Solution 4.1:9

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Current revision (11:50, 8 October 2008) (edit) (undo)
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<math>\text{1}0</math>
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10 seconds corresponds to 1/6 minutes, so that during that time period, the second hand sweeps over 1/6 of a turn, i.e. the sector of a circle with angle
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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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{{Displayed math||<math>\alpha = \frac{1}{6}\cdot 2\pi\ \text{radians} = \frac{\pi}{3}\ \text{radians.}</math>}}
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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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{{NAVCONTENT_START}}
 
<center> [[Image:4_1_9_.gif]] </center>
<center> [[Image:4_1_9_.gif]] </center>
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{{NAVCONTENT_STOP}}
 
The area of the sector is
The area of the sector is
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Area
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{{Displayed math||<math>\text{Area} = \frac{1}{2}\alpha r^{2} = \frac{1}{2}\cdot \frac{\pi}{3}\cdot (8\ \text{cm})^2 = \frac{32\pi}{3}\ \text{cm}^{2} \approx 33\textrm{.}5\ \text{cm}^{2}\,\textrm{.}</math>}}
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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>
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Current revision

10 seconds corresponds to 1/6 minutes, so that during that time period, the second hand sweeps over 1/6 of a turn, i.e. the sector of a circle with angle

\displaystyle \alpha = \frac{1}{6}\cdot 2\pi\ \text{radians} = \frac{\pi}{3}\ \text{radians.}
Image:4_1_9_.gif

The area of the sector is

\displaystyle \text{Area} = \frac{1}{2}\alpha r^{2} = \frac{1}{2}\cdot \frac{\pi}{3}\cdot (8\ \text{cm})^2 = \frac{32\pi}{3}\ \text{cm}^{2} \approx 33\textrm{.}5\ \text{cm}^{2}\,\textrm{.}
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