Let the altitude dropped from $N$ to $RK$ intersect the line at $A$. Likewise, denote the same for point $M$ as $B$. Now since $\triangle NAK$ ~ $\triangle MBK$ we can set $NK=7k$ and $MK=5k$. Continuing on, we can drop the height of the trapezoid from $R$ onto side $MK$ (and denote the intersection as $C$). We know that $\triangle KRC$ ~ $\triangle KMB$ as well. Thus we get the equality $\frac{3}{h}=\frac{5k}{5}$ and $h=\frac{3}{k}$. Finally, we can get the area of the quadrilateral by $\frac{7k+5k}{2} \cdot \frac{3}{k}=18$