Homework Help

Calculate the perimeter of the triangle whose vertices are (1,2) (-1,4) and * (-2,3).

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totoo | Student, Undergraduate | (Level 1) Honors

Posted December 4, 2010 at 1:05 PM via web

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Calculate the perimeter of the triangle whose vertices are (1,2) (-1,4) and * (-2,3).

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted December 4, 2010 at 1:09 PM (Answer #1)

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To find the perimeter we need the distances between the points (1,2) and  (-1,4), and the points (-1, 4) and (-2,3) and the points (-2,3) and (1,2).

The distance between (x1, y1) and (x2, y2) is given by sqrt[(x1 - x2)^2 + (y1 - y2)^2]

The distance between (1,2) and  (-1,4)

=> sqrt [(1 +1)^2 + (2 - 4)^2]

=> sqrt ( 2^2 + 2^2)

=> sqrt 8

The distance between (-1, 4) and (-2,3)

=> sqrt [(-1 + 2)^2 + (4 - 3)^2]

=> sqrt (1^2 + 1^2)

=> sqrt 2

The distance between (-2,3) and (1,2)

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

=> sqrt (3^2 + 1^2)

=> sqrt 10

Therefore the perimeter is sqrt 10 + sqrt 8 + sqrt 2.

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

Posted December 4, 2010 at 1:05 PM (Answer #2)

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Let ABC be a triangle.

Given the vertices's of a triangle are:

A(1,2) , B(-1,4), and C(-2,3).

We need to find the perimeter of the triangle.

First we need to find the length of the sides.

==> AB = sqrt[( 1+1)^2 + ( 2-4)^2)

            = sqrt( 2^2 + 2^2 )

            = sqrt( 4+4)

             = sqrt8

==> AB = 2sqrt2.

==> AC = sqrt( 1+ 2)^2 + ( 2-3)^2

              = sqrt( 3^2 + 1^2)

                = sqrt(9+ 1)

                = sqrt10.

==> AC= sqrt10.

==> BC = sqrt( -1+2)^2 +(4-3)^2]

             = sqrt( 1^2 + 1^2 )

            = sqrt2

==> BC = sqrt2.

==> The perimeter of the triangle is:

P = AB + AC + BC

   = 2sqrt2 + sqrt10 + sqrt2

   = 7.4 units. ( approx.)

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted December 4, 2010 at 4:38 PM (Answer #3)

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To determine the perimeter of a triangle, we have to know the lengths of the sides of the triangle.

We don't know the sides but we know the vertices of the triangle.

We'll calculate the lengths of the sides, using the distance formula:

We'll compute the length a:

a = sqrt[(-1-1)^2 + (4-2)^2]

a = sqrt(4 + 4)

a = 2sqrt2

We'll compute the length b:

b = sqrt[(-2+1)^2 + (3-4)^2]

b = sqrt(1+1)

b = sqrt2

We'll compute the length c:

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

c = sqrt(9 + 1)

c = sqrt10

The perimeter of the triangle is the sum of the lengths of the sides a,b,c:

P = a+b+c

P = 2sqrt2 + sqrt2 + sqrt10

We'll combine like terms:

P = 3sqrt2 + sqrt10

We'll factorize by sqrt2:

P = sqrt2(3 + sqrt5) units

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