The figure obtained by joining the mid-points | vedclass

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(B) Let the rhombus be $ABCD$. Let $P, Q, R,$ and $S$ be the mid-points of sides $AB, BC, CD,$ and $DA$ respectively.By the Mid-point Theorem,in $	ri

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a rectangle

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(B) Let the rhombus be $ABCD$. Let $P, Q, R,$ and $S$ be the mid-points of sides $AB, BC, CD,$ and $DA$ respectively.By the Mid-point Theorem,in $	riangle ABC$,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC$.Similarly,in $	riangle ADC$,$SR \parallel AC$ and $SR = \frac{1}{2} AC$.Thus,$PQ \parallel SR$ and $PQ = SR$,so $PQRS$ is a parallelogram.Since the diagonals of a rhombus bisect each other at $90^\circ$,the sides of the quadrilateral $PQRS$ will be parallel to the diagonals $AC$ and $BD$.Because the diagonals of a rhombus are perpendicular $(AC \perp BD)$,the adjacent sides of the quadrilateral $PQRS$ are perpendicular to each other.$A$ parallelogram with one right angle is a rectangle. Therefore,$PQRS$ is a rectangle.

The figure obtained by joining the mid-points of the sides of a rhombus,taken in order,is

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