Geometry Word Problem I've been trying to find the solution for this problem for a long time... but I can't seem to do it. I'm not asking for the full solution, maybe just a hint or where to start off from. 
Question:
Let ABCD be a trapezium with AB||CD. M and N are the midpoints of AB and CD respectively. If AC=6 BD=8 and MN=4 Find the area of ABCD.

HINT.
Extend base $AB$ by $BB'=CD$, so that $CB'=BD=8$. Triangle $AB'C$ has the same area as the trapezium and its median $CM'$ is congruent to $MN$. 
So the problem amounts at finding the area of a triangle, with two sides of length $6$ and $8$, and the median relative to the third side of length $4$.
SECOND HINT.
To find the area of such a triangle, let $AM'=B'M'=x$, $\angle AM'C=\alpha$ and $\angle B'M'C=\pi-\alpha$ (see diagram below). By the cosine rule you then get
$$
\begin{align}
x^2+16-8x\cos\alpha=36,\\
x^2+16+8x\cos\alpha=64.\\
\end{align}
$$
Summing these you immediately obtain $x=\sqrt{34}$.

