Sets, Relations & Functions
Sets and their properties
Grade None
Question:
<p>Let <em>F</em><sub>1</sub> be the set of parallelograms, <em>F</em><sub>2</sub> the set of rectangles, <em>F</em><sub>3</sub> be the set of rhombuses, <em>F</em><sub>4</sub> be the set of squares and <em>F</em><sub>5</sub> be the set of trapeziums in a plane. Then <em>F</em><sub>1</sub> may be equal to</p>
<p>\(F_2 \cap F_3\)</p>
<p>\(F_3 \cap F_4\)</p>
<p>\(F_2 \cup F_5\)</p>
<p>\(F_2 \cup F_3 \cup F_4 \cup F_1\)</p>
Step-by-Step Solution
Key Concept: A parallelogram can be expressed as the union of rectangles and rhombuses when we partition based on angle and side properties. Specifically, F1 = F2 ∪ F3 because every parallelogram is either a rectangle (equal angles) or rhombus (equal sides) or both (square), but squares are already counted in both F2 and F3.
<p><strong>Step 1:</strong> Identify set relationships by definition.</p><p>F2 (rectangles): Parallelograms with all angles = 90°</p><p>F3 (rhombuses): Parallelograms with all sides equal</p><p>F4 (squares): Parallelograms with all sides equal AND all angles = 90° (F4 ⊂ F2 ∩ F3)</p><p><strong>Step 2:</strong> Verify if every parallelogram in F1 belongs to F2 ∪ F3.</p><p>Any parallelogram must satisfy: opposite sides equal and opposite angles equal. By the contrapositive logic—if a parallelogram is NOT a rectangle (unequal angles) it MUST have unequal adjacent sides, making it approach rhombus properties, or vice versa. The complete classification is: F1 = F2 ∪ F3</p><p><strong>Step 3:</strong> Confirm using set theory: F2 ∪ F3 includes all rectangles, all rhombuses, and their intersection (squares). This covers all possible parallelograms.</p><p>∴ Answer: D (F1 = F2 ∪ F3)
Correct Answer: D