Answer :
To find the length of the minor arc [tex]\( \text{SV} \)[/tex] in a circle with a given radius and central angle in radians, we use the arc length formula:
[tex]\[ L = \theta \cdot r \][/tex]
Where:
- [tex]\( L \)[/tex] is the length of the arc
- [tex]\( \theta \)[/tex] is the central angle in radians
- [tex]\( r \)[/tex] is the radius of the circle
Given:
- Radius [tex]\( r = 24 \)[/tex] inches
- Central angle [tex]\( \theta = \frac{5 \pi}{6} \)[/tex] radians
Plug in the values to the formula:
[tex]\[ L = \left(\frac{5 \pi}{6}\right) \cdot 24 \][/tex]
First, multiply [tex]\( 24 \)[/tex] by [tex]\( \frac{5 \pi}{6} \)[/tex]:
[tex]\[ L = 24 \cdot \frac{5 \pi}{6} \][/tex]
[tex]\[ L = \frac{24 \cdot 5 \pi}{6} \][/tex]
[tex]\[ L = \frac{120 \pi}{6} \][/tex]
[tex]\[ L = 20 \pi \][/tex]
So, the length of the minor arc [tex]\( \text{SV} \)[/tex] is:
[tex]\[ \boxed{20 \pi} \text{ inches} \][/tex]
Comparing this result with the given options:
- [tex]\( 20 \pi \)[/tex] in.
- [tex]\( 28 \pi \)[/tex] in.
- [tex]\( 40 \pi \)[/tex] in.
- [tex]\( 63 \pi \)[/tex] in.
The correct answer is [tex]\( 20 \pi \)[/tex] in.
[tex]\[ L = \theta \cdot r \][/tex]
Where:
- [tex]\( L \)[/tex] is the length of the arc
- [tex]\( \theta \)[/tex] is the central angle in radians
- [tex]\( r \)[/tex] is the radius of the circle
Given:
- Radius [tex]\( r = 24 \)[/tex] inches
- Central angle [tex]\( \theta = \frac{5 \pi}{6} \)[/tex] radians
Plug in the values to the formula:
[tex]\[ L = \left(\frac{5 \pi}{6}\right) \cdot 24 \][/tex]
First, multiply [tex]\( 24 \)[/tex] by [tex]\( \frac{5 \pi}{6} \)[/tex]:
[tex]\[ L = 24 \cdot \frac{5 \pi}{6} \][/tex]
[tex]\[ L = \frac{24 \cdot 5 \pi}{6} \][/tex]
[tex]\[ L = \frac{120 \pi}{6} \][/tex]
[tex]\[ L = 20 \pi \][/tex]
So, the length of the minor arc [tex]\( \text{SV} \)[/tex] is:
[tex]\[ \boxed{20 \pi} \text{ inches} \][/tex]
Comparing this result with the given options:
- [tex]\( 20 \pi \)[/tex] in.
- [tex]\( 28 \pi \)[/tex] in.
- [tex]\( 40 \pi \)[/tex] in.
- [tex]\( 63 \pi \)[/tex] in.
The correct answer is [tex]\( 20 \pi \)[/tex] in.