Question

If an analog clock has hands at 12 and 2 and a radius of 9 inches, then what is the length of the major arc created by the central angle to the nearest tenth of an inch?(1 point)
Responses

51.8 inches
51.8 inches

47.1 inches
47.1 inches

4.7 inches
4.7 inches

9.4 inches

Answer 1

To find the length of the major arc created by the central angle, we first need to determine the central angle formed by the clock hands at 12 and 2.

On a clock, each number represents a 30-degree increment (360 degrees divided by 12 hours). The hands are at 12 and 2:

- From 12 to 2 is 2 intervals of 30 degrees.
- Therefore, the angle between the hands is \(2 \times 30° = 60°\).

The major arc will be the remaining angle around the circle:

\[
\text{Major angle} = 360° - 60° = 300°
\]

Next, we will calculate the length of the major arc using the formula for the length of an arc:

\[
\text{Length of arc} = \frac{\theta}{360°} \times 2\pi r
\]

Where:
- \(\theta\) is the central angle (300 degrees),
- \(r\) is the radius (9 inches).

Substituting the values into the formula:

\[
\text{Length of major arc} = \frac{300}{360} \times 2\pi(9)
\]

\[
= \frac{5}{6} \times 2\pi(9)
\]

\[
= \frac{5 \times 18\pi}{6}
\]

\[
= 15\pi
\]

Now calculating \(15\pi\):

\[
15\pi \approx 15 \times 3.1416 \approx 47.1 \text{ inches}
\]

Thus, the length of the major arc is approximately 47.1 inches.

The answer is **47.1 inches**.