Answer :
To determine where Alex can place point [tex]\( H \)[/tex] so that he does not need more fencing, we need to follow several steps.
1. Calculate the distances between points [tex]\( E, F, \)[/tex] and [tex]\( G \)[/tex]:
- [tex]\( E(1, 5) \)[/tex] to [tex]\( F(3, 5) \)[/tex]
- [tex]\( F(3, 5) \)[/tex] to [tex]\( G(6, 1) \)[/tex]
- [tex]\( E(1, 5) \)[/tex] to [tex]\( G(6, 1) \)[/tex]
2. Determine the total fencing already used.
3. Subtract this amount from the total fencing available (16 units) to find the remaining fencing available for side [tex]\( GH \)[/tex].
4. Check if point [tex]\( H \)[/tex] at the given coordinates can be fenced without exceeding the remaining available fencing.
### Step-by-Step Solution:
1. Distance [tex]\( EF \)[/tex]:
- Coordinates: [tex]\( E(1,5) \)[/tex] and [tex]\( F(3,5) \)[/tex]
- Using the distance formula: [tex]\( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)[/tex]
- [tex]\( d(EF) = \sqrt{(3-1)^2 + (5-5)^2} \)[/tex]
- [tex]\( d(EF) = \sqrt{4} = 2 \)[/tex]
2. Distance [tex]\( FG \)[/tex]:
- Coordinates: [tex]\( F(3,5) \)[/tex] and [tex]\( G(6,1) \)[/tex]
- Using the distance formula:
- [tex]\( d(FG) = \sqrt{(6-3)^2 + (1-5)^2} \)[/tex]
- [tex]\( d(FG) = \sqrt{9 + 16} = \sqrt{25} = 5 \)[/tex]
3. Distance [tex]\( EG \)[/tex]:
- Coordinates: [tex]\( E(1,5) \)[/tex] and [tex]\( G(6,1) \)[/tex]
- Using the distance formula:
- [tex]\( d(EG) = \sqrt{(6-1)^2 + (1-5)^2} \)[/tex]
- [tex]\( d(EG) = \sqrt{25 + 16} = \sqrt{41} \approx 6.4 \)[/tex]
4. Calculate the total fencing used so far:
- [tex]\( Total\ fencing\ used = d(EF) + d(FG) + d(EG) \)[/tex]
- [tex]\( Total\ fencing\ used = 2 + 5 + 6.4 = 13.4 \)[/tex]
5. Determine the remaining fencing for side [tex]\( GH \)[/tex]:
- [tex]\( Remaining\ fencing = Total\ fencing\ available - Total\ fencing\ used \)[/tex]
- [tex]\( Remaining\ fencing = 16 - 13.4 = 2.6 \)[/tex]
6. Evaluate the possible points for [tex]\( H \)[/tex]:
- For each proposed point, we will calculate the distance [tex]\( GH \)[/tex] and check if it is less than or equal to 2.6 units.
- For [tex]\( H(0, 1) \)[/tex]:
- [tex]\( d(GH) = \sqrt{(0-6)^2 + (1-1)^2} = \sqrt{36} = 6 \)[/tex]
- [tex]\( 6 > 2.6 \)[/tex]: Not possible.
- For [tex]\( H(0, -2) \)[/tex]:
- [tex]\( d(GH) = \sqrt{(0-6)^2 + (-2-1)^2} = \sqrt{36 + 9} = \sqrt{45} \approx 6.7 \)[/tex]
- [tex]\( 6.7 > 2.6 \)[/tex]: Not possible.
- For [tex]\( H(1, 1) \)[/tex]:
- [tex]\( d(GH) = \sqrt{(1-6)^2 + (1-1)^2} = \sqrt{25} = 5 \)[/tex]
- [tex]\( 5 > 2.6 \)[/tex]: Not possible.
- For [tex]\( H(1, -2) \)[/tex]:
- [tex]\( d(GH) = \sqrt{(1-6)^2 + (-2-1)^2} = \sqrt{25 + 9} = \sqrt{34} \approx 5.8 \)[/tex]
- [tex]\( 5.8 > 2.6 \)[/tex]: Not possible.
### Conclusion:
None of the proposed points [tex]\( (0,1) \)[/tex], [tex]\( (0,-2) \)[/tex], [tex]\( (1,1) \)[/tex], or [tex]\( (1,-2) \)[/tex] can be chosen as point [tex]\( H \)[/tex] without exceeding the remaining 2.6 units of fencing.
Thus, Alex cannot place point [tex]\( H \)[/tex] in any of the given proposed locations without having to buy more fencing.
1. Calculate the distances between points [tex]\( E, F, \)[/tex] and [tex]\( G \)[/tex]:
- [tex]\( E(1, 5) \)[/tex] to [tex]\( F(3, 5) \)[/tex]
- [tex]\( F(3, 5) \)[/tex] to [tex]\( G(6, 1) \)[/tex]
- [tex]\( E(1, 5) \)[/tex] to [tex]\( G(6, 1) \)[/tex]
2. Determine the total fencing already used.
3. Subtract this amount from the total fencing available (16 units) to find the remaining fencing available for side [tex]\( GH \)[/tex].
4. Check if point [tex]\( H \)[/tex] at the given coordinates can be fenced without exceeding the remaining available fencing.
### Step-by-Step Solution:
1. Distance [tex]\( EF \)[/tex]:
- Coordinates: [tex]\( E(1,5) \)[/tex] and [tex]\( F(3,5) \)[/tex]
- Using the distance formula: [tex]\( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)[/tex]
- [tex]\( d(EF) = \sqrt{(3-1)^2 + (5-5)^2} \)[/tex]
- [tex]\( d(EF) = \sqrt{4} = 2 \)[/tex]
2. Distance [tex]\( FG \)[/tex]:
- Coordinates: [tex]\( F(3,5) \)[/tex] and [tex]\( G(6,1) \)[/tex]
- Using the distance formula:
- [tex]\( d(FG) = \sqrt{(6-3)^2 + (1-5)^2} \)[/tex]
- [tex]\( d(FG) = \sqrt{9 + 16} = \sqrt{25} = 5 \)[/tex]
3. Distance [tex]\( EG \)[/tex]:
- Coordinates: [tex]\( E(1,5) \)[/tex] and [tex]\( G(6,1) \)[/tex]
- Using the distance formula:
- [tex]\( d(EG) = \sqrt{(6-1)^2 + (1-5)^2} \)[/tex]
- [tex]\( d(EG) = \sqrt{25 + 16} = \sqrt{41} \approx 6.4 \)[/tex]
4. Calculate the total fencing used so far:
- [tex]\( Total\ fencing\ used = d(EF) + d(FG) + d(EG) \)[/tex]
- [tex]\( Total\ fencing\ used = 2 + 5 + 6.4 = 13.4 \)[/tex]
5. Determine the remaining fencing for side [tex]\( GH \)[/tex]:
- [tex]\( Remaining\ fencing = Total\ fencing\ available - Total\ fencing\ used \)[/tex]
- [tex]\( Remaining\ fencing = 16 - 13.4 = 2.6 \)[/tex]
6. Evaluate the possible points for [tex]\( H \)[/tex]:
- For each proposed point, we will calculate the distance [tex]\( GH \)[/tex] and check if it is less than or equal to 2.6 units.
- For [tex]\( H(0, 1) \)[/tex]:
- [tex]\( d(GH) = \sqrt{(0-6)^2 + (1-1)^2} = \sqrt{36} = 6 \)[/tex]
- [tex]\( 6 > 2.6 \)[/tex]: Not possible.
- For [tex]\( H(0, -2) \)[/tex]:
- [tex]\( d(GH) = \sqrt{(0-6)^2 + (-2-1)^2} = \sqrt{36 + 9} = \sqrt{45} \approx 6.7 \)[/tex]
- [tex]\( 6.7 > 2.6 \)[/tex]: Not possible.
- For [tex]\( H(1, 1) \)[/tex]:
- [tex]\( d(GH) = \sqrt{(1-6)^2 + (1-1)^2} = \sqrt{25} = 5 \)[/tex]
- [tex]\( 5 > 2.6 \)[/tex]: Not possible.
- For [tex]\( H(1, -2) \)[/tex]:
- [tex]\( d(GH) = \sqrt{(1-6)^2 + (-2-1)^2} = \sqrt{25 + 9} = \sqrt{34} \approx 5.8 \)[/tex]
- [tex]\( 5.8 > 2.6 \)[/tex]: Not possible.
### Conclusion:
None of the proposed points [tex]\( (0,1) \)[/tex], [tex]\( (0,-2) \)[/tex], [tex]\( (1,1) \)[/tex], or [tex]\( (1,-2) \)[/tex] can be chosen as point [tex]\( H \)[/tex] without exceeding the remaining 2.6 units of fencing.
Thus, Alex cannot place point [tex]\( H \)[/tex] in any of the given proposed locations without having to buy more fencing.
