[tex]\displaystyle\bf\\\text{\bf Folosim formulele:}\\\\A_\Delta=\frac{b\times c\times \sin A }{2}~~~\text{(Aria triunghiului)}\\\\a^2=b^2+c^2-2bc\cos A~~~\text{(Teorema lui Pitagora generalizata)}\\\\\text{\bf Rezolvare:}\\\text{\bf Vezi desenul atasat.}[/tex]
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[tex]\displaystyle\bf\\a)\\\text{\bf Diagonala BD imparte paralelogramul in 2 triunghiuri congruente.}\\\\Aria~paralelogramului~ABCD=2\times A_{\Delta ABD}=\\\\=2\times \frac{AB\cdot AD\cdot\sin A}{2}=AB\cdot AD\cdot\sin A=\\\\=6\cdot4\cdot \sin60^o=24\cdot\frac{\sqrt{3}}{2}=\boxed{\bf12\sqrt{3}~cm^2}[/tex]
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[tex]\displaystyle\bf\\b)\\BD^2=AB^2+AD^2-2\cdot~AB\cdot~AD\cdot\cos A\\\\BD^2=6^2+4^2-2\cdot6\cdot4\cdot\cos 60^o\\\\BD^2=36+16-48\cdot\frac{1}{2}\\\\BD^2=36+16-24\\\\BD^2=36+16-24\\\\BD^2=28\\\\BD=\sqrt{28}=\sqrt{4\cdot7}=\boxed{\bf2\sqrt{7}~cm}[/tex]
