5. (3 points) In a square $ABCD$ with a side length of 1 cm, draw quarter circles with a radius of 1 cm, centered at $A, B, C, D$, intersecting at points $E, F, G, H$, as shown in the figure. The perimeter of the shaded area in the middle is $\qquad$ cm. (Take $\pi=3.141$)
【Analysis】As shown in the figure: It is easy to conclude from the problem that $\triangle \triangle_{A B F}$ is an equilateral triangle, so arc $\widehat{\mathrm{AGF}}$ is $\frac{1}{6}$ of the circle. Similarly, arc $\widehat{\mathrm{FGC}}$ is also $\frac{1}{6}$ of the circle. Therefore, arc $\widehat{\mathrm{GF}}=\widehat{\mathrm{AGF}}+\widehat{\mathrm{FGC}}-\widehat{\mathrm{AGC}}=\frac{1}{12}$ of the circle. Similarly, the other three segments are also $\frac{1}{12}$ of the circle. Hence, the perimeter $=\frac{1}{3}$ of the circle, and the perimeter can be solved using the formula for the circumference of a circle.
【Solution】From the problem, it is easy to know that $\triangle A B F$ is an equilateral triangle, so arc $\widehat{\mathrm{AGF}}$ is $\frac{1}{6}$ of the circle, and similarly, arc $\widehat{\mathrm{FGC}}$ is also $\frac{1}{6}$ of the circle,
so arc $\widehat{\mathrm{GF}}=\widehat{\mathrm{AGF}}+\widehat{\mathrm{FGC}}-\widehat{\mathrm{AGC}}=\frac{1}{12}$ of the circle,
Similarly, the other three segments are also $\frac{1}{12}$ of the circle,
Therefore, the perimeter of the shaded part
$$
\begin{array}{l}
=\frac{1}{12} \text { of the circle } \times 4 \\
=\frac{1}{3} \text { of the circle } \\
=\frac{2 \times 1 \times \pi}{3} \\
=2.094 \text { (cm); }
\end{array}
$$

Answer: The perimeter of the shaded part in the middle is 2.094 cm.