question_answer

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

A)  \[\frac{4}{5}\]             

B)  \[\frac{5}{4}\]

C)  \[\frac{3}{4}\]             

D)  \[\frac{3}{5}\]

Correct Answer: B

Solution :

 BP and CQ are medians of right angle\[\Delta ABC\], right angled at A. Now, \[B{{P}^{2}}={{\left( \frac{1}{2}AC \right)}^{2}}~+\text{ }A{{B}^{2}}\] and\[C{{Q}^{2}}=A{{C}^{2}}+{{\left( \frac{1}{2}AB \right)}^{2}}\] \[\Rightarrow \]   \[B{{P}^{2}}~+\text{ }C{{Q}^{2}}=\frac{5}{4}(A{{B}^{2}}~+\text{ }A{{C}^{2}})\] \[\Rightarrow \]   \[B{{P}^{2}}+\text{ }C{{Q}^{2}}=\frac{5}{4}B{{C}^{2}}\] \[(\therefore A{{C}^{2}}+A{{B}^{2}}~=\text{ }B{{C}^{2}})\] \[\Rightarrow \]\[\]