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Question Number 103872 by mikolo last updated on 17/Jul/20

Commented by bramlex last updated on 18/Jul/20

Commented by bramlex last updated on 18/Jul/20

$${AB}^2={AD}^2+{BD}^2$$$${(2+r)}^2=4+{(4-r)}^2$$$${(2+r)}^2-{(4-r)}^2=4$$$$\left(6\right)(2r-2)=4\Rightarrow r-1=\frac{1}{3}$$$$\therefore r=\frac{4}{3}.$$

Commented by bramlex last updated on 18/Jul/20

$$theshadedareaA=\frac{1}{2}\pi {\left(4\right)}^2-\pi {\left(2\right)}^2-\pi {\left(\frac{4}{3}\right)}^2$$$$A=8\pi -4\pi -\frac{16\pi }{9}=\frac{20\pi }{9}.$$

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