# The Area of a Parallelogram

A parallelogram is a quadrilateral whose opposite sides are of equal length and parallel. The diagonals of a parallelogram are unequal and bisect each other. If $ABCD$ is a parallelogram, then its area can be calculated in two ways.

(a) When the base and height are given:

Since $ABCD$ is a parallelogram with base $AB = b$ and height, $h = DL$

$\therefore$ the area of the parallelogram $ABCD$ $=$ the area of the rectangle $DLMC$.

The area of the parallelogram $= LM \times DL$.
The area of the parallelogram $= AB \times h$ (as $LM = AB$).
The area of the parallelogram $= b \times h$.

$\therefore$ the area of the parallelogram $=$ base $\times$ height.

(b) When the adjacent sides and their included angle is given:

Since $ABCD$ is a parallelogram, take $AB$ and $CD$ as its two adjacent sides with the included angle $\theta$.

$\therefore$ the area of the parallelogram $= AB \times DL = b \times h$
But $\frac{h}{{AD}} = \sin \theta$
Or $h = AD\sin \theta$
Or $h = c\sin \theta$.

$\therefore$ the area of $ABCD = h \times c\sin \theta$
$\therefore$ the area of the parallelogram $=$ the product of the adjacent sides$\times \sin \theta$

Example:

Find the base of a parallelogram whose area is $256$ square cm and height $32$ cm.

Solution:
Since the area $= 256$ square cm
The height $= 32$ cm
$\therefore$ the area of the parallelogram $=$ base $\times$ height.
$\therefore$ $256 =$base $\times 32$.

The base $= \frac{{256}}{{32}} = 8$ cm.

Example:

Find the area of a parallelogram, two adjacent sides of which are $17$ cm and $20$ cm and their included angle is${60^ \circ }$.

Solution:
Here, one side $b = 17$cm, the other side $c = 20$cm, and $\theta = {60^ \circ }$

$\therefore$ the area $= bc\sin \theta$
$= 17 \times 20 \times \sin {60^ \circ }$
$= 340 \times 0.866$
$= 294.44$ square cm.