A) \[\text{22 c}{{\text{m}}^{2}}\]
B) \[\text{11 c}{{\text{m}}^{2}}\]
C) \[\text{45 c}{{\text{m}}^{2}}\]
D) \[\text{31 c}{{\text{m}}^{2}}\]
Correct Answer: B
Solution :
| [b] Lenth of the minute hand = radius of the circle \[(r)=\sqrt{21}\,cm\] |
| Angle swept by minute hand in 60 minutes \[=360{}^\circ \] |
| \[\therefore \]Angle swept by minute hand in 10 minutes |
| \[=\frac{360{}^\circ }{60}\times 10=60{}^\circ \] |
| Now area of sector with \[r=\sqrt{21}\,cm\]and \[\theta =60{}^\circ \] is |
| \[\frac{\theta }{360{}^\circ }\times \pi {{r}^{2}}=\frac{60{}^\circ }{360{}^\circ }\times \frac{22}{7}\times {{(\sqrt{21})}^{2}}\] |
| \[=\frac{1}{6}\times \frac{22}{7}\times 21=11c{{m}^{2}}\] |
| Hence, the area swept by the minute hand in 10 minutes is \[11c{{m}^{2}}\]. |
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