ABC26GN4656 · Area

Let the original circumference be $x$ units. Then, new circumference $=120 \%$ of $x=\left(\frac{6 x}{5}\right)$. Let original radius $=r$ and new radius $=R$. $$2 \pi r=x \Rightarrow r=\frac{7 x}{2 \times 22}=\frac{7 x}{44}$$ And, $2 \pi R=\frac{6 x}{5} \Rightarrow R=\frac{6 x}{5} \times \frac{7}{2 \times 22}=\frac{21 x}{110}$ 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 $=\pi R^{2}=\left(\frac{22}{7} \times \frac{21 x}{110} \times \frac{21 x}{110}\right)=\frac{63 x^{2}}{550}$ Increase in area $=\left(\frac{63 x^{2}}{550}-\frac{7 x^{2}}{88}\right)=\frac{77 x^{2}}{2200}$. $$\text { ∴ } \text { Increase } \%=\left(\frac{77 x^{2}}{2200} \times \frac{88}{7 x^{2}} \times 100\right) \%=44 \% \text {. }$$