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

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samerrima | Student, Grade 9

Posted August 4, 2010 at 8:22 AM via web

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

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giorgiana1976 | College Teacher

Posted August 4, 2010 at 3:05 PM (Answer #1)

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The perimeter of a geometric shape is a sum of the lengths of the sides of the shape.

In our case, we have to determine the perimeter of a triangle, so we have to calculate first the lengths of the sides of the triangle.

The formula for calculating the length of the side AB is:

AB = sqrt [(xB-xA)^2 + (yB-yA)^2]

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

AB = sqrt(4+4)

AB = 2 sqrt 2

AC = sqrt [(xC-xA)^2 + (yC-yA)^2]

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

AC = sqrt 2

BC = sqrt [(xC-xB)^2 + (yC-yB)^2]

BC = sqrt [(-1-2)^2 + (2-5)^2]

BC = sqrt (18)

BC = 3sqrt 2

P = AB+AC+BC

P = 2 sqrt 2 + sqrt 2 + 3sqrt 2

P = 6 sqrt 2

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neela | High School Teacher

Posted August 4, 2010 at 3:31 PM (Answer #2)

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We find the sides of the triangle by distance formula.The distance d between (x1,y1) and (x2,y2) is given by:

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

So AB = sqrt{(2-0)^2+(5-3)^2} = sqrt(4+4) = sqrt8 = 2sqrt2.

BC = sqrt{((-1-2)^2+(2-5)^2 } = sqrt{9+9} = 3sqrt2.

CA = sqrt{(0- -1)^2+(3-2)^2 }= sqrt(1+1) = sqrt2.

Therefore the primeter of the triangle= AB+BC+CA = 2sqrt2+3sqrt2+sqrt2 = 6sqrt2 =   8.4853 nearly

AC = sqrt{(

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted August 4, 2010 at 8:26 AM (Answer #3)

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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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