AMC8 2018

Answer (B): To see what fraction of the square is occupied by quadrilateral $AFED$, first note that $\triangle ADC$ occupies half of the area. Next note that because $AB$ and $CD$ are parallel, it follows that $\triangle ABF$ and $\triangle CEF$ are similar, and $CE=\frac{1}{2}AB$. Draw $PQ$ through $F$ such that $PF$ and $FQ$ are altitudes of $\triangle ABF$ and $\triangle CEF$, respectively. Then $FQ=\frac{1}{2}PF$, so $FQ=\frac{1}{3}PQ=\frac{1}{3}BC$. Therefore the area of $\triangle CEF$ is $\frac{1}{2}(CE)(FQ)=\frac{1}{2}\left(\frac{1}{2}AB\right)\left(\frac{1}{3}BC\right)=\frac{1}{12}(AB)(BC)$, which is $\frac{1}{12}$ of the area of the square. Because the area of quadrilateral $AFED$ equals the area of $\triangle ADC$ minus the area of $\triangle CEF$, the fraction of the square occupied by quadrilateral $AFED$ is $\frac{1}{2}-\frac{1}{12}=\frac{5}{12}$. Because the area of $AFED$ is $45$, the area of $ABCD$ is $\left(\frac{12}{5}\right)(45)=108$.