The problem provided in the image is related to the area of a sector of a circle, where the minute hand of a clock sweeps through a specific angle in one minute.

Problem Analysis:

1. The minute hand of the clock completes one full revolution (360°) in 60 minutes.
2. Therefore, in one minute, the angle swept by the minute hand is: $\theta = \frac{360^\circ}{60} = 6^\circ$
3. The area of a sector of a circle is given by the formula: $\text{Sector Area} = \frac{\theta}{360^\circ} \cdot \pi r^2$ where $r$ is the length of the radius (minute hand), and $\theta$ is the angle of the sector.

Substituting the values:

• $\theta = 6^\circ$,
• $r^2$ is the square of the radius,
• The formula becomes: $\text{Sector Area} = \frac{6}{360} \cdot \pi r^2 = \frac{\pi r^2}{60}$

Final Answer:

The area of the sector formed by the minute hand in one minute is: $\boxed{\frac{\pi r^2}{60}}$

This matches Option (A).

Would you like further clarification on the formula or the steps? Here are five related questions to explore:

1. How is the formula for the area of a sector derived?
2. What would be the area swept in 15 minutes instead of 1 minute?
3. How is the angle swept by the hour hand calculated, and how would the formula differ for it?
4. What is the area swept in a full revolution of the clock's minute hand?
5. Can you calculate the perimeter of the sector in this problem, including the arc length?

Tip: Always remember, for sector problems, knowing the angle is key to determining the fraction of the circle being considered.