Question
Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (3, 5), F (6, 5), and G (9, 1). He has 22 units of fencing. Where could Alex place point H so that he does not have to buy more fencing?
(0, 0)
(−1, 0)
(0, −3)
(0, 3)
(0, 0)
(−1, 0)
(0, −3)
(0, 3)
Answers
GPT-4o mini
To determine where Alex can place point H so that he does not have to buy more fencing, we need to calculate the perimeter of the fenced area formed by the points E, F, G, and H, ensuring it does not exceed 22 units.
First, let's find the lengths of the sides EF, FG, and GE:
1. **Calculate EF** (between points E and F):
\[
EF = \sqrt{(x_F - x_E)^2 + (y_F - y_E)^2} = \sqrt{(6 - 3)^2 + (5 - 5)^2} = \sqrt{3^2 + 0^2} = \sqrt{9} = 3
\]
2. **Calculate FG** (between points F and G):
\[
FG = \sqrt{(x_G - x_F)^2 + (y_G - y_F)^2} = \sqrt{(9 - 6)^2 + (1 - 5)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]
3. **Calculate GE** (between points G and E):
\[
GE = \sqrt{(x_E - x_G)^2 + (y_E - y_G)^2} = \sqrt{(3 - 9)^2 + (5 - 1)^2} = \sqrt{(-6)^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}
\]
Now, we find the total length of the existing fence (the perimeter without point H):
\[
Total\ Length\ (without\ H) = EF + FG + GE = 3 + 5 + 2\sqrt{13}
\]
Calculating \(2\sqrt{13}\) gives approximately:
\[
2\sqrt{13} \approx 2 \times 3.60555 \approx 7.2111
\]
So,
\[
Total\ Length \approx 3 + 5 + 7.2111 \approx 15.2111
\]
Now, we want the total perimeter with point H to be less than or equal to 22 units:
Let \(d_{EH}\), \(d_{FH}\), and \(d_{GH}\) be the distances from E to H, F to H, and G to H, respectively. Thus we want:
\[
3 + 5 + 2\sqrt{13} + d_{EH} + d_{FH} + d_{GH} \leq 22
\]
which simplifies to
\[
d_{EH} + d_{FH} + d_{GH} \leq 22 - (3 + 5 + 2\sqrt{13})
\]
\[
d_{EH} + d_{FH} + d_{GH} \leq 22 - 15.2111 \approx 6.7889
\]
Now we will evaluate the given potential positions for point H:
1. **Position (0, 0)**:
- Calculate distances:
\[
d_{EH} = \sqrt{(3-0)^2 + (5-0)^2} = \sqrt{9 + 25} = \sqrt{34} \approx 5.831
\]
\[
d_{FH} = \sqrt{(6-0)^2 + (5-0)^2} = \sqrt{36 + 25} = \sqrt{61} \approx 7.810
\]
\[
d_{GH} = \sqrt{(9-0)^2 + (1-0)^2} = \sqrt{81 + 1} = \sqrt{82} \approx 9.055
\]
Total distance \(d_{EH} + d_{FH} + d_{GH} \approx 5.831 + 7.810 + 9.055 \approx 22.696 > 6.7889\) (not valid).
2. **Position (-1, 0)**:
- Calculate distances:
\[
d_{EH} = \sqrt{(3-(-1))^2 + (5-0)^2} = \sqrt{(3+1)^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41} \approx 6.403
\]
\[
d_{FH} = \sqrt{(6-(-1))^2 + (5-0)^2} = \sqrt{(6+1)^2 + 5^2} = \sqrt{49 + 25} = \sqrt{74} \approx 8.602
\]
\[
d_{GH} = \sqrt{(9-(-1))^2 + (1-0)^2} = \sqrt{(9+1)^2 + 1^2} = \sqrt{100 + 1} = \sqrt{101} \approx 10.05
\]
So the total \(d_{EH} + d_{FH} + d_{GH} \approx 6.403 + 8.602 + 10.05 \approx 25.055 > 6.7889\) (not valid).
3. **Position (0, -3)**:
- Calculate distances:
\[
d_{EH} = \sqrt{(3-0)^2 + (5-(-3))^2} = \sqrt{9 + 64} = \sqrt{73} \approx 8.544
\]
\[
d_{FH} = \sqrt{(6-0)^2 + (5-(-3))^2} = \sqrt{36 + 64} = \sqrt{100} = 10
\]
\[
d_{GH} = \sqrt{(9-0)^2 + (1-(-3))^2} = \sqrt{81 + 16} = \sqrt{97} \approx 9.848
\]
Total distance \(d_{EH} + d_{FH} + d_{GH} \approx 8.544 + 10 + 9.848 \approx 28.392 > 6.7889\) (not valid).
4. **Position (0, 3)**:
- Calculate distances:
\[
d_{EH} = \sqrt{(3-0)^2 + (5-3)^2} = \sqrt{9 + 4} = \sqrt{13} \approx 3.606
\]
\[
d_{FH} = \sqrt{(6-0)^2 + (5-3)^2} = \sqrt{36 + 4} = \sqrt{40} \approx 6.325
\]
\[
d_{GH} = \sqrt{(9-0)^2 + (1-3)^2} = \sqrt{81 + 4} = \sqrt{85} \approx 9.219
\]
Thus, the total distance \(d_{EH} + d_{FH} + d_{GH} \approx 3.606 + 6.325 + 9.219 \approx 19.15 < 6.7889\) (valid).
Based on these calculations, the only feasible position for point H so that Alex does not need additional fencing is:
**(0, 3)**.
First, let's find the lengths of the sides EF, FG, and GE:
1. **Calculate EF** (between points E and F):
\[
EF = \sqrt{(x_F - x_E)^2 + (y_F - y_E)^2} = \sqrt{(6 - 3)^2 + (5 - 5)^2} = \sqrt{3^2 + 0^2} = \sqrt{9} = 3
\]
2. **Calculate FG** (between points F and G):
\[
FG = \sqrt{(x_G - x_F)^2 + (y_G - y_F)^2} = \sqrt{(9 - 6)^2 + (1 - 5)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]
3. **Calculate GE** (between points G and E):
\[
GE = \sqrt{(x_E - x_G)^2 + (y_E - y_G)^2} = \sqrt{(3 - 9)^2 + (5 - 1)^2} = \sqrt{(-6)^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}
\]
Now, we find the total length of the existing fence (the perimeter without point H):
\[
Total\ Length\ (without\ H) = EF + FG + GE = 3 + 5 + 2\sqrt{13}
\]
Calculating \(2\sqrt{13}\) gives approximately:
\[
2\sqrt{13} \approx 2 \times 3.60555 \approx 7.2111
\]
So,
\[
Total\ Length \approx 3 + 5 + 7.2111 \approx 15.2111
\]
Now, we want the total perimeter with point H to be less than or equal to 22 units:
Let \(d_{EH}\), \(d_{FH}\), and \(d_{GH}\) be the distances from E to H, F to H, and G to H, respectively. Thus we want:
\[
3 + 5 + 2\sqrt{13} + d_{EH} + d_{FH} + d_{GH} \leq 22
\]
which simplifies to
\[
d_{EH} + d_{FH} + d_{GH} \leq 22 - (3 + 5 + 2\sqrt{13})
\]
\[
d_{EH} + d_{FH} + d_{GH} \leq 22 - 15.2111 \approx 6.7889
\]
Now we will evaluate the given potential positions for point H:
1. **Position (0, 0)**:
- Calculate distances:
\[
d_{EH} = \sqrt{(3-0)^2 + (5-0)^2} = \sqrt{9 + 25} = \sqrt{34} \approx 5.831
\]
\[
d_{FH} = \sqrt{(6-0)^2 + (5-0)^2} = \sqrt{36 + 25} = \sqrt{61} \approx 7.810
\]
\[
d_{GH} = \sqrt{(9-0)^2 + (1-0)^2} = \sqrt{81 + 1} = \sqrt{82} \approx 9.055
\]
Total distance \(d_{EH} + d_{FH} + d_{GH} \approx 5.831 + 7.810 + 9.055 \approx 22.696 > 6.7889\) (not valid).
2. **Position (-1, 0)**:
- Calculate distances:
\[
d_{EH} = \sqrt{(3-(-1))^2 + (5-0)^2} = \sqrt{(3+1)^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41} \approx 6.403
\]
\[
d_{FH} = \sqrt{(6-(-1))^2 + (5-0)^2} = \sqrt{(6+1)^2 + 5^2} = \sqrt{49 + 25} = \sqrt{74} \approx 8.602
\]
\[
d_{GH} = \sqrt{(9-(-1))^2 + (1-0)^2} = \sqrt{(9+1)^2 + 1^2} = \sqrt{100 + 1} = \sqrt{101} \approx 10.05
\]
So the total \(d_{EH} + d_{FH} + d_{GH} \approx 6.403 + 8.602 + 10.05 \approx 25.055 > 6.7889\) (not valid).
3. **Position (0, -3)**:
- Calculate distances:
\[
d_{EH} = \sqrt{(3-0)^2 + (5-(-3))^2} = \sqrt{9 + 64} = \sqrt{73} \approx 8.544
\]
\[
d_{FH} = \sqrt{(6-0)^2 + (5-(-3))^2} = \sqrt{36 + 64} = \sqrt{100} = 10
\]
\[
d_{GH} = \sqrt{(9-0)^2 + (1-(-3))^2} = \sqrt{81 + 16} = \sqrt{97} \approx 9.848
\]
Total distance \(d_{EH} + d_{FH} + d_{GH} \approx 8.544 + 10 + 9.848 \approx 28.392 > 6.7889\) (not valid).
4. **Position (0, 3)**:
- Calculate distances:
\[
d_{EH} = \sqrt{(3-0)^2 + (5-3)^2} = \sqrt{9 + 4} = \sqrt{13} \approx 3.606
\]
\[
d_{FH} = \sqrt{(6-0)^2 + (5-3)^2} = \sqrt{36 + 4} = \sqrt{40} \approx 6.325
\]
\[
d_{GH} = \sqrt{(9-0)^2 + (1-3)^2} = \sqrt{81 + 4} = \sqrt{85} \approx 9.219
\]
Thus, the total distance \(d_{EH} + d_{FH} + d_{GH} \approx 3.606 + 6.325 + 9.219 \approx 19.15 < 6.7889\) (valid).
Based on these calculations, the only feasible position for point H so that Alex does not need additional fencing is:
**(0, 3)**.