Question 9.

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  ($\sqrt{37} - \sqrt{13}$)