2025 Cayley Problem 19

In the diagram, ABCD is a square with area k. Point E is on side AB with AE = \frac{1}{3}AB. Point G is on side BC with BG = \frac{1}{4}BC. Point F is on ED so that GF is perpendicular to BC. The area of \triangle FGC is

\textbf{(A)}\ \frac{9}{32}k\quad \textbf{(B)}\ \frac{13}{48}k\quad \textbf{(C)}\ \frac{3}{14}k\quad \textbf{(D)}\ \frac{7}{24}k\quad \textbf{(E)}\ \frac{2}{7}k

If there are no answer choices shown, enter a numerical answer.

Full credit to this problem is given to the CEMC, you may view all Cayley contests here.