Formula ariei trapezulu:
(B + b) * h / 2
[tex]A_{ABCD}=\frac{(AB+CD)AD}{2}=576\ cm^2[/tex]
Fie ME ⊥ BC si E ∈ BC
Vom imparti trapezul in triunghiurile dreptunghice: ΔMAB, ΔMDC, ΔMEC, ΔMEB, si vom scrie aria trapezului in functie de ariile lor:
[tex]A_{ABCD}= \frac{MA\cdot AB}{2}+\frac{MD\cdot CD}{2} +\frac{ME\cdot CE}{2} + \frac{ME\cdot EB}{2}\\
2A_{ABCD}=MA(AB+CD)+ME(CE+EB)\\
2A_{ABCD}=\frac{AD(AB+CD)}{2}+ME\cdot CB\\
2A_{ABCD}=A_{ABCD+ME\cdot CB}\\
ME\cdot CB =A_{ABCD}\\\\
ME=\frac{A_{ABCD}}{CB}=\frac{576}{18}=32\ cm[/tex]
