Perim ABCD=2AB+2BC=2*80+2*60=160+120=280m, cerinta
b) NC=BM=60cm , ipoteza
NC||BM⇒BMCN paralelogram
MN⊥BC (ipoteza)⇒BMCN romb⇒NB≡NC=60m (ipoteza)
dar BC≡AB=60m (ipoteza)
⇒ΔNBC echilateral⇒m∡(NCB)=60°
darABCD paralelogram (ipoteza)⇒m∡(DAB)=m∡(NCB)=60°, cerinta
c)ME|| CF, (BEFC paralelogram ipoteza)⇒MEFC tyrapez
⇒Arie MEFC= (ME+CF)*d(ME, FC)/2
dar d ( ME, FC)=d(B,NC) distanta intre 2 drepte paralerle est constantasi se masoara pe perpendiculara comuna
d(B, NC) inaltimea truunghi echilateral lat 60 =(60√3)/2=30√3
Arie MEFC=[(80+20)2]*30√3=50*30√3=1500√3
tebuie sa aratam ca 1500√3<2600
adica
15√3<26
sau
(15√3)²<26²
adica
225*3<676
675<676, Adevarat, cerinta
