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.