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This image shows a geometry problem involving similar right triangles. We are asked to find the length \( x \). 

From the diagram, we can observe two right triangles: a smaller one and a larger one. The sides of the smaller triangle are labeled 3 (height) and 4.5 (base). The larger triangle has the same height (6) and an unknown base \( x \). The triangles are similar because they share a common angle and have right angles.

Since the triangles are similar, we can set up a proportion using the corresponding sides:

\[
\frac{\text{Height of smaller triangle}}{\text{Base of smaller triangle}} = \frac{\text{Height of larger triangle}}{\text{Base of larger triangle}}
\]

Substituting the given values:

\[
\frac{3}{4.5} = \frac{6}{x}
\]

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

\[
\frac{3}{4.5} = \frac{6}{x} \quad \Rightarrow \quad x = \frac{6 \times 4.5}{3}
\]

I'll calculate the value for \( x \).It seems like I can’t do more advanced data analysis right now. Please try again later.

You can easily solve for \( x \) by calculating:

\[
x = \frac{6 \times 4.5}{3} = 9
\]

So, the length of \( x \) is 9 units.

Would you like further details or have any other questions? Here are some related questions for deeper understanding:

1. What are the properties of similar triangles?
2. How do you identify corresponding sides in similar triangles?
3. What other methods can be used to find missing sides in right triangles?
4. What is the significance of the Pythagorean theorem in solving right triangles?
5. How would this problem change if the triangles were not similar?

**Tip:** When dealing with similar triangles, always match the corresponding sides and angles carefully before setting up proportions.

Find the length x in a geometry problem involving two similar right triangles. The smaller triangle has a height of 3 and a base of 4.5. The larger triangle has a height of 6, and we need to find its base (x).

Learn how to find the missing side of a larger right triangle using the properties of similar triangles. This problem demonstrates how to set up and solve proportions based on corresponding sides of similar triangles. Ideal for students in Grades 8-10.

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