Find the perimeter of the triangle whose vertices are
(-2, 1), (4, 6) and (6, -3).​

2 thoughts on “Find the perimeter of the triangle whose vertices are<br /> (-2, 1), (4, 6) and (6, -3).​”

1. Answer:

   [tex] \sqrt{80} + \sqrt{61} + \sqrt{85} [/tex]

   Step-by-step explanation:

   Perimeter means sum of all Sides.

   So, distance between two points (a,b) and (c,d) is defined as

   [tex] \sqrt{ {(c – a)}^{2} + \: {(d – b)}^{2} } [/tex]

   Assumptions:

   Let A = (-2,1) , B = (4,6) , C = (6,-3) and AB , BC, CA be the Sides of the Triangle

   Finding distance between them:

   AB = [tex] \sqrt{ {(4 + 2)}^{2} + {(6 – 1)}^{2} } [/tex]

   = [tex] \sqrt{36 + 25} [/tex]

   = [tex] \sqrt{61} [/tex]

   BC = [tex] \sqrt{ {(6 – 4)}^{2} + {( – 3 – 6)}^{2} } [/tex]

   = [tex] \sqrt{4 + 81} [/tex]

   = [tex] \sqrt{85} [/tex]

   CA = [tex] \sqrt{ {(6 + 2)}^{2} + {( – 3 – 1)}^{2} } [/tex]

   = [tex] \sqrt{64 + 16} [/tex]

   = [tex] \sqrt{80} [/tex]

   Result:

   Perimeter = Sum of all sides

   = [tex] \sqrt{80} + \sqrt{61} + \sqrt{85} [/tex]

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2. Answer:

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

   To find the length of each side, we use distance formula.

   Let A = (-2, 1)

   Let B = (4, 6)

   Let C = (6, -3)

   Distance AB = [tex]\sqrt{(4+2)^{2} +(6-1)^{2} } = \sqrt{36+25} = \sqrt{61}[/tex] units

   Distance BC = [tex]\sqrt{(6-4)^{2}+(-3+6)^{2}} = \sqrt{4+9} = \sqrt{13}[/tex] units

   Distance CA = [tex]\sqrt{(6+2)^2+(-3-1)^2} = \sqrt{64+16} = \sqrt{80}[/tex] units

   The required answer is the sum of these three values:

   [tex]\sqrt{61} + \sqrt{13} + \sqrt{80}[/tex] units

   Do mark as brainliest if it helped!

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