smc

Note: In the image AB and CD have been swapped from their given lengths in the problem. However, this doesn't affect any of the solving. Drop perpendiculars to $CD$ from $A$ and $B$, and call the intersections $X,Y$ respectively. Now, $D{A}^2-B{C}^2=(7-5)(7+5)=D{X}^2-C{Y}^2$ and $DX+CY=19-11=8$. Thus, $DX-CY=3$. We conclude $DX=\frac{11}{2}$ and $CY=\frac{5}{2}$. To simplify things even more, notice that $90^{\circ }=\frac{\angle D+\angle A}{2}=180^{\circ }-\angle APD$, so $\angle P=\angle Q=90^{\circ }$. Also, $$\sin (\angle PAD)=\sin (\frac{1}{2}\angle XDA)=\sqrt{\frac{1-\cos (\angle XDA)}{2}}=\sqrt{\frac{3}{28}}$$ So the area of $△APD$ is: $$R\cdot c\sin a\sin b=\frac{7\cdot 7}{2}\sqrt{\frac{3}{28}}\sqrt{1-\frac{3}{28}}=\frac{35}{8}\sqrt{3}$$ Over to the other side: $△BCY$ is $30-60-90$, and is therefore congruent to $△BCQ$. So $[BCQ]=\frac{5\cdot 5\sqrt{3}}{8}$. The area of the hexagon is clearly $$\begin{array}{rl}[ABCD]-([BCQ]+[APD])& =\frac{15\cdot 5\sqrt{3}}{2}-\frac{60\sqrt{3}}{8}\\ & =30\sqrt{3}⟹\boxed{30\sqrt{3}}\end{array}$$ Note: Once $DY$ is found, there is no need to do the trig. Notice that the hexagon consists of two trapezoids, $ABPQ$ and $CDPQ$. $PQ=\frac{19-7-5+11}{2}=9$. The height is one half of $BY$ which is $\frac{5\sqrt{3}}{4}$. So the area is $$\frac{1}{2}\cdot \frac{5\sqrt{3}}{4}(19+9+11+9)=30\sqrt{3}$$