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The image shows two right triangles, \( \triangle STU \) and \( \triangle ABC \), and the problem states that these triangles are similar: \( \triangle STU \sim \triangle ABC \).

### Step 1: Use Similarity of Triangles
Since the triangles are similar, the corresponding sides of the triangles are proportional:
\[
\frac{ST}{AB} = \frac{TU}{BC} = \frac{SU}{AC}
\]
In the given triangles:
- \( ST = 6 \), \( TU = 3 \), and \( SU \) is the hypotenuse of triangle \( \triangle STU \), which can be calculated using the Pythagorean theorem.
- \( AB = 36 \), \( BC = 2x - 6 \), and \( AC = 24 \).

### Step 2: Find the Length of \( SU \)
Apply the Pythagorean theorem in \( \triangle STU \):
\[
SU^2 = ST^2 + TU^2 = 6^2 + 3^2 = 36 + 9 = 45
\]
\[
SU = \sqrt{45} = 3\sqrt{5}
\]

### Step 3: Set up Proportions
Now use the proportion between corresponding sides:
\[
\frac{ST}{AB} = \frac{TU}{BC} = \frac{SU}{AC}
\]
Starting with \( \frac{TU}{BC} \):
\[
\frac{3}{2x - 6} = \frac{6}{36}
\]
Simplify \( \frac{6}{36} \) to \( \frac{1}{6} \), so:
\[
\frac{3}{2x - 6} = \frac{1}{6}
\]

### Step 4: Solve for \( x \)
Cross-multiply to solve for \( x \):
\[
3 \times 6 = (2x - 6) \times 1
\]
\[
18 = 2x - 6
\]
Add 6 to both sides:
\[
24 = 2x
\]
Divide by 2:
\[
x = 12
\]

### Final Answer: 
The value of \( x \) is \( 12 \).

Would you like any further details or have any questions?

Here are 5 related questions to explore:
1. How do you verify that two triangles are similar?
2. What is the importance of proportionality in similar triangles?
3. Can the Pythagorean theorem be applied in non-right triangles?
4. What would happen if the triangles weren't similar but still shared some side ratios?
5. How does changing one side affect the corresponding sides in similar triangles?

**Tip:** Always check that your proportionality ratios match between all corresponding sides in similar triangles!

Learn how to solve for x in a problem involving similar triangles, triangle STU and triangle ABC, using the principles of triangle similarity and proportions. This solution includes step-by-step application of the Pythagorean theorem and algebra. Suitable for students in Grades 9-10.

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