Quantitative Aptitude Ques 1354

Question: In $ \Delta ABC, $ $ \angle A=90{}^\circ , $ $ BP $ and $ CQ $ are two medians. Then, the value of $ \frac{BP^{2}+CQ^{2}}{BC^{2}} $ is

Options:

A) $ \frac{4}{5} $

B) $ \frac{5}{4} $

C) $ \frac{3}{4} $

D) $ \frac{3}{5} $

Show Answer

Answer:

Correct Answer: B

Solution:

  • In $ \Delta ABC, $ $ AQ=BQ $ and $ AP=PC $ From $ \Delta BAP, $ we have
    $ BP^{2}=AB^{2}+AP^{2} $ … (i) From $ \Delta CAQ, $ we have
    $ CQ^{2}=AQ^{2}+AC^{2} $ … (ii) From $ \Delta ABC, $ we have
    $ BC^{2}=AB^{2}+AC^{2} $ … (iii)
    $ \because $ $ \frac{BP^{2}+CQ^{2}}{BC^{2}}=\frac{AB^{2}+AP^{2}+AQ^{2}+AC^{2}}{BC^{2}} $ [from Eqs. (1) and (ii)] $ =\frac{AB^{2}+AC^{2}+{{( \frac{1}{2}AB )}^{2}}+{{( \frac{1}{2}AC )}^{2}}}{BC^{2}} $ $ =\frac{BC^{2}+\frac{1}{4}(AB^{2}+AC^{2})}{BC^{2}} $

$ \Rightarrow $ $ \frac{BC^{2}+\frac{1}{4}BC^{2}}{BC^{2}} $
$ \Rightarrow $ $ \frac{\frac{5}{4}BC^{2}}{BC^{2}}=\frac{5}{4} $

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