To the nearest tenth, find the perimeter of ABC with vertices A(-2-2), B(0,5) and C(3,-1).

Answer 1

AB=sqrt{(x₂-x₁)²+(y₂-y₁)²} =
=sqrt{(0-(-2))²+(5-(-2)) ²} =
=sqrt(4+49)=7.3

BC = sqrt{(x₃-x₂ )²+(y₃-y₂)²} =
=sqrt{(3-0)²+(-1-5) ²} =
=sqrt(9+36)=6.7

CA = sqrt{(x₁-x₃ )²+(y₁-y₃)²} =
=sqrt{(-2-3)²+(-2+1) ²} =
=sqrt(25+1)=5.1

P=AB+BC+CA = 7.3+6.7+5.1 =14.1

Answer 2

19.1 units

AB= 7.3
BC=6.8
CA=5