Deoarece N apartine lui AD astfel incat AN = 8 cm => NB = 4 cm (AN+NB=8+4=12=AD)
M este mijlocul lui [AB] => AM = MB = 12 : 2 = 6 cm.
AΔMNC=A ABCD - (AΔNAM + AΔDNC + AΔCBM)
A ABCD = l² => A ABCD = 12² = 144 cm²
ΔNAM dreptunghic => AΔNAM = (c1 * c2) supra 2 =>
AΔNAM=(AN * AM) supra 2 = 8 * 6 supra 2 => AΔNAM=24 cm²
ΔDNC dreptunghic => AΔDNC = 4*12 supra 2 => AΔDNC=24 cm²
ΔCBM dreptunghic => AΔCBM = 12*6 supra 2 => AΔCBM=36 cm²
=> AΔMNC=144-(24+24+36)=144-(48+36)=144-84=>
=>AΔMNC=60 cm²
Fie CP_|_MN => d(C,MN)=CP
In ΔNAM aplicam T.P.:
MN²=AN²+AM² => MN² = 64 + 36 = 100 => MN=10 cm
AΔMNC = (b * h) supra 2 => 60 = (MN * CP) supra 2 => 60 = (10*CP) supra 2 => 10 * CP = 60 * 2 => 10 * CP = 120 => CP = 120 : 10 => CP = 12 cm
Bafta.
