very hard geometry problem

The area of rectangle ABCD is 73/2.

Find the area of rectangle  ABCD

\(\text{Let area of rectangle $ABCD=A$} \\ A=wh\quad \text{or}\quad \mathbf{h=\dfrac{A}{w}}\)

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{\dfrac{w(h-x)}{2}} &=& \mathbf{5} \\\\ & wh-wx &=& 10 \quad | \quad wh = A \\ & A-wx &=& 10 \\ & \mathbf{wx} &=& \mathbf{A-10} \\ \hline \end{array} \begin{array}{|lrcll|} \hline (2) & \mathbf{\dfrac{h(w-y)}{2}} &=& \mathbf{8} \\\\ & wh-hy &=& 16 \quad | \quad wh = A \\ & A-hy &=& 16 \\ & \mathbf{A} &=& \mathbf{16+hy} \\ \hline \end{array} \begin{array}{|lrcll|} \hline (3) & \mathbf{\dfrac{xy}{2}} &=& \mathbf{9} \\\\ & xy &=& 18 \\ & \mathbf{y} &=& \mathbf{\dfrac{18}{x}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{A} &=& \mathbf{16+hy} \quad | \quad \mathbf{h=\dfrac{A}{w}},\ \mathbf{y=\dfrac{18}{x}} \\\\ A &=& 16+\dfrac{A}{w}*\dfrac{18}{x} \\\\ A &=& 16+\dfrac{18A}{wx} \\\\ A-16 &=& \dfrac{18A}{wx} \quad | \quad \mathbf{wx=A-10} \\\\ A-16 &=& \dfrac{18A}{A-10} \\\\ (A-16)(A-10) &=& 18A \\ A^2-26A+160 &=& 18A \\ \mathbf{A^2-44A+160} &=& \mathbf{0} \\\\ A &=& \dfrac{44\pm \sqrt{44^2-4*160}} {2} \\\\ A &=& \dfrac{44\pm \sqrt{1296}} {2} \\\\ A &=& \dfrac{44\pm 36} {2} \\\\ A = \dfrac{44+ 36} {2} &\text{or}& A = \dfrac{44- 36} {2} \\\\ A = \dfrac{80} {2} && A = \dfrac{8} {2} \\\\ \mathbf{A = 40} && A = 4 \quad \text{no solution!} \\ \hline \end{array}\)

The area of rectangle  ABCD is \(\mathbf{40}\)