Triunghiul BMn este dreptunghic.Determini ipotenuzaNM
NM=[tex] \sqrt({ 4^{2} } + 3^{2}) = \sqrt{25} =5[/tex]
tΔBNM~ΔBCA conf teorema fundamentala a asemanarii
[tex] \frac{BN}{Bc} = \frac{5}{15= \frac{1}{3} } [/tex]
BC=5*BN=3*3=9
[tex] \frac{BM}{AB} = \frac{4}{AB}=\ \textgreater \ [/tex]
=[tex] \frac{1}{3}
[/tex]
AB=4*3=12
P=15+12+9=36
