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Q: 7     P and  Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that
              (ii)    \small ar(RQC)=\frac{3}{8}ar(ABC)

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In \DeltaRBC, RQ is the median
Therefore ar(\DeltaRQC) = ar(\DeltaRBQ) 
                                   = ar (PRQ) + ar (BPQ)
                                   = 1/8 (ar \DeltaABC) + ar(\DeltaBPQ)  [from eq (vi) & eq (A) in part (i)]
                                   = 1/8 (ar \DeltaABC) + 1/2 (ar \Delta PBC)  [ since PQ is the median of \DeltaBPC]
                                   = 1/8 (ar \DeltaABC) + (1/2).(1/2)(ar \DeltaABC) [CP is the medain of \DeltaABC]
                                   = 3/8 (ar \DeltaABC)

Hence proved.

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manish

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