To determine the value of \( x \) that makes \( (x, 45, 51) \) a Pythagorean triple, we need to satisfy the Pythagorean theorem, which states that:
\[ a^2 + b^2 = c^2 \]
In our case, we can assume \( c = 51 \) (the largest side), \( b = 45 \), and \( a = x \). The equation thus becomes:
\[ x^2 + 45^2 = 51^2 \]
Calculating \( 45^2 \) and \( 51^2 \):
\[ 45^2 = 2025 \] \[ 51^2 = 2601 \]
Now, substituting these values into the equation gives:
\[ x^2 + 2025 = 2601 \]
To find \( x^2 \), we rearrange the equation:
\[ x^2 = 2601 - 2025 \] \[ x^2 = 576 \]
Taking the square root of both sides:
\[ x = \sqrt{576} = 24 \]
Since \( 24 \) is less than \( 51 \), the solution is valid.
Thus, the value of \( x \) that makes \( (x, 45, 51) \) a Pythagorean triple and is less than 51 is:
\[ \boxed{24} \]
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