7. (10 points) As shown in the figure, in isosceles trapezoid $A B C D$, the upper base $A B$ is 4 cm, the lower base $C D$ is 12 cm, and the angle between the leg $A D$ and the lower base $D C$ is $45^{\circ}$. If $A F=F E=E D$ and $B C=4 B G$, then the area of $\triangle E F G$ is $\qquad$ square centimeters.
【Solution】Solution: According to the analysis, draw the height of the trapezoid and mark the relevant data, as shown in the figure:
From the characteristics of the isosceles trapezoid, we know that $D M=A M=B N=C N, A B=M N$, so the length of $D M$ is:
$(12 - 4) \div 2=4$ (cm) Therefore, $A M=4$ (cm),
The area of trapezoid $A B C D$ is $=(4+12) \times 4 \div 2=16 \times 4 \div 2=32$ (square cm)
Connect $D G$ and intersect the extension of $A B$ at point $P$, as shown in the figure below:
Since $B C=3 D G, C D=3 B P$.
According to the scaling rule of the figure, we can know: $D G=3 P G, C D=3 B P$.
Since $C D=12$ cm, thus $B P=12 \div 3=4$ cm,
The area of triangle $A D P$ is $=(4+4) \times 4 \div 2=8 \times 4 \div 2=32 \div 2=16$ (square cm);
Since $D G=3 P G$, the area of triangle $A D G$ is: $16 \div(3+1) \times 3=16 \div 4 \times 3=4 \times 3=12$ (square
cm);
Since $A F=F E=E D$, the area of triangle $E F G$ is $=12 \div 3=4$ (square cm)
Therefore, the answer is: 4