Question

ABCD is rectangle and P, Q, R and S are mid point point of the sides AB, BC, CD and DA resp. then the quadrilateral PQRS in rhombus.

• A. True
• B. False

Answer 1

The correct option is A True

$ABCD$ is a rectangle where $P,Q,R$ and $S$ are mid-point of the sides $AB,BC,CD$ and $DA$ respectively.

In $△ABC,$

$P$ is mid-point of $AB$, $Q$ is mid-point of $BC$

Line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.

$\therefore$ $PQ\parallel AC$ and $PQ=\frac{1}{2}AC$ ----- ( 1 )

In $△ADC,$

$R$ is mid-point of $CD,$ $S$ is mid-point of $AD$ respectively.

Line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.

$\therefore$ $RS\parallel AC$ and $RS=\frac{1}{2}AC$ ----- ( 2 )

From ( 1 ) and ( 2 ),

$PQ\parallel RS$ and $PQ=RS$

In $PQRS,$

One pair of opposite side is parallel and equal.

Hence, $PQRS$ is a parallelogram.

In $△APS$ and $△BPQ$

$\Rightarrow$ $AP=BP$ [ $P$ is the mid-point of $AB$ ]

$\Rightarrow$ $\angle PAS=\angle PBQ$ [ All angles of rectangle are ${90}^o.$ ]

$\Rightarrow$ $AS=BQ$ [ $AD=BC,\Rightarrow \frac{1}{2}AD=\frac{1}{2}BC,\Rightarrow$ AS=BQ$ ]

$\therefore$ $△APS\cong △BPQ$ [ $SAS$ congruence rule ]

$\therefore$ $PS=PQ$ [ CPCT ]

But $PS=RQ$ and $PQ=RS$ [ Opposite sides of parallelogram are equal ]

$\therefore$ $PQ=RS=PS=RQ$

Hence, all sides are equal.

Thus, $PQRS$ is a parallelogram with all sides equal.

$\therefore$ $PQRS$ is a rhombus.