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Question
In the following figure, ABCD to a trapezium with AB ‖ DC. If AB = 9 cm, DC = 18 cm, CF = 13.5 cm, AP = 6 cm and BE = 15 cm, Calculate: PE
Solution
Given: ABCD is trapezium, AB || DC
AB = 9 cm, DC = 18 cm, CF = 13.5 cm, AP = 6 cm, BE = 15 cm
Consider ΔAPB and ΔFPD,
`=>` ∠APB = ∠FPD ...(Vertically opposite angles)
`=>` ∠BAP = ∠DFP ...(Since AB || DF)
ΔAPB ~ ΔFPD ...(AA criterion for similarity)
`=> (AP)/(FP) = (AB)/(FD)`
`=> 6/(FP) = 9/31.5`
`=>` FP = 21 cm
So, AF = AP + PF
= 6 + 21
= 27 cm
In ΔAEB and ΔFEC,
∠AEB = ∠FEC ...(Vertically opposite angles)
∠BAF = ∠CFE ...(Since AB || DC)
ΔAEB ~ ΔFEC ...(AA criterion for Similarity)
`=> (AF)/(FE) = (AB)/(FC)`
`=> (AP + PE)/(FE) = 9/13.5`
`=> (AP + PE)/(FE) = 90/135`
`=> (6 + PE)/(FE) = 30/45 = 2/3`
`=>` 3(6 + PE) = 2FE
`=> (3(6 + PE))/2 = FE`
Now, PF = PE + EF
`21 = PE + (3 (6 + PE))/2`
`21 = (2PE + 18 + 3PE)/2`
42 = 5PE + 18
42 – 18 = 5PE
24 = 5PE
`24/5` = PE
PE = 4.8
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