Question

What value of x that is less than 51 will make (x,45,51) a Pythagorean triple?(1 point)
x=

Answer 1

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:

\[
\boxed{24}
\]