The length of one side of a rhombus is 41 cm and its area is 720 $$cm^2$$. What is the sum of the lengths of its diagonals?

Let A and B be the 2 diagonals  of the rhombus.

$$Area of rhombus = Product of diagonals  \diagup 2$$

720 = P.Q \diagup 2

\Rightarrow P.Q =1440

using Pythagorean theorem  

$$\left(\begin{array}{c}P\\ 2\end{array}\right)^2  + \left(\begin{array}{c}Q\\ 2\end{array}\right)^2 = 41^{2}$$

$$p^{2} + Q^{2} = 4 \star 1681 =6724$$

$$Using perfect square formula P+Q^{2} = P^{2} + Q^{2} + 2PQ$$

$$P+Q^{2} = 6784 + 2 \star 1440$$

$$P+Q = \sqrt{9604}$$

P+q = 98

 Explanation not proper