Question

$P,Q,R$ and $S$ are the mid-points of sides $AB,BC,CD$ and $DA$ respectively of rhombus $ABCD$. Quadrilateral $PQRS$ 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.