Question

find the exact value of x. x = do the side lengths form a pythagorean triple? yes no

Explanation:

Step1: Apply Pythagorean theorem

For a right - triangle with legs \(a = 16\), \(b=30\) and hypotenuse \(x\), the Pythagorean theorem states that \(x^{2}=a^{2}+b^{2}\).
Substitute \(a = 16\) and \(b = 30\) into the formula:
\(x^{2}=16^{2}+30^{2}\)
\(16^{2}=256\) and \(30^{2} = 900\), so \(x^{2}=256 + 900=1156\)

Step2: Solve for \(x\)

Take the square root of both sides: \(x=\sqrt{1156}\)
Since \(34\times34 = 1156\), we have \(x = 34\)

Now, check if \(16\), \(30\), \(34\) form a Pythagorean triple. A Pythagorean triple consists of three positive integers \(a\), \(b\), \(c\) such that \(a^{2}+b^{2}=c^{2}\). Here, \(16\), \(30\), \(34\) are all positive integers and \(16^{2}+30^{2}=256 + 900 = 1156=34^{2}\), so they do form a Pythagorean triple. But in the original problem, the selected option was wrong. Let's re - evaluate the Pythagorean triple part.

Wait, maybe there was a miscalculation. Wait, \(16^{2}=256\), \(30^{2}=900\), \(256 + 900=1156\), and \(\sqrt{1156}=34\). Now, check if \(16\), \(30\), \(34\) have a common factor. The greatest common divisor of \(16\), \(30\), \(34\) is \(2\). Dividing each by \(2\) gives \(8\), \(15\), \(17\). And \(8^{2}+15^{2}=64 + 225=289 = 17^{2}\), so \(16\), \(30\), \(34\) is a Pythagorean triple (since they are positive integers satisfying \(a^{2}+b^{2}=c^{2}\)).

Answer:

For the value of \(x\): \(x = 34\)

For the Pythagorean triple question, the correct answer is "Yes" (the original selected "No" was incorrect). So the correct options are:
\(x=\boldsymbol{34}\)
Do the side lengths form a Pythagorean triple? \(\boldsymbol{\text{Yes}}\)