Lösung 4.1:9

Aus Online Mathematik Brückenkurs 1

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Zeile 1: Zeile 1:
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
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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{{Displayed math||<math>\alpha = \frac{1}{6}\cdot 2\pi\ \text{radians} = \frac{\pi}{3}\ \text{radians.}</math>}}
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{{Abgesetzte Formel||<math>\alpha = \frac{1}{6}\cdot 2\pi\ \text{radians} = \frac{\pi}{3}\ \text{radians.}</math>}}
<center> [[Image:4_1_9_.gif]] </center>
<center> [[Image:4_1_9_.gif]] </center>
Zeile 7: Zeile 7:
The area of the sector is
The area of the sector is
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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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{{Abgesetzte Formel||<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>}}

Version vom 08:49, 22. Okt. 2008

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{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.1:9“