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, \[C{{E}^{2}}=B{{E}^{2}}+B{{C}^{2}}={{\left( \frac{1}{2}AB \right)}^{2}}+B{{C}^{2}}\] |
\[=\frac{1}{4}A{{B}^{2}}+B{{C}^{2}}\] \[[\because A{{B}^{2}}=25\cdot B{{C}^{2}}]\] |
\[=\frac{5}{3}+\frac{55}{3}=\frac{60}{3}=20\] |
\[\therefore \] \[CE=2\sqrt{5}\,cm\] |
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