A) 26%
B) 32%
C) 36%
D) 53%
Correct Answer: C
Solution :
Let initial radius of the circle be r, then new radius will be 80% of \[r=r\times \frac{80}{100}=\frac{4r}{5}\] Initial area \[=\pi {{r}^{2}}\]and new area \[=\pi {{\left( \frac{4r}{5} \right)}^{2}}=\frac{16}{25}\pi {{r}^{2}}\] \[\therefore \]Decrease in area\[=\left( \pi {{r}^{2}}-\frac{16}{25}\pi {{r}^{2}} \right)\] \[=\frac{9}{25}\pi {{r}^{2}}\] \[\therefore \]Percentage decrease in area \[=\frac{9}{25}\pi {{r}^{2}}\times \frac{1}{\pi {{r}^{2}}}\times 100=36%\]
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