To determine if the sides \(a = 6\) and \(c = 9\) of a right-angled triangle form part of a Pythagorean triple, we first need to recognize the notation for a right-angled triangle, where:
- \(a\) is one of the legs,
- \(b\) is the other leg,
- \(c\) is the hypotenuse (the longest side opposite the right angle).
In a Pythagorean triple, the sides of the triangle must satisfy the equation:
\[ a^2 + b^2 = c^2 \]
From the values given, we know:
- \(a = 6\)
- \(c = 9\)
Now we need to find \(b\) and verify if there exists an integer \(b\) such that the Pythagorean theorem holds.
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Calculate \(c^2\): \[ c^2 = 9^2 = 81 \]
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Substitute \(a\) into the Pythagorean theorem and solve for \(b\): \[ a^2 + b^2 = c^2 \] \[ 6^2 + b^2 = 81 \] \[ 36 + b^2 = 81 \]
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Rearranging the equation to solve for \(b^2\): \[ b^2 = 81 - 36 \] \[ b^2 = 45 \]
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Taking the square root to find \(b\): \[ b = \sqrt{45} = \sqrt{9 \cdot 5} = 3\sqrt{5} \] Since \(3\sqrt{5}\) is not an integer, \(b\) is not an integer.
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Conclusion: The sides \(a = 6\), \(b = 3\sqrt{5}\), and \(c = 9\) do not form a Pythagorean triple because for a set of Pythagorean triples, all sides must be integers. Thus, \(a = 6\), \(b = 3\sqrt{5}\), and \(c = 9\) is not an integer solution.
Therefore, the values \(6\) and \(9\) cannot be considered parts of a Pythagorean triple as \(b\) is not an integer.
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