question_answer

In figure, ABC is a right triangle, right angled at B. AD and CE are the two medians drawn from A and C, respectively. If AC = 5 cm and \[AD=\frac{3\sqrt{5}}{2}cm.\] Find the length of CE.

A) \[2\sqrt{5}\,\text{cm}\]              

B) 2.5 cm

C) 5 cm    

D) \[4\sqrt{2}\,\text{cm}\]

Correct Answer: C

Solution :

\[AC=5cm\]\[\Rightarrow \]\[AD=\frac{3\sqrt{5}}{5}cm\]

\[AE=BE\] and \[BD=CD\]

\[A{{B}^{2}}=A{{C}^{2}}-B{{C}^{2}}\]

\[=25-B{{C}^{2}}\]                 ... (i)

and \[A{{B}^{2}}=A{{D}^{2}}-B{{D}^{2}}\]

\[={{\left( \frac{3\sqrt{5}}{2} \right)}^{2}}-B{{D}^{2}}\]

\[=\frac{45}{2}-\frac{B{{C}^{2}}}{4}\]

From Eqs. (i) and (ii), we get

\[B{{C}^{2}}=\frac{55}{3}\]

Now, from Eqs. (i)

\[A{{B}^{2}}=25-\frac{55}{3}=\frac{20}{3}\]

Also,     \[C{{E}^{2}}=B{{E}^{2}}+B{{C}^{2}}={{\left( \frac{1}{2}AB \right)}^{2}}+B{{C}^{2}}\]

\[=\frac{1}{4}A{{B}^{2}}+B{{C}^{2}}\]      \[[\because A{{B}^{2}}=25\cdot B{{C}^{2}}]\]

\[=\frac{5}{3}+\frac{55}{3}=\frac{60}{3}=20\]

\[\therefore \]      \[CE=2\sqrt{5}\,cm\]