Area of Parallelogram

Area of Parallelogram is the region bounded by the parallelogram in a given two-dimension space. A parallelogram is a special type of quadrilateral in which it has four sides.  Both pair of opposite sides are parallel.  In a parallelogram, the opposite sides are of equal length and opposite angles are of equal measures. Since the rectangle and the parallelogram have similar properties, the area of the rectangle is equal to the area of a parallelogram.

Area of Parallelogram

b × h 
Perimeter of Parallelogram 

2 (a+b)

Area of a Parallelogram

Area of Parallelogram Formula

To find the area of the parallelogram, multiply the base by the height. It should be noted that the base and the height of the parallelogram are perpendicular to each other whereas the lateral side of the parallelogram is not perpendicular to the base.

area of parallelogram formula


Area = b × h Square units 

 where b is the base and h is the height of the parallelogram.

Perimeter of a Parallelogram

Perimeter of any shape is the total distance of the covered around the shape or its the total length of any shape. Similarly, the perimeter of a parallelogram is the total distance of the boundaries of the parallelogram. To calculate the perimeter value we have to know the values of its length and breadth. The parallelogram has its opposite sides equal in length. Therefore, the formula of the perimeter could be written as;


Perimeter = 2 (a+b) 

Where a and b are the length of the equal sides of the parallelogram.

Properties of Parallelogram

Some of the properties of a parallelogram are as follows:

  • Opposite sides are equal. AB = DC
  • Opposite angles are equal. A = C
  • The diagonals of the parallelogram bisect each other
  • Each diagonal bisects the parallelogram into two congruent triangles
  • Consecutive angles are supplementary. i.e ∠A + ∠D = 180°

Parallelogram Theorems

Theorem: Parallelograms on the same base and between the same parallel sides are equal in area.

Proof: Two parallelograms ABCD and ABEF, on the same base DC and between the same parallel line AB and FC.

To prove that area (ABCD) = area (ABEF).

Parallelogram Theorems

In ∆ADF and ∆BCE,

AD=BC (∴ABCD is a parallelogram ∴ AD=BC)

AF=BE (∴ABEF is a parallelogram ∴AF=BE)

∠ADF=∠BCE (Corresponding Angles)

∠AFD=∠BEC (Corresponding Angles)

∠DAF =∠CBE (Angle Sum Property)

∆ADE ≅ ∆BCF (From SAS-rule)

Area(ADF) = Area(BCE) (By congruence area axiom)

Area(ABCD)=Area(ABED) + Area(BCE)



Hence, the area of parallelograms on the same base and between the same parallel sides is equal.

Corollary: A parallelogram and a rectangle on the same base and between the same parallels are equal in area.

Proof: Since a rectangle is also a parallelogram so, the result is a direct consequence of the above theorem.

Theorem: The area of a parallelogram is the product of its base and the corresponding altitude.

Given: In a parallelogram ABCD, AB is the base.

To prove that Area(||gmABCD) = AB×AL

Construction: Complete the rectangle ALMB by Drawing BM perpendicular to CD.

Triangles in Parallelogram

Proof: Since (||gmABCD) and rectangle ALMB are on the same base and between the same parallels.

∴ Area(||gmABCD) =Area(rect.ALMB)

= AB×AL (Area of rectangle= base ×X height)

Hence, Area(||gm ABCD)= AB×AL

Solved Example


Find the area of the parallelogram with the base of 4 cm and height of 5 cm?



Base, b = 4 cm

h = 5 cm

We know that,

Area of Parallelogram = b×h Square units

= 4 × 5 = 20

Therefore, the area of a parallelogram = 20 cm2

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