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In this problem, we are given two similar figures: a red parallelogram with a side length of 11 and a blue parallelogram with a side length of 6. The task is to find:

1. The ratio of the perimeters.
2. The ratio of the areas.

### Step 1: Ratio of the Perimeters
Since the figures are similar, the ratio of their perimeters is the same as the ratio of their corresponding side lengths. The side length ratio (red to blue) is:

\[
\text{Ratio of perimeters} = \frac{11}{6}
\]

This ratio is already given in the image.

### Step 2: Ratio of the Areas
The ratio of the areas of two similar figures is the square of the ratio of their corresponding side lengths. Therefore, we square the ratio of the side lengths:

\[
\text{Ratio of areas} = \left(\frac{11}{6}\right)^2 = \frac{11^2}{6^2} = \frac{121}{36}
\]

Thus, the ratio of the areas is:

\[
\frac{121}{36}
\]

Would you like further details on this? Also, here are five related questions you might find interesting:

1. How do the perimeters of two similar shapes change if the side length ratio is doubled?
2. Can the ratio of the areas ever be equal to the ratio of the perimeters?
3. How does the ratio of volumes behave for three-dimensional similar figures?
4. How do different types of transformations affect the ratios of perimeters and areas?
5. What happens to the area ratio if one side length ratio is 1:1?

**Tip:** For any two similar figures, the area ratio is always the square of the perimeter (or side length) ratio.

Find the values of the ratios (red to blue) of the perimeters and areas of the similar figures. Write the ratios as fractions in simplest form.

This problem focuses on finding the ratio of perimeters and areas of two similar parallelograms. It demonstrates the concepts of similarity in geometry and how the ratios of side lengths, perimeters, and areas are related. Suitable for students in Grades 6-8.

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