The side of the regular hexagon $A B C D E F A$ is $a$. Connect the vertex $A$ to the midpoint $G$ of the opposite side $C D$ and calculate the area of the quadrilateral $A B C G$.

Let's proceed step by step:

1. **Identify the coordinates of the vertices:**
   - Place the hexagon in the coordinate plane with the center at the origin $(0, 0)$.
   - The vertices of the hexagon can be represented as follows:
     - $A = (a, 0)$
     - $B = \left(\frac{a}{2}, \frac{a\sqrt{3}}{2}\right)$
     - $C = \left(-\frac{a}{2}, \frac{a\sqrt{3}}{2}\right)$
     - $D = (-a, 0)$
     - $E = \left(-\frac{a}{2}, -\frac{a\sqrt{3}}{2}\right)$
     - $F = \left(\frac{a}{2}, -\frac{a\sqrt{3}}{2}\right)$

2. **Find the midpoint $G$ of the side $CD$:**
   - The coordinates of $C$ are $\left(-\frac{a}{2}, \frac{a\sqrt{3}}{2}\right)$.
   - The coordinates of $D$ are $(-a, 0)$.
   - The midpoint $G$ is given by:
     \[
     G = \left(\frac{-\frac{a}{2} + (-a)}{2}, \frac{\frac{a\sqrt{3}}{2} + 0}{2}\right) = \left(-\frac{3a}{4}, \frac{a\sqrt{3}}{4}\right)
     \]

3. **Calculate the area of the quadrilateral $ABCG$:**
   - The vertices of the quadrilateral $ABCG$ are $A = (a, 0)$, $B = \left(\frac{a}{2}, \frac{a\sqrt{3}}{2}\right)$, $C = \left(-\frac{a}{2}, \frac{a\sqrt{3}}{2}\right)$, and $G = \left(-\frac{3a}{4}, \frac{a\sqrt{3}}{4}\right)$.
   - Use the Shoelace formula to find the area:
     \[
     \text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right|
     \]
     Substituting the coordinates:
     \[
     \text{Area} = \frac{1}{2} \left| a \cdot \frac{a\sqrt{3}}{2} + \frac{a}{2} \cdot \frac{a\sqrt{3}}{2} + \left(-\frac{a}{2}\right) \cdot \frac{a\sqrt{3}}{4} + \left(-\frac{3a}{4}\right) \cdot 0 - \left(0 \cdot \frac{a}{2} + \frac{a\sqrt{3}}{2} \cdot \left(-\frac{a}{2}\right) + \frac{a\sqrt{3}}{2} \cdot \left(-\frac{3a}{4}\right) + \frac{a\sqrt{3}}{4} \cdot a \right) \right|
     \]
     Simplifying inside the absolute value:
     \[
     \text{Area} = \frac{1}{2} \left| \frac{a^2\sqrt{3}}{2} + \frac{a^2\sqrt{3}}{4} - \frac{a^2\sqrt{3}}{8} - \left( -\frac{a^2\sqrt{3}}{4} - \frac{3a^2\sqrt{3}}{8} + \frac{a^2\sqrt{3}}{4} \right) \right|
     \]
     \[
     \text{Area} = \frac{1}{2} \left| \frac{4a^2\sqrt{3}}{8} + \frac{2a^2\sqrt{3}}{8} - \frac{a^2\sqrt{3}}{8} + \frac{a^2\sqrt{3}}{4} + \frac{3a^2\sqrt{3}}{8} - \frac{a^2\sqrt{3}}{4} \right|
     \]
     \[
     \text{Area} = \frac{1}{2} \left| \frac{4a^2\sqrt{3}}{8} + \frac{2a^2\sqrt{3}}{8} - \frac{a^2\sqrt{3}}{8} + \frac{2a^2\sqrt{3}}{8} + \frac{3a^2\sqrt{3}}{8} - \frac{2a^2\sqrt{3}}{8} \right|
     \]
     \[
     \text{Area} = \frac{1}{2} \left| \frac{6a^2\sqrt{3}}{8} \right| = \frac{1}{2} \cdot \frac{3a^2\sqrt{3}}{4} = \frac{3a^2\sqrt{3}}{8}
     \]

Thus, the area of the quadrilateral $ABCG$ is:
\[
\boxed{\frac{3a^2\sqrt{3}}{8}}
\]
The quadrilateral $ABCG$ is divided into two triangles by the diagonal $AC$. These triangles have equal areas because their base $AC$ is common, and their height is equal to $\frac{a}{2}$. The area of one triangle is

$$
\frac{a}{4} \cdot AC = \frac{a}{4} \cdot a \sqrt{3} = \frac{a^2}{4} \cdot \sqrt{3}
$$

Thus, the area we are looking for is

$$
\frac{a^2}{2} \sqrt{3}
$$

(Ernó Freund, Budapest.)

The problem was also solved by: J. Felhőssy, P. Füstös, P. Heimlich, Gy. Jánosy, J. Kiss, A. Paunz, Z. Reich, Gy. Schilhanek, Gy. Schuster, J. Steiger, J. Szabó.