1.
Ducem înălțimea AD, care este și mediană, deci BD=DC=30cm.
(30, 40, 50) - triplet pitagoreic ⇒ AD = 40cm
[tex]\it \mathcal{A}=\dfrac{BC\cdot AD}{2}=\dfrac{60\cdot40}{2}=1200\ cm^2[/tex]
2.
[tex]\it \mathcal{A}=\dfrac{\ell ^2\sqrt3}{4}=\dfrac{(3\sqrt2)^2\cdot\sqrt3}{4}=\dfrac{9\cdot2\cdot\sqrt3}{4}=\dfrac{9\sqrt3}{2}=4,5\cdot\sqrt3\ cm^2[/tex]
3.
[tex]\it \mathcal{A}=AB\cdot BD\cdot sin150^o=20\cdot10\cdot\dfrac{1}{2}=100\ cm^2[/tex]
