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The problem involves a right triangle, and we need to find the value of \( x \). The sides of the triangle are labeled as follows:

- The hypotenuse (the side opposite the right angle) is \( x \).
- One leg of the triangle is \( 8 \).
- The other leg of the triangle is \( x - 4 \).

We can use the **Pythagorean theorem** to solve for \( x \). The Pythagorean theorem states:

\[
\text{(Hypotenuse)}^2 = \text{(Leg 1)}^2 + \text{(Leg 2)}^2
\]

Substituting the given values into the equation:

\[
x^2 = 8^2 + (x - 4)^2
\]

Now, let's solve for \( x \):

1. Expand the equation:
   \[
   x^2 = 64 + (x^2 - 8x + 16)
   \]

2. Simplify the equation:
   \[
   x^2 = x^2 - 8x + 80
   \]

3. Subtract \( x^2 \) from both sides:
   \[
   0 = -8x + 80
   \]

4. Solve for \( x \):
   \[
   8x = 80
   \]
   \[
   x = \frac{80}{8} = 10
   \]

So, the value of \( x \) is **10**.

Would you like more details on this solution or have any other questions?

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**Related Questions:**

1. How does the Pythagorean theorem work in non-right triangles?
2. What happens if both legs of the triangle were given and we needed to find the hypotenuse?
3. Can the value of \( x \) ever be negative in such problems?
4. How can we verify the solution in the context of the triangle's dimensions?
5. What if the triangle's dimensions were based on real-life measurements? How would rounding affect the answer?

**Tip:** Always check the context of the problem to ensure the triangle is right-angled before applying the Pythagorean theorem.

Learn how to solve for the value of x in a right triangle using the Pythagorean Theorem. The triangle's sides include one leg of 8 units and another leg labeled as x - 4, with the hypotenuse labeled as x. This problem is suitable for students in Grades 8-10.

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