Answer
$\left(4\sqrt{2}+2\sqrt{106}\right)$ units
Work Step by Step
The Distance Formula, $d$, between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $
d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}
$.
Using the Distance Formula, the distance, $d_1$, between $(2,6)$ and $(6,2)$
is
\begin{align*}\require{cancel}
d_1&=
\sqrt{(2-6)^2+(6-2)^2}
\\&=
\sqrt{(-4)^2+(4)^2}
\\&=
\sqrt{16+16}
\\&=
\sqrt{32}
\\&=
\sqrt{16\cdot2}
\\&=
\sqrt{16}\cdot\sqrt{2}
\\&=
4\sqrt{2}
.\end{align*}
Using the Distance Formula, the distance, $d_2$, between $(6,2)$ and $(-3,-3)$
is
\begin{align*}\require{cancel}
d_2&=
\sqrt{(6-(-3))^2+(2-(-3))^2}
\\&=
\sqrt{(6+3)^2+(2+3)^2}
\\&=
\sqrt{9^2+5^2}
\\&=
\sqrt{81+25}
\\&=
\sqrt{106}
.\end{align*}
Using the Distance Formula, the distance, $d_3$, between $(-3,-3)$ and $(2,6)$
is
\begin{align*}\require{cancel}
d_3&=
\sqrt{(-3-2)^2+(-3-6)^2}
\\&=
\sqrt{(-5)^2+(-9)^2}
\\&=
\sqrt{25+81}
\\&=
\sqrt{106}
.\end{align*}
Adding all the distances, then the perimeter, $P$, of the given triangle is
\begin{align*}\require{cancel}
P&=
d_1+d_2+d_3
\\&=
4\sqrt{2}+\sqrt{106}+\sqrt{106}
\\&=
4\sqrt{2}+2\sqrt{106}
.\end{align*}
Hence, the perimeter of the triangle is $\left(4\sqrt{2}+2\sqrt{106}\right)$ units.
