A) \[6\pi \,c{{m}^{2}}\]
B) \[6\,c{{m}^{2}}\]
C) \[12\pi \,c{{m}^{2}}\]
D) \[20\,c{{m}^{2}}\]
Correct Answer: A
Solution :
| [a] Here, diameter \[=8cm\] and \[=135{}^\circ \] |
| \[\therefore \] Area of sector \[\frac{\theta }{360{}^\circ }\times \pi {{r}^{2}}=\frac{135{}^\circ }{360{}^\circ }\times \pi {{\left( \frac{8}{2} \right)}^{2}}=6\pi c{{m}^{2}}.\] |
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