a) MN||BC din Teorema fundamentală a asemănării=>ΔAMN~ΔABC
=>MN/BC=AM/AB=AN/AC
MN=BC*AM/AB
MN=30 m*x/40 m
MN=3x/4
ΔAMN este dreptunghic=> AΔAMN=(c1*c2)/2=(AM*MN)/2=(x*3x/4)/2=3x²/4*1/2=3x²/8
b) BCNM e trapez=>A BCNM=(B+b)*h/2=(BC+MN)*BM/2=(30+3x/4)*(40-x)/2
A BCNM=(30+3x/4)*(40/2-x/2)
A BCNM=(30+3x/4)*(20-x/2)
A BCNM=30*20-30x/2+3x*20/4-3x²/8
A BCNM=600-15x+3x*5-3x²/8
A BCNM=600-15x+15x-3x²/8
A BCNM=600-3x²/8
BM=AB-AM=40 m-x
