Question

If the circumference of a circle is reduced by $50\%$, then by how much percent is the area of the circle reduced?

• A. $25\%$
• B. $50\%$
• C. $65\%$
• D. $75\%$

Answer 1

The correct option is D $75\%$

Let circumference of circle be 44cm.

$\therefore 2\pi r=44cm$

$r=\frac{44\times 7}{44}=7cm$

area $A_1=\pi r^2=\frac{22}{7}\times 7\times 7=154c{m}^2$

After reducing circumference $50\%,=\frac{50}{100}\times \frac{44}{7}=22cm$

$\therefore 2\pi R=22cm$

$R=\frac{22\times 7}{44}=3.5cm$

$\therefore$ area, $A_2=\pi r^2=\frac{22}{7}\times 3.5\times 3.5=38.5cm$

Reduction % in area$=\frac{A_1-A_2}{A_1}\times 100=\frac{154-38.5}{154}\times 100=75\%$