Question

The minute hand of a clock is √21 cm long. Find the area described by the minute hand on the face of the clock between 7.00 AM and 7.05 AM.

Answer 1

Answer:

The angle described by the minute hand on the face of the clock between 7:00 AM and 7:05 AM is 5.5 cm²

Step-by-step explanation:

Given :

Radius of circle,r = √21 cm

Minutes between 7:00 AM and 7:05 AM = 5 min  

Angle described by the minute hand in one minute = 6°

Angle described by the minute hand in 5 minutes ,θ = 6° ×5 = 30°

Angle swept by the minute hand in 5 minutes = Area of sector of angle 30° in a circle of radius √21 cm

 

Area of the sector of a circle, A = (θ/360) × πr²

A = (30°/360°) × π ×√21²

A = 1/12  × 22/7 × 21  

A = (22 × 3)/12

A = 22/4

A = 5.5 cm²

Angle described by the minute hand in 5 minutes , A = 5.5 cm²

 Hence, the angle described by the minute hand on the face of the clock between 7:00 AM and 7:05 AM is 5.5 cm²

HOPE THIS ANSWER WILL HELP YOU….

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Answer 2

Answer:

Step-by-step explanation:

If we divide the parts of a clock based on it's time interval we get,

1-2

2-3

3-4

4-5

5-6

6-7

7-8

8-9

9-10

10-11

11-12

so that gives us 12 equal parts

now, between 7:00 and 7:05 we have 1/12 part

so that is 360*/12= 30*

so area of that particular sector will be

22/7 *21*30/360 ( root 21* root21 is 21)

=5.5cm cube