To the nearest tenth, find the perimeter of ABC with vertices A(-1,4), B(-2,1) and C(2,1). show your work
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we have to firstly apply the distance formula to find the length of sides of the triangle
distance formula = [tex] \sqrt{(x_{2} -x_{1})^2 +(y_{2}- y_{1})^2} [/tex]
So length AB = [tex] \sqrt{(-2-(-1))^2+(1-4)^2}= \sqrt{1 +9}=\sqrt{10} [/tex]
BC= [tex] \sqrt{(2-(-2))^2+(1-1)^2}= \sqrt{16+0}=4 [/tex]
AC = [tex] \sqrt{(2-(-1))^2+ (1-4)^2}= \sqrt{9+9}=\sqrt{18}=3\sqrt{2} [/tex]
Now perimeter = AB+BC+AC = [tex] \sqrt{10}+4+3\sqrt{2} [/tex]
Plug in calculator
perimeter= 11.4 units