Longer diagonal of a rhombus

Let the diagonals be $d_1$ and $d_2$. Area = $\frac{1}{2} d_1 d_2 = 12 \implies d_1 d_2 = 24$.

Side length $s = 5 \implies \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = 5^2 = 25$. $d_1^2 + d_2^2 = 100$

Now: $(d_1 + d_2)^2 = d_1^2 + d_2^2 + 2 d_1 d_2 = 100 + 48 = 148 \implies d_1 + d_2 = \sqrt{148} = 2\sqrt{37}$ $(d_1 - d_2)^2 = d_1^2 + d_2^2 - 2 d_1 d_2 = 100 - 48 = 52 \implies d_1 - d_2 = \sqrt{52} = 2\sqrt{13}$

Adding both equations: $2 d_1 = 2\sqrt{37} + 2\sqrt{13} \implies d_1 = \sqrt{37} + \sqrt{13}$.