Area Of Rhombus

Theorem: Area of a rhombus is equal to one half of the product of diagonals.

Prerequisites:
Properties of rhombus (proof)
Area of right triangle (proof)

Proof:

Let ABCD be a rhombus having diagonals AC and BD of lengths $d_1$ and $d_2$ respectively, intersecting at a point E.
Since in a rhombus, diagonals bisect each other at right angles. Hence,

$\qquad\quad\; AE = CE = \dfrac{d_1}{2}\\
\qquad\quad BE = DE = \dfrac{d_2}{2}$

Now consider $\triangle ABE$.  Since, $\angle AEB$ is a right angle, hence using the formula for area of a right triangle, area of $\triangle ABE$ is equal to:

$\qquad\quad\begin{align} ar (\triangle ABE) &= \dfrac{1}{2}\dfrac{d_1}{2}\dfrac{d_2}{2}\\[4pt]
&= \dfrac{d_1 d_2}{8}\end{align}$

Here $ar$ stands for area. Similarly, area of other three triangles, i.e., $\triangle BCE, \triangle CDE$ and $\triangle DAE = \dfrac{d_1 d_2}{8}$.

$\begin{align}\qquad\quad\text{Area of ABCD} &= ar (\triangle ABE) + ar (\triangle BCE) + ar (\triangle CDE) + ar (\triangle DAE)\\
&= 4\times\dfrac{d_1 d_2}{8}\\
&= \dfrac{d_1 d_2}{2}\end{align}$

Q.E.D.


Recommended:
Area of triangle
Area of rectangle
Properties of rhombus