Question

reasoning can 96 and 100 be part of a pythagorean triple? yes no explain your reasoning.

Explanation:

Step1: Recall Pythagorean triple formula

A Pythagorean triple consists of three positive integers \(a\), \(b\), and \(c\) such that \(a^{2}+b^{2}=c^{2}\) (or \(a^{2}+c^{2}=b^{2}\), \(b^{2}+c^{2}=a^{2}\) depending on which is the hypotenuse). Let's assume two cases: one where 100 is the hypotenuse (the largest side) and one where 100 is a leg.

Step2: Case 1: 100 is the hypotenuse (\(c = 100\), \(a=96\))

We need to find \(b\) such that \(96^{2}+b^{2}=100^{2}\). Calculate \(96^{2}=9216\) and \(100^{2} = 10000\). Then \(b^{2}=10000 - 9216=784\). Taking the square root of 784, we get \(b = \sqrt{784}=28\), which is an integer.

Step3: Verify the triple

So the triple would be 28, 96, 100 since \(28^{2}+96^{2}=784 + 9216=10000=100^{2}\).

Answer:

Yes, because if we consider 96 and 100 as two sides (with 100 as the hypotenuse), we can find the third side \(b\) using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\). For \(a = 96\) and \(c=100\), \(b^{2}=100^{2}-96^{2}=10000 - 9216 = 784\), and \(b=\sqrt{784}=28\) (an integer). Thus, 28, 96, 100 is a Pythagorean triple.