Question
Leon verified that the side lengths 21, 28, 35 form a Pythagorean triple using this procedure.
Step 1: Find the greatest common factor of the given lengths: 7
Step 2: Divide the given lengths by the greatest common factor: 3, 4, 5
Step 3: Verify that the lengths found in step 2 form a Pythagorean triple: 3 squared + 4 squared = 9 + 16 = 25 = 5 squared
Leon states that 21, 28, 35 is a Pythagorean triple because the lengths found in step 2 form a Pythagorean triple. Which explains whether or not Leon is correct? pick one
Yes, multiplying every length of a Pythagorean triple by the same whole number results in a Pythagorean triple.
Yes, any set of lengths with a common factor is a Pythagorean triple.
No, the lengths of Pythagorean triples cannot have any common factors.
No, the given side lengths can form a Pythagorean triple even if the lengths found in step 2 do not.
Step 1: Find the greatest common factor of the given lengths: 7
Step 2: Divide the given lengths by the greatest common factor: 3, 4, 5
Step 3: Verify that the lengths found in step 2 form a Pythagorean triple: 3 squared + 4 squared = 9 + 16 = 25 = 5 squared
Leon states that 21, 28, 35 is a Pythagorean triple because the lengths found in step 2 form a Pythagorean triple. Which explains whether or not Leon is correct? pick one
Yes, multiplying every length of a Pythagorean triple by the same whole number results in a Pythagorean triple.
Yes, any set of lengths with a common factor is a Pythagorean triple.
No, the lengths of Pythagorean triples cannot have any common factors.
No, the given side lengths can form a Pythagorean triple even if the lengths found in step 2 do not.
Answers
GPT-4o mini
Leon is correct in his reasoning, and the correct statement is:
**Yes, multiplying every length of a Pythagorean triple by the same whole number results in a Pythagorean triple.**
This is because if \( (a, b, c) \) is a Pythagorean triple that satisfies the equation \( a^2 + b^2 = c^2 \), then for any integer \( k \), the lengths \( (ka, kb, kc) \) will also satisfy \( (ka)^2 + (kb)^2 = (kc)^2 \). Since \( (3, 4, 5) \) is a Pythagorean triple, multiplying each side by 7 (the greatest common factor Leon found) gives the lengths of 21, 28, and 35, which confirms that these lengths can indeed form a Pythagorean triple.
**Yes, multiplying every length of a Pythagorean triple by the same whole number results in a Pythagorean triple.**
This is because if \( (a, b, c) \) is a Pythagorean triple that satisfies the equation \( a^2 + b^2 = c^2 \), then for any integer \( k \), the lengths \( (ka, kb, kc) \) will also satisfy \( (ka)^2 + (kb)^2 = (kc)^2 \). Since \( (3, 4, 5) \) is a Pythagorean triple, multiplying each side by 7 (the greatest common factor Leon found) gives the lengths of 21, 28, and 35, which confirms that these lengths can indeed form a Pythagorean triple.