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: [tex]\bold{4+3\sqrt2+\sqrt{10}}[/tex]Step-by-step explanation:Perimeter is the sum of the lengths of the sides.  Use the distance formula to find the lengths of each of the sides, then find their sum.[tex]d_{AB}=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}[/tex]      [tex]=\sqrt{(-2+1)^2+(1-4)^2}[/tex]      [tex]=\sqrt{(-1)^2+(-3)^2}[/tex]      [tex]=\sqrt{1+9}[/tex]      [tex]=\sqrt{10}[/tex][tex]d_{AC}=\sqrt{(x_C-x_A)^2+(y_C-y_A)^2}[/tex]      [tex]=\sqrt{(2+1)^2+(1-4)^2}[/tex]      [tex]=\sqrt{(3)^2+(-3)^2}[/tex]      [tex]=\sqrt{9+9}[/tex]      [tex]=\sqrt{18}[/tex]      [tex]=3\sqrt{2}[/tex][tex]d_{BC}=\sqrt{(x_C-x_B)^2+(y_C-y_B)^2}[/tex]      [tex]=\sqrt{(2+2)^2+(1-1)^2}[/tex]      [tex]=\sqrt{(4)^2+(0)^2}[/tex]      [tex]=\sqrt{16+0}[/tex]      [tex]=\sqrt{16}[/tex]      [tex]=4[/tex]