Area of Parallelogram

May 17, 2023

Table of Contents

A parallelogram is a special type of quadrilateral that is formed by parallel lines. In a parallelogram, both pairs of opposite sides are parallel and equal. Like other 2D shapes, there are two types of measurements associated with parallelogram – perimeter and area.

Let’s understand what is area of parallelogram and what are different formulas to find the area of a parallelogram with examples.

The area of a parallelogram is the region or space covered by a parallelogram in a 2D plane and it refers to the total number of unit squares that can fit into it. It is measured in square units such as $\text{cm}^2$, $\text{m}^2$, $\text{ft}^2$, $\text{in}^2$, etc. Base (one of the sides) and height (also called altitude) are used to calculate the area of a parallelogram.

Area of Parallelogram Formula

The area of the parallelogram can be calculated using different formulas using either the sides or the diagonals. Some of the most commonly used formulas are

Using Side and Height
Using Diagonals
Using Sides

Area of Parallelogram Using Side and Height

Consider a parallelogram $\text{ABCD}$, such that the adjacent sides are $a$ and $b$ and $h$ is the height (perpendicular distance of side $b$ from the opposite vertex), then the area of a parallelogram is given by the formula $\text{Area} = b \times h$.

Examples on Area of Parallelogram Using Side and Height

Example 1: Find the area of the parallelogram with a base of $5$ cm and height of $7$ cm.

Base of parallelogram $b = 5$ cm

Height of parallelogram $h = 7$ cm

Area of a parallelogram = $b \times h = 5 \times 7 = 35 \text{ cm}^2$.

Example 2: Find the area of a parallelogram whose breadth is $8$ cm and height is $11$ cm.

Breadth (base) of parallelogram $b = 8$ cm

Height of parallelogram $h = 11$ cm

Area of a parallelogram = $b \times h = 8 \times 11 = 88 \text{ cm}^2$.

Example 3: The base of the parallelogram is thrice its height. If the area is $192 \text{ cm}^2$, find the base and height.

Let height of parallelogram = $x$ cm

Therefore base of parallelogram = $3x$ cm

Area of a parallelogram = $\text{Base} \times \text{Height}$

Therefore $3x \times x = 192 => 3x^2 = 192$

$ => x^2 = \frac{192}{3} => x^2 = 64$

Taking the square root of both sides

$x = \sqrt{64} => x = 8$

Base of parallelogram = $3 \times 8 = 24$ cm

And height of parallelogram = $8$ cm

Example 4: The area of a parallelogram is $1250$ sq. cm. Its height is twice its base. Find the height and base.

Let base of parallelogram = $x$ cm

Therefore height of parallelogram = $2x$ cm

Area of a parallelogram = $\text{Base} \times \text{Height}$

Therefore $x \times 2x = 1250 => 2x^2 = 1250$

$=> x^2 = \frac{1250}{2} => x^2 = 625$

Taking the square root of both sides

$x = \sqrt{625} => x = 25$

Therefore height of parallelogram = $2 \times 25 = 50$ cm

And base of parallelogram = $25$ cm

Example 5: The area of a playground which is in the shape of a parallelogram is $2500 \text{ ft}^2$, with one side measuring $250$ ft. Find the corresponding altitude.

Area of a parallelogram-shaped playground = $2500 \text{ ft}^2$

Length of one side $b = 250$ ft

Let the height of the corresponding altitude = $h$

Area of a parallelogram = $bh$

Therefore $250 \times h = 2500 => h = \frac{2500}{250}$

$=> h = 10$ ft

Therefore altitude (height) = $10$ ft.

Area of Parallelogram Using Diagonals

Consider a parallelogram $\text{ABCD}$, such that the adjacent sides are $a$ and $b$ and $\text{AC}$ and $\text{BD}$ are the diagonals intersecting at $\text{O}$ making an angle $\alpha$. If the length of the diagonals is $\text{d}_1$ and $\text{d}_2$, then area of a parallelogram is given by the formula $\text{Area} = \frac{1}{2} d_1 d_2 \sin \alpha$.

Examples Area of Parallelogram Using Diagonals

Example 1: Find the area of a parallelogram whose diagonals measure $8$ cm and $6$ cm and the angle between them is $30^{\circ}$.

Length of first diagonal of a parallelogram $\text{d}_1 = 8$ cm

Length of second diagonal of a parallelogram $\text{d}_2 = 6$ cm

Angle between the diagonals $\alpha = 30^{\circ}$

Area of a parallelogram = $\frac{1}{2} \text{d}_1 \text{d}_2 \sin \alpha$

$= \frac{1}{2} \times 8 \times 6 \times \sin 30^{\circ}$

$= 24 \times \frac{1}{2} = 12 \text{cm}^2$

Example 2: Area of a parallelogram whose diagonals are $5$ cm and $7$ cm respectively is $\frac{35 \sqrt{3}}{4} \text{ cm}^2$. Find the angle between the diagonals.

Area of a parallelogram = $\frac{1}{2} d_1 d_2 \sin \alpha$

Here $d_1 = 5$ cm, $d_2 = 7$ cm, and Area = $\frac{35 \sqrt{3}}{4} \text{ cm}^2$

Therefore $\frac{1}{2} \times 5 \times 7 \times \sin \alpha = \frac{35 \sqrt{3}}{4} \text{ cm}^2$

$=> \frac{1}{2} \times 35 \times \sin \alpha = \frac{35 \sqrt{3}}{4}$

$=> \frac{1}{2} \times \sin \alpha = \frac{\sqrt{3}}{4}$

$=> \sin \alpha = \frac{\sqrt{3}}{2}$

$=> \alpha = 60^{\circ}$

Thus angle between the diagonals = $60^{\circ}$.

Area of Parallelogram Using Sides

Consider a parallelogram $\text{ABCD}$, such that the adjacent sides are $a$ and $b$. Further, if $\alpha$ is the angle between the sides $a$ and $b$, then the area of a parallelogram is given by the formula $\text{Area} = a b \sin \alpha$.

Examples on Area of Parallelogram Using Sides

Example 1: Find the area of a parallelogram whose adjacent sides are $10$ cm and $8$ cm and the angle between them is $30^{\circ}$.

Area of a parallelogram = $a b \sin \alpha$

Here $a = 10$ cm, $b = 8$ cm, and $\alpha = 30^{\circ}$

Area of a parallelogram = $10 \times 8 \times \sin 30^{\circ}$

$= 10 \times 8 \times \frac{1}{2} = 40 \text{ cm}^2$.

Practice Problems

Find the area of a parallelogram whose adjacent sides are
- $a = 9$ cm and $b = 12$ cm
- $a = 16$ in and $b = 20$ in
Find the area of a parallelogram whose two diagonals and the angle between them is
- $d_1 = 4$ cm, $d_2 = 6$ cm and $\alpha = 45^{\circ}$
- $d_1 = 10$ cm, $d_2 = 14$ cm and $\alpha = 90^{\circ}$
Find the area of a parallelogram whose two sides and the angle between the sides is
- $a = 12$ cm, $b = 18$ cm, $\alpha = 90^{\circ}$
- $a = 8$ cm, $b = 14$ cm, $\alpha = 60^{\circ}$

FAQs

What is the area of a parallelogram?

The area of a parallelogram is defined as the region enclosed by it in two-dimensional space. It is measured in square units such as $\text{cm}^2$, $\text{m}^2$, $\text{in}^2$, etc.

What is the area formula for parallelograms?

There are three commonly used formulas to find the area of a parallelogram. These are
a. $b \times h$, where $b$ is the length of one of the sides and $h$ is its perpendicular distance from the opposite vertex
b. $\frac{1}{2} \text{d}_1 \text{d}_2 \sin \alpha$, where $\text{d}_1$, $\text{d}_2$ are the diagonals and $\alpha$ is the angle between them
c. $a b \sin \alpha$, where $a$, $b$ are the adjacent sides and $\alpha$ is the angle between them

What is the area of a parallelogram when diagonals are given?

The formula used to find the area of a parallelogram when diagonals are given is $\frac{1}{2} \times \text{d}_1 \times $\text{d}_2 \times \sin \alpha$, where $\text{d}_1$ and $\text{d}_2$ are lengths of diagonals of the parallelogram, and $\alpha$ is the angle between them.

What is the perimeter of a parallelogram?

To find the perimeter of a parallelogram, add all the sides together. The formula used to find the perimeter of a parallelogram is $2 (a + b)$, where $a$, and $b$ are the adjacent sides of a parallelogram.

Conclusion

A parallelogram is a special type of quadrilateral that is formed by parallel lines. The three most commonly used formulas for finding the area of a parallelogram are

$b \times h$, where $b$ is length of one of the sides and $h$ is its perpendicular distance from the opposite vertex
$\frac{1}{2} \text{d}_1 \text{d}_2 \sin \alpha$, where $\text{d}_1$, $\text{d}_2$ are the diagonals and $\alpha$ is the angle between them
$a b \sin \alpha$, where $a$, $b$ are the adjacent sides and $\alpha$ is the angle between them

Area of Parallelogram – Formulas & Examples