5. Lucio has two identical rectangular cards with sides measuring 6 and 9; he places them on the table so that they share a diagonal but are not perfectly overlapped. What is the area of the portion of the first card that is covered by the second?
(A) 27
(B) 36
(C) 37
(D) 38
(E) 39
## 5. The answer is $\mathbf{( E )}$.

The problem asks to calculate the area of $A G C H$. Let's prove that this quadrilateral is a rhombus. Let $O$ be the midpoint of $A C$. This is the center of symmetry for both the rectangle $A B C D$ and the rectangle $A E C F$, so it is also the center of symmetry for $A G C H$, which is therefore a parallelogram. It is enough to prove that two adjacent sides of $A G C H$ are congruent. Consider the triangles $A F H$ and $H D C$. These are two right triangles whose angles at $H$ are vertical angles, and thus congruent. By difference, all three angles are congruent, so the triangles are similar.

![](https://cdn.mathpix.com/cropped/2024_04_17_c5c03d45c82457ddb3cag-13.jpg?height=520&width=594&top_left_y=731&top_left_x=1368)

Moreover, $F A$ and $D C$ are congruent by hypothesis, so the two triangles are similar with a corresponding side congruent, and therefore they are congruent, and consequently $\overline{H C}=\overline{H A}$. Let $l$ be the length of $H C$. Applying the Pythagorean theorem to the triangle $D C H$ we get:

$$
l^{2}=(9-l)^{2}+36
$$

Expanding this, we get $l^{2}=81-18 l+l^{2}+36$, from which $l=\frac{13}{2}$. The area of $A G C H$ is therefore given by $\overline{A B} \cdot \overline{A H}=6 \cdot \frac{13}{2}=39$.