Q. 214.8( 10 Votes )

P.Q.R.S are respectively the midpoints of the sides AB, BC, CD and DA of gm ABCD. Show that PQRS is a parallelogram and also show that

Ar(gm PQRS) = x ar(gm ABCD).


Answer :


Given : ABCD is a parallelogram and P,Q,R,S are the midpoints of AB,BC,CD,AD respectively


To prove: (i) PQRS is a parallelogram


(ii) Area(gm PQRS) = x area(gm ABCD)


Construction : Join AC ,BD,SQ


Proof:


(i)


As S and R are midpoints of AD and CD respectively, in ∆ACD


SR || AC [By midpoint theorem] ------------------- (1)


Similarly in ∆ABC , P and Q are midpoints of AB and BC respectively


PQ || AC [By midpoint theorem] ------------------ (2)


From (1) and (2)


SR || AC || PQ


SR || PQ ------------------- (3)


Again in ∆ACD as S and P are midpoints of AD and CB respectively


SP || BD [By midpoint theorem] ------------------ (4)


Similarly in ∆ABC , R and Q are midpoints of CD and BC respectively


RQ || BD [By midpoint theorem] -------------------- (5)


From (4) and (5)


SP || BD || RQ


SP || RQ ----------- (6)


From (3) and (6)


We can say that opposite sides are Parallel in PQRS


Hence we can conclude that PQRS is a parallelogram.


(ii)


Here ABCD is a parallelogram


S and Q are midpoints of AD and BC respectively


SQ || AB


SQAB is a parallelogram


Now, area(∆SQP) = x area of (SQAB) -------------- 1


[Since ∆SQP and ||gm SQAB have same base and lie between same parallel lines]


Similarly


S and Q are midpoints of AD and BC respectively


SQ || CD


SQCD is a parallelogram


Now, area(∆SQR) = x area of (SQCD) ------------------- 2


[Since ∆SQR and ||gm SQCD have same base and lie between same parallel lines]


Adding (1) and (2) we get


area(∆SQP) + area(∆SQR) = x area of (SQAB) + x area of (SQCD)


area(PQRS) = (area of (SQAB) + area of (SQCD))


Area(gm PQRS) = x area(gm ABCD)


Hence proved


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