If the circumference of a circle is decreased by 50% then the percentage of decrease in its area is :

Solution(By Examveda Team)

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