Answer :
To determine a counterexample for the conditional statement, "If a square has side length [tex]\( s \)[/tex], then the perimeter is less than the area," we'll analyze the perimeters and areas of squares with given side lengths [tex]\( s = 3, 5, 7, \)[/tex] and [tex]\( 9 \)[/tex].
1. Determine the perimeter and area for each side length:
For a square with side length [tex]\( s \)[/tex]:
- The perimeter [tex]\( P \)[/tex] is given by [tex]\( P = 4s \)[/tex].
- The area [tex]\( A \)[/tex] is given by [tex]\( A = s^2 \)[/tex].
2. Calculate the perimeters and areas:
- For [tex]\( s = 3 \)[/tex]:
[tex]\[ P = 4 \times 3 = 12 \][/tex]
[tex]\[ A = 3^2 = 9 \][/tex]
- For [tex]\( s = 5 \)[/tex]:
[tex]\[ P = 4 \times 5 = 20 \][/tex]
[tex]\[ A = 5^2 = 25 \][/tex]
- For [tex]\( s = 7 \)[/tex]:
[tex]\[ P = 4 \times 7 = 28 \][/tex]
[tex]\[ A = 7^2 = 49 \][/tex]
- For [tex]\( s = 9 \)[/tex]:
[tex]\[ P = 4 \times 9 = 36 \][/tex]
[tex]\[ A = 9^2 = 81 \][/tex]
3. Compare the perimeter and area for each side length:
- For [tex]\( s = 3 \)[/tex]:
[tex]\[ P = 12 \quad \text{and} \quad A = 9 \][/tex]
Here, [tex]\( P \geq A \)[/tex]. (In fact, [tex]\( P > A \)[/tex])
- For [tex]\( s = 5 \)[/tex]:
[tex]\[ P = 20 \quad \text{and} \quad A = 25 \][/tex]
Here, [tex]\( P < A \)[/tex].
- For [tex]\( s = 7 \)[/tex]:
[tex]\[ P = 28 \quad \text{and} \quad A = 49 \][/tex]
Here, [tex]\( P < A \)[/tex].
- For [tex]\( s = 9 \)[/tex]:
[tex]\[ P = 36 \quad \text{and} \quad A = 81 \][/tex]
Here, [tex]\( P < A \)[/tex].
4. Identify the counterexample:
We need to find a value of [tex]\( s \)[/tex] for which the perimeter [tex]\( P \)[/tex] is not less than the area [tex]\( A \)[/tex]. From the calculations above, we see that for [tex]\( s = 3 \)[/tex], the perimeter [tex]\( P = 12 \)[/tex] is greater than the area [tex]\( A = 9 \)[/tex].
Therefore, the counterexample to the conditional statement is:
[tex]\[ \boxed{s = 3} \][/tex]
1. Determine the perimeter and area for each side length:
For a square with side length [tex]\( s \)[/tex]:
- The perimeter [tex]\( P \)[/tex] is given by [tex]\( P = 4s \)[/tex].
- The area [tex]\( A \)[/tex] is given by [tex]\( A = s^2 \)[/tex].
2. Calculate the perimeters and areas:
- For [tex]\( s = 3 \)[/tex]:
[tex]\[ P = 4 \times 3 = 12 \][/tex]
[tex]\[ A = 3^2 = 9 \][/tex]
- For [tex]\( s = 5 \)[/tex]:
[tex]\[ P = 4 \times 5 = 20 \][/tex]
[tex]\[ A = 5^2 = 25 \][/tex]
- For [tex]\( s = 7 \)[/tex]:
[tex]\[ P = 4 \times 7 = 28 \][/tex]
[tex]\[ A = 7^2 = 49 \][/tex]
- For [tex]\( s = 9 \)[/tex]:
[tex]\[ P = 4 \times 9 = 36 \][/tex]
[tex]\[ A = 9^2 = 81 \][/tex]
3. Compare the perimeter and area for each side length:
- For [tex]\( s = 3 \)[/tex]:
[tex]\[ P = 12 \quad \text{and} \quad A = 9 \][/tex]
Here, [tex]\( P \geq A \)[/tex]. (In fact, [tex]\( P > A \)[/tex])
- For [tex]\( s = 5 \)[/tex]:
[tex]\[ P = 20 \quad \text{and} \quad A = 25 \][/tex]
Here, [tex]\( P < A \)[/tex].
- For [tex]\( s = 7 \)[/tex]:
[tex]\[ P = 28 \quad \text{and} \quad A = 49 \][/tex]
Here, [tex]\( P < A \)[/tex].
- For [tex]\( s = 9 \)[/tex]:
[tex]\[ P = 36 \quad \text{and} \quad A = 81 \][/tex]
Here, [tex]\( P < A \)[/tex].
4. Identify the counterexample:
We need to find a value of [tex]\( s \)[/tex] for which the perimeter [tex]\( P \)[/tex] is not less than the area [tex]\( A \)[/tex]. From the calculations above, we see that for [tex]\( s = 3 \)[/tex], the perimeter [tex]\( P = 12 \)[/tex] is greater than the area [tex]\( A = 9 \)[/tex].
Therefore, the counterexample to the conditional statement is:
[tex]\[ \boxed{s = 3} \][/tex]
