1. **State the problem:** We are given points A(2, -2), B(2, 3), C(-1, 5), and D(-1, 0) which form a parallelogram ABCD. We need to find the perimeter of this parallelogram. 2. **Recall the formula for distance between two points:** The distance between points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ 3. **Calculate the lengths of the sides:** - Side AB: $$AB = \sqrt{(2 - 2)^2 + (3 - (-2))^2} = \sqrt{0^2 + 5^2} = \sqrt{25} = 5$$ - Side BC: $$BC = \sqrt{(-1 - 2)^2 + (5 - 3)^2} = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$$ - Side CD: $$CD = \sqrt{(-1 - (-1))^2 + (0 - 5)^2} = \sqrt{0^2 + (-5)^2} = \sqrt{25} = 5$$ - Side DA: $$DA = \sqrt{(2 - (-1))^2 + (-2 - 0)^2} = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}$$ 4. **Check that opposite sides are equal:** - $AB = CD = 5$ - $BC = DA = \sqrt{13}$ 5. **Calculate the perimeter:** $$\text{Perimeter} = AB + BC + CD + DA = 5 + \sqrt{13} + 5 + \sqrt{13} = 10 + 2\sqrt{13}$$ 6. **Approximate the value:** $$\sqrt{13} \approx 3.6055$$ $$\text{Perimeter} \approx 10 + 2 \times 3.6055 = 10 + 7.211 = 17.211$$ Rounded to the nearest tenth: $$17.2$$ **Final answer:** The perimeter of parallelogram ABCD is approximately $17.2$ units.