ΔABC oarecare
Fie BE_|_ AC, E ∈ [AC]
Avem ΔBEA, mas<E= 90°
mas<A= 30° => mas<ABE= 60° => BE= AB:2= 6cm
ArieΔ=[tex] \frac{b*h}{2} [/tex]
AΔ= [tex] \frac{BE*AC}{2} [/tex]= [tex] \frac{6*8}{2} [/tex]= [tex]24 cm^{2} [/tex]
b) ΔABC oarecare
Fie BE_|_AC, E∈ [AC]
ΔBEC, mas<E= 90°
mas<C= 45° => ΔBEC dreptunghic isoscel⇒ BE=CE= 6√2cm
AΔ= 6√2·6√2:2= 24:2= 12cm²
c) ΔABC oarecare
Fie AE_|_ BC, unde E∈[BC]
Avem ΔAEB, mas<E= 90°
mas<B= 60°=>mas<BAE= 30°=> BE= AB/2= 4√3cm
AE= hΔ
h²= (8√3)²- (4√3)²
h²= 192- 48
h= √144= 12cm
AΔ= [tex] \frac{12*5}{2} = \frac{60}{2} = 30 cm^{2} [/tex]
