[tex]\displaystyle\\
\text{\^In triunghiul ABC avem:}\\
D \in [BC]\\
E\text{ este mijlocul lui }[AC]\\
F\text{ este miljocul lui }[AB]\\
\text{Aria }\Delta ABC=18~cm^2 \\
\text{Aflati aria lui AEDF.} \\\\
\text{Rezolvare:}\\
\text{AP este inaltimea }\Delta ABC\\
\text{AM este inaltimea }\Delta AFE\\
\text{DN este inaltimea }\Delta DFE\\\\
\text{EF este linie mijlocie in }\Delta ABC\\\\
=\ \textgreater \ ~EF =\frac{BC}{2}\\
\text{EF imparte inaltimea AP in doua parti egale.}
=\ \textgreater \ ~AM =\frac{AP}{2}\\
[/tex]
[tex]\displaystyle\\
Aria~\Delta AEF = \frac{FE\times AM}{2} =\frac{ \frac{BC}{2} \times \frac{AP}{2}}{2} =\frac{ \frac{BC\times AP}{4} }{2} =\\\\\\
=\frac{BC\times AP}{4\times 2}=\frac{ \frac{BC\times AP}{2} }{4} = \frac{\text{aria}~\Delta ABC}{4} = \frac{18}{4} = \frac{9}{2} =\boxed{4,\!5~cm^2}\\\\
{Aria }\Delta DE\! F = \frac{EF\times DN}{2} \\
\text{EF este baza comuna a triunghiurilor AEF si DE\! F.}\\
DN = MP = AM\\
\text{Rezulta ca inaltimile triunghiurilor AEF si DE\!\! F sunt egale.} \\
[/tex]
[tex]\displaystyle\\
\Longrightarrow~~\text{Ariile triunghiurilor AEF si DE\!\! F sunt egale.}\\\\
\Longrightarrow~~\text{Aria }\Delta DE\! F = \text{Aria }\Delta AEF = \boxed{4,\!5~cm^2 } \\\\
\text{Aria patrulaterului }AEDF = \text{Aria }\Delta AEF + \text{Aria }\Delta DE\! F\\\\
\text{Aria patrulaterului }AEDF =4,5+4,5 = \boxed{\boxed{\bf 9~cm^2}}[/tex]
