\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}

Question

UpStudy AI · Accepted Answer

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.

$$
\boxed{(1,1)}
$$
Alex should place point H at (1,1) to use exactly 16 units of fencing.

Upstudy AI Solution

Answer

Solution

Beyond the Answer

Related Questions

Latest Geometry Questions