a: Xét ΔABC có MN//BC
nên AM/MB=AN/NC
=>8/NC=6/4=3/2
=>NC=8:3/2=16/3(cm)
AC=AN+NC=8+16/3=40/3(cm)
AB=6+4=10(cm)
\(BC=\sqrt{AB^2+AC^2}=\dfrac{50}{3}\left(cm\right)\)
Xét ΔBAC có MN//BC
nên MN/BC=AM/AB
=>\(MN=\dfrac{3}{5}\cdot\dfrac{50}{3}=10\left(cm\right)\)
b: \(S_{BMND}=MN\cdot MB=10\cdot4=40\left(cm^2\right)\)