Question

if radius of circle reduce by 10%,then how much area of circle be reduced?​

Answer 1

Answer:

19% is the answer of given question.

Answer 2

$Let \: the \: radius \: of \: original \: circle = R$

$Area \: of \: Original \: circle = \pi R^{2} \: --(1)$

/* According to the problem given */

$After \: reducing \:the \: radius \:10\% , \\Measure \: of \: new \: radius (r)\\= R\left(\frac{100-10}{100}\right)\\= \frac{9R}{10} \:--(2)$

$Area \:of \:new \:circle \\= \pi \left(\frac{9R}{10}\right)^{2} \\= \pi \times \frac{81R^{2}}{100}\:--(3)$

$Reduced \: Area = (3) - (2) \\= \pi R^{2} - \pi \times \frac{81R^{2}}{100} \\= \pi R^{2} [1- \frac{81}{100} ]\\= \frac{100-81}{100} \pi R^{2} \\= \frac{19}{100} \pi R^{2} \: -- (4)$

$Percentage \:of \: reduced \:area \\= \frac{Reduced \:area }{Original \:area } \times 100\\= \frac{ \frac{19}{100}\pi R^{2}}{\pi R^{2}} \times 100\\= 19\%$

Therefore.,

$\red {Percentage \:of \: reduced \:area}\green {= 19\% }$

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