By definition, the perimeter of the triangle is the sum of its sides.
We must then use the formula of distance between points:
[tex]d = \sqrt{(x2-x1) ^ 2 + (y2-y1) ^ 2} [/tex]
We now look for the longitus for each of the sides:
For L1:
[tex]L1 = \sqrt{(6-2)^2 + (6-2)^2} [/tex]
[tex]L1 = \sqrt{32}
[/tex]
[tex]L1 = 4\sqrt{2} [/tex]
For L2:
[tex]L2 = \sqrt{(7-2)^2 + (-3-2)^2} [/tex]
[tex]L2 = \sqrt{50} [/tex]
[tex]L2 = 5\sqrt{2} [/tex]
For L3:
[tex]L3 = \sqrt{(7-6)^2 + (-3-6)^2} [/tex]
[tex]L3 = \sqrt{82} [/tex]
Then, the perimeter is given by:
P = L1 + L2 + L3
Substituting values we have:
[tex]P = 4\sqrt{2} + 5\sqrt{2} + \sqrt{82}
P = 9\sqrt{2} + \sqrt{82} [/tex]
Answer:
the perimeter of triangle ABC is:
none of the above.
