【Question 14】
As shown in the figure, in rectangle $A B C D$, $A B=6, B C=8$. After folding rectangle $A B C D$ along $C E$, point $D$ lands on point $F$ on diagonal $A C$. What is the area of trapezoid $A B C E$? $\qquad$
【Analysis and Solution】Geometry, Pythagorean Theorem.

Since $\angle B=90^{\circ}$;

Therefore, $A B^{2}+B C^{2}=A C^{2}$; Since $A B=6, B C=8$; Therefore, $A C=10$;

Since $\triangle C D E$ is symmetric to $\triangle C F E$ about $C E$; Therefore, $\triangle C D E \cong \triangle C F E$;

Therefore, $C D=C F=6$, Let $D E=F E=x, \angle D=\angle C F E=90^{\circ}$;
Then $A F=A C-C F=10-6=4, A E=A D-D E=8-x$;

Since $\angle C F E=90^{\circ}$; Therefore, $\angle A F E=90^{\circ}$; Therefore, $A F^{2}+E F^{2}=A E^{2}$;

Therefore, $4^{2}+x^{2}=(8-x)^{2}$; Solving for $x$ gives $x=3$; Therefore, $A E=8-3=5$;
Therefore, $S_{\text {trapezoid } A B C E}=(A E+B C) \times A B \div 2=(5+8) \times 6 \div 2=39$.