Let $F_{1}$ be the set all of parallelograms, $F_{2}$ be the set of the rectangles, $F_{3}$ be the set of rhombuses, $F_{4}$ be the set of squares $F_{5}$ be the set of trapezium in a plane then $F_{1}$ =

F_{2} \cap F_{3}

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F_{2} \cup F_{3} \cup F_{4}

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F_{3} \cup F_{4} \cup F_{5}

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F_{3} \cap F_{1}

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Solution

The correct option is B $F_{2} \cup F_{3} \cup F_{4}$
As we know that all rectangles , rhombuses , squares are parallelogram with opposite sides parallel and opposite angles equal while for trapezium it does has only one opposite sides parallel. Hence
$F_{1} = F_{2} \cup F_{3} \cup F_{4}$

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Let $F_{1}$ be the set of all 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 $F_{1}$ is equal to

Let $F_{1}$ be the set of all parallelograms, $F_{2}$ the set of all reactangles, $F_{3}$ the set of all rhombuses, $F_{4}$ the set of all sqauares and $F_{5}$ the set of trapeziums in a plane. Then $F_{1}$ may be equal to