https://file.aiquickdraw.com/mathbot/file/1736855715043ft0qeu32.jpg

The diagram shows a right triangle. Based on the image:

- One leg is labeled as \( x \).
- The other leg is labeled as \( 2x \).
- The hypotenuse is labeled as \( 18 \).

This appears to be a problem requiring the Pythagorean theorem, which states:
\[
a^2 + b^2 = c^2
\]
where \( a \) and \( b \) are the legs, and \( c \) is the hypotenuse.

### Solution:
Using the Pythagorean theorem:
\[
x^2 + (2x)^2 = 18^2
\]

1. Expand \( (2x)^2 \):
   \[
   x^2 + 4x^2 = 324
   \]

2. Combine like terms:
   \[
   5x^2 = 324
   \]

3. Solve for \( x^2 \):
   \[
   x^2 = \frac{324}{5} = 64.8
   \]

4. Take the square root to find \( x \):
   \[
   x = \sqrt{64.8} \approx 8.05
   \]

Thus:
- \( x \approx 8.05 \)
- The other leg \( 2x \approx 16.1 \).

Let me know if you'd like a detailed explanation or if you have any questions!

---

**Follow-up Questions:**
1. How is the Pythagorean theorem applied to solve right triangles?
2. Why is the hypotenuse always the longest side in a right triangle?
3. Can the values of \( x \) and \( 2x \) be verified further using trigonometric ratios?
4. What happens if the hypotenuse is replaced with another value?
5. How can similar triangles help solve right-triangle problems?

**Tip:** Always double-check units and ensure the triangle follows right-angle properties before applying the Pythagorean theorem.

Solve for x in a right triangle where one leg is x, the other leg is 2x, and the hypotenuse is 18.

Learn how to solve for x in a right triangle where the legs are x and 2x, and the hypotenuse is 18. This problem involves applying the Pythagorean theorem and solving a quadratic equation. Suitable for students in Grades 8-10.

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