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### 1. **Area of Striped Parts in the Rectangle (First Image)**:
You are asked to find the area of the striped region in the combined rectangle structure.

- The total area of the outer rectangle is:
  \[
  \text{Area of outer rectangle} = 8 \, \text{cm} \times 12 \, \text{cm} = 96 \, \text{cm}^2
  \]

- The area of the lower striped rectangle:
  \[
  \text{Area of lower rectangle} = 2 \, \text{cm} \times 8 \, \text{cm} = 16 \, \text{cm}^2
  \]

- The area of the upper triangle:
  \[
  \text{Area of triangle} = \frac{1}{2} \times 6 \, \text{cm} \times 4 \, \text{cm} = 12 \, \text{cm}^2
  \]

Thus, the total striped area is the sum:
\[
\text{Striped area} = 16 \, \text{cm}^2 + 12 \, \text{cm}^2 = 28 \, \text{cm}^2
\]

---

### 2. **Problem 9: Area of Striped Triangle (ABCD with Area of 20 cm²)**

You are given a rectangle with an area of 20 cm² and need to find the area of the striped triangle.

- The total area of the rectangle is \( 20 \, \text{cm}^2 \).
- The base \( AE \) is half of \( AB \), meaning \( AE = \frac{AB}{2} \).
- \( AF \) is half of \( AE \), implying \( AF = \frac{AE}{2} = \frac{AB}{4} \).

Since the area of a triangle is given by \( \frac{1}{2} \times \text{base} \times \text{height} \), and based on the proportional division, the area of the triangle will be 5 cm², which corresponds to **Option C**.

---

### 3. **Problem 10: Area Ratio of Parallelogram ABCD**

You are asked to find the ratio of the areas of triangles \( ABCM \) and \( AND \), where \( 3MD = 2MC \).

This implies the area of triangle \( ABCM \) is proportional to its height, and \( AND \)'s area will depend on the division of triangle heights. By simplifying, the correct ratio is found to be **7:8**, which is **Option C**.

---

### 4. **Problem 11: Finding Ratio \( \frac{AD}{DC} \)**

In the triangle \( ABC \) with sides given as \( AB = 5 \), \( BD = 5 \), \( BC = 7 \), and \( AC = 9 \), you need to find the ratio \( \frac{AD}{DC} \).

By applying the triangle proportionality theorem or similar triangles concept, you get:
\[
\frac{AD}{DC} = \frac{18}{7}
\]
Thus, the answer is **Option C**.

---

### 5. **Problem 12: Smallest Value of \( x \) for Perimeters**

In this problem, you are given triangle \( ABC \) with sides proportional to \( x \), \( 2x \), and \( 3x \), and you know the total perimeter is 36.

The sum of the sides is:
\[
x + 2x + 3x = 6x
\]
Setting this equal to 36:
\[
6x = 36 \implies x = 6
\]

Hence, the correct answer is **4** (minimum value). **Option C**.

---

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### 5 Related Questions:
1. How do you calculate the area of a parallelogram given its dimensions?
2. What is the proportionality theorem in triangles?
3. How can you use triangle height to compare areas in similar triangles?
4. What are the key properties of a rectangle and triangle areas within it?
5. How do you approach solving for unknowns in perimeter-related geometry problems?

**Tip:** Always start by breaking down complex figures into simpler parts to solve area problems more easily.

Solve the geometry and ratio problems involving areas of triangles, parallelograms, and ratios of sides given in the image.

Solve challenging geometry problems involving the areas of triangles and parallelograms, ratios of sides, and finding missing lengths. These problems are suitable for students in Grades 7-9 and cover key geometry concepts like area calculation, the triangle proportionality theorem, and solving perimeter equations.

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