Find the perimeter of triangle ABC.

Answer:
21.79 units
Step-by-step explanation:
We have to use the distance formula, which says that for two points (x_1, y_1) and (x_2, y_2), the distance between them is: [tex]\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}[/tex]
First up is (2, 2) and (6, 6): [tex]\sqrt{(2-6)^2+(2-6)^2}=\sqrt{(-4)^2+(-4)^2}=\sqrt{16+16}=\sqrt{32} =4\sqrt{2}[/tex] ≈ 5.66
Next is (6, 6) and (7, -3):
[tex]\sqrt{(7-6)^2+(-3-6)^2}=\sqrt{(1)^2+(-9)^2}=\sqrt{1+81}=\sqrt{82}[/tex] ≈ 9.06
Finally is (2, 2) and (7, -3):
[tex]\sqrt{(2-7)^2+(2-(-3))^2}=\sqrt{(-5)^2+(5)^2}=\sqrt{25+25}=\sqrt{50} =5\sqrt{2}[/tex] ≈ 7.07
Add these up:
5.66 + 9.06 + 7.07 = 21.79 units
Hope this helps!