Question

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

Answer 1

we have to firstly apply the distance formula to find the length of sides of the triangle

distance formula = $\sqrt{(x_{2} -x_{1})^2 +(y_{2}- y_{1})^2}$

So length AB = $\sqrt{(-2-(-1))^2+(1-4)^2}= \sqrt{1 +9}=\sqrt{10}$

BC= $\sqrt{(2-(-2))^2+(1-1)^2}= \sqrt{16+0}=4$

AC = $\sqrt{(2-(-1))^2+ (1-4)^2}= \sqrt{9+9}=\sqrt{18}=3\sqrt{2}$

Now perimeter = AB+BC+AC = $\sqrt{10}+4+3\sqrt{2}$

Plug in calculator

perimeter= 11.4 units