In a rectangle $\mathrm{ABCD}$, points $\mathrm{X}$ and $\mathrm{Y}$ are the midpoints of $\mathrm{AD}$ and $\mathrm{DC}$, respectively. Lines $\mathrm{BX}$ and $\mathrm{CD}$ when extended intersect at $\mathrm{E}$, lines $\mathrm{BY}$ and $\mathrm{AD}$ when extended intersect at $\mathrm{F}$. If the area of $\mathrm{ABCD}$ is $60$ then the area of $\mathrm{BEF}$ is

• A

  $60$
• B

  $80$
• Correct answer
  C

  $90$
• D

  $120$

Solution

Let $A B=x$ $B C=y$ $x y=60$ $\Delta \mathrm{XED} \simeq \triangle \mathrm{XBA}$ and $\triangle D F Y \simeq \Delta C B Y$ $\Delta B E F=\Delta B X D Y+\Delta X E D+\Delta D F Y+\triangle D E F$ $=60+\triangle D E F$ $60+\frac{1}{2} \times x y$ $60+30=90$

Hence, the correct answer is option: 'C'.