, 2025

The area of $\triangle FGC$ is expressed as a fraction of the area of square $ABCD$, and so we begin by letting the side length of $ABCD$ be equal to $12$, as shown, and its area $k=12^2=144$. [[IMAGE0]] Since $AE=\dfrac13AB=\dfrac13(12)=4$, then $EB=12-4=8$. Since $BG=\dfrac14BC=\dfrac14(12)=3$, then $GC=12-3=9$. Next, we position point $H$ on $FG$ so that $EH$ is perpendicular to $FG$, and so $EBGH$ is a rectangle with $GH=EB=8$ and $EH=BG=3$. The slope of $ED=\dfrac{AE}{AD}=\dfrac{4}{12}=\dfrac{1}{3}$, and so the slope of $EF$ is also equal to $\dfrac13$. The slope of $EF=\dfrac{FH}{EH}$ or $\dfrac13=\dfrac{FH}{3}$, and so $FH=3\times\dfrac13=1$. Since $GH=8$ and $FH=1$, then $FG=8+1=9$. The area of $\triangle FGC$ is $\dfrac12(GC)(FG)=\dfrac12(9)(9)=\dfrac{81}{2}$. At this point, we could substitute $k=144$ into each of the given answers to determine which is equal to $\dfrac{81}{2}$. Doing so, we get $\dfrac{9}{32}k=\dfrac{9}{32}(144)=\dfrac{9}{2}(9)=\dfrac{81}{2}$, and so the answer is (A). Alternately, we could express $\dfrac{81}{2}$ in terms of $k$ since $\dfrac{81}{2}=\dfrac{81}{2}\times\dfrac{k}{k}=\dfrac{81}{2}\times\dfrac{k}{144}=\dfrac{81}{2\times144}k=\dfrac{9}{2\times16}k=\dfrac{9}{32}k$.