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

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] Reply

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! Reply

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

Findingdistancebetweenthem:AB= [tex] \sqrt{ {(4 + 2)}^{2} + {(6 – 1)}^{2} } [/tex]= [tex] \sqrt{36 + 25} [/tex]

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

BC== [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]

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!