Let ABCD be a rectangle. Let P, Q, R, S be the mid-points of sides AB, BC, CD, DA respectively. Then the quadrilateral PQRS is a

- Square
- Rectangle, but need not be a square
- Rhombus, but need not be a square
- Parallelogram, but need not be a rhombus

A rhombus is a parallelogram with all sides equal.

In ΔABC, P and Q are the mid-points of AB and BC respectively.

∴ PQ || AC and PQ = ½AC (Mid-point theorem)

Similarly in ΔADC, SR || AC and SR = ½AC (Mid-point theorem)

So, PQ || SR and PQ = SR

Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a parallelogram.

∴ PS || QR and PS = QR (Opposite sides of parallelogram)

In ΔBCD, Q and R are the mid-points of side BC and CD respectively.

∴ QR || BD and QR = ½BD (Mid-point theorem)

However, the diagonals of a rectangle are equal.

∴ AC = BD

PQ = QR = SR = PS

Therefore, PQRS is a rhombus.

**The correct option is C.**

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