[tex]\displaystyle\\
\text{Avem trapezul }ABCD\text{ cu }AB||CD\text{ si }AB\ \textgreater \ CD.\\
m(\sphericalangle B)=45^o\\
BC=40~cm\\
\text{Linia mijlocie a trapezului }=\frac{AB+CD}{2}=42~cm\\\\
\text{Rezolvare:}\\\\
\text{Din C ducem perpendiculara: }CE~\bot~AB},~~E\in AB.\\
\text{Se formeaza triunghiul dreptunghic isoscel}~\Delta EBC\text{cu ipotenuza }BC\\
\text{Inaltimea trapezului este cateta}CE \text{ in }\Delta EBC.\\\\
CE=BC\times\sin45^o=BC\times\frac{\sqrt{2}}{2}=40\times\frac{\sqrt{2}}{2}=20\sqrt{2}~cm[/tex]
[tex]\displaystyle\\
A_{trapez}= \frac{(AB+CD)\times CE}{2}=\\\\
= \frac{(AB+CD)}{2}\times CE= 42 \times 20\sqrt{2}= \boxed{\bf 840\sqrt{2}~cm^2}[/tex]
