question_answer

In \[\Delta ABC,\]\[\angle A=90{}^\circ ,\]\[BP\]and \[CQ\] are two medians. Then, the value of \[\frac{B{{P}^{2}}+C{{Q}^{2}}}{B{{C}^{2}}}\]is  [SSC (10+2) 2014]

A) \[\frac{4}{5}\]                          

B) \[\frac{5}{4}\]

C) \[\frac{3}{4}\]                          

D) \[\frac{3}{5}\]

Correct Answer: B

Solution :

In \[\Delta ABC,\]\[AQ=BQ\]

and \[AP=PC\]

From \[\Delta BAP,\]we have

\[B{{P}^{2}}=A{{B}^{2}}+A{{P}^{2}}\]               (i)

From \[\Delta CAQ,\]we have

\[C{{Q}^{2}}=A{{Q}^{2}}+A{{C}^{2}}\]              (ii)

From \[\Delta ABC,\] we have

\[B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}\]               ... (iii)

\[\because \]\[\frac{B{{P}^{2}}+C{{Q}^{2}}}{B{{C}^{2}}}=\frac{A{{B}^{2}}+A{{P}^{2}}+A{{Q}^{2}}+A{{C}^{2}}}{B{{C}^{2}}}\]

[from Eqs. (1) and (ii)]

\[=\frac{A{{B}^{2}}+A{{C}^{2}}+{{\left( \frac{1}{2}AB \right)}^{2}}+{{\left( \frac{1}{2}AC \right)}^{2}}}{B{{C}^{2}}}\]

\[=\frac{B{{C}^{2}}+\frac{1}{4}(A{{B}^{2}}+A{{C}^{2}})}{B{{C}^{2}}}\]

\[\Rightarrow \]\[\frac{B{{C}^{2}}+\frac{1}{4}B{{C}^{2}}}{B{{C}^{2}}}\]\[\Rightarrow \]\[\frac{\frac{5}{4}B{{C}^{2}}}{B{{C}^{2}}}=\frac{5}{4}\]