Area of a rhombus $ABCD$ is a rhombus. We are given the the circumradius of triangles $ABD$ and $ACD$. So how do we compute the area and the side and area of the rhombus? I have tried some properties of the circumcenter, but have failed. If there was a geometric solution, I would like to prefer it.    
 A: Hint: The area of a triangle is equal to $\frac{abc}{4R}$.
This gives you the ratio of the diagonals of the rhombus, which are perpendicular to each other.
This determines the rhombus, up to a scaling factor.

Seeing that you're really young, let me expand on the above.
Let the circumradius of $ABD$ be $R_1$ and the circumradius of $ACD$ be $R_2$.
Since $AD \parallel BC$, hence $[ABD] = [ACD]$. Applying the formula for area given circumradius, we get that
$$ \frac{AC}{BD} = \frac{R_2}{R_1}$$
Now, consider an isosceles triangle $XYZ$ with $XY=YZ$, $ZX = R_1$ and height of $\frac{R_2}{2}$. It has area $\frac{R_1 R_2}{4}$, and hence circumradius of $\frac{ R_1 ( R_1^2 + R_2^2)}{4 \times 2 \times \frac{R_1 R_2}{4}} = \frac{R_1^2 + R_2^2}{2 R_2}$. Hence, in order for it to scale and have a circumradius of $R_1$, we need to scale it by $\frac{ 2 R_1R_2}{R_1^2 + R_2^2}$.
This tells us that $AB = XY \times \frac{ 2 R_1R_2}{R_1^2 + R_2^2}  = \frac{R_1 R_2 } { \sqrt{R_1^2 + R_2^2}}$, and that the area of the rhombus is $2[XYZ] \times \left( \frac{ 2 R_1R_2}{R_1^2 + R_2^2} \right)^2  = \frac{ 2 R_1^3 R_2^3 } { (R_1^2 + R_2^2)^2} $.
