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.

To find the perimeter of a triangle, you need to calculate the distance between all pairs of vertices and then sum them up.

Let's start by finding the distance between points A and B.

Distance between two points (x1, y1) and (x2, y2) can be calculated using the distance formula:

Distance = √((x2 - x1)^2 + (y2 - y1)^2)

For points A(0, 8) and B(6, -2), the distance is:

AB = √((6 - 0)^2 + (-2 - 8)^2)
= √(6^2 + (-10)^2)
= √(36 + 100)
= √136
≈ 11.66 (rounded to two decimal places)

Now let's find the distance between points A and C.

AC = √((-4 - 0)^2 + (4 - 8)^2)
= √((-4)^2 + (-4)^2)
= √(16 + 16)
= √32
≈ 5.66 (rounded to two decimal places)

Finally, let's calculate the distance between points B and C.

BC = √((-4 - 6)^2 + (4 - (-2))^2)
= √((-4 - 6)^2 + (4 + 2)^2)
= √((-10)^2 + 6^2)
= √(100 + 36)
= √136
≈ 11.66 (rounded to two decimal places)

Now, we can sum up all these distances to find the perimeter of the triangle ABC:

Perimeter = AB + AC + BC
≈ 11.66 + 5.66 + 11.66
≈ 28.98 (rounded to two decimal places)

Therefore, the approximate perimeter of triangle ABC is 28.98 units.