Question

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

Answer 1

Since $△ABD$ is a right-angled triangle at B. Therefore,

${AD}^2={AB}^2+{BD}^2$

$\Rightarrow$ ${AD}^2={AB}^2+{\left(\frac{BC}{2}\right)}^2$ [$\because BD=DC$]

$\Rightarrow$ ${AD}^2={AB}^2+\frac{1}{4}{BC}^2$.......(i)

Again, $△BCE$ is right-angled triangle at B

$\therefore$ ${CE}^2={BC}^2+{BE}^2$

$\Rightarrow$ ${CE}^2={BC}^2+{\left(\frac{AB}{2}\right)}^2$ [$\because BE=EA$]

$\Rightarrow$ ${CE}^2={BC}^2+\frac{1}{4}{AB}^2$.........(ii)

Adding (i) and (ii), we get

${AD}^2+{CE}^2={AB}^2+\frac{1}{4}{BC}^2+{BC}^2+\frac{1}{4}{AB}^2$

$\Rightarrow$ ${AD}^2+{CE}^2=\frac{5}{4}({AB}^2+{BC}^2)$

$\Rightarrow$ ${AD}^2+{CE}^2=\frac{5}{4}{AC}^2$ [$\because △ABC$ is right triangle $\therefore {AC}^2={AB}^2+{BC}^2$]

$\Rightarrow$ ${\left(\frac{3\sqrt{5}}{2}\right)}^2+{CE}^2=\frac{5}{4}\times 25$

$\Rightarrow$ ${CE}^2=\frac{125}{4}-\frac{45}{4}=20$

$\Rightarrow$ $CE=\sqrt{20}cm=2\sqrt{5}cm$