geometry
posted by Tina on .
Find the approximate perimeter of triangle abc with vertices a(0,8 b(6,2 c(4,4

A(0 , 8), B(6 , 2), C(4 , 4).
(AB)^2 = (6  0)^2 + (2  8)^2 = 136.
AB  sqrt(136) = 11.66.
(BC)^2 = (4 6)^2 + (4 + 2)^2 = 136,
BC = sqrt(136) = 11.66.
(AC)^2 = (4  0)^2 + (4  8)^2 = 32,
AC = sqrt(32) = 5.66.
P = 11.66 + 11.66 + 5.66 = 29.