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Find the approximate perimeter of triangle abc with vertices a(0,8 b(6,-2 c(-4,4

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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.
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