[tex]\it ABCD-tapez\ dreptunghic,\ AB||CD,\ AB>CD,\ DA\perp AB,\ m(\hat B)=60^o\\ \\ Ducem\ CF\perpAB,\ F\in AB\\ \\ AFCD-dreptunghi\ \Rightarrow AF=CD=8cm \Rightarrow FB=12-8=4cm\\ \\ Din\ \Delta CFB \Rightarrow m(\widehat{ BCF}) =30^o\ (complementul\ lui\ 60^o) \stackrel{T.\angle30^o}{\Longrightarrow} \\ \\ \Rightarrow BC=2BF=2\cdot4=8cm[/tex]
Cu teorema lui Pitagora, în triunghiul FBC, obținem:
[tex]\it CF = 4\sqrt3\Rightarrow DA=CF=4\sqrt3cm\\ \\ \mathcal{P}=AB+BC+CD+DA=12+8+8+4\sqrt3=28+4\sqrt3cm\\ \\ \mathcal{A}=\dfrac{AB+CD}{2}\cdot DA= \dfrac{12+8}{2}\cdot4\sqrt3=10\cdot4\sqrt3=40\sqrt3cm^2[/tex]
