Question

$P,Q,R$ and $S$ are the mid-points of sides $AB.BC,CD$ and $DA$ respectively of rhombus $ABCD$. Show that $PQRS$ is a rectangle.
Under what condition will $PQRS$ be a square ?

• A. When $ABCD$ is a square.
• B. When $ABCD$ is a parallelogram
• C. When $ABCD$ is a rectangle
• D. When $ABCD$ is a square or a rectangle

Answer 1

The correct option is A When $ABCD$ is a square.

Given: $ABCD$ is a rhombus. $P,Q,R,S$ are mid points of $AB,BC,CD,DA$ respectively.
Join: $AC$ and $BD$
In $△ABC$
$P$ is mid point of $AB$ and $Q$ is mid point of $AC$.
Thus, by mid point theorem, $PQ\parallel AC$ and $PQ=\frac{1}{2}AC$
Similarly, In $△ACD$,
$S$ is mid point of $AD$ and $R$ is mid point of $CD$.
Thus, by mid point theorem, $SR\parallel AC$ and $SR=\frac{1}{2}AC$
Hence, $PQ\parallel SR$ and $PQ=SR$
Similarly, $PS=QR$ and $PS\parallel QR$
Thus, the opposite sides of $PQRS$ are equal and parallel.
We know the diagonals of a rhombus bisect each other at right angles.
Now, since $AC\perp BD$ thus, $PS\perp PQ$ (Angle between two lines is same as the angle between their corresponding parallel lines)
Now, the opposite sides of $PQRS$ are equal and parallel and the sides meet each other at right angles. Hence, $PQRS$ is a rectangle.
If $PQRS$ had to be a square, the diagonals must be equal and bisect at right angles. It is possible only if $ABCD$ is a square.