To find the perimeter of the parallelogram with the given vertices \( (2,1) \), \( (2,3) \), \( (4,4) \), and \( (4,2) \), we first identify the pairs of vertices that form the sides of the parallelogram.
The vertices are as follows:
- Vertex A: \( (2, 1) \)
- Vertex B: \( (2, 3) \)
- Vertex C: \( (4, 4) \)
- Vertex D: \( (4, 2) \)
Let's compute the lengths of the sides AB and BC.
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Length of side AB (between points \( A(2,1) \) and \( B(2,3) \)): \[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(2 - 2)^2 + (3 - 1)^2} = \sqrt{0 + 4} = \sqrt{4} = 2 \]
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Length of side BC (between points \( B(2,3) \) and \( C(4,4) \)): \[ \text{Distance} = \sqrt{(4 - 2)^2 + (4 - 3)^2} = \sqrt{(2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5} \approx 2.236 \]
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Length of side CD (between points \( C(4,4) \) and \( D(4,2) \)): \[ \text{Distance} = \sqrt{(4 - 4)^2 + (2 - 4)^2} = \sqrt{0 + 4} = \sqrt{4} = 2 \]
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Length of side DA (between points \( D(4,2) \) and \( A(2,1) \)): \[ \text{Distance} = \sqrt{(2 - 4)^2 + (1 - 2)^2} = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \approx 2.236 \]
Now, we can find the perimeter \( P \) of the parallelogram by adding the lengths of all four sides: \[ P = AB + BC + CD + DA = 2 + 2.236 + 2 + 2.236 = 8.472 \]
Rounding to the nearest tenth, we get: \[ P \approx 8.5 \text{ units} \]
Thus, the perimeter of the parallelogram is \( \boxed{8.5} \text{ units} \).