Q.Let $F_1$ be the set of parallelograms, $F_2$ the set of rectangles, $F_3$ 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
(A) $F_2 \cap F_3$
(B) $F_3 \cap F_4$
(C) $F_2 \cup F_5$
(D) $F_2 \cup F_3 \cup F_4 \cup F_1$

Concept: Set Operations and Hierarchy of Quadrilaterals

We need to understand the containment relationships among these sets of quadrilaterals.

Every rectangle, rhombus, and square is a parallelogram, so $F_2 \subset F_1$, $F_3 \subset F_1$, and $F_4 \subset F_1$. A square is both a rectangle and a rhombus, so $F_4 \subset F_2$ and $F_4 \subset F_3$. Trapeziums have only one pair of parallel sides, while parallelograms have both pairs parallel, so $F_5$ and $F_1$ are disjoint except for the degenerate case.

Now check each option:

(A) $F_2 \cap F_3$ = rectangles AND rhombuses = squares = $F_4 \neq F_1$

(B) $F_3 \cap F_4$ = rhombuses AND squares = squares = $F_4 \neq F_1$ …

The question asks us to understand the relationships between different quadrilateral families through set operations. Let me first clarify what each set contains and how they relate to one another.

Understanding the hierarchy of quadrilaterals:

A parallelogram is a quadrilateral with opposite sides parallel. From this basic definition, we get special cases:

• A rectangle ($F_2$) is a parallelogram with all angles equal to $90°$
• A rhombus ($F_3$) is a parallelogram with all sides equal
• A square ($F_4$) is both a rectangle AND a rhombus (right angles + equal sides)
• A trapezium ($F_5$) has only one pair of parallel sides, so it's NOT a parallelogram

This means: $F_4 \subset F_2 \subset F_1$, $F_4 \subset F_3 \subset F_1$, and $F_5 \cap F_1 = \emptyset$.

Now let's examine each option systematically:

1. Option (A): $F_2 \cap F_3$

   The intersection of rectangles and rhombuses gives us quadrilaterals that are BOTH rectangles AND rhombuses. A shape with right angles and equal sides is precisely a square. So $F_2 \cap F_3 = F_4$, which is a proper subset of $F_1$, not equal to it.

2. Option (B): $F_3 \cap F_4$

   Since every square is already a rhombus, we have $F_4 \subset F_3$. Therefore $F_3 \cap F_4 = F_4$. Again, this is just the set of squares, not all parallelograms.

3. Option (C): $F_2 \cup F_5$

   This union combines rectangles with trapeziums. But trapeziums are not parallelograms at all—they have only one pair of parallel sides. Meanwhile, this union misses all the rhombuses that aren't rectangles (like a "diamond" shape). So $F_2 \cup F_5 \neq F_1$.

4. Option (D): $F_2 \cup F_3 \cup F_4 \cup F_1$ …