Question

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 1

Answer: $\bold{4+3\sqrt2+\sqrt{10}}$

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.

$d_{AB}=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$

     $=\sqrt{(-2+1)^2+(1-4)^2}$

     $=\sqrt{(-1)^2+(-3)^2}$

     $=\sqrt{1+9}$

     $=\sqrt{10}$

$d_{AC}=\sqrt{(x_C-x_A)^2+(y_C-y_A)^2}$

     $=\sqrt{(2+1)^2+(1-4)^2}$

     $=\sqrt{(3)^2+(-3)^2}$

     $=\sqrt{9+9}$

     $=\sqrt{18}$

     $=3\sqrt{2}$

$d_{BC}=\sqrt{(x_C-x_B)^2+(y_C-y_B)^2}$

     $=\sqrt{(2+2)^2+(1-1)^2}$

     $=\sqrt{(4)^2+(0)^2}$

     $=\sqrt{16+0}$

     $=\sqrt{16}$

     $=4$