ABC26GN4657 · Area

Let the original circumference be $x$. Then, new circumference $=50 \%$ of $x=\frac{x}{2}$. Let original radius $=r$ and new radius $=R$. $$\begin{aligned} & 2 \pi r=x \Rightarrow r=\frac{x \times 7}{2 \times 22}=\frac{7 x}{44} . \\ & 2 \pi R=\frac{x}{2} \Rightarrow R=\frac{x}{2} \times \frac{7}{2 \times 22}=\frac{7 x}{88} . \end{aligned}$$ Original area $=\pi r^{2}=\left(\frac{22}{7} \times \frac{7 x}{44} \times \frac{7 x}{44}\right)=\frac{7 x^{2}}{88}$. New area $=\pi R^{2}=\left(\frac{22}{7} \times \frac{7 x}{88} \times \frac{7 x}{88}\right)=\frac{7 x^{2}}{352}$. Decrease in area $=\left(\frac{7 x^{2}}{88}-\frac{7 x^{2}}{352}\right)=\frac{21 x^{2}}{352}$. ∴ Decrease\% $=\left(\frac{21 x^{2}}{352} \times \frac{88}{7 x^{2}} \times 100\right) \%=75 \%$.