Question

What is the perimeter of the triangle shown on the coordinate plane, to the nearest tenth of a unit?

Answer 1

A(-4,-1), B(-1,7), C(6,3)

$AB = \sqrt{(-1 - -4)^2 + (7--1)^2}=\sqrt{3^2+8^2}=\sqrt{73}$

$BC = \sqrt{(6 - -1)^2+(7-3)^2}=\sqrt{49+16}=\sqrt{65}$

$AC=\sqrt{(6 - -4)^2+(3--1)^2} = \sqrt{116}= 2 \sqrt{29}$

$\textrm{perimeter} = \sqrt{73} + \sqrt{65} + 2 \sqrt{29} = 27.3766...$

Answer: 27.4