Question

Find the perimeter of triangle ABC.

Answer 1

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: $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

First up is (2, 2) and (6, 6): $\sqrt{(2-6)^2+(2-6)^2}=\sqrt{(-4)^2+(-4)^2}=\sqrt{16+16}=\sqrt{32} =4\sqrt{2}$ ≈ 5.66

Next is (6, 6) and (7, -3):

$\sqrt{(7-6)^2+(-3-6)^2}=\sqrt{(1)^2+(-9)^2}=\sqrt{1+81}=\sqrt{82}$ ≈ 9.06

Finally is (2, 2) and (7, -3):

$\sqrt{(2-7)^2+(2-(-3))^2}=\sqrt{(-5)^2+(5)^2}=\sqrt{25+25}=\sqrt{50} =5\sqrt{2}$ ≈ 7.07

Add these up:

5.66 + 9.06 + 7.07 = 21.79 units

Hope this helps!