Salut!
1. Aria unui romb este de 120 m², iar una dintre diagonale are 10 m. Aflati latura rombului.
Voi nota rombul ABCD, diagonalele sunt AC, BD. Laturi: AB,BC,CD,DA. Fie {O}=AC∩BD. Si BD=10.
[tex]Aria=\frac{d_1*d_2}{3} \\\\120=\frac{BD*AC}{2}\\\\120=\frac{10*AC}{2}\\\\120=5AC|:5\\\\AC=24[/tex]
Deci AO=OC=12 si BO=OD=5
Pitagora in ΔAOB:
AB²=AO²+OB²
AB²=12²+5²
AB²=144+25
AB²=169
AB=√169
AB=13
2. Aria unui triunghi dreptunghic ABC, cu ∡A-drept este de 24 m², iar AB=8 m. Calculati sin B.
[tex]Aria=\frac{c_1*c_2}{2} \\\\24=\frac{AB*AC}{2} \\\\24=\frac{8*AC}{2} \\\\24=4*AC|:4\\\\AC=6[/tex]
In ΔABC-Pitagora
BC²=AC²+AB²
BC²=6²+8²
BC²=36+64
BC²=100
BC=√100
BC=10
[tex]sin B=\frac{cat.op.}{ip.}\\\\sinB=\frac{AC}{BC} \\\\sinB=\frac{6}{10} \\\\sinB=\frac{3}{5}[/tex]
Succes!
