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Updated on : 2022-09-05

Solution

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Correct option is B)

ABCD is a rhombus and P, Q, R, S are mid-point of AB, BC, CD and DA.

In $△$ABD,

Since, P and S are mid-point of AB and AD

$∴$ By Mid-point theorem,

P $∣∣$BD; PS = $21 $BD ----- (1)

In $△$BCD,

Since, R and Q are mid-point of sides CD & CB

$∴$ By Mid-point theorem,

RQ$∣∣$BD; RQ = $21 $BD ------(2)

From (1) & (2),

PS$∣∣$QR and PS = QR -------(3)

Similarly,

PQ$∣∣$RS and PQ = RS -------(4)

From (3) and (4) we get,

Quadrilateral PQRS is a parallelogram.

The diagonals of a rhombus bisect each other at right angles.

In quadrilateral OMQL,

since, QR$∣∣$PS

$∴$ OM$∣∣$QL; OM = QL

As PQ$∣∣$RS

$∴$ QM$∣∣$OL; QM = OL

$∴$ Quadrilateral OMQL is a parallelogram.

As, $∠$O = 90$_{∘}$

$∠$MQL = $∠$MOL = 90$_{∘}$ ---(opposite angles of parallelogram are equal)

$∴$ PQRS is a rectangle.

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