Как найти площадь циферблата часов

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 $9$ am and $9:35$ am.

$begin{align}&color{green}{a.)183.3quad cm^2}\
&b.)366.6quad cm^2\
&c.)244.4quad cm^2\
&d.)188.39quad cm^2\
end{align}$

I tried

$Area=dfrac{30}{360} times pitimes 10^2=26.17quad cm^2$

what is the mistake ?

asked May 17, 2015 at 20:56

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When the minute hand moved from 9 am to 9:35 am,

It moved $frac7{12}$ cicrle,that is,

it moved $frac{7}{12}cdot360=210$ degree NOT $30$ degree.

answered May 17, 2015 at 21:05

In a 12 hour clock, we know that $$1 space minute=frac{360}{60}=6^o=frac{6times pi}{180}=frac{pi}{30} space radian$$

Hence, the area swept by the minute hand between 9 am to 9:35 am is $$=frac{1}{2}times(text{total angle swept})times(text{length of minute hand})^2=frac{1}{2}timesleft(35times frac{pi}{30}right)times(10)^2=frac{175pi}{3}approx183.2595715 space cm^2$$

answered May 17, 2015 at 23:45

The fraction of full angle in minutes time =$ dfrac{35}{60} $

The fraction of swept area =$ dfrac{35}{60} pi 10^{2} approx 183.26. $

answered May 18, 2015 at 0:04

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