Question

question 1 of 28 which of the following represents side lengths that form a pythagorean triple? a. 3,4,5 b. 1,2,3 c. 4,5,6 d. 7,8,9

Explanation:

Step1: Recall Pythagorean theorem

For a right - triangle with side lengths \(a\), \(b\), and hypotenuse \(c\), \(a^{2}+b^{2}=c^{2}\). A Pythagorean triple satisfies this equation.

Step2: Check option A

For \(a = 3\), \(b = 4\), \(c = 5\). Calculate \(a^{2}+b^{2}\) and \(c^{2}\). \(a^{2}=3^{2}=9\), \(b^{2}=4^{2}=16\), so \(a^{2}+b^{2}=9 + 16=25\). And \(c^{2}=5^{2}=25\). Since \(a^{2}+b^{2}=c^{2}\), \((3,4,5)\) is a Pythagorean triple.

Step3: Check option B

For \(a = 1\), \(b = 2\), \(c = 3\). \(a^{2}=1^{2}=1\), \(b^{2}=2^{2}=4\), \(a^{2}+b^{2}=1 + 4=5\), \(c^{2}=3^{2}=9\). Since \(a^{2}+b^{2}
eq c^{2}\), \((1,2,3)\) is not a Pythagorean triple.

Step4: Check option C

For \(a = 4\), \(b = 5\), \(c = 6\). \(a^{2}=4^{2}=16\), \(b^{2}=5^{2}=25\), \(a^{2}+b^{2}=16+25 = 41\), \(c^{2}=6^{2}=36\). Since \(a^{2}+b^{2}
eq c^{2}\), \((4,5,6)\) is not a Pythagorean triple.

Step5: Check option D

For \(a = 7\), \(b = 8\), \(c = 9\). \(a^{2}=7^{2}=49\), \(b^{2}=8^{2}=64\), \(a^{2}+b^{2}=49 + 64=113\), \(c^{2}=9^{2}=81\). Since \(a^{2}+b^{2}
eq c^{2}\), \((7,8,9)\) is not a Pythagorean triple.

Answer:

A. 3, 4, 5