Answer
$10+6\sqrt 5$
Work Step by Step
The perimeter of the triangle $AMN$ is:
$$P_{\triangle AMN}=AM+MN+NA.\tag1$$
Determine $AM$ using the Pythagorean theorem in the triangle $ABM$:
$$\begin{align*}
AM^2&=AB^2+BM^2\\
AM&=\sqrt{8^2+4^2}=\sqrt{80}=4\sqrt 5.
\end{align*}$$
Determine $MN$ using the Pythagorean theorem in the triangle $MCN$:
$$\begin{align*}
MN^2&=CM^2+CN^2\\
MN&=\sqrt{4^2+2^2}=\sqrt{20}=2\sqrt 5.
\end{align*}$$
Determine $NA$ using the Pythagorean theorem in the triangle $NDA$:
$$\begin{align*}
NA^2&=ND^2+DA^2\\
NA&=\sqrt{(8-2)^2+8^2}=\sqrt{100}=10.
\end{align*}$$
We substitute the values of $AM$, $MN$ and $NA$ in equation$(1)$:
$$P_{\triangle AMN}=4\sqrt 5+2\sqrt 5+10=10+6\sqrt 5.$$
