Question:

In trapezium ABCD, side AB ||side PQ ||side DC, AP = 15, PD = 12, QC = 14, find BQ.

Answer:

Given ABPQDC.
AP = 15
PD = 12
QC = 14
AP/PD = BQ/QC [ Property of three parallel lines and their transversals]
15/12 = BQ /14
BQ = 15×14/12
BQ = 17.5 units.
Hence measure of BQ is 17.5 units.

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In trapezium ABCD, side AB ||side PQ ||side DC, AP = 15, PD = 12, QC = 14, find BQ.

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