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The quadrilateral ABCD has right angles at A and D.

The numbers show the areas of the colored triangles.

Find the area of ABC

\(\begin{array}{|rcll|} \hline \dfrac{hb}{2} &=& 16 + 48 \quad \text{or} \quad \mathbf{\dfrac{hb}{2}=64} \qquad (1) \\\\ \mathbf{\dfrac{hb}{2}} &=& \mathbf{16+A_1} \qquad (2) \\ \hline \dfrac{hb}{2} = 64 &=& 16+A_1 \\ 64 &=& 16+A_1 \\ A_1 &=& 64-16 \\ \mathbf{A_1} &=& \mathbf{48} \\ \hline \end{array} \begin{array}{|rcll|} \hline \dfrac{ha}{2} &=& A_1+A_2 \quad |\quad \mathbf{A_1=48} \\\\ \mathbf{\dfrac{ha}{2}} &=& \mathbf{48+A_2} \qquad (3) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{bh_1}{2} &=& 16 \quad \text{or} \quad \mathbf{h_1=\dfrac{32}{b}} \qquad (4) \\ \hline \dfrac{ah_2}{2} &=& A_2 \quad | \quad h = h_1+h_2 \quad \text{or} \quad \mathbf{h_2=h-h_1} \\\\ \dfrac{a(h-h_1)}{2} &=& A_2 \\\\ \dfrac{ah}{2} - \dfrac{ah_1}{2} &=& A_2\quad | \quad \mathbf{h_1=\dfrac{32}{b}} \\\\ \dfrac{ah}{2} - \dfrac{a\dfrac{32}{b}}{2} &=& A_2 \\\\ \dfrac{ah}{2} - \dfrac{16a}{b} &=& A_2 \quad | \quad \mathbf{\dfrac{ha}{2}= 48+A_2} \\\\ 48+A_2 - \dfrac{16a}{b} &=& A_2 \\\\ 48 - \dfrac{16a}{b} &=& 0 \\\\ \dfrac{16a}{b} &=& 48 \\\\ \dfrac{a}{b} &=& \dfrac{48}{16} \\\\ \mathbf{\dfrac{a}{b}} &=& \mathbf{3} \qquad (5) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \dfrac{(3)}{(1)}: & \dfrac{ \dfrac{ah}{2} } { \dfrac{bh}{2} } &=& \dfrac{48+A_2}{64} \\\\ & \dfrac{ah}{2} \cdot \dfrac{2}{bh} &=& \dfrac{48+A_2}{64} \\\\ & \dfrac{a}{b} &=& \dfrac{48+A_2}{64} \quad | \quad \mathbf{\dfrac{a}{b}=3} \\\\ & 3 &=& \dfrac{48+A_2}{64} \\\\ & 192 &=& 48+A_2 \\ & A_2 &=& 192-48 \\ & \mathbf{A_2} &=& \mathbf{144} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{The area of $ABCD$} &=& 16+48+A_1+A_2 \\ &=& 16+48+48+144 \\ \mathbf{\text{The area of $ABCD$}} &=& \mathbf{256} \\ \hline \end{array} \)