Question

The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). what is the perimeter and area in square units

Answer 1

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,

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &A&(~ 2 &,& 8~) % (c,d) &C&(~ 6 &,& 2~) \end{array}~~~ % distance value d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AC=\sqrt{(6-2)^2+(2-8)^2}\implies AC=\sqrt{4^2+(-6)^2} \\\\\\ AC=\sqrt{16+36}\implies AC=\sqrt{52}\implies AC=\sqrt{4\cdot 13} \\\\\\ AC=\sqrt{2^2\cdot 13}\implies AC=2\sqrt{13}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &A&(~ 2 &,& 8~) % (c,d) &B&(~ 16 &,& 2~) \end{array}\\\\\\ AB=\sqrt{(16-2)^2+(2-8)^2}\implies AB=\sqrt{14^2+(-6)^2} \\\\\\ AB=\sqrt{196+36}\implies AB=\sqrt{232}\implies AB=\sqrt{4\cdot 58} \\\\\\ AB=\sqrt{2^2\cdot 58}\implies AB=2\sqrt{58}$

so, add AC + AB + CB, and that's the perimeter of the triangle.