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Question
The length of the minute hand of a clock is 5 cm. Find the area swept by the minute hand during the time period 6 : 05 am and 6 : 40 am.
Solution
We know that, in 60 mins, minute hand revolving = 360°
In 1 min, minute hand revolving = `360^circ/60^circ`
∴ In (6 : 05 am to 6 : 40 am) = 35 mins,
minute hand revolving = `360^circ/60 xx 35` = (6 × 35)°
Given that, length of minute hand (r) = 5 cm
∴ Area of sector AOBA with angle ∠O
= `(π"r"^2)/360^circ xx ∠"O"`
= `22/7 xx (5)^2/360^circ xx (6 xx 35)^circ`
= `22/7 xx (5 xx 5)/360^circ xx (6 xx 35)^circ`
= `(22 xx 5 xx 5 xx 5)/60^circ`
= `(22 xx 5 xx 5)/12`
= `(11 xx 5 xx 5)/6`
= `275/6`
= `45 5/6 "cm"^2`
Hence, the required area swept by the minute hand is `45 5/6 "cm"^2`.
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