A) \[26\,\pi \,c{{m}^{2}}\]
B) \[16\,\pi \,c{{m}^{2}}\]
C) \[28\,\pi \,c{{m}^{2}}\]
D) \[30\,\pi \,c{{m}^{2}}\]
Correct Answer: C
Solution :
[c] We have, \[360{}^\circ -80{}^\circ =280{}^\circ \] \[\therefore \] Required ratio \[=\frac{280{}^\circ }{360{}^\circ }\times \pi \times {{(6)}^{2}}\] \[=\frac{7}{9}\times \pi \times 36\] \[=28\,\pi \,c{{m}^{2}}\]
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