what is the answer of the following question:-

in a parallelogram ABCD, P,Q,R,S are mid- points of the sides AB, CD, DA, and BC respectively. AS, BQ, CR, and DP are joined.Find the ratio of the area of the shaded region to the area of parellogram ABCD.

figure is given Below in picture

Answer :

From given information , We form our figure As :



Here , from given information and the property of parallelogram , we get

AP  =  PB  =  DQ  =  QC  = $\frac{1}{2}$  AB  =  $\frac{1}{2}$ CD

And

DR  =  RA  =  CS  =  SB  = $\frac{1}{2}$ DA  =  $\frac{1}{2}$ BC

Here we join P and  Q , Whose are the mid point of sides AB and CD respectively ,  And get

$∆$ ADP  =  $∆$ DPQ  =  $∆$ PQB  =  $∆$ QBC  =  $\frac{1}{4}$ (  Area of parallelogram ABCD )  ------------  ( 1 )

In $∆$ ADH 

DR  =  RA                                 ( As given )

RE  | |  AH  (  From construction )

So,

DE =  EH  (  From B.P.T. )

And

In $∆$ ABG

AP  =  PB  = $\frac{1}{2}$  AB                         ( As given )

PH  | |  BG                                       (  From construction )

So,

PH =  $\frac{1}{2}$ BG   (  From B.P.T. )

But

BG  =  DE  ( From symmetry )
So,

DE  =  EH  =  2 PH
So,

DE  = $\frac{2}{5}$ DP  ------------------ ( 2 )
And

In $∆$ ADP

DR  =  $\frac{1}{2}$ DA                                ( As given )

DE  = $\frac{2}{5}$ DP                ( From equation 2  )

So,

Area of $∆$ RDE  = $\frac{1}{2}$ $\times$ $\frac{2}{5}$ $\times$ Area of $∆$ ADP

From equation 1 , we get

Area of $∆$ RDE  = $\frac{1}{2}$ $\times$ $\frac{2}{5}$ $\times$   $\frac{1}{4}$ (  Area of parallelogram ABCD ) 

So, Area of shaded region  = 4 [ $\frac{1}{2}$ $\times$ $\frac{2}{5}$ $\times$   $\frac{1}{4}$ (  Area of parallelogram ABCD )  ]

Therefore

$\frac{Area of shaded region}{Area of paralle\log ram ABCD} = \frac{4 \left[ {\displaystyle \frac{1}{20}} \times Area of paralle\log ram ABCD \right]}{Area of paralle\log ram ABCD} = \frac{1}{5}$

So,

Area of shaded region  :  Area of parallelogram ABCD  =  1  :  5  ( Ans )