# The Area of a Rhombus

A rhombus is a quadrilateral having all sides with unequal diagonals which bisect each other.

Note: If a square is pressed from two opposite corners, a rhombus is formed.

Let $ABCD$ be a rhombus, then its area can be evaluated in two ways.

(1) When one side and the included angle is given:

Let one side be equal to $a$ with includes angle $\theta$. Since the diagonal $AC$ divides the rhombus into two equal triangles, therefore

The area of the rhombus $= 2 \times {\text{area of }}\Delta ABC$.
The area of the rhombus $= 2\left( {\frac{1}{2}a \times a \times \sin \theta } \right) = 2\left( {\frac{1}{2}{a^2}} \right)\sin \theta = {a^2}\sin \theta$.
The area of the rhombus $= {({\text{one side}})^2}\sin \theta$.

(2) When two diagonal are given:

Let $d$ and $d'$ be the length of the diagonals $AC$ and $BD$ respectively, and since the rhombus is divided into four equal triangles, therefore

The area of the rhombus $= 4 \times \frac{1}{2} \times \frac{{BD}}{2} \times \frac{{CA}}{2} = 4 \times \frac{{BD \times CA}}{8} = \frac{{BD \times CA}}{2}$.

The area of the rhombus $= \frac{{AC \times BD}}{2} = \frac{{d \times d'}}{2}$.

The area of the rhombus $= \frac{{{\text{Product of two diagonals}}}}{2}$.

Example:

The length of each side of a rhombus is $120$ cm and two of its opposite angles are $60^\circ$ each. Find the area.

Solution:

Given that each side $a = 120$ and $\theta = 60^\circ$,
the area of the rhombus $= {({\text{side}})^2} \times \sin \theta$.
The area of the rhombus $= 120 \times 120 \times \sin 60^\circ = 120 \times 120 \times 0.866 = 12470.4$ square cm.

Example:

The diagonals of a rhombus are $40$ m and $30$ m. Find its area.

Solution:

Given that the diagonals are $d = 40$ m and $d' = 30$ m,

the area of the rhombus $= \frac{{d \times d'}}{2} = \frac{{40 \times 30}}{2} = 600$ square m.