Question

Consider a right-angled triangle with sides a = 6 and c = 9. Determine whether these values from a Pythagorean triple. Show your steps and explain your answer.

Answer 1

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.

1. **Calculate \(c^2\)**:
\[
c^2 = 9^2 = 81
\]

2. **Substitute \(a\) into the Pythagorean theorem and solve for \(b\)**:
\[
a^2 + b^2 = c^2
\]
\[
6^2 + b^2 = 81
\]
\[
36 + b^2 = 81
\]

3. **Rearranging the equation to solve for \(b^2\)**:
\[
b^2 = 81 - 36
\]
\[
b^2 = 45
\]

4. **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.

5. **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.