jmc

The diagonals of the rhombus perpendicularly bisect each other. Call half of diagonal BD $a$ and half of diagonal AC $b$. The length of the four sides of the rhombus is $\sqrt{a^2+b^2}$. The area of any triangle can be expressed as $\frac{a\cdot b\cdot c}{4R}$, where $a$, $b$, and $c$ are the sides and $R$ is the circumradius. Thus, the area of $△ABD$ is $ab=2a(a^2+b^2)/(4\cdot 12.5)$. Also, the area of $△ABC$ is $ab=2b(a^2+b^2)/(4\cdot 25)$. Setting these two expressions equal to each other and simplifying gives $b=2a$. Substitution yields $a=10$ and $b=20$, so the area of the rhombus is $20\cdot 40/2=\boxed{400}$.