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# 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) ar (RQC) = 3/8 ar (ABC)

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 $\Delta$RBC, RQ is the median
Therefore ar($\Delta$RQC) = ar($\Delta$RBQ)
= ar (PRQ) + ar (BPQ)
= 1/8 (ar $\Delta$ABC) + ar($\Delta$BPQ)  [from eq (vi) & eq (A) in part (i)]
= 1/8 (ar $\Delta$ABC) + 1/2 (ar $\Delta$ PBC)  [ since PQ is the median of $\Delta$BPC]
= 1/8 (ar $\Delta$ABC) + (1/2).(1/2)(ar $\Delta$ABC) [CP is the medain of $\Delta$ABC]
= 3/8 (ar $\Delta$ABC)

Hence proved.

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