11. Given point $A(m, n)$ lies on the line $x+3 y=41$, where $0<n<m$. If the symmetric point of point $A$ with respect to the line $y=x$ is $B$, the symmetric point of point $B$ with respect to the $y$-axis is $C$, the symmetric point of point $C$ with respect to the $x$-axis is $D$, and the symmetric point of point $D$ with respect to the $y$-axis is $E$, and the area of pentagon $A B C D E$ is 451, then the coordinates of point $A$ are
11. $(11,10)$.

By symmetry, we get $B(n, m)$, $C(-n, m)$, $D(-n, -m)$, and $E(n, -m)$.
The area of pentagon $ABCDE$ is
$$
\begin{array}{l}
S_{\text {pentagon } ABCDE}=2 m \times 2 n+\frac{1}{2} \times 2 m \times(m-n) \\
=m(m+3 n)=41 m,
\end{array}
$$

Thus, $41 m=451$. Therefore, $m=11, n=10$.
Hence, the coordinates of point $A$ are $(11,10)$.