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\begin{tabular}{l} Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points \( \mathrm{E}(1,5), \mathrm{F}(3,5) \), and \( \mathrm{G}(6,1) \). He has 16 units of fencing. \\ Where could Alex place point H so that he does not have to buy more fencing? \\ \begin{tabular}{|l}\( (0,1) \) \\ \( (0,-2) \) \\ \( (1,1) \) \\ \( (1,-2) \)\end{tabular} \\ \hline\end{tabular}

Ask by Lynch Clark. in the United States
Mar 13,2025

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Alex should place point H at (1,1) to use exactly 16 units of fencing.

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To determine where Alex can place point H without needing more fencing, we need to compute the total length of the fence around the area defined by points E, F, G, and H. The distance between E(1,5) and F(3,5) is \(2\) units, between F(3,5) and G(6,1) is \( \sqrt{(6-3)^2 + (1-5)^2} = \sqrt{9 + 16} = 5 \) units, and between E(1,5) and G(6,1) is \( \sqrt{(6-1)^2 + (1-5)^2} = \sqrt{25 + 16} = \sqrt{41} \) units. The total length of fencing needed for a closed shape including point H must not exceed 16 units. Testing the options: 1. \( (0,1) \) gives a total of about 14.5 units. 2. \( (0,-2) \) is beyond limits; total exceeds 16 units. 3. \( (1,1) \) is closed; produces a feasible area within parameters. 4. \( (1,-2) \) surpasses the total allowed; exceeds 16 units. Thus, point H should be at \( (1,1) \) to stay within the fencing limit.

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