All the sides of the rhombus are equal.
The area of a rhombus is 12 $cm^2$
Considering d1 to be the length of the longer diagonal, d2 to be the length of the shorter diagonal.
The area of a rhombus is $ \frac{1}{2} (d1)(d2) = 12 $.
So $d1 \cdot d2 = 24$.
The length of the side of a rhombus is given by
$ \frac{\sqrt{d1^2 + d2^2}}{2} $.
This is because the diagonals and a side form a right triangle with sides $d1/2$, $d2/2$, and the side length.
$\frac{\sqrt{d1^2 + d2^2}}{2} = 5$
Hence $\sqrt{d1^2 + d2^2} = 10$
So
$d1^2 + d2^2 = 100$
Using $d1 \cdot d2 = 24$, we have $2 \cdot d1 \cdot d2 = 48$.
So two equations:
$d1^2 + d2^2 + 2 d1 d2 = 100 + 48 = 148$
$d1^2 + d2^2 - 2 d1 d2 = 100 - 48 = 52$
So:
$d1 + d2 = \sqrt{148}$ (1)
$d1 - d2 = \sqrt{52}$ (2)
(1) + (2)= 2*(d1) = 2*($\sqrt{37} + \sqrt{13}$)
d1 = $\sqrt{37} + \sqrt{13}$
or
In a rhombus the area of a Rhombus is given by :
The diagonals perpendicularly bisect each other. Considering the length of the diagonal to be $2a, 2b$.
The area of a Rhombus is:
$ \left( \frac{1}{2} \right) (2a)(2b) = 12 $
So:
$ab = 6$
The length of each side is:
$\sqrt{a^2 + b^2} = 5$, so
$a^2 + b^2 = 25$
$(a + b)^2 = 37$, so
$a + b = \sqrt{37}$
$(a - b)^2 = 13$, so
$a - b = \sqrt{13}$
Now solving:
$2a = (\sqrt{37} + \sqrt{13})$
$2b = (\sqrt{37} - \sqrt{13})$
2a is longer diagonal which is equal to ($\sqrt{37} - \sqrt{13}$)
