Question

If $AB>AC$ and length of median from $C$ is $4$ units, then the length of median from $B$ can be-

Answer 1

$CN=\frac{1}{2}\sqrt{2({AC}^2+{BC}^2)-{AB}^2}$

$⟹Y=\frac{1}{2}\sqrt{2({AC}^2+{BC}^2)-{AB}^2}$

$⟹2({AC}^2+{BC}^2)-{AB}^2=64$ -(1)

$⟹2{BC}^2=64+{AB}^2-{2AC}^2$-(2)

$BP=\frac{1}{2}\sqrt{2({AB}^2+{BC}^2)-{AC}^2}$

$⟹BP=\frac{1}{2}\sqrt{2({AB}^2+{BC}^2)-{AC}^2}$

using(2)

$⟹BP=\frac{1}{2}\sqrt{2{AB}^2+64+{AB}^2-{3AC}^2}$

$⟹BP=\frac{1}{2}\sqrt{64+3({AB}^2-{AC}^2)}$

$\because$AB>AC

$\therefore$AB-AC>0 -(3)

$⟹BP=\frac{1}{2}\sqrt{64+3(AB-AC)(AB+AC)}$ using(3)

$BP=\frac{1}{2}\sqrt{64+3(AB-AC)(AB+AC)}>\frac{1}{2}\sqrt{64+0}$

$⟹BP>\frac{1}{2}\times 8$

$⟹$ BP>4 (BP=length of median from B)