\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}

We know three vertices of the quadrilateral: \[ E(1,5), \quad F(3,5), \quad G(6,1) \] and we want to choose \(H\) from the options so that the total fence length (perimeter) is 16 units. Assuming the vertices are connected in order \(E \to F \to G \to H \to E\), we first calculate the known side lengths. 1. From \(E\) to \(F\): \[ EF = \sqrt{(3-1)^2 + (5-5)^2} = \sqrt{2^2 + 0^2} = 2 \] 2. From \(F\) to \(G\): \[ FG = \sqrt{(6-3)^2 + (1-5)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = 5 \] The total used fencing for these two sides is: \[ EF + FG = 2 + 5 = 7 \] Since the entire perimeter is 16 units, the remaining two sides (from \(G\) to \(H\) and from \(H\) to \(E\)) must sum to: \[ 16 - 7 = 9 \] Now, we evaluate each candidate for \(H\): --- **Candidate 1: \(H(0,1)\)** - Distance from \(G\) to \(H\): \[ GH = \sqrt{(6-0)^2 + (1-1)^2} = \sqrt{6^2+0^2} = 6 \] - Distance from \(H\) to \(E\): \[ HE = \sqrt{(1-0)^2 + (5-1)^2} = \sqrt{1^2+4^2} = \sqrt{1+16} = \sqrt{17} \approx 4.12 \] - Sum: \[ 6 + \sqrt{17} \approx 6 + 4.12 = 10.12 \] This is more than 9. --- **Candidate 2: \(H(0,-2)\)** - Distance from \(G\) to \(H\): \[ GH = \sqrt{(6-0)^2 + (1-(-2))^2} = \sqrt{6^2+3^2} = \sqrt{36+9} = \sqrt{45} \approx 6.71 \] - Distance from \(H\) to \(E\): \[ HE = \sqrt{(1-0)^2 + (5-(-2))^2} = \sqrt{1^2+7^2} = \sqrt{1+49} = \sqrt{50} \approx 7.07 \] - Sum: \[ 6.71 + 7.07 \approx 13.78 \] This is more than 9. --- **Candidate 3: \(H(1,1)\)** - Distance from \(G\) to \(H\): \[ GH = \sqrt{(6-1)^2 + (1-1)^2} = \sqrt{5^2+0^2} = 5 \] - Distance from \(H\) to \(E\): \[ HE = \sqrt{(1-1)^2 + (5-1)^2} = \sqrt{0^2+4^2} = 4 \] - Sum: \[ 5 + 4 = 9 \] This exactly matches the required remaining length. --- **Candidate 4: \(H(1,-2)\)** - Distance from \(G\) to \(H\): \[ GH = \sqrt{(6-1)^2 + (1-(-2))^2} = \sqrt{5^2+3^2} = \sqrt{25+9} = \sqrt{34} \approx 5.83 \] - Distance from \(H\) to \(E\): \[ HE = \sqrt{(1-1)^2 + (5-(-2))^2} = \sqrt{0^2+7^2} = 7 \] - Sum: \[ 5.83 + 7 \approx 12.83 \] This is greater than 9. --- Since only the candidate \(H(1,1)\) gives a sum of \(GH + HE = 9\), it is the correct location where Alex can place point \(H\) without requiring extra fencing. \[ \boxed{(1,1)} \]