задача номер 15
BN=NA=4 tg ∠ ANT=0,5 ⇒ AT=2 TD=AD-AT=8-2=6 Δ АNT ∼ Δ TDM ⇒ DM=12 S_( Δ BMT)=S_(BTMC)-S_( Δ BMC)=S_(квадрата)- S_( Δ ABT) + S_( Δ TDM) -S_( Δ BMC)= =8*8-(1/2)*8*2+(1/2)*6*12-(1/2)*20*8=[b]12[/b]