Quantitative Aptitude Ques 2334

Question: In a trapezium ABCD, if E and F be the mid-points of the diagonals, AC and BD, respectively. Then, EF is equal to

Options:

A) $ \frac{1}{2}AB $

B) $ \frac{1}{2}CD $

C) $ \frac{1}{2}(AB+CD) $

D) $ \frac{1}{2}(AB-CD) $

Show Answer

Answer:

Correct Answer: D

Solution:

  • Join CF and produce it to meet AB at G. Then, $ \Delta CDF\cong \Delta GBF $ [by AA criterion] CD = GB and CF = GF Since, E and F are mid-points of CA and CG, we have $ EF=\frac{1}{2}AG=\frac{1}{2}(AB,-GB)=\frac{1}{2}(AB-CD) $
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