Question

ABCD is a rectangle and P, Q, R, S are mid-points of the sides AB, BC, CD and DA respectively. The quadrilateral PQRS is a rhombus.

• A. True
• B. False

Answer 1

The correct option is A True


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