Here is a question I came across

> 
> Find the area of rhombus ABCD given that the radii of the circles
>  circumscribed around triangles ABD and ACD are 12.5 and 25,
>  respectively.
> 
> 
> 

With the diagonals of the rhombus perpendicularly bisecting each other, let $a$ be half of diagonal BD and $b$ half of diagonal AC. Thus, the sides of the rhombus are $\sqrt{a^2 + b^2}$ long. 

The area of any triangle can be expressed as $\frac{\text{product of side lengths}}{\text{4 multiplied by the circumradius}}$. Thus, the area $\Delta ABC$ is $\frac{2b (a^2 + b^2)}{4(12.5)}$, while the area of $\Delta ABD$ is $\frac{2a (a^2 + b^2)}{4(25)}$. These two areas are equal since both of the said triangles are half of the rhombus. Setting these two expressions equal to each other and simplifying gives $b = 2a$. 

I've been stuck here for the past few hours. I'm blatantly missing something here. What's the next step?