11. Consider an equilateral triangle with a side length of 1, denoted as $A_{1}$ (as shown in Figure 3).
Trisect each side of $A_{1}$, and construct an equilateral triangle outward on the middle segment, then remove the middle segment to obtain the figure denoted as $A_{2}$ (as shown in Figure 4); trisect each side of $A_{2}$, and repeat the above process to obtain the figure denoted as $A_{3}$ (as shown in Figure 5); trisect each side of $A_{3}$, and repeat the above process to obtain the figure denoted as $A_{4}$. Then, the perimeter of $A_{4}$ is $\qquad$
11. $\frac{64}{9}$.

Starting from $A_{1}$, each operation results in a figure whose perimeter is $\frac{4}{3}$ times the perimeter of the original figure. Therefore,
the perimeter of $A_{2}$ is $\frac{4}{3} \times 3=4$,
the perimeter of $A_{3}$ is $\frac{4}{3} \times 4=\frac{16}{3}$,
the perimeter of $A_{4}$ is $\frac{4}{3} \times \frac{16}{3}=\frac{64}{9}$.