This problem involves a right triangle where the hypotenuse is 4, one leg is $\sqrt{2}$, and the other leg is labeled $x$. To solve for $x$, we can use the Pythagorean Theorem:

$a^2 + b^2 = c^2$

Here:

• $a = x$,
• $b = \sqrt{2}$,
• $c = 4$ (the hypotenuse).

Step 1: Substitute the values into the equation

$x^2 + (\sqrt{2})^2 = 4^2$

$x^2 + 2 = 16$

Step 2: Solve for $x^2$

$x^2 = 16 - 2$

$x^2 = 14$

Step 3: Take the square root of both sides

$x = \sqrt{14}$

Thus, the value of $x$ is:

$\boxed{\sqrt{14}}$

Would you like further clarification or detailed steps on this?

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Here are 5 related questions to expand your understanding:

1. How can the Pythagorean Theorem be used to check triangle properties?
2. What is the significance of $\sqrt{2}$ in geometry?
3. How would the problem change if the hypotenuse were given in terms of $\sqrt{k}$?
4. Can $x$ ever be negative in such a problem? Why or why not?
5. How do we interpret $x^2$ geometrically in this context?

Tip: Always double-check if the triangle is right-angled when using the Pythagorean Theorem!