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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) rhombus

It 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 ΔBPQ
AP=PB as P is mid point of AB.
PAS=PBQ as Each angle is right angle
AS=BQ .
Thus, By SAS congruence criteria, we get
ΔAPSΔBPQ
By corresponding parts of congruent triangle, we have
PS=PQ
Thus, the two adjacent sides are equal.
Therefore , PQRS is a rhombus.


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