RHOM

As the diagonals of a rhombus bisects each other at 90 degrees, so Let the diagonals be 2x & 2y respectively. Now, since 2x + 2y = 20, or x + y = 10.

again, x^2 + y^2 = 81 ( since x,y, 9 forms a right angled triangle with hypotenuse '9' which is also the side of the rhombus) but, (x + y)^2 - (x^2 + y^2) = 2xy = 10^2 - 81 = 19, which is the equal to the area of the rhombus i.e 1/2(4xy).

If the diagonals are $e$ and $f$ , the given information is describe by the equations

$\begin{cases} e+f=20 \\ \sqrt{\left( \frac e2 \right)^2 + \left( \frac e2 \right)^2} = 9 \end{cases}$

These equations have the symmetric solutions

$(e,f) = (10-\sqrt{62}, 10+\sqrt{62}) \lor (10+\sqrt{62}, 10-\sqrt{62})$

Since a rhombus consists of four right triangles with the diagonals as their base and height, the total area is given by

$A_{RHOM} = 4 \cdot \frac 12 \cdot \left( \frac e2 \right) \cdot \left( \frac f2 \right) = \frac 12 ef = \frac 12 \cdot (10-\sqrt{62})^2 \cdot (10+\sqrt{62})^2 = \boxed{19}$

I would write a solution, but it would be impossible to understand since my phone can't use pictures/diagrams.