Question #8f739

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Question #8f739

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Answer 1 · 2017-03-07T15:44:08

Area swept by minute hand: $\frac{7 π}{4}$ $(7pi)/4$ sq.cm.

Explanation:

I have assumed underoot 21 means $\sqrt{21}$ $sqrt(21)$

The surface area of the clock when the minute hand makes 1 full rotation is
${Area}_{\circ} \leq = π r^{2} = π {(\sqrt{21})}^{2} = 21 π$ $"Area"_circle = pir^2 =pi(sqrt(21))^2=21pi$

During the interval from 7:00 to 7:05, the minute hand sweeps $\frac{1}{12}$ $1/12$ of its full surface area.

${Area}_{swept 7:00 to 7:05} = \frac{1}{12} \times 21 π = \frac{7 π}{4}$ $"Area"_"swept 7:00 to 7:05"=1/12xx21pi = (7pi)/4$

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Question #8f739

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