Calculate the perimeter of triangle ABC. A(1,1) , B(0,3) , C(1,-1)

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hala718's profile pic

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

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A(1, 1)  B(0, 3)    C(1, -1)

First we will calculate the measure of the sides:

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

       = sqrt( 1 + 4)

        = sqrt(5)

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

       = sqrt(1+ 16)

        = sqrt(17)

AC = sqrt[(1-1)^2  + (-1-1)^2]

     = sqrt(0+ 4)

     = sqrt4 = 2

Then the perimeter is:

P = AB + BC + AC

   = sqrt5 + sqrt17 + 2

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The perimeter of a triangle is the sum of the lengths of the sides of the triangle.

Because we don't have the lengths of the sides of the triangle, we'll have to calculate them.

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

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

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

AB = sqrt(1+4)

AB = sqrt 5

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

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

AC = sqrt 4

AC = 2

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

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

BC = sqrt (17)

P = AB+AC+BC

P = sqrt 5 + 2 + sqrt 17

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

The distance between the point (x1,y1) and (x2,y2) is given as

sqrt [(x2-x1)^2 + (y2-y1)^2].

For the points:

  • A and B

Length of AB is sqrt [(0-1)^2 + (3-1)^2]

=sqrt [1+4]= sqrt 5

  • B and C

Length of BC is sqrt [(1-0)^2 + (-1-3)^2]

=sqrt [1^2 + 4^2]= sqrt 17

  • C and A

Length of CA is sqrt [(1-1)^2 + (-1-1)^2]

=sqrt [0 + 4]=2

Therefore the perimeter is 2 + sqrt 5 + sqrt 17

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The perimeter of the triangle ABC = AB+BC+CA.

A(1,1) , B(0,3) , C(1,-1)

We know that the distance  d between the the points (x1,y1) and x2,y2) is given by:

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

AB = sqrt((0-1)^2+(3-1)^2} = sqrt(1+2^2 ) = sqrt5.

BC= sqrt{(1-0)^2+(-1-3)^2} = sqrt(1+(-4)^2} = sqrt17.

CA = sqrt{(1-1)^2+(1- -1)^2} = sqrt 4 = 2.

Therefore Perimeter of the triangle ABC = sqrt5+sqrt17 +2  = 8.26 units.

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