a) P= Ab+BC+CD+AD= 10+15+10+15
=25+25=50
b) Aplicam Teorema lui Pitagora in DNC=> NC^2= DN^2+DC^2=> NC^2=5^2+10^2=> NC^2=25+100=> NC^2= 125=> NC=√125= 5√5
AN=AD-DN=15-5=10
Aplicam teorema lui pitagora in AMN=> MN^2=AN^2+AM^2= 10^2+5^2= 125=> MN=√125=5√5
=> MN=NC=> MNC-isoscel
fie NP perpendicular pe MC
pt ca MNC-isoscel=> NP este si mediana
apl teorema lui pitagora in MBC=> MC^2= MB^2+BC^2= 5^2+15^2= 25+225=250=> MC=√250=5√10=>MP= MC/2=5√10/2
Apl t.p in MNP=> NP^2=MN^2-MP^2= (5√5)^2-(5√10/2)^2=125-250/4 (acum il amplifici pe 125 cu 4)=> NP^2=500/4-250/4=250/4=> NP=√250/4=5√10/2
Aria= (MC•NP)/2=(5√10/2)•2/2= 5√10/2
