The minute hand of a clock is 10 cm long. Find the area of the face of the clock described by the minute hand between 8 AM and 8.25 AM.
Answers
Answer:
The angle described by the minute hand on the face of the clock between 8:00 AM and 8:25 AM is 130.95 cm²
Step-by-step explanation:
Given :
Radius of circle,r = 10 cm
Minutes between 8:00 AM and 8:25 AM = 25 min
Angle described by the minute hand in one minute = 6°
Angle described by the minute hand in 25 minutes ,θ = 6° ×25 = 150°
Angle swept by the minute hand in 25 minutes = Area of sector of angle 150° in a circle of radius 10 cm
Area of the sector of a circle, A = (θ/360) × πr²
A = (150°/360°) × π ×10²
A = 5/12 × 22/7 × 100
A = (5 × 22 × 100) /(12×7)
A = (5 × 11 × 100) /(6×7)
A = (5 × 11 × 50) /(3 ×7)
A = (55 × 50)/21
A = 2750/21
A = 130.95 cm²
Angle described by the minute hand in 25 minutes , A = 130.95 cm²
Hence, the angle described by the minute hand on the face of the clock between 8:00 AM and 8:25 AM is 130.95 cm²
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Answer:
length of minute hand =10cm .
time taken =25 min.
1min=6degree
25min=25×6degree=150degree
therefore angle formed by sector =150degree
area =given angle\ 360degree ×πr2
therefore 130.95 cm2 is area