【Question 12】
As shown in the figure, in rectangle $A B C D$, $A D-A B=9$ cm, the area of trapezoid $A B C E$ is 5 times the area of triangle $A D E$, and the perimeter of triangle $A D E$ is 68 cm shorter than the perimeter of trapezoid $A B C E$. The area of rectangle $A B C D$ is $\qquad$ square cm.
Connect $A C$;
Since $S_{\text {trapezoid } A B C E}=S_{\triangle A D E} \times 5$; therefore $S_{\text {rectangle } A B C D}=S_{\triangle A D E} \times 6$;
Since $S_{\text {rectangle } A B C D}=S_{\triangle A D C} \times 2$; therefore $S_{\triangle A D C}=S_{\triangle A D E} \times 3$; so $D C=D E \times 3, D E: C E=1: 2$;
The perimeter of triangle $A D E$ is shorter than the perimeter of trapezoid $A B C E$ by $A B+C E-D E=A B+A B \times \frac{2}{3}-A B \times \frac{1}{3}=A B \times \frac{4}{3}$, which is 68 cm shorter; $A B \times \frac{4}{3}=68$ cm, $A B=68 \div \frac{4}{3}=51$ cm;
Since $A D-A B=9$; therefore $A D=51+9=60$ cm;
$S_{\text {rectangle } A B C D}=60 \times 51=3060$ square cm.