To determine the counterexample for the conditional statement, "If a square has side length [tex]\( s \)[/tex], then the perimeter is less than the area," consider the mathematical definitions and calculate them for each of the given side lengths.
1. Formulas:
- 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 the same square is [tex]\( A = s^2 \)[/tex].
2. Check each side length:
- For [tex]\( s = 3 \)[/tex]:
- Perimeter: [tex]\( P = 4 \times 3 = 12 \)[/tex]
- Area: [tex]\( A = 3^2 = 9 \)[/tex]
- Compare: [tex]\( 12 < 9 \)[/tex] which is false
- For [tex]\( s = 5 \)[/tex]:
- Perimeter: [tex]\( P = 4 \times 5 = 20 \)[/tex]
- Area: [tex]\( A = 5^2 = 25 \)[/tex]
- Compare: [tex]\( 20 < 25 \)[/tex] which is true
- For [tex]\( s = 7 \)[/tex]:
- Perimeter: [tex]\( P = 4 \times 7 = 28 \)[/tex]
- Area: [tex]\( A = 7^2 = 49 \)[/tex]
- Compare: [tex]\( 28 < 49 \)[/tex] which is true
- For [tex]\( s = 9 \)[/tex]:
- Perimeter: [tex]\( P = 4 \times 9 = 36 \)[/tex]
- Area: [tex]\( A = 9^2 = 81 \)[/tex]
- Compare: [tex]\( 36 < 81 \)[/tex] which is true
3. Conclusion:
- Out of the given side lengths, when [tex]\( s = 3 \)[/tex], the statement that the perimeter is less than the area is false.
- Hence, [tex]\( s = 3 \)[/tex] serves as a counterexample to the conditional statement.
So, 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].
