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To determine a counterexample for the statement "If a square has side length [tex]\( s \)[/tex], then the perimeter is less than the area," we need to compare the perimeter and the area of the square for different values of [tex]\( s \)[/tex].
### Perimeter and Area Calculation
The perimeter [tex]\( P \)[/tex] of a square with side length [tex]\( s \)[/tex] is given by:
[tex]\[ P = 4s \][/tex]
The area [tex]\( A \)[/tex] of a square with side length [tex]\( s \)[/tex] is given by:
[tex]\[ A = s^2 \][/tex]
We need to identify a value of [tex]\( s \)[/tex] where the perimeter is not less than the area, i.e., [tex]\( P \geq A \)[/tex].
### Step-by-Step Calculation
#### For [tex]\( s = 3 \)[/tex]:
1. Perimeter: [tex]\( P = 4 \times 3 = 12 \)[/tex]
2. Area: [tex]\( A = 3^2 = 9 \)[/tex]
Since [tex]\( 12 \geq 9 \)[/tex]:
- The perimeter (12) is greater than the area (9) for [tex]\( s = 3 \)[/tex].
Therefore, [tex]\( s = 3 \)[/tex] is a counterexample for the conditional statement.
##### Verification for other values is unnecessary because we only need one counterexample to disprove the statement. However, for completeness, let's quickly review the other options:
#### For [tex]\( s = 5 \)[/tex]:
1. Perimeter: [tex]\( P = 4 \times 5 = 20 \)[/tex]
2. Area: [tex]\( A = 5^2 = 25 \)[/tex]
Since [tex]\( 20 < 25 \)[/tex]:
- The perimeter (20) is less than the area (25) for [tex]\( s = 5 \)[/tex].
Thus, [tex]\( s = 5 \)[/tex] is not a counterexample.
#### For [tex]\( s = 7 \)[/tex]:
1. Perimeter: [tex]\( P = 4 \times 7 = 28 \)[/tex]
2. Area: [tex]\( A = 7^2 = 49 \)[/tex]
Since [tex]\( 28 < 49 \)[/tex]:
- The perimeter (28) is less than the area (49) for [tex]\( s = 7 \)[/tex].
Thus, [tex]\( s = 7 \)[/tex] is not a counterexample.
#### For [tex]\( s = 9 \)[/tex]:
1. Perimeter: [tex]\( P = 4 \times 9 = 36 \)[/tex]
2. Area: [tex]\( A = 9^2 = 81 \)[/tex]
Since [tex]\( 36 < 81 \)[/tex]:
- The perimeter (36) is less than the area (81) for [tex]\( s = 9 \)[/tex].
Thus, [tex]\( s = 9 \)[/tex] is not a counterexample.
### Conclusion
The counterexample for the conditional statement "If a square has side length [tex]\( s \)[/tex], then the perimeter is less than the area" is:
- [tex]\( s = 3 \)[/tex]
For [tex]\( s = 3 \)[/tex], the perimeter (12) is greater than the area (9), disproving the statement.
### Perimeter and Area Calculation
The perimeter [tex]\( P \)[/tex] of a square with side length [tex]\( s \)[/tex] is given by:
[tex]\[ P = 4s \][/tex]
The area [tex]\( A \)[/tex] of a square with side length [tex]\( s \)[/tex] is given by:
[tex]\[ A = s^2 \][/tex]
We need to identify a value of [tex]\( s \)[/tex] where the perimeter is not less than the area, i.e., [tex]\( P \geq A \)[/tex].
### Step-by-Step Calculation
#### For [tex]\( s = 3 \)[/tex]:
1. Perimeter: [tex]\( P = 4 \times 3 = 12 \)[/tex]
2. Area: [tex]\( A = 3^2 = 9 \)[/tex]
Since [tex]\( 12 \geq 9 \)[/tex]:
- The perimeter (12) is greater than the area (9) for [tex]\( s = 3 \)[/tex].
Therefore, [tex]\( s = 3 \)[/tex] is a counterexample for the conditional statement.
##### Verification for other values is unnecessary because we only need one counterexample to disprove the statement. However, for completeness, let's quickly review the other options:
#### For [tex]\( s = 5 \)[/tex]:
1. Perimeter: [tex]\( P = 4 \times 5 = 20 \)[/tex]
2. Area: [tex]\( A = 5^2 = 25 \)[/tex]
Since [tex]\( 20 < 25 \)[/tex]:
- The perimeter (20) is less than the area (25) for [tex]\( s = 5 \)[/tex].
Thus, [tex]\( s = 5 \)[/tex] is not a counterexample.
#### For [tex]\( s = 7 \)[/tex]:
1. Perimeter: [tex]\( P = 4 \times 7 = 28 \)[/tex]
2. Area: [tex]\( A = 7^2 = 49 \)[/tex]
Since [tex]\( 28 < 49 \)[/tex]:
- The perimeter (28) is less than the area (49) for [tex]\( s = 7 \)[/tex].
Thus, [tex]\( s = 7 \)[/tex] is not a counterexample.
#### For [tex]\( s = 9 \)[/tex]:
1. Perimeter: [tex]\( P = 4 \times 9 = 36 \)[/tex]
2. Area: [tex]\( A = 9^2 = 81 \)[/tex]
Since [tex]\( 36 < 81 \)[/tex]:
- The perimeter (36) is less than the area (81) for [tex]\( s = 9 \)[/tex].
Thus, [tex]\( s = 9 \)[/tex] is not a counterexample.
### Conclusion
The counterexample for the conditional statement "If a square has side length [tex]\( s \)[/tex], then the perimeter is less than the area" is:
- [tex]\( s = 3 \)[/tex]
For [tex]\( s = 3 \)[/tex], the perimeter (12) is greater than the area (9), disproving the statement.
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