Pretty Rhombic

Let, AC and BD are intersect at O. So, area of right triangle AOB is= 384/4=96. While AC:BD = 3:4 then AO:BO=3:4, AO= $\frac{3BO}{4}$

Now from right triangle AOB,

$\frac{1}{2} \times BO \times AO=96$

$\frac{1}{2} \times BO \times \frac{3 \times BO}{4} = 96$

So, BO=16 and then AO=12

so, side $AB=\sqrt{AO^2+BO^2}$

$AB=\sqrt{12^2+16^2}$

$AB=\sqrt{144+256}$

$AB=\sqrt{400}$

$AB= 20$