The quadrilateral formed by joining the mid-points of consecutive sides of a rectangle ABCD, taken in order, is a rhombus. Options: A . &nbsp PQRS is a rectangle B . &nbsp PQRS is a parallelogram C . &nbsp diagonals of PQRS are perpendicular and bisect each other D . &nbsp diagonals of PQRS are equal and bisect each other

The quadrilateral formed by joining the mid-points of consecutive sides of a re

Question

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

Options:

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: Option C
:
C

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 $Δ$ ABC,
P and Q are the mid-points of AB and BC respectively.
$∴$ PQ || AC and PQ = $\frac{1}{2}$ AC (Midpoint theorem)...........(1)
Similarly in $Δ$ 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
$∴$ PS || QR and PS = QR (opposite sides of parallelogram).........(3)
In $Δ$ BCD, Q and R are the mid-points of sides BC and CD respectively.
$∴$ QR || BD and QR= $\frac{1}{2}$ BD (Midpoint theorem)..........(4)
However, the diagonals of a rectangle are equal.
$∴$ 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.