Given that (20, x, 29) is a Pythagorean triple and x < 29, what is the value of x?

(1 point)
Responses

21
21

18
18

19
19

22

1 answer

To determine the value of \( x \) in the Pythagorean triple \( (20, x, 29) \), we use the Pythagorean theorem, which states:

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

where \( c \) is the hypotenuse. In this case, \( c = 29 \), \( a = 20 \), and \( b = x \), so we can write the equation as:

\[ 20^2 + x^2 = 29^2 \]

Calculating \( 20^2 \) and \( 29^2 \):

\[ 20^2 = 400 \] \[ 29^2 = 841 \]

Now substituting these values into the equation:

\[ 400 + x^2 = 841 \]

Subtracting 400 from both sides gives:

\[ x^2 = 841 - 400 \] \[ x^2 = 441 \]

Now take the square root of both sides:

\[ x = \sqrt{441} \] \[ x = 21 \]

Since \( x < 29 \), the value of \( x \) is \( 21 \).

Thus, the value of \( x \) is \(\boxed{21}\).