## Task B-3.3.

The length of the side of the rhombus $A B C D$ is $6 \, \text{cm}$, and the measure of the angle at vertex $B$ is $120^{\circ}$. All points inside the rhombus that are closer to vertex $B$ than to the other vertices lie within the pentagon $P$. Determine its area.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_8cadc17e7912dc7108beg-13.jpg?height=677&width=971&top_left_y=1863&top_left_x=491)

Let's consider the area bounded by the perpendicular bisectors of segments $\overline{B A}, \overline{B C}$, and $\overline{B D}$. Points on the perpendicular bisector of a segment are equidistant from its endpoints, which in this case are vertices $A, C, D$, so the points closer to vertex $B$ are inside the pentagon $B E F G H$.

The area of this pentagon is obtained by subtracting the areas of two triangles $A E F$ from half the area of the rhombus.

We calculate step by step:

$$
\begin{array}{ll}
P_{\text {rhombus }}=a^{2} \sin \alpha & 1 \text{ point} \\
|E F|=3 \cdot \operatorname{tg} 30^{\circ}=\sqrt{3} \text{ cm} & 1 \text{ point} \\
P_{A E F}=\frac{1}{2} \cdot 3 \cdot|E F|=\frac{3 \sqrt{3}}{2} \text{ cm}^{2} & 1 \text{ point} \\
P=\frac{1}{2} \cdot 6^{2} \cdot \sin 120^{\circ}-2 \cdot \frac{3 \sqrt{3}}{2}=6 \sqrt{3} \text{ cm}^{2} . & 1 \text{ point}
\end{array}
$$