To solve the problem, we need to understand the relationship between the arc length and the radius of a circle involving the sector of the circle.
Given:
[tex]\[
\frac{\text{length of }\hat{AB}}{\text{radius}} = \frac{\pi}{10}
\][/tex]
We denote:
- [tex]\(l\)[/tex] as the length of arc [tex]\(\hat{AB}\)[/tex]
- [tex]\(r\)[/tex] as the radius of the circle
The formula connecting the arc length [tex]\(l\)[/tex] to the central angle [tex]\(\theta\)[/tex] in radians is:
[tex]\[
l = r \theta
\][/tex]
From the given relation, we can equate:
[tex]\[
\frac{l}{r} = \frac{\pi}{10}
\][/tex]
So,
[tex]\[
l = r \cdot \frac{\pi}{10}
\][/tex]
This implies that the central angle [tex]\(\theta\)[/tex] subtended by the arc [tex]\(\hat{AB}\)[/tex] is:
[tex]\[
\theta = \frac{\pi}{10} \text{ radians}
\][/tex]
Next, we calculate the area of sector [tex]\(AOB\)[/tex]. The area [tex]\(A_{\text{sector}}\)[/tex] of a sector with a central angle [tex]\(\theta\)[/tex] (in radians) in a circle of radius [tex]\(r\)[/tex] is given by:
[tex]\[
A_{\text{sector}} = \frac{1}{2} r^2 \theta
\][/tex]
Substituting [tex]\(\theta = \frac{\pi}{10}\)[/tex]:
[tex]\[
A_{\text{sector}} = \frac{1}{2} r^2 \cdot \frac{\pi}{10} = \frac{\pi r^2}{20}
\][/tex]
Now, the area [tex]\(A_{\text{circle}}\)[/tex] of the entire circle is:
[tex]\[
A_{\text{circle}} = \pi r^2
\][/tex]
Finally, the ratio of the area of the sector [tex]\(AOB\)[/tex] to the area of the whole circle is:
[tex]\[
\text{Ratio} = \frac{A_{\text{sector}}}{A_{\text{circle}}} = \frac{\frac{\pi r^2}{20}}{\pi r^2} = \frac{1}{20}
\][/tex]
Therefore, the correct answer is:
[tex]\[
\boxed{\frac{1}{20}}
\][/tex]
