Find the perimeter of a triangle with vertices A(2,5) B(2,-2) C(5,-2). Round your answer to the nearest tenth and show your work.​ Edit: Nvm I got it:)

Step-by-step explanation:

Question

Find the perimeter of a triangle with vertices A(2,5) B(2,-2) C(5,-2). Round your answer to the nearest tenth and show your work.​

perimeter of a triangle = AB+AC+BC

Using the distance formula

AB = sqrt(-2-5)²+(2-2)²

AB = sqrt(-7)²

AB =sqrt(49)

AB =7

BC = sqrt(-2+2)²+(2-5)²

BC = sqrt(0+3²)

BC =sqrt(9)

BC =3

AC= sqrt(-2-5)²+(2-5)²

AC= sqrt(-7)²+3²

AC =sqrt(49+9)

AC =sqrt58

Perimeter = 10+sqrt58

y = x + 95

y = 42 + 95

y=137 mm hg