解：在$Rt\triangle ABC$中，$AB = 5，$$AC = 3，$根據(jù)勾股定理$BC=\sqrt{AB^{2}-AC^{2}}=\sqrt{5^{2}-3^{2}} = 4。$
以$AC$為直徑的半圓面積$S_{1}=\frac{1}{2}\pi(\frac{AC}{2})^{2}=\frac{1}{2}\pi(\frac{3}{2})^{2}=\frac{9\pi}{8}；$
以$BC$為直徑的半圓面積$S_{2}=\frac{1}{2}\pi(\frac{BC}{2})^{2}=\frac{1}{2}\pi(\frac{4}{2})^{2}=2\pi；$
以$AB$為直徑的半圓面積$S_{3}=\frac{1}{2}\pi(\frac{AB}{2})^{2}=\frac{1}{2}\pi(\frac{5}{2})^{2}=\frac{25\pi}{8}；$
$\triangle ABC$的面積$S_{\triangle ABC}=\frac{1}{2}AC\cdot BC=\frac{1}{2}\times3\times4 = 6。$
陰影部分面積$S = S_{1}+S_{2}+S_{\triangle ABC}-S_{3}$
$=\frac{9\pi}{8}+2\pi + 6-\frac{25\pi}{8}$
$=6。$
