Question

Find the perimeter of the triangle whose vertices are the following specified (0,3), (-10,-4) (-9,-7) points in the plane.

Answer 1

The perimeter of the triangle is 28.82

Explanation:

Please check the image uploaded for the diagram.

Perimeter = d₁ + d₂ + d₃

`d_1 = √((-10-0)^2 + (-4-3)^2) = √(100+49) =√(149) = 12.21\\\\d_2 = √((-9-0)^2 + (-7-3)^2) = √(81+100) =√(181) = 13.45\\\\d_3 = √((-10+9)^2 + (-4+7)^2) = √(1+9) =√(10) = 3.16\\\\Perimeter = 12.21 + 13.45 + 3.16 = 28.82`