Question

To the nearest tenth, find the perimeter of ∆ABC with vertices A(-1,4), B(-2,1) and C(2,1).

Answer 1

First we need to find the length of each side of the triangle , and for that we need to use the distance formula, that is
$d =\sqrt{(x_{2} -x_{1})^2 +(y_{2} -y_{1})^2}$
$AB = \sqrt{(1-4)^2 +(-2+1)^2}= \sqrt{10}$
$\sqrt{BC} = \sqrt{(1-1)^2 + (2+2)^2}=4$
Perimeter is the sum of all sides.
$Perimeter = AB + BC + AC = \sqrt{10}+4 +\sqrt{18}$
$Perimeter= 11.4$