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Question

ABCD is a parallelogram. P is the mid point of AB. BD and CP intersect at Q such that CQ:QP=3:1.Ifar(PBQ)=10cm2
Find the ara of parallelogram ABCD.

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Solution

In ||gm ABCD, P is mid point on AB, PC and BD intersect each otehr at Q

CQ: QP = 3 :1
ar(PBQ)=10cm2
BD is its diagonal
ar(ABD)=ar(BCD)=12ar||gmABCDar(||gmABCD)=2ar(ABD)InPBCCQ:QP=3:1PBQandCQB have same vertice B
3×areaPBQ=ar(CBQ)Area(CBQ)=3×10=30cm2ar(PBC)=30+10=40cm2
Now ABDandPBC are between the same parallel but base PB=12AB
ar(ABD)=2ar(PBC)=2×40=80cm2Butar(||gmABCD)=2ar(ABD)=2×80=160cm2


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