prove that area of the quadrilateral formed by joining the mid points of the sides of a parallelogram is half the area of the parallelogram

Answer :

Let P , Q , R , S are the mid point of parallelogram ABCD .

And

Diagonals of parallelogram ABCD meet at  "  O "  , And we join OP , OQ  , OR and OS  .

And

We know AB =  CD  , AB | |  CD  and BC  =  DA  , BC | |  DA  (  Sides of parallelogram  )
And

AP  =  PB   =  12 AB
And
RC  =  RD  =  12 CD  And we know AB  =  CD  , SO 

AP  =  PB =  RC  =  RD 

Similarly

BQ  =  QC  =  DS  =  SA 

We also know diagonal of parallelogram bisect each other , SO  "  O  "  is mid point of AC and BD  .

Now In ABC 

O is mid point of AC and Q is mid point of BC  , So

Line OQ is parallel to AB  , As we know from basic proportionality theorem  , SO

OQ  | |  AB  |  |  CD Then

OQ  | | RC                                (  RC is a part of line CD )                     ------------- ( 1 )

So ,
QCOQ = BCAB12BCOQ = BCABOQ= 12AB

So,

OQ  =  RC                                   ----------------( 2 ) 

Similarly we can show

QC  | | OR  , QC  =  OR            ---------------- ( 3 )

So, From equation 1 , 2 and 3 , we get 

OQCR is a parallelogram  , So OF  = FC As we know diagonals of parallelogram bisect each other .

So ,

Area of  OFQ  =  Area of  CFQ                      --------------- ( 4 )   (  As they have same base (  OF  =  FC )  And same height as they have same vertex " Q "  )

Similarly

Area of  OEQ  =  Area of  BEQ                      --------------- ( 5 ) 

Add equation 4 and 5 , we get

Area of  OFQ  + Area of  OEQ  = Area of  CFQ + Area of  BEQ      

Area of OEQF = Area of  CFQ + Area of  BEQ                            --------------- ( 6 )

Similarly 

Area of OFRG = Area of  CFR + Area of  DGR                            --------------- ( 7 )

And

Area of OGSH = Area of  DGS + Area of  AHS                            --------------- ( 8 )

And

Area of OHPE = Area of  AHP + Area of  BPE                            --------------- ( 9 )

Now we add equation  6   ,  7   ,  8  and  9  and get

Area of OEQF +  Area of OFRG + Area of OGSH + Area of OHPE = Area of  CFQ + Area of  BEQ +  Area of  CFR + Area of  DGR + Area of  DGS + Area of  AHS   + Area of  AHP + Area of  BPE

Area of quadrilateral PQRS  = Area of  QCR + Area of  RDS +  Area of  SAP + Area of  PBQ

Area of quadrilateral PQRS  =  Area of parallelogram ABCD  -  Area of quadrilateral PQRS 

2 ( Area of quadrilateral PQRS ) =  Area of parallelogram ABCD

Area of quadrilateral PQRS  =  Area of parallelogram ABCD2                                                  ( Hence proved )


 

  • 1
What are you looking for?