In trapezoid ABCD segments AB and CD are parallel. Point P is the intersection of diagonals AC and BD. The area of APAB is 16 square units, and the area of ▲PCD is 25 square units. What is the area of trapezoid ABCD?
Triangles $PAB$ and $PCD$ are similar, so
\[\frac{[PAB]}{[PCD]} = \left( \frac{AB}{CD} \right)^2 = \frac{x^2}{y^2}.\]
But we are given that $[PAB] = 8$ and $[PCD] = 18$, so
\[\frac{x^2}{y^2} = \frac{8}{18} = \frac{4}{9},\]
which means $x/y = 2/3$.
Then $AP/PC = AB/CD = 2/3$. Triangles $ABP$ and $BCP$ have the same height with respect to base $\overline{AC}$, so
\[\frac{[BCP]}{[ABP]} = \frac{CP}{AP} = \frac{3}{2},\]
which means $[BCP] = 3/2 \cdot [ABP] = 3/2 \cdot 8 = 12$.
Also, $BP/PD = AB/CD = 2/3$. Triangles $ABP$ and $ADP$ have the same height with respect to base $\overline{BD}$, so
\[\frac{[ADP]}{[ABP]} = \frac{DP}{BP} = \frac{3}{2},\]
which means $[ADP] = 3/2 \cdot [ABP] = 3/2 \cdot 8 = 12$.
Therefore, the area of trapezoid $ABCD$ is $[ABP] + [BCP] + [CDP] + [DAP] = 8 + 12 + 18 + 12 = 50
