Final answer:
The area described by the minute hand on a clock face between 7:00 am and 7:05 am, with a hand length of 21 cm, is computed by finding the area of a sector that represents 1/12th of a full circle, resulting in 11/12 π cm². Therefore, the correct answer is (b) 11/12 π cm².
Step-by-step explanation:
The student is asking about the area swept by the minute hand on the face of a clock in a given amount of time. This is a classic problem in geometry concerning circular motion and involves calculating a sector of a circle's area.
Specifically, the minute hand of a clock that is 21 cm long covers a certain angle as it moves from 7:00 am to 7:05 am. Since there are 60 minutes in a full revolution of 360 degrees, 5 minutes represent 5/60 of the full rotation, which is 1/12th of the entire circle. Since the minute hand is 21 cm long, it serves as the radius (r) of the circle.
The area (A) of a sector of a circle can be calculated using the formula A = 1/2 × r² × θ, where θ is the angle in radians. In this case, the angle covered is (360/12)= 30 degrees, which when converted to radians is θ = π/6. Hence, the area that the minute hand sweeps between 7:00 am and 7:05 am is A = 1/2 × (21)^2 × (π/6) which simplifies to 11/12 π cm².
The area described by the minute hand on the face of the clock between 7:00 am and 7:05 am can be found by calculating the sector of the circle formed by the minute hand. The length of the minute hand is given as 21 cm, which is also the radius of the clock.
In 5 minutes, the minute hand moves from the 12 o'clock position to the 1 o'clock position. This is a fraction of 1/12 of a complete revolution.
Therefore, the area described by the minute hand can be calculated using the formula for the area of a sector, A = (θ/360) * π * r^2, where θ is the angle in degrees, r is the radius, and π is a constant. Plugging in the values, we get (1/12) * π * (21 cm)^2 = 7/12 π cm².
Therefore, the correct answer is (b) 11/12 π cm².