2019-II-16

Ans: D
Since $ABCD$ is a parallelogram, $BC=AD$ (opp. sides of //gram).

Since $\mathrm{\Delta }BEX\sim \mathrm{\Delta }DAX$,

$\begin{array}{rcl}{\displaystyle \frac{BX}{DX}}& =& {\displaystyle \frac{BE}{DA}}\\ & =& {\displaystyle \frac{BE}{BC}}\\ & =& {\displaystyle \frac{2}{12}}\\ & =& {\displaystyle \frac{1}{6}}\end{array}$

Consider $\mathrm{\Delta }ABX$ and $\mathrm{\Delta }ADX$. They have the same height if $BX$ and $DX$ are the bases respectively. Therefore, we have

$\begin{array}{rcl}{\displaystyle \frac{\text{the area of }\mathrm{\Delta }ABX}{\text{the area of }\mathrm{\Delta }ADX}}& =& {\displaystyle \frac{BX}{DX}}\\ {\displaystyle \frac{24}{\text{the area of }\mathrm{\Delta }ADX}}& =& {\displaystyle \frac{1}{6}}\\ \text{the area of }\mathrm{\Delta }ADX& =& 144{\text{ cm}}^2\end{array}$

Since $\mathrm{\Delta }ADX\sim \mathrm{\Delta }FBY$, we have

$\begin{array}{rcl}{\displaystyle \frac{\text{the area of }\mathrm{\Delta }ADX}{\text{the area of }\mathrm{\Delta }FBY}}& =& {\left({\displaystyle \frac{AD}{BY}}\right)}^2\\ {\displaystyle \frac{144}{\text{the area of }\mathrm{\Delta }FBY}}& =& {\left({\displaystyle \frac{12}{9}}\right)}^2\\ \text{the area of }\mathrm{\Delta }FBY& =& 81{\text{ cm}}^2\end{array}$

Therefore, the area of the quadrilateral $CDYF$

$\begin{array}{cl}=& \text{the area of }\mathrm{\Delta }BCD–\text{the area of }\mathrm{\Delta }BFY\\ =& \text{the area of }\mathrm{\Delta }ABD–\text{the area of }\mathrm{\Delta }BFY\\ =& \text{the area of }\mathrm{\Delta }ABX+\text{the area of }\mathrm{\Delta }ADX–\text{the area of }\mathrm{\Delta }BFY\\ =& 24+144–81\\ =& 87{\text{ cm}}^2\end{array}$