Question

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

Answer 1

Let us find the co ordinates of each vertex of the triangle .

Vertex A ( in firs second quadrant) = ( -5 ,3)

vertex B in third quadrant = ( -5, -5)

vertex C in fourth quadrant = ( 4, -2)

let us use distance formula AB^2 = ( -5 - 3)^2 + (-5 - -5 )^2 = 64 + 0

AB= 8

BC^2 = ( -2 - -5 )^2 + ( 4 - - 5)^2 = 9 + 81 = 90

BC = 9.48

AC^2 = ( -2 -3)^2 +( 4- -5)^2 = 25 + 81 = 106

AC= 10.29

Perimeter = sum of length of AB+ BC+ Ac = 8 + 10.29 + 9.48= 22.77