2. In AABC, DE || BC, AD = 4 cm, DB = 6 cm and AE = 5 cm. The length of EC is
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Answer:
Given DE∥BC
∴△ADE∼△ABC
Thus, by property of similar triangles,
ABAD=ACAE
∴AD+DBAD=AE+ECAE
∴4+64=5+EC5
∴104=5+EC5
∴4(5+EC)=50
∴20+4EC=50
∴4EC=30
∴EC=7.5cm
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