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Question
ABCD is a trapezium in which AB || DC. P and Q are points on sides AD and BC such that PQ || AB. If PD = 18, BQ = 35 and QC = 15, find AD.
Solution
In trapezium ABCD, AB || DC. P and Q are points on sides AD and BC such that PQ || AB.
Join AC. Suppose AC intersects PQ in O.
In Δ ACD , OP || CD
`∴ (AP)/(PD)=(AO)/(OC)`.......(1) (BPT)
`In Δ Abc , OQ||AB`
`∴ (BQ)/(QC)=(AO)/(OC)`.......(2) (BPT)
From (1) and (2), we get
`(AP)/(DP)=(BQ)/(QC)`
`(AP)/(18)=(35)/(15)`
`AP=(35xx18)/(5xx3)`
`AP=(7xx5xx3xx6)/(5xx3)`
`AP=(7xxcancel5xxcancel3xx6)/(cancel5xxcancel3)`
`AP = 42`
`AD=AP+PD`
`AD=42+18`
`AD=60`
Hence, the value of AD is 60 .
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