# Area of Parallelogram

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 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.

Therefore, the Area of Parallelogram = b×h Square units

where,

b is the base and h is the height 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).

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

∠AFD=∠BEC (Corresponding Angles)

∠DAF =∠CBE (Angle Sum Property)

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

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

Area(ABCD)=Area(ABEF)

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($||^{gm}$ ABCD) = AB×AL

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

Proof: Since ($||^{gm}$ABCD) and rectangle ALMB are on the same base and between the same parallels.

∴ Area($||^{gm}$ ABCD) =Area(rect.ALMB)

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

Hence, Area($||^{gm}$ ABCD)= AB×AL

### Solved Problem

Question:

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

Solution:

Given:

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

Learn about the area of parallelograms and some other quadrilaterals in a simple manner with detailed information along with step by step solutions to all questions of this chapter in an NCERT Books only at BYJU’S.

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A line segment is a part of the line that has a fixed length