Question

In a coordinate plane, triangle ABC has vertices A(- 4, 5), B(- 4, - 4) , and C(3,-4), . Find the perimeter of triangle ABC to the nearest tenth.

Answer 1

Answer: perimeter of triangle ABC to the nearest tenth = 27.4

Step-by-step explanation: Given that

vertices are

A(- 4, 5), B(- 4, - 4) , and C(3,-4)

Length AB = root( [-4-5]^2 + [-4+4]^2)

Length AB = sqr(9^2) = 9

Length AC = root( [-4-5]^2 + [3+4]^2)

Length AC = sqr( 9^2 + 7^2)

Length AC = sqr(130) = 11.4

Length BC = root( [-4+4]^2 + [3+4]^2)

Length BC = sqr( 7^2) = 7

Perimeter = 9 + 11.4 + 7 = 27.4