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The vertices of triangle ABC are A(-2, 2), B(6, 2), and C(0, 8). What is its perimeter?

1 Answer

Jan 8, 2018

$8+6sqrt2+2sqrt10$

Explanation:

We are essentially finding the distance between these points, and then add them up.
Remember the distance formula: $d=sqrt ((x_2-x_1)^2+(y_2-y_1)^2)$

The distance from points A to B is:
$d=sqrt ((6-(-2))^2+(2-2)^2)$
$d=sqrt (64+0)$
$d=8$

The distance from points B to C is:
$d=sqrt ((0-6)^2+(8-2)^2)$
$d=sqrt (36+36)$
$d=sqrt 72$
$d=6sqrt 2$

The distance from points A to C is:
$d=sqrt ((0-(-2))^2+(8-2)^2)$
$d=sqrt (4+36)$
$d=sqrt 40$
$d=2sqrt 10$

We add'em up.
$8+6sqrt2+2sqrt10$ That is your answer!

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