Giúppp emmmmmm vsssssss

Gọi trung tuyến kẻ từ $A$ là $AM$

$A{{M}^{2}}=\dfrac{A{{B}^{2}}+A{{C}^{2}}}{2}-\dfrac{B{{C}^{2}}}{4}$

$\to {{\left( \dfrac{\sqrt{673}}{2} \right)}^{2}}=\dfrac{{{13}^{2}}+{{14}^{2}}}{2}-\dfrac{B{{C}^{2}}}{4}$

$\to BC=\sqrt{57}$

Nữa chu vi $\Delta ABC$

$p=\dfrac{AB+AC+BC}{2}=\dfrac{27+\sqrt{57}}{2}$

Diện tích $\Delta ABC$

$S=\sqrt{p\left( p-AB \right)\left( p-AC \right)\left( p-BC \right)}=28\sqrt{3}$