1. Consider square ABCD. You know that
[tex]A_{ABCD}=AD^2=200,[/tex]
then
[tex]AB=BC=CD=AD=10\sqrt{2}.[/tex]
2. Consider traiangle AED. F is mipoint of AE and G is midpoint of DE, then FG is midline of triangle AED. This means that
[tex]FG=\dfrac{AD}{2}=\dfrac{10\sqrt{2} }{2}=5\sqrt{2}.[/tex]
3. Consider trapezoid BFGC. Its area is
[tex]A_{BFGC}=\dfrac{FG+BC}{2}\cdot h,[/tex] where h is the height of trapezoid and is equal to half of AB. Thus,
[tex]A_{BFGC}=\dfrac{FG+BC}{2}\cdot \dfrac{AB}{2}=\dfrac{5\sqrt{2}+10\sqrt{2}}{2}\cdot \dfrac{10\sqrt{2}}{2}=75.[/tex]
4.
[tex]A_{BFGC}=A_{BFGE}+A_{EGC},\\A_{EGC}=A_{BFGC}-A_{BFGE}=75-34=41.[/tex]
5. Note that angles EGC and CGD are supplementary and
[tex]\sin \angle CGD=\sin \angle EGC.[/tex]
Then
[tex]A_{CGD}=\dfrac{1}{2}CG\cdot CD\cdot \sin \angle CGD=\dfrac{1}{2}CG\cdot EG\cdot \sin \angle CGE=A_{ACG}=41.[/tex]
Answer: [tex]A_{CGD}=41.[/tex]