check the picture below.
the triangle has that base and that height, recall that A = 1/2 bh.
now as for the perimeter, you can pretty much count the units off the grid for the segment CB, so let's just find the lengths of AC and AB,
[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~)
% (c,d)
&C&(~ 6 &,& 2~)
\end{array}~~~
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{(6-2)^2+(2-8)^2}\implies AC=\sqrt{4^2+(-6)^2}
\\\\\\
AC=\sqrt{16+36}\implies AC=\sqrt{52}\implies AC=\sqrt{4\cdot 13}
\\\\\\
AC=\sqrt{2^2\cdot 13}\implies AC=2\sqrt{13}[/tex]
[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~)
% (c,d)
&B&(~ 16 &,& 2~)
\end{array}\\\\\\
AB=\sqrt{(16-2)^2+(2-8)^2}\implies AB=\sqrt{14^2+(-6)^2}
\\\\\\
AB=\sqrt{196+36}\implies AB=\sqrt{232}\implies AB=\sqrt{4\cdot 58}
\\\\\\
AB=\sqrt{2^2\cdot 58}\implies AB=2\sqrt{58}[/tex]
so, add AC + AB + CB, and that's the perimeter of the triangle.
