1. **State the problem:** We need to find the area and perimeter of a parallelogram with vertices A(-8,0), B(5,5), C(9,0), and D(-4,-5). Each unit on the graph corresponds to 5 meters. 2. **Find the side lengths for the perimeter:** Use the distance formula between points: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ Calculate AB: $$AB = \sqrt{(5 - (-8))^2 + (5 - 0)^2} = \sqrt{13^2 + 5^2} = \sqrt{169 + 25} = \sqrt{194}$$ Calculate BC: $$BC = \sqrt{(9 - 5)^2 + (0 - 5)^2} = \sqrt{4^2 + (-5)^2} = \sqrt{16 + 25} = \sqrt{41}$$ Since ABCD is a parallelogram, opposite sides are equal: $$CD = AB = \sqrt{194}, \quad DA = BC = \sqrt{41}$$ 3. **Calculate perimeter in graph units:** $$P = 2(AB + BC) = 2(\sqrt{194} + \sqrt{41})$$ 4. **Convert perimeter to meters:** Each unit = 5 meters, so $$P_{meters} = 5 \times 2(\sqrt{194} + \sqrt{41}) = 10(\sqrt{194} + \sqrt{41})$$ Calculate approximate values: $$\sqrt{194} \approx 13.9284, \quad \sqrt{41} \approx 6.4031$$ So, $$P_{meters} \approx 10(13.9284 + 6.4031) = 10(20.3315) = 203.315$$ Rounded to nearest hundredth: $$P_{meters} \approx 203.32$$ meters 5. **Find the area:** Area of parallelogram = base \times height. Use vector method: area = magnitude of cross product of adjacent sides. Vectors: $$\vec{AB} = (5 - (-8), 5 - 0) = (13, 5)$$ $$\vec{AD} = (-4 - (-8), -5 - 0) = (4, -5)$$ Area in graph units: $$|\vec{AB} \times \vec{AD}| = |13 \times (-5) - 5 \times 4| = |-65 - 20| = 85$$ 6. **Convert area to square meters:** Each unit length = 5 meters, so area scales by $5^2 = 25$: $$Area_{meters} = 85 \times 25 = 2125$$ 7. **Final answers:** - Area = 2125 square meters - Perimeter = 203.32 meters