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Question

The following figure shows a trapezium ABCD in which AB || DC.

P is the mid-point of AD and PR || AB. Then:

194305_44cf5ad7019d488daad7467a99ea31a7.png

A
PR=12(AB+CD)
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B
PR=(AB+CD)
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C
PR=13(AB+CD)
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D
PR=14(AB+CD)
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Solution

The correct option is A PR=12(AB+CD)
In , BDC and BQR,
B=B (Common)
BDC=BQR (Corresponding angles of parallel lines)
BCD=BRQ (Corresponding angles of parallel lines)
thus, BDCBQR
Thus, BDQB=DCQR
2=DCQR (Q is the mid point of BD)
QR=12DC
Similarly, QP=12AB
Hence, QR+QP=12(AB+DC)
PR=12(AB+CD)

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