If the radius of a circle is decreased by 20 | vedclass

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(C) Let the initial radius of the circle be $r$.The new radius after a $20 \%$ decrease is $80 \%$ of $r$,which is $r 	imes \frac{80}{100} = \frac{4r

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(C) Let the initial radius of the circle be $r$.The new radius after a $20 \%$ decrease is $80 \%$ of $r$,which is $r 	imes \frac{80}{100} = \frac{4r}{5}$.The initial area of the circle is $A_1 = \pi r^2$.The new area of the circle is $A_2 = \pi \left(\frac{4r}{5}ight)^2 = \frac{16}{25} \pi r^2$.The decrease in area is $A_1 - A_2 = \pi r^2 - \frac{16}{25} \pi r^2 = \frac{9}{25} \pi r^2$.The percentage decrease in area is $\frac{	ext{Decrease in Area}}{	ext{Initial Area}} 	imes 100 = \frac{\frac{9}{25} \pi r^2}{\pi r^2} 	imes 100$.Calculating this gives $\frac{9}{25} 	imes 100 = 9 	imes 4 = 36 \%$.

If the radius of a circle is decreased by $20 \%$,then the percentage decrease in its area is (in $\%$)?

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