Răspuns:
a) A ABCD= AB×BC=50×60=3000 cm pătrați.
A GFEA= A HIJC=100cm²
A hasurata = A ABCD -A GFEA -A HIJC = 3000cm² -200cm²=2800cm²
b) M mijloc AB=> AM=MB=30cm
HI _|_ DC
DC||AB => HI_|_ AB
Fie HI intersecteaza AB=N => IJBM dreptunghi => HJ = BN=10cm => MN=20cm si IN=JB=40cm
EM=30-10=20cm
In triunghiul FEM: sinFME=10/20=1/2
In triunghiul MIN: sin MIN=20/40=1/2 => mFME=mMIN
mMIN+mNMI=90° => mFME+mNMI=90° => FM_|_IM
c)FE=HI
HI||BC
FE||BC => HI||FE => FEIH paralelogram (Fie FI intersecteaza HE in O)
GF=IJ
GF||AB
IJ||AB => GF||IJ =>GFJI paralelogram dar cum O mijloc FI => O mijloc GF => GF intersecteaza FI => GF,FI si EH concurente
