11. Given that a regular $n$-sided polygon has $n+3$ diagonals, its perimeter is $x$, and the sum of the lengths of the diagonals is $y$. Then $\frac{y}{x}=$ $\qquad$
$$
=.11 .1+\sqrt{3} \text {. }
$$

The number of diagonals is $\frac{n(n-3)}{2}=n+3$. Therefore,
$$
n^{2}-5 n-6=0 \text {. }
$$

Solving gives $n=6$ or $n=-1$ (discard).
$$
\text { Hence } \frac{y}{x}=1+\sqrt{3} \text {. }
$$