27. Three semicircular protractors with radii of $1 \mathrm{~cm}, 3 \mathrm{~cm}, 5 \mathrm{~cm}$ have their centers coinciding at $O$, and their diameters also coincide on a straight line, as shown in the figure. Let the areas of the shaded segments in the semicircles be $S_{\text {甲, }} S_{\text {乙, }}, S_{\text {丙, }}$ respectively. Find $S_{\text {甲: }} S_{\text {乙: }} S_{\text {丙. }}$.
27. Three semicircular protractors with radii of $1 \mathrm{~cm}, 3 \mathrm{~cm}, 5 \mathrm{~cm}$ respectively have their centers coinciding at $O$, and their diameters also coincide on a straight line, as shown in the figure. Let the areas of the shaded segments in the sectors and the semicircle be $S_{\text {甲 }}, S_{\text {乙 }}, S_{\text {丙, }}$ respectively. Find $S_{\text {甲 }}: S_{\text {乙: }} S_{\text {丙. }}$.

【Analysis】Clearly, the ratio of the areas of 甲, 乙, and 丙 can be found by first calculating the areas of 甲, 乙, and 丙. 甲, 乙, and 丙 are sectors with different central angles, and their areas can be easily calculated based on the indicated arcs in the figure.

【Solution】According to the analysis, since 丙 is a semicircle, the area of 丙 $S$ 丙 $=\frac{180^{\circ}}{360^{\circ}} \times \pi \times 1^{2}=\frac{1}{2} \pi$; the central angle of 乙 is: $120^{\circ}-60^{\circ}=60^{\circ}$, so the area of 乙 $S_{\text {乙 }}=\frac{60^{\circ}}{360^{\circ}} \times \pi \times\left(3^{2}-1^{2}\right)$ $=\frac{8}{6} \pi=\frac{4}{3} \pi$; the central angle of 甲 is: $108^{\circ}-72^{\circ}=36^{\circ}$, so the area of 甲 $S_{\text {甲 }}=\frac{36^{\circ}}{360^{\circ}} \times \pi \times\left(5^{2}-3^{2}\right)=$
$$
\begin{array}{l}
\frac{1}{10} \times \pi \times 16=\frac{8}{5} \pi, \\
\therefore S_{\text {甲: }} S_{\text {乙: }} S_{\text {丙 }}=\left(\frac{8}{5} \pi\right):\left(\frac{4}{3} \pi\right):\left(\frac{1}{2} \pi\right)=48: 40: 15,
\end{array}
$$

Therefore, the answer is: 48: $40: 15$.