Example: 31 red right-angled triangular paper pieces with a hypotenuse of 29, 1 blue right-angled triangular paper piece with a hypotenuse of 49, and 1 yellow square paper piece, are used to form a right-angled triangle as shown in Figure 4. Question: What is the sum of the areas of the red and blue triangular paper pieces? Please explain your reasoning.
(7th Hua Luogeng Cup Junior Mathematics Invitational Team Final (Oral))
Solution: As shown in Figure 5, rotate the right triangle $\triangle BDE$ counterclockwise around point $D$ by $90^{\circ}$. Since $DE = DF$, point $E$ clearly coincides with point $F$. Given $\angle DEB = 90^{\circ} = \angle DFC$, $EB$ falls along $FC$, and point $B$ lands on point $G$ on $FC$. At this moment, the right triangle $\triangle BDE$ moves to the position of the right triangle $\triangle DFG$, and $DB$ moves to the position of $DG$. Therefore,
$$
\begin{array}{l}
DG = DB = 29, \\
\angle ADG = \angle ADF + \angle FDG \\
= \angle ADF + \angle BDE = 180^{\circ} - \angle EDF \\
= 180^{\circ} - 90^{\circ} = 90^{\circ}.
\end{array}
$$

At this point, the area of the right triangle $\triangle ADG$ is equal to the sum of the areas of the red and blue right triangles.

Since $S_{\text{ADG}} = \frac{1}{2} AD \cdot DG = \frac{1}{2} \times 49 \times 29 = \frac{1421}{2} = 710.5$, the sum of the areas of the red and blue right triangles in the figure is 710.5.