Question

ABCD is a rhombus, P, Q, R, S are midpoints of AB, BC, CD, DA respectively. PQRS is a

• A. rectangle
• B. trapezium
• C. rhombus
• D. parallelogram

Answer 1

Answer: (A)

rectangle

According to the given statement, the figure will be a shown alongside; using mid-point theorem:

In $△ABC,PQ\parallel AC$ and $PQ=\frac{1}{2}AC$ .......$\left(1\right)$

In $△ADC,SR\parallel AC$ and $SR=\frac{1}{2}AC$ .......$\left(2\right)$

$\therefore PQ=SR$ and $PQ\parallel SR$ from $\left(1\right)$ and $\left(2\right)$

$\Rightarrow PQRS$ is a parallelogram.

Now,$PQRS$ will be a rectangle if any angle of the parallelogram $PQRS$ is ${90}^{\circ }$

$PQ\parallel AC$(by midpoint theorem)

$QR=BD$(by midpoint theorem)

But $AC\perp BD$(diagonals of a rhombus are perpendicular to each other)

$\therefore PQ\perp QR$(angle between two lines $=$ angle between their parallels)

Hence $PQRS$ is a rectangle.