Question

The quadrilateral formed by joining the mid-points of consecutive sides of a rectangle ABCD, taken in order, is a rhombus.

• A. PQRS is a rectangle
• B. PQRS is a parallelogram
• C. diagonals of PQRS are perpendicular and bisect each other
• D. diagonals of PQRS are equal and bisect each other

Answer 1

The correct option is C

diagonals of PQRS are perpendicular and bisect each other

Let ABCD be a rectangle such as AB = CD and BC = DA. P,Q,R and S are the midpoints of the sides AB, BC, CD and DA respectively.

Let us join AC and BD
In $\Delta$ABC,
P and Q are the mid-points of AB and BC respectively.
$\therefore$ PQ || AC and PQ = $\frac{1}{2}$AC (Midpoint theorem)...........(1)
Similarly in $\Delta$ ADC,
SR || AC and SR=$\frac{1}{2}$AC (Midpoint theorem)..........(2)
Clearly, PQ || SR and PQ = SR
Since, in quadrilateral PQRS, one pair of opposite side is equal and parallel to each other, it is parallelogram
$\therefore$ PS || QR and PS = QR (opposite sides of parallelogram).........(3)
In $\Delta$ BCD, Q and R are the mid-points of sides BC and CD respectively.
$\therefore$ QR || BD and QR=$\frac{1}{2}$BD (Midpoint theorem)..........(4)
However, the diagonals of a rectangle are equal.
$\therefore$ AC = BD ...........(5)
By using equation (1), (2), (3), (4), (5), we obtain
PQ = QR = SR = PS
Therefore, PQRS is a rhombus and hence, the given statement is true.