, 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 $\dfrac{81}{2}=\dfrac{81}{2}\times\dfrac{k}{k}=\dfrac{81}{2}\times\dfrac{k}{144}=\dfrac{81}{2\times144}k=\dfrac{9}{2\times16}k=\dfrac{9}{32}k$.