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

Respuesta :

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]