# Area of a Parallelogram

A parallelogram is a quadrilateral whose opposite sides are 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, $h = DL$

$\therefore$ Area of the parallelogram $ABCD$ $=$ Area of the rectangle $DLMC$
Area of the parallelogram $= LM \times DL$
Area of the parallelogram $= AB \times h$ (as $LM = AB$)
Area of the parallelogram $= b \times h$
$\therefore$ Area of parallelogram $=$ base $\times$ height

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

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

$\therefore$ Area of parallelogram $= AB \times DL = b \times h$
But $\frac{h}{{AD}} = \sin \theta$
Or $h = AD\sin \theta$
Or $h = c\sin \theta$
$\therefore$ Area of $ABCD = h \times c\sin \theta$
$\therefore$ Area of Parallelogram $=$ Product of adjacent sides$\times \sin \theta$

Example:

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

Solution:
Since, area $= 256$square cm
height $= 32$cm
$\therefore$ Area of parallelogram $=$ base $\times$ height
$\therefore$ $256 =$base $\times 32$
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, other side, $c = 20$cm, $\theta = {60^ \circ }$
$\therefore$ Area $= bc\sin \theta$
$= 17 \times 20 \times \sin {60^ \circ }$
$= 340 \times 0.866$
$= 294.44$square cm