Solution:

The answer is $(E)$.

The problem asks us to compute the area of $AGCH$. We show that this quadrilateral is a rhombus. Let $O$ be the midpoint of $AC$; this is a center of symmetry both for the rectangle $ABCD$ and for the rectangle $AECF$, hence it is also a center of symmetry for $AGCH$, which is therefore a parallelogram. It thus suffices to show that two adjacent sides of $AGCH$ are congruent. Consider the triangles $AFH$ and $HDC$. These are two right triangles whose angles at $H$ are vertical angles, hence congruent. By subtraction, all 3 angles turn out to be congruent, so the triangles are similar.

Moreover $FA$ and $DC$ are congruent by hypothesis, so the two triangles are similar with one corresponding side congruent, hence they are congruent, and consequently $\overline{HC}=\overline{HA}$. Letting $l$ be the length of $HC$, applying the Pythagorean theorem to triangle $DCH$ we get:
$$l^2 = (9-l)^2 + 36$$
Expanding, we obtain $l^2 = 81 - 18l + l^2 + 36$, from which $l = \frac{13}{2}$. The area of $AGCH$ is therefore given by $\overline{AB} \cdot \overline{AH} = 6 \cdot \frac{13}{2} = 39$.