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


parallelogram-01

\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 .


parallelogram-02

\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 = 17cm, the other side c = 20cm, 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.