BC² = AD² +CE² = 160² + ( 160 -40 )²= 160² +120² = 25600+14400=40000
BC = √40000=200 m
lungimea gardului = perimetrul = AB+BC+CD+DA =160 +200 +40+160 =560m
b. ΔDMC drept din ipoteza
Δ ABM drept din ipoteza
MB² = AB² +MA² = 160² +80² = 25600 +6400=32000
MB =√32000=80√5m
MC²=40²+80²=1600+6400=8000 , MC =√8000=40√5
daca in ΔMCB avem BC²=MC² +MB²
200²=(40√5)² + (80√5)²
40000=8000+32000 adevarat ⇒ Δ BMC cu mas <( BMC) =90°
c. aria Δ MDC = MD·DC /2 = 80·40 /2 = 1600 m²
aria ΔMAB = AB·MA /2 = 80·160 /2 =80²m²= 6400m²
aria ΔCMB = MC·MB /2 = 40√5 ·80√5 /2 = 40²·5 m²=8000 m²
