Question

$P,Q,R,S$ are respectively the midpoints of the sides $AB,BC,CD$ and $DA$ of $1gm$ $ABCD.$ Show that $PQRS$ is a parallelogram and also show that
at $(1gm$ $PQRS)=\frac{1}{2}\times ar(1gm$ $ABCD).$

Answer 1

REF.Image.

In $△ADC$, by mid point

theorem, $SR\parallel AC,SR=\frac{1}{2}\left(AC\right)$

Similarly $PQ\parallel AC,PQ=\frac{1}{2}\left(AC\right)$

$\Rightarrow PQ\parallel RS,$ similarly $PS\parallel QR$

$\Rightarrow$ opposite sides parallel

$\Rightarrow PQRS$ is a parallelogram

As $△ALS\sim △AOD$ (AA similarity)

$\Rightarrow \frac{SL}{OD}=\frac{AS}{AD}=\frac{1}{2}$

$\Rightarrow$ Area of rectangle $SL\times R$

$=SL\times L\times =\frac{1}{2}\left(OD\right)\left(\frac{AC}{2}\right)$

$\{LX\times =SR=\frac{AC}{2}\}=\frac{1}{2}\left(ar△ACD\right)$

$\Rightarrow$ Area of $PQRS=\frac{1}{2}\left(arACD\right)$

$+\frac{1}{2}\left(arACB\right)=\frac{1}{2}\left(arABCD\right)$