Since $ABCD$ is a parallelogram, $BC = AD$ (opp. sides of //gram).
Since $\Delta BEX \sim \Delta DAX$,
$\begin{array}{rcl}
\dfrac{BX}{DX} & = & \dfrac{BE}{DA} \\
& = & \dfrac{BE}{BC} \\
& = & \dfrac{2}{12} \\
& = & \dfrac{1}{6}
\end{array}$
Consider $\Delta ABX$ and $\Delta ADX$. They have the same height if $BX$ and $DX$ are the bases respectively. Therefore, we have
$\begin{array}{rcl}
\dfrac{\text{the area of $\Delta ABX$}}{\text{the area of $\Delta ADX$}} & = & \dfrac{BX}{DX} \\
\dfrac{24}{\text{the area of $\Delta ADX$}} & = & \dfrac{1}{6} \\
\text{the area of $\Delta ADX$} & = & 144\text{ cm}^2
\end{array}$
Since $\Delta ADX \sim \Delta FBY$, we have
$\begin{array}{rcl}
\dfrac{\text{the area of $\Delta ADX$}}{\text{the area of $\Delta FBY$}} & = & \left(\dfrac{AD}{BY}\right)^2 \\
\dfrac{144}{\text{the area of $\Delta FBY$}} & = & \left(\dfrac{12}{9}\right)^2 \\
\text{the area of $\Delta FBY$} & = & 81 \text{ cm}^2
\end{array}$
Therefore, the area of the quadrilateral $CDYF$
$\begin{array}{cl}
= & \text{the area of $\Delta BCD$} – \text{the area of $\Delta BFY$} \\
= & \text{the area of $\Delta ABD$} – \text{the area of $\Delta BFY$} \\
= & \text{the area of $\Delta ABX$} + \text{the area of $\Delta ADX$} – \text{the area of $\Delta BFY$} \\
= & 24 + 144 – 81 \\
= & 87\text{ cm}^2
\end{array}$
