Question

The coordinates of the vertices of △ABC are A(−2,2), B(5,−3), and C(−4,−1). Identify the perimeter of △ABC. Round each side length to the nearest tenth before adding.

Answer 1

The perimeter of the triangle is 21.4 units

How to determine the perimeter?

The points are given as:

A(−2,2), B(5,−3), and C(−4,−1)

Calculate the side lengths using:

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

So, we have:

$AB = \sqrt{(-2 -5)^2 + (2 + 3)^2} = 8.6$

$AC = \sqrt{(-2 +4)^2 + (2 + 1)^2} = 3.6$

$BC = \sqrt{(5+4)^2 + (-3 + 1)^2} = 9.2$

The perimeter is then calculated as:

P = AB + AC + BC

This gives

P = 8.6 + 3.6 + 9.2

Evaluate

P = 21.4

Hence, the perimeter of the triangle is 21.4 units

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