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,
![\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 = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ AC=√((6-2)^2+(2-8)^2)\implies AC=√(4^2+(-6)^2) \\\\\\ AC=√(16+36)\implies AC=√(52)\implies AC=√(4\cdot 13) \\\\\\ AC=√(2^2\cdot 13)\implies AC=2√(13)]()
![\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=√((16-2)^2+(2-8)^2)\implies AB=√(14^2+(-6)^2) \\\\\\ AB=√(196+36)\implies AB=√(232)\implies AB=√(4\cdot 58) \\\\\\ AB=√(2^2\cdot 58)\implies AB=2√(58)]()
so, add AC + AB + CB, and that's the perimeter of the triangle.