If the length of a side of a rhombus is 36 cm and the area of the rhombus is 396 sq. cm, then the absolute value of the difference between the lengths, in cm, of the diagonals of the rhombus is

Let the diagonals of the rhombus be $$d_1$$ and $$d_2$$

We know that the $$\text{Area} = \frac{1}{2} d_1 d_2 = 396 \implies d_1 d_2 = 792$$

For a rhombus with side a=36, the diagonals intersect at right angle. Giving a right-angle triangle with the side as hypotenuse.

$$\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = a^2 \implies \frac{d_1^2 + d_2^2}{4} = 36^2 = 1296$$

$$d_1^2 + d_2^2 = 5184$$

We want $$|d_1 - d_2|$$ 

$$(d_1 - d_2)^2 = d_1^2 + d_2^2 - 2d_1 d_2 = 5184 - 2 \cdot 792 = 5184 - 1584 = 3600$$

$$|d_1 - d_2| = \sqrt{3600} = 60$$