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Question

# The adjoining figure shows a trapezium ABCD in which side AB is parallel to side DC. P and Q are mid-points of diagonals AC and BD respectively, CQ joined and produced meets AB at point R. Then:

A
PQ=12(ABDC)
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B
PQ=(ABDC)
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C
PQ=13(ABDC)
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D
PQ=14(ABDC)
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Solution

## The correct option is A PQ=12(AB−DC)Given: ABCD is a trapezium. AB∥CD, P is mid point of AC and Q is mid point of BDNow, In △CDQ and △RBQ∠QDC=∠QBR .....(Alternate angles)∠CQD=∠BQR ....(Vertically Opposite angles)DQ=BQ .....(Q is mid point of BD)Thus, △DQC≅△BQR ....(ASA rule)Hence, by CPCT, CQ=QR and DC=RBNow, In ΔACR, P is mid-point of AC and Q is mid-point of RCBy Mid Point Theorem, we havePQ=12ARPQ=12(AB−BR)PQ=12(AB−DC) ....(Since, DC=RB)

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