a) Aria CMD =CM*CD:2=10*30:2
Aria CMD =150 cm²
b) Aria triunghiului decupat e : AriaΔDMN= Aria ABCD:3=(20*30:2):3=600:3=200
A ΔADN+AΔNBM=Aria ABCD-Aria ΔDCM-Aria AMN=600-150-200=250
Si avem:
A ΔADN=AD*AN:2=20*AN:2=10*AN
AΔNBM=MB*NB:2=10*(AB-AN):2=5*(30-AN)=150-5*AN
A ΔADN+AΔNBM=250=10*AN+150-5*AN=150+5*AN
5*AN=250-150=100
AN=20 cm²
c) DN²=AD²+AN²=20²*2=800
DN=20√2
NB=AB-NB=30-20=10
MN²=NB²+BM²=10²+10²=2*10²=200
MN=10√2
DM²=DC²+CM²=30²+10²=900+100=1000
DM=10√10
si avem 800+200=1000 adica:
(10√2)²+(20√2)²=(10√10)² , adica
MN²+DN²=DM²
=> Triunghiul decupat DMN este dreptunghic
d) AΔDAN=DA*AN:2=20*20:2=200
AΔBMN=BM*NB:2=10*10:2=50
200:50=4 => deci ΔBMN reprezinta o patrime din ΔDAN=>
ΔBMN reprezinta 25% din ΔDAN
