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 :

[b] In \[\Delta ABC,\]

\[AQ=QB\] 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. (i) and (ii)]

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

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

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