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Question
In the figure given below, AB ‖ EF ‖ CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm. Calculate : AC
Solution
In ΔPCD and ΔPEF,
∠CPD = ∠EPF .....(vertically opposite angles)
∠DCE = ∠FEP ...(since DC || EF)
ΔPCD ∼ ΔPEF ...(AA criterion for similarity)
`=> 27/(EF) = 15/(7.5)`
`=>` EF = 13.5 cm
Since EF || AB, ΔCEF ∼ ΔCAB
`=> (EC)/(AC) = (EF)/(AB)`
`=> 22.5/(AC) = 13.5/22.5`
`=>` AC = 37.5 cm
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