Question

ABCD is a rectangle.P and Q are points on side AD and AB respectively.Show that APOQ is a rectangle and find ar(APOQ):ar(ABCD),when it is given that BR=1/4BC and DS=1/4CD.
The intersection of PR and OS is O

Answer 1

 its given that
In rectangle ABCD, P,Q,R and S are points on sides AD,AB,BC and CD,
AQ=1/4AB,
AP=1/4AD
BR=1/4BC and
DS=1/4CD

To prove:
APOQ is a rectangle
ar(APOQ):ar(ABCD)=?

Proof:

AB∥CD ,AB=CD (Opposite sides of rectangle ABCD)
AQ∥DS 
AQ=DS 
this is a pair of opposite sides of quadrilateral ADSQ
ADSQ is a parallelogram
AD∥QS (Opposite sides of parallelogram ADSQ)
AP∥QO .....(i)
AD∥BC and AD=BC (Opposite sides of rectangle ABCD)
AP∥BR and
AP=BR
this is a pair of opposite sides of quadrilateral APRB
APRB is a parallelogram
AB∥PR (Opposite sides of parallelogram ADSQ)
AQ∥PO .....(ii)

From (i) and (ii),
APOQ is a parallelogram

∠A=90. (angle of rectangle is 90°)
this is an angle of parallelogram APOQ
APOQ is a rectangle


(2)
ar(APOQ)=AQ×AP
=1/4AB×1/4AD
=1/16(AB×AD)
=1/16ar(ABCD)

ar(APOQ)ar(ABCD)=1/16

∴ ar(APOQ):ar(ABCD)=1:16

Answer 2

Hi friend,

AB∥CD ,AB=CD (Opposite sides of rectangle ABCD)
AQ∥DS
AQ=DS
this is a pair of opposite sides of quadrilateral ADSQ
ADSQ is a parallelogram
AD∥QS (Opposite sides of parallelogram ADSQ)
AP∥QO .....(i)
AD∥BC and AD=BC (Opposite sides of rectangle ABCD)
AP∥BR and
AP=BR
this is a pair of opposite sides of quadrilateral APRB
APRB is a parallelogram
AB∥PR (Opposite sides of parallelogram ADSQ)
AQ∥PO .....(ii)

From (i) and (ii),
APOQ is a parallelogram

∠A=90. (angle of rectangle is 90°)
this is an angle of parallelogram APOQ
APOQ is a rectangle


(2)
ar(APOQ)=AQ×AP
=1/4AB×1/4AD
=1/16(AB×AD)
=1/16ar(ABCD)

ar(APOQ)ar(ABCD)=1/16

∴ ar(APOQ):ar(ABCD)=1:16

Hope this helps you...
Please mark it as brainliest answer....