A) \[4\text{ }cm\]
B) \[2\sqrt{5}\,cm\]
C) \[3\sqrt{5}\,cm\]
D) \[5\text{ }cm\]
Correct Answer: B
Solution :
Since \[\Delta ABD\]is a right angled triangle \[\therefore \] \[A{{D}^{2}}=A{{B}^{2}}+B{{D}^{2}}\] [By using Pythagoras theorem] \[\Rightarrow \] \[A{{D}^{2}}=A{{B}^{2}}+B{{\left( \frac{BC}{2} \right)}^{2}}\] [ \[\because \] EC is median on AB] \[\Rightarrow \] \[C{{E}^{2}}=B{{C}^{2}}+\frac{1}{4}.A{{B}^{2}}\] ?.(ii) Adding (i) and (ii), we get \[A{{D}^{2}}+C{{E}^{2}}=A{{B}^{2}}+\frac{1}{4}B{{C}^{2}}+B{{C}^{2}}+\frac{1}{4}A{{B}^{2}}\] \[\Rightarrow \] \[A{{D}^{2}}+C{{E}^{2}}=\frac{5}{4}\left( A{{B}^{2}}+B{{C}^{2}} \right)\] \[\Rightarrow \]\[A{{D}^{2}}+C{{E}^{2}}=\frac{5}{4}A{{C}^{2}}\] \[[\because \,A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}]\] \[\Rightarrow \]\[{{\left( \frac{3\sqrt{5}}{2} \right)}^{2}}+C{{E}^{2}}=\frac{5}{4}\times 25\Rightarrow C{{E}^{2}}=20\] \[\Rightarrow \] \[CE=\sqrt{20}\,cm=2\sqrt{5}\,cm\]
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