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

Answer 1

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 $\Delta ABC$, $P$ and $Q$ are the mid points of AB and BC respectively.
From mid point theorem, $PQ=\frac{1}{2}\times AC$ ...........(i)
Similarly, in $\Delta ACD$, we have $SR=\frac{1}{2}\times AC$ ........(ii)
From equation (i) and (ii), we have $PQ=SR$.
Therefore, PQRS is a parallelogram.
Since, $AD=BC$
$\Rightarrow \frac{AD}{2}=\frac{BC}{2}$
$\Rightarrow AS=BQ$.
Now, consider $\Delta APS$ and $\Delta BPQ$
$AP=PB$ as $P$ is mid point of AB.
$\angle PAS=\angle PBQ$ as Each angle is right angle
$AS=BQ$ .
Thus, By SAS congruence criteria, we get
$\Delta APS\cong \Delta BPQ$
By corresponding parts of congruent triangle, we have
$PS=PQ$
Thus, the two adjacent sides are equal.
Therefore , PQRS is a rhombus.