Question

$\Delta ABC$ is a right triangle right angled at B. AD and CE are the two medians drawn from A and C respectivley. If AC =5cm and AD $=\frac{3\sqrt{5}}{2}$ cm, then CE =??

• A. 4 cm
• B. $2\sqrt{5}cm$
• C. $3\sqrt{5}cm$
• D. 5 cm

Answer 1

The correct option is B

$2\sqrt{5}cm$

Let AE =BE =a cm

and BD=DC =b cm.

Using mid point theorem $DE=\frac{1}{2}AC.$

$=\frac{5}{2}cm.$

In $\Delta DBC,D{B}^2+B{E}^2=E{D}^2$

$\Rightarrow a^2+b^2=\frac{25}{4}---\left(I\right)$

In $\Delta ABD,A{B}^2+B{D}^2=A{D}^2$

$(2a{)}^2+b^2=(\frac{3\sqrt{3}}{2}{)}^2$

$\Rightarrow 4a^2+b^2=\frac{45}{4}---\left(II\right)$

Solving (I) and (II), we get, $a^2=\frac{5}{3}$ and $b^2=\frac{55}{12}.$

Now, in $\Delta BEC.$

$B{E}^2+B{C}^2=E{C}^2$ or,$E{C}^2=a^2+(2b{)}^2=a^2+4b^2$

$=\frac{5}{3}+\frac{55}{3}=\frac{60}{3}=20$

$\therefore EC=\sqrt{20}cm=2\sqrt{5}cm$