1. Consider square ABCD. You know that
![A_(ABCD)=AD^2=200,](https://img.qammunity.org/2019/formulas/mathematics/college/sy2l7zvxzzvpm7qx1rw723vwjmizf7vo5m.png)
then
![AB=BC=CD=AD=10√(2).](https://img.qammunity.org/2019/formulas/mathematics/college/5ps33fome5ttp8wikhprjpwz2rtkbdlezm.png)
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
![FG=(AD)/(2)=(10√(2) )/(2)=5√(2).](https://img.qammunity.org/2019/formulas/mathematics/college/nibml5u5tayrq21dxyoqyhj1p4gq5i6a6n.png)
3. Consider trapezoid BFGC. Its area is
where h is the height of trapezoid and is equal to half of AB. Thus,
![A_(BFGC)=(FG+BC)/(2)\cdot (AB)/(2)=(5√(2)+10√(2))/(2)\cdot (10√(2))/(2)=75.](https://img.qammunity.org/2019/formulas/mathematics/college/v5zw86t454d0y94syixwfpi0dne6f9pboi.png)
4.
![A_(BFGC)=A_(BFGE)+A_(EGC),\\A_(EGC)=A_(BFGC)-A_(BFGE)=75-34=41.](https://img.qammunity.org/2019/formulas/mathematics/college/wlz0yf2zmjxwv3uhwxxwm6e36k61p6lk9r.png)
5. Note that angles EGC and CGD are supplementary and
![\sin \angle CGD=\sin \angle EGC.](https://img.qammunity.org/2019/formulas/mathematics/college/ax669mf1s3dxdk2rjobto6civcv4t06xo6.png)
Then
![A_(CGD)=(1)/(2)CG\cdot CD\cdot \sin \angle CGD=(1)/(2)CG\cdot EG\cdot \sin \angle CGE=A_(ACG)=41.](https://img.qammunity.org/2019/formulas/mathematics/college/rxn4tcg46zpslb0fifz4wm868q78268j3q.png)
Answer:
![A_(CGD)=41.](https://img.qammunity.org/2019/formulas/mathematics/college/uee27p7me5nk4de6vpkh4dtmikw37nmbb5.png)