Solution 4.1:9

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m (Lösning 4.1:9 moved to Solution 4.1:9: Robot: moved page)
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<math>\text{1}0</math>
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seconds corresponds to
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<math>\frac{1}{6}</math>
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minutes, so that during that time period, the second hand sweeps over
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<math>\frac{1}{6}</math>
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of a turn, i.e. the sector of a circle with angle
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<math>\alpha =\frac{1}{6}\centerdot 2\pi </math>
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radians
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<math>=\frac{\pi }{3}</math>
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radians
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<center> [[Image:4_1_9_.gif]] </center>
<center> [[Image:4_1_9_.gif]] </center>
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<center> [[Image:4_1_9.gif]] </center>
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{{NAVCONTENT_STOP}}
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The area of the sector is
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Area
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<math>=\frac{1}{2}\alpha r^{2}=\frac{1}{2}\centerdot \frac{\pi }{3}\centerdot \left( 8\ \text{cm} \right)^{2}=\frac{32\pi }{3}\ \text{cm}^{2}\approx 33.5\ \text{cm}^{2}</math>

Revision as of 12:30, 27 September 2008

\displaystyle \text{1}0 seconds corresponds to \displaystyle \frac{1}{6} minutes, so that during that time period, the second hand sweeps over \displaystyle \frac{1}{6} of a turn, i.e. the sector of a circle with angle


\displaystyle \alpha =\frac{1}{6}\centerdot 2\pi radians \displaystyle =\frac{\pi }{3} radians


Image:4_1_9_.gif


The area of the sector is

Area \displaystyle =\frac{1}{2}\alpha r^{2}=\frac{1}{2}\centerdot \frac{\pi }{3}\centerdot \left( 8\ \text{cm} \right)^{2}=\frac{32\pi }{3}\ \text{cm}^{2}\approx 33.5\ \text{cm}^{2}

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