In trapezoid $A B C D$, $A B / / C D, A D$ $=49, B C=60, C D=B D=50$. Find the area of trapezoid $A B C D$.
Solve as shown in Figure 3, construct $D E \perp B C, B F \perp C D$, $A P \perp C D$, with the feet of the perpendiculars being $E, F, P$ respectively.

In the isosceles $\triangle B D C$, by the "three lines coincide" property, we have
$$
C E=\frac{1}{2} B C=30.
$$

In the right $\triangle D E C$, by the Pythagorean theorem, we get
$$
D E=\sqrt{C D^{2}-C E^{2}}=40.
$$

From the area of $\triangle B D C$, we know
$$
C D \cdot B F=2 S_{\triangle B D C}=B C \cdot D E.
$$

Thus, $B F=\frac{B C \cdot D E}{C D}=48$.
It is easy to see that quadrilateral $A B F P$ is a rectangle.
Therefore, $A P=B F=48, A B=P F$.
In the right $\triangle B D F$ and right $\triangle A D P$, by the Pythagorean theorem, we respectively get
$$
\begin{array}{l}
D F=\sqrt{B D^{2}-B F^{2}}=14, \\
D P=\sqrt{A D^{2}-A P^{2}}=\sqrt{97}.
\end{array}
$$

Thus, $A B=P F=D F-D P=14-\sqrt{97}$.
Therefore, $S_{\text {trapezoid } A B C D}=\frac{1}{2}(A B+C D) B F$
$$
=1536-24 \sqrt{97}.
$$

Additionally, there is another scenario, as shown in Figure 4.
Similarly,
$$
\begin{array}{l}
B F=48, \\
D F=14, \\
D P=\sqrt{97}.
\end{array}
$$

Thus, $A B=P F$
$$
=D F+D P=14+\sqrt{97}.
$$

Therefore, $S_{\text {trapezoid } B C D}=\frac{1}{2}(A B+C D) B F$
$$
=1536+24 \sqrt{97}.
$$

In summary, the area of trapezoid $A B C D$ is
$$
1536 \pm 24 \sqrt{97}.
$$
(From the 29th Middle School of Shuzhuang City, Shandong Province, 277000)