Correct Answer: Option C
Let original radius = R.
New radius = \(\frac{50}{100}R\) = \(\frac{50}{100}R\)
Original area = \(\frac R2\) and new area = \(\mathrm{πR}^2\)
\(\frac{3\mathrm{πR}^2}4\ast\frac1{\mathrm{πR}^2}\ast100\)
Decrease in area = \(\mathrm\pi\left(\frac{\mathrm R}2\right)^2=\frac{\mathrm{πR}^2}4\)
= 75%
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