To the nearest tenth, find the perimeter of ABC with vertices A(-1,4), B(-2,1) and C(2,1). show your work
![To the nearest tenth find the perimeter of ABC with vertices A14 B21 and C21 show your work class=](https://us-static.z-dn.net/files/df2/9caf26c63d9f72aea14c2edba5eaf2cf.jpg)
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