A) \[8\,c{{m}^{2}}\]
B) \[8\,\sqrt{\pi \,}c{{m}^{2}}\]
C) \[16\,c{{m}^{2}}\]
D) \[16\sqrt{\pi }\,c{{m}^{2}}\]
Correct Answer: C
Solution :
Given figure is Area of \[{{C}_{1}}=4\,c{{m}^{2}}\] \[\Rightarrow \] \[4=\pi {{r}_{1}}^{2}\] \[\Rightarrow \] \[\sqrt{\frac{4}{\pi }}={{r}_{1}}\] Now, \[{{r}_{2}}=2{{r}_{1}}\] \[\Rightarrow \] \[{{r}_{2}}=\frac{2\times 2}{\sqrt{\pi }}=\frac{4}{\sqrt{\pi }}\] Area of \[{{C}_{2}}=\pi \times {{r}_{2}}^{2}=\pi \times \frac{4\times 4}{\pi }=16c{{m}^{2}}\]You need to login to perform this action.
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