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Question Number 161424 by cherokeesay last updated on 17/Dec/21

Answered by aleks041103 last updated on 17/Dec/21

$${(R-2)}^2+{(R-4)}^2=R^2$$$$R^2-4R+4+R^2-8R+16=R^2$$$$R^2-12R+20=0$$$$R_{1;2}=\frac{1}{2}(12\pm \sqrt{144-4.20})=$$$$=\frac{1}{2}(12\pm 8)=6\pm 4=2;10$$$$R=2\to impossible,sinceR-4>0$$$$\Rightarrow R=10$$$$redarea:S$$$$\frac{S}{R^2}=1-\frac{(2.4)}{R^2}-\frac{1}{4}\pi =\frac{92}{100}-\frac{\pi }{4}$$$$\frac{S_{red}}{S_{square}}=\frac{92-25\pi }{100}\approx 0,135$$

Commented by cherokeesay last updated on 17/Dec/21

$$verynice!$$$$thankyou!$$

Commented by Tawa11 last updated on 18/Dec/21

$$\mathrm{Great}\mathrm{sir}.$$

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