Solution:

Let $s$ be the side length and $\alpha = \angle BAC$ (the diagonal $AC$ bisects angle $A$). The diagonals then have lengths $AC = 2s\cos\alpha$ and $BD = 2s\sin\alpha.$ In triangle $ABD,$ side $BD$ subtends the angle $\angle BAD = 2\alpha,$ so the extended law of sines gives $$12.5 = R_1 = \frac{BD}{2\sin 2\alpha} = \frac{2s\sin\alpha}{4\sin\alpha\cos\alpha} = \frac{s}{2\cos\alpha}.$$ In triangle $ACD,$ side $AC$ subtends $\angle ADC = 180^\circ - 2\alpha,$ so similarly $25 = R_2 = \frac{s}{2\sin\alpha}.$

Dividing, $\tan\alpha = \frac{R_1}{R_2} = \frac{1}{2},$ so $\sin\alpha = \frac{1}{\sqrt{5}}$ and $\cos\alpha = \frac{2}{\sqrt{5}}.$ Then $s = 2R_2\sin\alpha = \frac{50}{\sqrt{5}} = 10\sqrt{5}.$

The area is half the product of the diagonals: $\frac{1}{2} \cdot 2s\cos\alpha \cdot 2s\sin\alpha = 2s^2\sin\alpha\cos\alpha = 2 \cdot 500 \cdot \frac{2}{5} = 400.$