Question

Geometrie
Va rog frumos...!!!!

Answer 1

   
[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]