Question

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

Answers

GPT-4o mini
To determine the value of \( x \) in the Pythagorean triple \( (20, x, 29) \) where \( x < 29 \), we will use the Pythagorean theorem, which states that in a right triangle with legs \( a \) and \( b \), and hypotenuse \( c \), the following relationship holds:

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

In our case, we will set \( a = 20 \), \( b = x \), and \( c = 29 \). Thus, we have:

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

Now we need to calculate \( 20^2 \) and \( 29^2 \):

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

Now substituting these values into the equation gives:

\[
400 + x^2 = 841
\]

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

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

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

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

Since we have the condition \( x < 29 \), and \( 21 < 29 \) holds true, the value of \( x \) is:

\[
\boxed{21}
\]