Select the correct answer.

Points [tex]$A$[/tex] and [tex]$B$[/tex] lie on a circle centered at point [tex]$O$[/tex]. If [tex]$\frac{\text{length of }\hat{AB}}{\text{radius}} = \frac{\pi}{10}$[/tex], what is the ratio of the area of sector [tex]$AOB$[/tex] to the area of the circle?

A. [tex]$\frac{1}{10}$[/tex]

B. [tex]$\frac{\pi}{10}$[/tex]

C. [tex]$\frac{1}{20}$[/tex]

D. [tex]$\frac{\pi}{20}$[/tex]



Answer :

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]