# Area of Rhombus Shortcut

An edge of a Math enthusiasts than the normal the people is they can create their own formula. Last year while we were reviewing for elimination round of MMC, I came across with this question, “The diagonal of a rhombus differ by 4. If its perimeter is 40, find its area”.

This problem could be easy for everyone but not for beginners, let me show how to derive a shortcut formula for this problem.

Before we do that, we need first to know some important properties of Rhombus that will help us along the way.

• Each side of a rhombus is equal
• The intersection of the diagonals of rhombus forms a right angle

We will base our derivation using the figure below

Let $ABCD$ be a rhombus with side $s$ and diagonals $d_1$  and  $d_2$. The diagonals intersect at point $O$.

Objective: Find a shortcut formula to solve for the area of the rhombus given a side s and the difference of the length of diagonals.

Area of Rhombus:

$A=\displaystyle\frac{d_1d_2}{2}$

Rearranging

$2A=d_1d_2$  (i)

Observe in the figure that $\triangle AOD$ is a right triangle. By Pythagorean theorem, we have the following relations.

$(\displaystyle\frac{d_1}{2})^2+(\displaystyle\frac{d_2}{2})^2=s^2$

$\displaystyle\frac{d_1^2}{4}+\displaystyle\frac{d_2^2}{4}=s^2$

$d_1^2+d_2^2=4s^2$

Consider the identity $(d_1-d_2)^2=d_1^2-2d_1d_2+d_2^2$.

This can be written in $d_1^2+d_2^2=(d_1-d_2)^2+2d_1d_2$

By substitution we have,

$d_1^2+d_2^2=4s^2$

$(d_1-d_2)^2+2d_1d_2=4s^2$

From (i) we have $2A=d_1d_2$

By another substitution,

$(d_1-d_2)^2+2(2A)=4s^2$

$4A=4s^2-(d_1-d_2)^2$

$\boxed{A=s^2-(\displaystyle\frac{d_1-d_2}{2})^2}$

What about if the given is the sum of the lengths of diagonal?

Then we have to consider the following identity

$(d_1+d_2)^2=d_1^2+2d_1d_2+d_2^2$

Do the same derivation technique until you get the following formula.

$\boxed{A=(\displaystyle\frac{d_1+d_2}{2})^2-s^2}$

Going back to the problem posted above, using the derived formula the answer must be 96 sq. units.