Task B-1.3. On the sides $A C$ and $B C$ of an equilateral triangle $A B C$, squares $A C D E$ and $B C F G$ are constructed outward. If $|A B|=4 \mathrm{~cm}$, what is the perimeter of the resulting hexagon $A B G F D E$?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_06_03_ade005c4283eff439485g-02.jpg?height=585&width=814&top_left_y=583&top_left_x=623)

The sides of the hexagon (and the square) $B G, G F, D E, E A$ have the same length as the side $A B$ of the triangle $A B C$. Therefore, the perimeter of the hexagon $A B G F D E$ is calculated as

$$
o=5 \cdot|A B|+|D F|
$$

The length of the side $D F$ is calculated from the isosceles triangle $C F D$.

The angle $\measuredangle D C F=360^{\circ}-90^{\circ}-90^{\circ}-60^{\circ}=120^{\circ}$, and

$\measuredangle P C F=60^{\circ}$, so the triangle $C F C^{\prime}$ is equilateral, with side length $4 \mathrm{~cm}$.

Then the length $|P F|$ is equal to the height of the triangle $C F C^{\prime}$, and $|D F|=2|P F|$. $|D F|=2 \cdot \sqrt{4^{2}-2^{2}}=4 \sqrt{3}$.

Thus, the desired perimeter is $o=5 \cdot 4+4 \sqrt{3}=(20+4 \sqrt{3}) \mathrm{cm}$.