Area of Parallelogram (Definition, Formulas & Examples)

A variety of quadrilaterals created by parallel lines is called a parallelogram. In a parallelogram, the opposite sides must be parallel, however, the angle between them may vary. Opposite sides of a parallelogram must be parallel and congruent. Consequently, a quadrilateral in which both pairs of opposite sides are parallel and equal is referred to as a parallelogram. There are three basic categories of parallelograms: square, rectangle, and rhombus. The region or space a parallelogram occupies in a two-dimensional plane is known as the area of a parallelogram. The total number of unit squares that can fit inside a parallelogram is its area, which is measured in square units.

Area Of A Parallelogram Using Base And Height

Consider a parallelogram with ‘b’ as the length of the set of one of its parallel sides and height equal to ‘h’.

Using the base length ‘b’ and height ‘h’, the area of the parallelogram is given by the following formula:

Area = base ⨯ height

A = bh

Area Of A Parallelogram Using Side Lengths

Consider a parallelogram whose height is not known but the length of its sides is given. Let the lengths of the set of parallel sides of the parallelogram be equal to “a” and “b”. Let the measurement of the angle between adjacent sides of the parallelogram be equal to \alpha .

The area of the parallelogram can be calculated using concepts of trigonometry.

We know, the area of a parallelogram is equal to base times height.

Using trigonometry, the height can be written as:

Height = b\sin\alpha

Hence, the area using the side lengths is given by the following formula:

A=ab\sin\alpha

It can be noted that if the angle between the adjacent sides of a parallelogram is equal to 90^{\circ} then, it becomes a rectangle.

Since, \sin90^{\circ}=1

The area of the parallelogram which is a rectangle becomes equal to ab.

Area Of A Parallelogram Using Diagonal Lengths

Let ABCD be a parallelogram with the diagonal lengths BD equal to d_1 and AC equal to d_2 . Suppose the angle between these intersecting diagonals is equal to \theta .

The area of the parallelogram with the given specifications is calculated using the following formula:

A = \frac{1}{2} d_1 d_2 \sin\theta

Area Of A Parallelogram In Vector Form

If the sides of a parallelogram are given in vector form then different formulas can be used to determine the area of a parallelogram. Take into consideration a parallelogram with vector ‘a’ and vector ‘b’ as its two sides. Let us say ABCD is a parallelogram with AB and DC equal to vector ‘a’ and AD and BC equal to vector ‘b’. Then,

Area = Mod of cross-product of vector a and vector b

A=|a \times b|

Let the diagonal vectors AC equal to vector d_1 and BD equal to vector d_2 . Then,

\begin{aligned}

& \vec{a}+\vec{b}=\overrightarrow{d_1} \\

& \vec{b}-\vec{a}=\overrightarrow{d_2} \\

& \Rightarrow \overrightarrow{d_1} \times \overrightarrow{d_2}=(\vec{a}+\vec{b}) \times(\vec{b}-\vec{a}) \\

& =\vec{a} \times(\vec{b}-\vec{a})+\vec{b}(\vec{b}-\vec{a}) \\

& =\vec{a} \times \vec{b}-\vec{a} \times \vec{a}+\vec{b} \times \vec{b}-\vec{b} \times \vec{a} \\

& \because \vec{a} \times \vec{a}=0, \vec{b} \times \vec{b}=0 \\

& =\vec{a} \times \vec{b}-0+0-\vec{b} \times \vec{a} \\

& \because \vec{a} \times \vec{b}=-\vec{b} \times \vec{a} \\

& \overrightarrow{d_1} \times \overrightarrow{d_2}=\vec{a} \times \vec{b}+\vec{a} \times \vec{b} \\

& =2(\vec{a} \times \vec{b})=2 A \\

& A=\frac{1}{2}\left|\overrightarrow{d_1} \times \overrightarrow{d_2}\right|

\end{aligned}

Consequently, the area of the parallelogram when diagonals are given in vector form is equal to half the mod of the vector product of the diagonals.

Examples

1. Given a parallelogram with a base equal to 4m and height equal to 7m, find the area of the parallelogram.

Solution: Given base = 4m and height = 7m

Area = base ⨯ height

Area = 4m ⨯ 7m = 28 sq. m

2. If the angle between adjacent sides of a parallelogram is 30 degrees and the length of two parallel sides is 6cm and 5cm respectively, then find the area of the parallelogram.

Solution: The area of a parallelogram using sides is given by:

A=ab\sin\alpha

Given, a = 6cm and b = 5cm

\sin\alpha = \sin30^{\circ} = 0.5

Area = 6 ⨯ 5 ⨯ 0.5 = 15 sq. cm

3. If the base of a parallelogram is twice its height and the area of the parallelogram is equal to 32 sq. units, then find the base and height of the parallelogram.

Solution: Let the height = h units

The base will be = 2h units

Given, area = 32 sq. units

This gives,

Base ⨯ height = 32

h ⨯ 2h = 32

h ⨯ h = 16

Hence, h = 4 units

So, the height is equal to 4 units and the base is equal to twice the height which is 8 units.