The area of parallelogram 1 is 4 square units greater than the area of parallelogram 2
<h3>How to compare the areas?</h3>
The coordinates are given as:
Parallelogram 1: (0, 2), (2, 6), (6, 4), and (4, 0).
Parallelogram 2: (2, 0), (4, -6), (2, -8), and (0, -2)
Next, calculate the base and the height
For parallelogram 1, we have:
Base = (0, 2) and (2, 6)
Height = (6, 4), and (4, 0).
So, we have:
![Base = \sqrt{(0 - 2)^2 + (2 -6)^2} = \sqrt{20}](https://tex.z-dn.net/?f=Base%20%3D%20%5Csqrt%7B%280%20-%202%29%5E2%20%2B%20%282%20-6%29%5E2%7D%20%3D%20%5Csqrt%7B20%7D)
![Height = \sqrt{(6 - 4)^2 + (4 -0)^2} = \sqrt{20}](https://tex.z-dn.net/?f=Height%20%3D%20%5Csqrt%7B%286%20-%204%29%5E2%20%2B%20%284%20-0%29%5E2%7D%20%3D%20%5Csqrt%7B20%7D)
The area is:
Area = Base * Height
Area = √20 * √20
Area = 20
For parallelogram 2, we have:
Base = (0, -2) and (4, -6)
Height = (4, -6) and (2, -8)
So, we have:
![Base = \sqrt{(0 -4)^2 + (-2 +6)^2} = \sqrt{32}](https://tex.z-dn.net/?f=Base%20%3D%20%5Csqrt%7B%280%20-4%29%5E2%20%2B%20%28-2%20%2B6%29%5E2%7D%20%3D%20%5Csqrt%7B32%7D)
![Height = \sqrt{(4 - 2)^2 + (-6 +8)^2} = \sqrt{8}](https://tex.z-dn.net/?f=Height%20%3D%20%5Csqrt%7B%284%20-%202%29%5E2%20%2B%20%28-6%20%2B8%29%5E2%7D%20%3D%20%5Csqrt%7B8%7D)
The area is:
Area = Base * Height
Area = √32 * √8
Area = 16
By comparing both areas, we have:
20 is greater than 16 by 4
This means that the area of parallelogram 1 is greater by 4 units
Read more about parallelogram at:
brainly.com/question/3050890
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