i dont have rd sharma so please solve it for me. In a trapezium ABCD, AB ll DC and DC = 2AB. EF ll AB where E and F lie on BC and AD respectively such that BE /EC = 4/3. Diagonal DB intersects EF at G . Prove that 7EF = 11AB.
RAHUL , 6 Years ago
Grade 10
1 Answers
Arun
Last Activity: 6 Years ago
Dear student
given DC = 2 AB
Draw BH parallel to AD. H will be the middle point of DC as AB = DH. Sin EF || AB, FI = AB
The ΔBHC and ΔBIE are similar as the corresponding sides are parallel. So IE / HC = BE / BC IE = HC * BE / BC = y * 4x/7x = 4 y / 7
EF = FI + IE= y + 4 y /7 = 11 y / 7
so 7 EF = 11 AB
Regards Arun (askIITians forum expert)
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