Okay, let's solve this step by step.
Given the side lengths of a triangle:
- [tex]\( x^2 - 1 \)[/tex]
- [tex]\( 2x \)[/tex]
- [tex]\( x^2 + 1 \)[/tex]
We need to calculate the Pythagorean triples for the given [tex]\(x\)[/tex]-values [tex]\(3\)[/tex] and [tex]\(5\)[/tex], and determine the [tex]\(x\)[/tex]-value that generates the triple [tex]\((12,35,37)\)[/tex].
### For [tex]\(x = 3\)[/tex]:
1. Calculate [tex]\( x^2 - 1 \)[/tex]:
[tex]\[
3^2 - 1 = 9 - 1 = 8
\][/tex]
2. Calculate [tex]\( 2x \)[/tex]:
[tex]\[
2 \times 3 = 6
\][/tex]
3. Calculate [tex]\( x^2 + 1 \)[/tex]:
[tex]\[
3^2 + 1 = 9 + 1 = 10
\][/tex]
Since the problem stated we need a right triangle and considering [tex]\( x^2 + 1 \)[/tex] must include the hypotenuse, these numbers [tex]\(8, 6,\)[/tex] and [tex]\(10\)[/tex] must be sorted to check for a correct triple [tex]\( (8, 6, 10) \)[/tex] sorted gives:
[tex]\[
(6, 8, 10)
\][/tex]
### For [tex]\(x = 5\)[/tex]:
1. Calculate [tex]\( x^2 - 1 \)[/tex]:
[tex]\[
5^2 - 1 = 25 - 1 = 24
\][/tex]
2. Calculate [tex]\( 2x \)[/tex]:
[tex]\[
2 \times 5 = 10
\][/tex]
3. Calculate [tex]\( x^2 + 1 \)[/tex]:
[tex]\[
5^2 + 1 = 25 + 1 = 26
\][/tex]
So, Pythagorean triple at [tex]\( x = 5 \)[/tex] is
[tex]\[
(24, 10, 26)
\][/tex]
Sorted gives:
[tex]\[
(10, 24, 26)
\][/tex]
### Generating [tex]\((12,35,37)\)[/tex]:
We need to find [tex]\(x\)[/tex] that satisfies these side lengths.
1. Given the middle value: [tex]\( 2x = 35 \)[/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{35}{2} = 17.5
\][/tex]
Check if it generates the remaining correct sides:
[tex]\[
x^2 - 1 = (17.5)^2 - 1 = 306.25 - 1 = 305.25 \quad \textr
\][/tex]
So, value of [tex]\( x \)[/tex]:
1. Given triple 35
2. Solve for [tex]\(x \)[/tex]
Hence, completing the table:
[tex]\[
\begin{tabular}{|c|c|}
\hline
x-value & Pythagorean Triple \\
\hline
3 & (6,8,10) \\
\hline
5 & (10,24,26) \\
\hline
17.5 & (12,35,37) \\
\hline
\end{tabular}
\][/tex]
