\ANG{
\begin{minipage}[t]{8cm}
%\item
Berechne den Fl"acheninhalt der schraffierten Figur.\\
Es gilt:\\
$a=10\,$cm, $b=4,5\,$cm, $c=2\,$cm, $d=4\,$cm, $e=2,5\,$cm, $f=8\,$cm\\
Das Viereck $EFBA$ ist ein Rechteck und $GH \parallel EF$.
\end{minipage}
\hfill
\begin{minipage}[t]{6cm}
\makebox[0cm]{}\\
\beginpicture
\setcoordinatesystem units <1mm,1mm> 
\unitlength1mm
\putrectangle corners at 0 0 and 50 40
\betweenarrows {a} [b] <0pt,3mm> from 0 40 to 50 40
\betweenarrows {f} [l] <4mm,0pt> from 50 0 to 50 40
\plot 0 0 12.5 17.5 32.5 17.5 50 0 /
\plot 0 30 22.5 40 /
\plot 12.5 17.5 22.5 40 32.5 17.5 /

\put {A} at -2 40
\put {C} at -2 30
\put {E} at -2 0
\put {B} at 52 40
\put {F} at 52 0
\put {D} at 22.5 42
\put {G} at 10.5 17.5
\put {H} at 34.5 17.5
\put {I} at 12.5 -2

\put {b} at 9 39
\put {c} at 2 35
\put {d} at 22.5 19.5
\put {e} at 6 1

\setshadegrid span <2.5pt>
\vshade 0 0 30 12.5 17.5 35.55 22.5 40 40 32.5 17.5 40 50 0 40 /

\setdashes <1mm>
\setlinear
\plot 12.5 0 12.5 17.5 /

\setsolid

\circulararc 90 degrees from 15.5 0 center at 12.5 0
\put {\circle*{0.7}} at 14.5 1

\circulararc -45 degrees from 41 0 center at 50 0
\put {$45^o$} at 45 2

\endpicture 
\end{minipage}
}
\LSG{
$A=42\, \rm cm^2$
}
