Solution.

Let the side of the square be $x$, then the perimeter of the square

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-570.jpg?height=664&width=1154&top_left_y=94&top_left_x=93)

$P_{0}=4 x, \triangle F B M$ is isosceles, $\angle F B M=360^{\circ}-90^{\circ}-2 \cdot 60^{\circ}=150^{\circ}$, $F M=\sqrt{2 x^{2}-2 x^{2} \cdot \cos 150^{\circ}}=x \sqrt{2+\sqrt{3}}$ (Fig. 10.30). Thus,

$$
P_{E F M N}=4 x \sqrt{2+\sqrt{3}}, \frac{P}{P_{0}}=\sqrt{2+\sqrt{3}}=\frac{\sqrt{6}+\sqrt{2}}{2}
$$

Answer: $\frac{\sqrt{6}+\sqrt{2}}{2}$.
10.030. On the sides of a square, equilateral triangles are constructed outside it, and their vertices are sequentially connected. Determine the ratio of the perimeter of the resulting quadrilateral to the perimeter of the given square.