Answer
$L_A= \sqrt {29}$
$L_B= 2\sqrt {5}$
$L_C= \sqrt {17}$
Work Step by Step
Step $1$.
Recall that the midpoint $M$ of the points $A=(x_1,y_1)$ and $B=(x_2,y_2)$ is $$M=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$$
Use the Midpoint Formula above to find the midpoint of each side.
The midpoint of $BC$ is $\left(\frac{6+4}{2},\frac{0+4}{2}\right)=(5,2)$.
The midpoint of $AC$ is $\left(\frac{0+4}{2},\frac{0+4}{2}\right)=(2,2)$.
The midpoint of $AB$ is $\left(\frac{0+6}{2},\frac{0+0}{2}\right)=(3,0)$.
Step $2$.
Recall that the distance $d$ between the points $A=(x_1,y_1)$ and $B=(x_2,y_2)$ is given by the formula
$$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$
Use the formula above to find the length of each median:
The length of the median from vertex $A$ (let us name it as $L_A$) is equal to the distance between $(5,2)$ and $(0,0)$:
$L_A=\sqrt {(5-0)^2+(2-0)^2}=\sqrt{25+4}=\sqrt {29}$
The length of the median from vertex $B$ (let us name it as $L_B$) is equal to the distance between $(2,2)$ and $(6,0)$:
$L_B=\sqrt {(2-6)^2+(2-0)^2}=\sqrt{16+4}=\sqrt{20}=\sqrt{4(5)}=2\sqrt {5}$
The length of the median from vertex $C$ (let us name it as $L_C$) is equal to the distance between $(3,0)$ and $(4,4)$:
$L_C=\sqrt {(3-4)^2+(0-4)^2}=\sqrt{1+16}=\sqrt {17}$
