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Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points [tex]\(E(0,5)\)[/tex], [tex]\(F(5,5)\)[/tex], and [tex]\(G(1,1)\)[/tex]. He has 20 units of fencing. Where could Alex place point [tex]\(H\)[/tex] so that he does not have to buy more fencing?

A. [tex]\((-3,1)\)[/tex]
B. [tex]\((-3,-1)\)[/tex]
C. [tex]\((-5,1)\)[/tex]
D. [tex]\((-5,-1)\)[/tex]


Sagot :

Let's determine the correct position for point [tex]\( H \)[/tex] that ensures Alex uses no more than 20 units of fencing by considering the given points and the available fencing.

First, calculate the distances between the given points [tex]\( E \)[/tex], [tex]\( F \)[/tex], and [tex]\( G \)[/tex]:

1. Distance [tex]\( EF \)[/tex] between [tex]\( E(0, 5) \)[/tex] and [tex]\( F(5, 5) \)[/tex]:

[tex]\[ EF = \sqrt{(5 - 0)^2 + (5 - 5)^2} = \sqrt{5^2 + 0^2} = \sqrt{25} = 5 \text{ units} \][/tex]

2. Distance [tex]\( FG \)[/tex] between [tex]\( F(5, 5) \)[/tex] and [tex]\( G(1, 1) \)[/tex]:

[tex]\[ FG = \sqrt{(1 - 5)^2 + (1 - 5)^2} = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \text{ units} \][/tex]

3. Distance [tex]\( GE \)[/tex] between [tex]\( G(1, 1) \)[/tex] and [tex]\( E(0, 5) \)[/tex]:

[tex]\[ GE = \sqrt{(0 - 1)^2 + (5 - 1)^2} = \sqrt{(-1)^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17} \text{ units} \][/tex]

Next, calculate the total fencing used by summing these distances:

[tex]\[ \text{Total fencing used} = EF + FG + GE = 5 + 4\sqrt{2} + \sqrt{17} \][/tex]

Given that Alex has 20 units of fencing, the remaining fencing available for the fourth side [tex]\( GH \)[/tex] is:

[tex]\[ \text{Remaining fencing} = 20 - (5 + 4\sqrt{2} + \sqrt{17}) \][/tex]

To decide where point [tex]\( H \)[/tex] could be placed, we evaluate the distances from point [tex]\( G \)[/tex] to each of the four potential points for [tex]\( H \)[/tex]:

1. Distance [tex]\( GH1 \)[/tex] between [tex]\( G (1, 1) \)[/tex] and [tex]\( H1 (-3, 1) \)[/tex]:

[tex]\[ GH1 = \sqrt{(-3 - 1)^2 + (1 - 1)^2} = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4 \text{ units} \][/tex]

2. Distance [tex]\( GH2 \)[/tex] between [tex]\( G (1, 1) \)[/tex] and [tex]\( H2 (-3, -1) \)[/tex]:

[tex]\[ GH2 = \sqrt{(-3 - 1)^2 + (-1 - 1)^2} = \sqrt{(-4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \text{ units} \][/tex]

3. Distance [tex]\( GH3 \)[/tex] between [tex]\( G (1, 1) \)[/tex] and [tex]\( H3 (-5, 1) \)[/tex]:

[tex]\[ GH3 = \sqrt{(-5 - 1)^2 + (1 - 1)^2} = \sqrt{(-6)^2 + 0^2} = \sqrt{36} = 6 \text{ units} \][/tex]

4. Distance [tex]\( GH4 \)[/tex] between [tex]\( G (1, 1) \)[/tex] and [tex]\( H4 (-5, -1) \)[/tex]:

[tex]\[ GH4 = \sqrt{(-5 - 1)^2 + (-1 - 1)^2} = \sqrt{(-6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \text{ units} \][/tex]

Now, check which of these distances for points [tex]\( H1, H2, H3, H4 \)[/tex] satisfies the remaining fencing condition:

[tex]\[ 4 \leq \text{remaining fencing} \][/tex]

After checking the distances, it turns out that:

[tex]\[ GH1 = 4 \text{ units} \][/tex]

Thus, point [tex]\( H \)[/tex] could be placed at [tex]\( (-3,1) \)[/tex] so that Alex does not have to buy more fencing. Therefore, the correct answer is:

[tex]\[ \boxed{(-3, 1)} \][/tex]