Răspuns:
Explicație pas cu pas:
b)ΔOMB asemenea cu ΔAOD O1=O2 opuse la varf
∠DAO=∠OMB alterne int
DO/OB=AO/OM=AD/BM=2
AO+OM/OM=2+1/1 AM/OM=3 AM=13 dam OM=13/3 dam
AO+OM/OA=3/2 AM/OA=3/2 OA=13×2/3 dam
c) A dcmo=Aabcd-(Aabd-+Abom)
Aabcd=10×12=120 dam²
Aabd=10×12/2=10×6=60 dam²
Abom=OH×BM/2
In ΔOMH OH²=OM²-(BM/2)²= (13/3)²- (5/2)²=(13/3-5/2)(13/3+5/2)==(26-15)12×(26+15)12=11×41/144 dam²
Adcmo=120-(60-11×41/144)=120-(8640-451)144=120-8189/144=120-56,86=63,14 dam²
Adcmo=
