Find the perimeter of the triangle whose vertices A(2,3)  B(4,8)  and C ( -1,-6)

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

Posted on

Given the points A(2,3)  B(4,8)  and C ( -1,-6) are the vertices's of the triangle.

We need to find the perimeter of the triangle ABC.

First, we will determine the length of the sides:

AB = sqrt( 4-2)^2( + ( 8-3)^2

      = sqrt(4 + 25) = sqrt29

==> AB = sqrt29

AC = sqrt(2+1)^2 + ( 3+6)^2

     = sqrt(3 + 81) = sqrt84 = 2sqrt21

==> AC= 2sqrt21

BC = sqrt(4+1)^2 ( 8+6)^2

    = sqrt(25+ 196) = sqrt221

BC = sqrt221

Then the perimeter is:

P = sqrt221 + sqrt29 + 2sqrt21

     = 14.9 + 5.4 + 9.2 = 29.5

Then, the perimeter of the triangle = 29.5

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

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

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

Therefore the distance between A(2,3) and B(4,8) is given by:

AB = sqrt{ (4-2)^2 +(8-3)^2} = sqrt(2^2+5^2) = sqrt29.

The distance between B(4,8) and C(-1, -6) is given by:

BC = sqrt{((-1-4)^2 +(-6-8)^2} = sqrt{(-5)^2+(-14)^2} = sqrt221.

The distance between C(-,1, -6) and A(2,3) is given by:

CA = {(-1-2)^2 + (3-(-6)^2} = sqrt{9+81 = sqrt90.

Therefore the perimeter of  the triangle ABC = AB+BC+CA = sqrt29+sqrt221+sqrt90 = 29.738 nearly.

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