Let `F_(1)` be the set of parallelograms, `F_(2)` the set of rectangle , `F_(3)` the set of rhombuses, `F_(4)` the set of squares and `F_(5)` the set of trapeziums in a plane. Then, `F_(1)` may be equal to

Options

A. `F_(2) cap F_(3)`
B. `F_(3) cap F_(4)`
C. `F_(2) cup F_(5)`
D. `F_(2)cup F_(3) cup F_(4) cup F_(1)`

Explanation

To solve the problem, we need to analyze the relationships between the sets defined in the question.

1. **Identify the Sets**:
- Let \( F_1 \) be the set of parallelograms.
- Let \( F_2 \) be the set of rectangles.
- Let \( F_3 \) be the set of rhombuses.
- Let \( F_4 \) be the set of squares.
- Let \( F_5 \) be the set of trapeziums.

2. **Understand the Relationships**:
- Every rectangle is a parallelogram, so \( F_2 \subseteq F_1 \).
- Every rhombus is also a parallelogram, so \( F_3 \subseteq F_1 \).
- Every square is a special type of rectangle and rhombus, hence \( F_4 \subseteq F_2 \) and \( F_4 \subseteq F_3 \). Therefore, \( F_4 \subseteq F_1 \).
- Trapeziums are not parallelograms, so \( F_5 \) does not contribute to \( F_1 \).

3. **Combine the Sets**:
- Since \( F_1 \) includes all rectangles, rhombuses, and squares, we can express this as:
\[ F_1 = F_2 \cup F_3 \cup F_4 \]
- However, since \( F_4 \) is already included in both \( F_2 \) and \( F_3 \), we can simplify this to:
\[ F_1 = F_2 \cup F_3 \cup F_4 \]

4. **Evaluate the Options**:
- \( F_2 \cap F_3 \): This would represent the set of shapes that are both rectangles and rhombuses, which are squares. This is not equal to \( F_1 \).
- \( F_3 \cap F_4 \): This would also yield squares, which is not equal to \( F_1 \).
- \( F_2 \cup F_5 \): This includes rectangles and trapeziums, which does not encompass all parallelograms.
- \( F_2 \cup F_3 \cup F_4 \): This includes all rectangles, rhombuses, and squares, which are all types of parallelograms.

5. **Conclusion**:
- Therefore, the correct answer is:
\[ F_1 = F_2 \cup F_3 \cup F_4 \]
- This corresponds to option number four.

### Final Answer:
The correct answer is option number four: \( F_1 = F_2 \cup F_3 \cup F_4 \).