Find the perimeter of the triangle whose vertices are A(0,3)  B(2, 5)  C (-1,2)

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We have the triangle ABC where A(0,3) B(2,5) and C(-1,2)

We need to calculate the length of the sides AB, AC, and BC

We knwo that the distance between two points is:

d = sqrt(x2-x1)^2 + (y2-y1)^2

AB = sqrt[(2-0)^2 + (5-3)^2]= sqrt(4+4) = sqrt8 = 2sqrt2

AC = sqrt[(-1-0)^2...

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We have the triangle ABC where A(0,3) B(2,5) and C(-1,2)

We need to calculate the length of the sides AB, AC, and BC

We knwo that the distance between two points is:

d = sqrt(x2-x1)^2 + (y2-y1)^2

AB = sqrt[(2-0)^2 + (5-3)^2]= sqrt(4+4) = sqrt8 = 2sqrt2

AC = sqrt[(-1-0)^2 + (2-3)^2] = sqrt(1+1) = sqrt2

BC = sqrt[(-1-2)^2 + (2-5)^2] = sqrt(9+9) = sqrt18 = 3sqrt2

Then the perimeter P is:

P = AB + AC + BC

  = 2sqrt2 + sqrt2 + 3sqrt2

  = 6sqrt2

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