A) \[15.1\text{ }c{{m}^{2}}\]
B) \[15.5\text{ }c{{m}^{2}}\]
C) \[15.6\,c{{m}^{2}}\]
D) \[15.9\,c{{m}^{2}}\]
Correct Answer: C
Solution :
Perimeter of a sector of circle \[=\frac{\theta }{{{360}^{o}}}\times 2\pi r+2r\] \[\Rightarrow \] \[\left( \frac{\theta }{{{360}^{o}}}\times 2\pi \times 5.2 \right)+(2\times 5.2)=16.4\] \[\Rightarrow \] \[\frac{\theta }{{{360}^{o}}}\pi =\frac{16.4-10.4}{10.4}=\frac{6}{10.4}\] Area of sector of circle \[\frac{\theta }{{{360}^{o}}}\times \pi {{r}^{2}}=\frac{6}{10.4}\times {{(5.2)}^{2}}=15.6\,c{{m}^{2}}\]
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