Quadrilateral $ABCD$ is a rhombus with a side length of $6 \, \text{cm}$ and a height of $4 \, \text{cm}$. Point $E$ is the midpoint of side $AD$, point $G$ is the midpoint of side $BC$, point $F$ is the intersection of segments $AG$ and $BE$, and point $H$ is the intersection of segments $CE$ and $DG$.

Determine the area of quadrilateral $EFGH$.

(K. Pazourek)

Hint. Use a diagram with the added segment $EG$.
Segment $E G$ divides the rhombus $A B C D$ into two congruent trapezoids $A B G E$ and $E G C D$ (points $E$ and $G$ are the midpoints of the sides, so segments $A E, B G, E D, G C$ are congruent and lines $A B, E G, C D$ are parallel). The remaining segments from the problem are the diagonals of these trapezoids, and points $F, H$ are their intersections:

![](https://cdn.mathpix.com/cropped/2024_04_17_1df09591dc133e92eb1dg-2.jpg?height=485&width=1102&top_left_y=1048&top_left_x=477)

Each trapezoid is divided into four triangles of equal area by its diagonals (each pair of adjacent triangles have congruent sides lying on the same line and a common height relative to that line).

The rhombus $A B C D$ is thus divided into eight triangles of equal area, and quadrilateral $E F G H$ is formed by two of these eight triangles. This quadrilateral therefore occupies two eighths, i.e., one quarter of the area of the given rhombus.

The area of the rhombus $A B C D$ is $6 \cdot 4 = 24 \left(\mathrm{~cm}^{2}\right)$, so

$$
S_{E F G H} = \frac{1}{4} S_{A B C D} = 6 \left(\mathrm{~cm}^{2}\right).
$$

Notes. In the given solution, it is not essential which of the diagonals of the rhombus is shorter and which is longer. Since we are only concerned with areas, we can even work with any trapezoid that has the same side and height as the original rhombus. Therefore, we can solve the problem in a rectangle with sides $6 \mathrm{~cm}$ and $4 \mathrm{~cm}$:

![](https://cdn.mathpix.com/cropped/2024_04_17_1df09591dc133e92eb1dg-3.jpg?height=482&width=688&top_left_y=290&top_left_x=684)

Additional division using lines parallel to the sides divides the rectangle $A B C D$ into 16 congruent triangles, four of which form quadrilateral $E F G H$. This quadrilateral therefore occupies $\frac{4}{16}$, i.e., one quarter of the area of the rectangle.