5. (10 points) In a square $ABCD$ with side length of 2 cm, draw quarter circles with radius 2 cm, centered at $A, B, C, D$, intersecting at points $E, F, G, H$, as shown in the figure. Then 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 $\triangle A B F$ is an equilateral triangle, so arc $\widehat{\mathrm{AGF}}$ is $\frac{1}{6}$ of the circumference of the circle. Similarly, arc $\widehat{\mathrm{FGC}}$ is also $\frac{1}{6}$ of the circumference of the circle. Therefore, arc $\widehat{\mathrm{GF}}=\widehat{\mathrm{AGF}}+\widehat{\mathrm{FGC}}-\widehat{\mathrm{AGC}}=\frac{1}{12}$ of the circumference of the circle. Similarly, the other three segments are also $\frac{1}{12}$ of the circumference of the circle. Hence, the perimeter of the shaded area $=\frac{1}{3}$ of the circumference of the circle, and the solution can be obtained using the formula for the circumference of a circle.

【Solution】Solution: It is easy to know from the problem that $\triangle A B F$ is an equilateral triangle, so arc $\widehat{\mathrm{AGF}}$ is $\frac{1}{6}$ of the circumference of the circle. Similarly, arc $\widehat{\mathrm{FGC}}$ is also $\frac{1}{6}$ of the circumference of the circle. Therefore, arc $\widehat{\mathrm{GF}}=\widehat{\mathrm{AGF}}+\widehat{\mathrm{FGC}}-\widehat{\mathrm{AGC}}=\frac{1}{12}$ of the circumference of the circle. Similarly, the other three segments are also $\frac{1}{12}$ of the circumference of the circle.
Hence, the perimeter of the shaded area $=\frac{1}{3}$ of the circumference of the circle $=\frac{2 \times 2 \times \pi}{3}=4.188$ (cm); Answer: The perimeter of the shaded area in the middle is 4.188 cm.

Therefore, the answer is: 4.188.