Example 1 According to the data marked in Figure 1, the area of the shaded part is $\qquad$ [1]
(2009, Nanchang City Mathematics Competition (Grade 8))
By symmetry,
we get

and $\frac{S_{0}}{S_{1}}=\frac{3}{2}$.
Assume $S_{0}=S_{0}^{\prime}=3 t, S_{1}=S_{1}^{\prime}=2 t$.
Then $3 t+3 t+2 t=8 t=\frac{3 \times 5}{2} \Rightarrow t=\frac{15}{16}$.
Thus, $S_{2}=\frac{5 \times 5}{2}-(3 t+3 t+2 t+2 t)$
$=\frac{25}{2}-10 t=\frac{25}{2}-\frac{75}{8}=\frac{25}{8}$.