Let the areas of $\triangle ABX$ and $CDX$ be $25x$ and $49x$ respectively.
Since $\triangle ABX\sim\triangle CDX$,
$$\displaystyle \frac{AX}{CX}=\frac{BX}{DX}=\sqrt{\frac{25}{49}}=\frac{5}{7}$$
We have
$$\frac{\textrm{area of }\triangle ABX}{\textrm{area of }\triangle ADX}=\frac{5}{7}$$
$$\textrm{area of }\triangle ADX=35x$$
and
$$\frac{\textrm{area of }\triangle ABX}{\textrm{area of }\triangle BCX}=\frac{5}{7}$$
$$\textrm{area of }\triangle BCX=35x$$
The area of the trapezium $ABCD$ is $25x+35x+35x+49x=144x$.
$$\frac{\textrm{area of }\triangle ADX}{\textrm{area of trapezium }ABCD}=\frac{35}{144}$$
In general, we have the following result:
Let $ABCD$ be a convex quadrilateral and $X$ be the point of intersection of its diagonals. Then we have
$$\textrm{area of }\triangle ABX\times\textrm{area of }\triangle CDX=\textrm{area of }\triangle BCX\times\textrm{area of }\triangle DAX$$
Furthermore, if $AB=CD$, then
$$\textrm{area of }\triangle BCX=\textrm{area of }\triangle DAX$$
and therefore,
$$\textrm{area of }\triangle ABX\times\textrm{area of }\triangle CDX=(\textrm{area of }\triangle DAX)^2$$
An overkill solution:
We know that the area of kite is $ef/2$ where $e,f$ are diagonals. By Ptolomey theorem we have:
$$ 2A =ef = ac+bd = 2ac = 2\cdot 104$$ so $A= 104$.
Best Answer
Note that since $EB \parallel DC$, we have $A(EBC) = A(EBD)$ (both triangles have base $EB$ and since $EB \parallel DC$, they have the same height for that base). Rest is just to use the given ratio.