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)] |
\[=\frac{A{{B}^{2}}+A{{C}^{2}}+{{\left( \frac{1}{2}AB \right)}^{2}}+{{\left( \frac{1}{2}AC \right)}^{2}}}{B{{C}^{2}}}\] |
\[=\frac{B{{C}^{2}}+\frac{1}{4}(A{{B}^{2}}+A{{C}^{2}})}{B{{C}^{2}}}\] |
\[\Rightarrow \]\[\frac{B{{C}^{2}}+\frac{1}{4}B{{C}^{2}}}{B{{C}^{2}}}\]\[\Rightarrow \]\[\frac{\frac{5}{4}B{{C}^{2}}}{B{{C}^{2}}}=\frac{5}{4}\] |
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