* THE LENGTH OF THE MINUTE HAND OF A CLOCK IS 6.3CM . FIND THE AREA SWEPT BY THE MINUTE HAND DURING THE TIME PERIOD 5:45AM TO 6:10 AM AD ALSO FIND THE DISTANCE DISTANCE TRAVELLED BY THE MINUTE HAND DURING THIS PERIOD. *

We know that in 1 hour (i.e., 60 minutes), the minute hand rotates 360°.

In 25 minutes, minute hand will rotate = (360^{0}/ 60^{0}) × 25 = 150^{0}

Therefore, the area swept by the minute hand in 25 minutes will be the area of a sector of 150° in a circle of 6.3 cm radius.

Area of sector of angle θ =( θ/360^{0}) × πr^{2}

Area of sector of 150° = (150^{0}/360^{0}) × (22/7) × 6.3 ×6.3 = (10395/200) = 51.975 cm^{2}

Therefore, the area swept by the minute hand in 25 minutes is 51.975 cm^{2}

Distance travelled by minute hand to cover 25 minutes is the length of the arc AB:

Therefore,

Length of the arc AB = ( θ/360^{0}) × 2πr = (150^{0}/360^{0}) × 2× (22/7) × 6.3 = (33/2) cm.

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