Question

using the identity for generating pythagorean triples, find three positive integers that could be the sides of a right triangle when one side is 24 feet

Answer 1

To generate Pythagorean triples, we can use the formula:

\[
a = m^2 - n^2, \quad b = 2mn, \quad c = m^2 + n^2
\]

where \(m\) and \(n\) are positive integers with \(m > n\).

Since one of the sides of the triangle is given as 24 feet, we'll check if we can express 24 as either \(a\) or \(b\).

### Case 1: Let \(b = 24\)

Set \(b = 2mn\):
\[
2mn = 24
\]
\[
mn = 12
\]

Now, we need to find pairs of positive integers \((m, n)\) such that \(mn = 12\). The factor pairs of 12 that satisfy \(m > n\) are:

- \( (4, 3) \)
- \( (6, 2) \)
- \( (12, 1) \)

#### For \( (m, n) = (4, 3) \)
\[
a = m^2 - n^2 = 4^2 - 3^2 = 16 - 9 = 7
\]
\[
c = m^2 + n^2 = 4^2 + 3^2 = 16 + 9 = 25
\]
Thus, one Pythagorean triple is \( (7, 24, 25) \).

#### For \( (m, n) = (6, 2) \)
\[
a = m^2 - n^2 = 6^2 - 2^2 = 36 - 4 = 32
\]
\[
c = m^2 + n^2 = 6^2 + 2^2 = 36 + 4 = 40
\]
Thus, another Pythagorean triple is \( (32, 24, 40) \).

#### For \( (m, n) = (12, 1) \)
\[
a = m^2 - n^2 = 12^2 - 1^2 = 144 - 1 = 143
\]
\[
c = m^2 + n^2 = 12^2 + 1^2 = 144 + 1 = 145
\]
Thus, another Pythagorean triple is \( (143, 24, 145) \).

### Conclusion
The Pythagorean triples we found that include the side of 24 feet are:

1. \( (7, 24, 25) \)
2. \( (32, 24, 40) \)
3. \( (143, 24, 145) \)

These integers can represent the sides of right triangles with one side being 24 feet.