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Question

# The figure formed by joining the midpoints of the adjacent sides of a rectangle is a (a) rhombus (b) square (c) rectangle (d) parallelogram

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Solution

## The correct option is (a) rhombusIt is given that rectangle ABCD, Where P, Q, R and S are the mid points of AB, BC, CD and DA respectively. Join PQ, QR, RS, SP .In triangle ΔABC, P and Q are the mid points of AB and BC respectively.From mid point theorem, PQ=12×AC ...........(i)Similarly, in ΔACD, we have SR=12×AC ........(ii)From equation (i) and (ii), we have PQ=SR.Therefore, PQRS is a parallelogram.Since, AD=BC⇒AD2=BC2⇒AS=BQ.Now, consider ΔAPS and ΔBPQAP=PB as P is mid point of AB.∠PAS=∠PBQ as Each angle is right angleAS=BQ .Thus, By SAS congruence criteria, we get ΔAPS≅ΔBPQ By corresponding parts of congruent triangle, we have PS=PQThus, the two adjacent sides are equal.Therefore , PQRS is a rhombus.

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