A hexagon has every angle $120^{\circ}$, and its sides are alternately 1 and $\sqrt{3} \mathrm{~cm}$ long. What is the area of the hexagon?
We attach a regular, $1 \mathrm{~cm}$ side triangle to each of the three $1 \mathrm{~cm}$ sides of the hexagon as shown in the figure. Since the angles of the hexagon are $120^{\circ}$ and the angles of the regular triangles are $60^{\circ}$, the sides of the regular triangles coincide with the sides of the hexagon. Therefore, the hexagon and the three small regular triangles together form a large regular triangle, each side of which is $2+\sqrt{3} \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_56b4160530e71090d297g-1.jpg?height=510&width=450&top_left_y=352&top_left_x=859)

The area $T$ of the hexagon is obtained by subtracting the area of the three small triangles from the area of the large triangle. The area of a regular triangle with side $a$ is $\frac{a^{2} \sqrt{3}}{4}$, so

$$
T=\frac{(2+\sqrt{3})^{2} \cdot \sqrt{3}}{4}-3 \cdot \frac{1^{2} \cdot \sqrt{3}}{4}=3+\sqrt{3} \mathrm{~cm}^{2}
$$