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प्रश्न
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: AF
योग
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उत्तर
In 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
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