Question

A triangle has vertices (-1, 2), (3, 1), and (7, 2). What is the approximate perimeter of the triangle? Round your answer to the nearest hundredth.

Answer 1

Use the distance formula to find the length of the sides, then add them up to find the perimeter.

$\sf d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

For points in the form of (x1, y1), (x2, y2).

(-1, 2), (3, 1)

$\sf d=\sqrt{(3+1)^2+(1-2)^2}$

$\sf d=\sqrt{(4)^2+(-1)^2}$

$\sf d=\sqrt{16+1}$

$\sf d=\sqrt{17}$

(3, 1), (7, 2)

$\sf d=\sqrt{(7-3)^2+(2-1)^2}$

$\sf d=\sqrt{(4)^2+(1)^2}$

$\sf d=\sqrt{16+1}$

$\sf d=\sqrt{17}$

(7, 2), (-1, 2)

$\sf d=\sqrt{(-1-7)^2+(2-2)^2}$

$\sf d=\sqrt{(-8)^2+(0)^2}$

$\sf d=\sqrt{64+0}$

$\sf d=\sqrt{64}=8$

So the perimeter will be:

$\sf 8+\sqrt{17}+\sqrt{17}\approx\boxed{\sf 16.25}$