacent angles are supplementary.
  • The sum of the interior angles is 360 degrees.
  • All the angles are a right angle if even one angle is a right angle.
  • The diagonals of a parallelogram bisect with each other.
  • The diagonals of a parallelogram separate each other into two congruent.

Area of a Parallelogram

The area of a Parallelogram is the region filled by the parallelogram. The area of a Parallelogram is equal to the area of a Rectangle. Also, the area of a parallelogram is twice the area of a triangle provided the triangle and the parallelogram are on the same base and between the same parallels.

How to Find the Area of a Parallelogram?

Finding the Area of a Parallelogram will require the measurements of its height/width and base/length. The height of the parallelogram is also known as the altitude of the parallelogram. The area can be easily found by multiplying the height with the length of the parallelogram. 

It should be noted that the parallelogram’s height should be perpendicular to its base on the opposite side. The base of the parallelogram is always its longer side.

Derivation of the Formula

Now that we know how to find a parallelogram area, let’s derive the formula for it.

Let us assume a Parallelogram PQRS with a base as RS and height as PT drawn from the vertices P forming a ∆PTS. To prove the formula for the area of a parallelogram, we will form another ∆QMR whose height is drawn from the vertices Q resulting in a rectangle PQMT.

Both the triangles formed are right-angled triangles where ∆PTS = ∆QMR. (Segment PS = QR and segment PT = QM in parallelogram and rectangle respectively. Hence, applying the SSS theorem justifies that both the triangles are congruent.)

Now that we know that both of our triangles are equal, it is proved that the parallelogram PQRS and the rectangle PQMT have the same area.

As we know, the Area of a rectangle = Width x Length.

Therefore, Area of a parallelogram = Width/Base x Length.

Let us understand more with an example:

Suppose there is a Parallelogram with a and b as its parallel sides and has its height. To find the area of the Parallelogram, we will multiply the base with the height giving,

Area = base x height (sq. unit.), or,

A = b x h (sq. unit)

Now, let us assume that the parallelogram’s base is 35 cm and the height is 11 cm.

We know that,

Area = b x h

Multiplying the base with the height of the parallelogram, we get,

Area = 35 cm x 11 cm

Area = 385 cm²

The area of a Parallelogram is equal to twice the Area of a Triangle.

Let us assume a Parallelogram has 4 sides as AB, BC, CD, and DA, and AC is the diagonal. The height h is drawn through CE perpendicular to the base b, which is side AB forming CE. We have to find the area of the parallelogram.

Given the Diagonal AC, we have 2 Right-angled Triangles ∆ADC and ∆ABC. 

Now, Area of parallelogram = base x height sq. units

                                         = b x h sq. units

                                         = 2 x ½ AB x CE sq. units (Given, AB & CE are b & h)

                                         = 2 x ½ b x h sq. units

                                         = 2 x Area of ∆ABC

Worked-out examples on Area of Parallelogram:

Question 1: Find the area of the parallelogram with a base of 5 cm and a height of 3 cm.

Solution: 

We know that,

Area of Parallelogram = b × h sq. units

Given,

Base, b = 5 cm

Height, h = 3 cm

Area of Parallelogram = 5 cm × 3 cm

                                 = 15 sq.cm

Therefore, the Area of Parallelogram = 15 cm2

Question 2: A parallelogram has sides 30 cm and 15 cm, with a height of 10 cm. What is its area?

Solution:

We know that the base of a parallelogram is always on its longer side.

Therefore, in the given problem,

Base b = 30 cm, and

Height h = 10 cm

Area of parallelogram = b x h

                                = 30 cm x 10 cm

                                = 300 sq. cm

Therefore, the Area of Parallelogram = 300 sq. cm

Question 3: The area of a parallelogram is 400 sq. cm. Its height is twice its base. Find the height and base.

Solution:

Given,

Area of the parallelogram = 400 sq. cm

Height = 2base, or

h = 2b

We know that,

Area of Parallelogram = b x h

Therefore, 400 = b x 2b

                          400 = 2b2

                          b2 = 200

                          b = 14.14 cm

Hence, height h = 2 x b

                        = 2 x 14.14

                        = 28.28 cm

Question 4: A parallelogram has sides of 9 cm and 6 cm. If the distance between its shorter sides is 4 cm, find the distance between its longer side.

Solution:

Given,

Sides of parallelogram = 9 cm and 6 cm

As the sides are not equal, these will be the adjacent sides.

Distance between the shorter sides = 4 cm

Area of parallelogram = b × h

                    = 6 × 4 cm²

                   = 24 cm²

 Now, area of parallelogram = b × h

                                    24 = 9 × h

         h = 24 / 9

                    h = 2.6 cm

Therefore, the distance between its longer side = 2.6 cm. 

Question 5: ABCD is a parallelogram in which AB = 15 cm, BC = 8 cm and the diagonal AC = 17 cm. Find the area of parallelogram ABCD.

Solution:

We know that,

Area of parallelogram = Twice area of a triangle

Now, with the diagonal being AC, we have ∆ABC and ∆ADC.

In ∆ ABC,

AB = 15 cm 

BC = 8 cm 

AC = 17 cm

Applying Heron’s Formula on ∆ABC,

s = ( a + b + c ) / 2, or

s = perimeter / 2

s = (15 + 8 + 17) / 2

s = 40 / 2

s = 20

 Therefore, area of ∆ABC = √ (s (s-a) (s-b) (s-c))

        = √ (20 (20 – 15) (20 – 8) (20 – 17))

                    = √ (20 × 5 × 12 × 3)

                    = √ (5 x 2 x 2 x 5 x 3 x 2 x 2 x 3)

                    = 5 x 2 x 3 x 2

                    = 60 cm²

 Area of parallelogram ABCD = 2 area of ∆ABC

                          = 2 × 60 cm²

                         = 120 cm²

Question 6: A parallelogram PQRS has its base of x+4 cm and a height of 3 cm. The area of the parallelogram is 30 cm2. Find the value of x.

Solution:

Given,

Area of Parallelogram = 30 cm2 

Also, we know that,

Area of Parallelogram = b x h, where,

b = base, and h = height

Therefore, if we substitute the area, base and height with their respective values, we get,

30 = (x+4) x 3 

30 = 3x + 12 

18 = 3x

x = 6 cm

Now, h = 6+4 =10 cm

Therefore, the height of the parallelogram is 10 cm. 

To Summarise

A parallelogram is a figure of 4 sides with two pairs of parallel lines.

The opposite sides and the opposite angles are equal in length and measurement, respectively.

A parallelogram area can be calculated by multiplying its base with its height, i.e., b x h.

The height of the parallelogram must always be perpendicular to its base.

The area of a parallelogram is expressed in square units.

Area of parallelogram = Area of Rectangle.

Area of parallelogram = Twice Area of Triangle.

Now that you have a fair idea of a parallelogram and finding its area, you can easily solve problems associated with it. But if you are still stuck in any parallelogram concerned problems, don’t worry! Build your basics strong with the utmost personalized guidance by Cuemath. Through their visual and application-oriented teaching methods, experts will ensure that you learn math the right way. Math will be memorized on tips because the measurement is now made easy with Cuemath!

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