Question

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

Answer 1

To find the value of \( x \) that is less than 51 and makes \( (x, 45, 51) \) a Pythagorean triple, we start with the Pythagorean theorem:

\[
a^2 + b^2 = c^2
\]

In our case, let's assign \( a = x \), \( b = 45 \), and \( c = 51 \). Thus, we have:

\[
x^2 + 45^2 = 51^2
\]

First, we calculate \( 45^2 \) and \( 51^2 \):

\[
45^2 = 2025
\]

\[
51^2 = 2601
\]

Now we can substitute these values into the equation:

\[
x^2 + 2025 = 2601
\]

Next, we will isolate \( x^2 \):

\[
x^2 = 2601 - 2025
\]

Calculating the right hand side gives us:

\[
x^2 = 576
\]

Now, we take the square root of both sides to find \( x \):

\[
x = \sqrt{576} = 24
\]

Since 24 is less than 51, it meets the criteria given in the problem.

Thus, the value of \( x \) that makes \( (x, 45, 51) \) a Pythagorean triple is:

\[
\boxed{24}
\]