Type the correct answer in each box. If necessary, round your answer(s) to the nearest hundredth.
The vertices of ∆ABC are A(-2, 2), B(6, 2), and C(0, 8). The perimeter of ∆ABC is units, and its area is square units.
To find the perimeter of ΔABC, we can use the distance formula to find the length of each side and then add them together.
The distance between A(-2, 2) and B(6, 2) is:
√((6 - (-2))^2 + (2 - 2)^2) = √((8)^2 + (0)^2) = √(64 + 0) = √64 = 8 units.
The distance between B(6, 2) and C(0, 8) is:
√((0 - 6)^2 + (8 - 2)^2) = √((-6)^2 + (6)^2) = √(36 + 36) = √72 ≈ 8.49 units.
The distance between C(0, 8) and A(-2, 2) is:
√((-2 - 0)^2 + (2 - 8)^2) = √((-2)^2 + (-6)^2) = √(4 + 36) = √40 ≈ 6.32 units.
Finally, to find the perimeter, we add all the side lengths together:
8 + 8.49 + 6.32 = 22.81 units.
The perimeter of ΔABC is approximately 22.81 units.
To find the area of ΔABC, we can use the Shoelace Formula.
Using the Shoelace Formula:
(((-2) * (2 + 8)) + (6 * (8 + 2)) + (0 * (2 + 2))) / 2 = ((-2 * 10) + ( 6 * 10) + (0 * 4)) / 2 = (-20 + 60 + 0) / 2 = 40 / 2 = 20 square units.
The area of ΔABC is 20 square units.