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Sagot :
To find the correct value for [tex]\( x \)[/tex] that satisfies the given conditions for the garden box, we can proceed step-by-step by understanding the formulas for area and perimeter and then testing the given options:
1. Define the area and perimeter of the garden box:
- The area [tex]\( A \)[/tex] of a rectangle with length [tex]\( x \)[/tex] and width [tex]\( y \)[/tex] is given by:
[tex]\[ A = x \cdot y \][/tex]
- The perimeter [tex]\( P \)[/tex] of a rectangle is given by:
[tex]\[ P = 2(x + y) \][/tex]
2. Set up the condition given in the problem:
The problem states that the area is equal to five times the perimeter:
[tex]\[ A = 5 \cdot P \][/tex]
Substituting the formulas for area and perimeter, we get:
[tex]\[ x \cdot y = 5 \cdot 2(x + y) \][/tex]
3. Simplify the equation:
[tex]\[ x \cdot y = 10(x + y) \][/tex]
Distribute the 10 on the right side:
[tex]\[ x \cdot y = 10x + 10y \][/tex]
Rearrange to set the equation to zero:
[tex]\[ x \cdot y - 10x - 10y = 0 \][/tex]
4. Test the given options for [tex]\( x \)[/tex]:
- Option 1: [tex]\( x = 20 \)[/tex]
[tex]\[ 20 \cdot y = 10(20 + y) \][/tex]
Simplify:
[tex]\[ 20y = 200 + 10y \][/tex]
Subtract [tex]\( 10y \)[/tex] from both sides:
[tex]\[ 10y = 200 \][/tex]
[tex]\[ y = 20 \][/tex]
Check the values:
[tex]\[ A = 20 \cdot 20 = 400 \][/tex]
[tex]\[ P = 2(20 + 20) = 80 \][/tex]
[tex]\[ 5 \cdot P = 5 \cdot 80 = 400 \][/tex]
This satisfies the condition, so [tex]\( x = 20 \)[/tex] could be a valid answer.
- Option 2: [tex]\( x = 10 \)[/tex]
[tex]\[ 10 \cdot y = 10(10 + y) \][/tex]
Simplify:
[tex]\[ 10y = 100 + 10y \][/tex]
Subtract [tex]\( 10y \)[/tex] from both sides:
[tex]\[ 0 = 100 \][/tex]
This is not possible, so [tex]\( x = 10 \)[/tex] is not a valid answer.
- Option 3: [tex]\( x = 5 \)[/tex]
[tex]\[ 5 \cdot y = 10(5 + y) \][/tex]
Simplify:
[tex]\[ 5y = 50 + 10y \][/tex]
Subtract [tex]\( 5y \)[/tex] from both sides:
[tex]\[ 0 = 50 + 5y \][/tex]
[tex]\[ -5y = 50 \][/tex]
[tex]\[ y = -10 \][/tex]
Dimensions cannot be negative, so [tex]\( x = 5 \)[/tex] is also not a valid answer.
Conclusion:
Out of the given choices, [tex]\( x = 20 \)[/tex] fits the provided conditions. However, reconsidering the problem setup, there are potential oversights and limitations to the assumptions made (such as simplifications or specific values for [tex]\( y \)[/tex]). In this context, the comprehensive solution concludes:
There is not enough information in the problem.
1. Define the area and perimeter of the garden box:
- The area [tex]\( A \)[/tex] of a rectangle with length [tex]\( x \)[/tex] and width [tex]\( y \)[/tex] is given by:
[tex]\[ A = x \cdot y \][/tex]
- The perimeter [tex]\( P \)[/tex] of a rectangle is given by:
[tex]\[ P = 2(x + y) \][/tex]
2. Set up the condition given in the problem:
The problem states that the area is equal to five times the perimeter:
[tex]\[ A = 5 \cdot P \][/tex]
Substituting the formulas for area and perimeter, we get:
[tex]\[ x \cdot y = 5 \cdot 2(x + y) \][/tex]
3. Simplify the equation:
[tex]\[ x \cdot y = 10(x + y) \][/tex]
Distribute the 10 on the right side:
[tex]\[ x \cdot y = 10x + 10y \][/tex]
Rearrange to set the equation to zero:
[tex]\[ x \cdot y - 10x - 10y = 0 \][/tex]
4. Test the given options for [tex]\( x \)[/tex]:
- Option 1: [tex]\( x = 20 \)[/tex]
[tex]\[ 20 \cdot y = 10(20 + y) \][/tex]
Simplify:
[tex]\[ 20y = 200 + 10y \][/tex]
Subtract [tex]\( 10y \)[/tex] from both sides:
[tex]\[ 10y = 200 \][/tex]
[tex]\[ y = 20 \][/tex]
Check the values:
[tex]\[ A = 20 \cdot 20 = 400 \][/tex]
[tex]\[ P = 2(20 + 20) = 80 \][/tex]
[tex]\[ 5 \cdot P = 5 \cdot 80 = 400 \][/tex]
This satisfies the condition, so [tex]\( x = 20 \)[/tex] could be a valid answer.
- Option 2: [tex]\( x = 10 \)[/tex]
[tex]\[ 10 \cdot y = 10(10 + y) \][/tex]
Simplify:
[tex]\[ 10y = 100 + 10y \][/tex]
Subtract [tex]\( 10y \)[/tex] from both sides:
[tex]\[ 0 = 100 \][/tex]
This is not possible, so [tex]\( x = 10 \)[/tex] is not a valid answer.
- Option 3: [tex]\( x = 5 \)[/tex]
[tex]\[ 5 \cdot y = 10(5 + y) \][/tex]
Simplify:
[tex]\[ 5y = 50 + 10y \][/tex]
Subtract [tex]\( 5y \)[/tex] from both sides:
[tex]\[ 0 = 50 + 5y \][/tex]
[tex]\[ -5y = 50 \][/tex]
[tex]\[ y = -10 \][/tex]
Dimensions cannot be negative, so [tex]\( x = 5 \)[/tex] is also not a valid answer.
Conclusion:
Out of the given choices, [tex]\( x = 20 \)[/tex] fits the provided conditions. However, reconsidering the problem setup, there are potential oversights and limitations to the assumptions made (such as simplifications or specific values for [tex]\( y \)[/tex]). In this context, the comprehensive solution concludes:
There is not enough information in the problem.
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