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Question Number 108331 by I want to learn more last updated on 16/Aug/20

Answered by mr W last updated on 16/Aug/20

Commented by mr W last updated on 16/Aug/20

$$R=a+b$$$$\frac{2a}{BD}=\frac{BD}{2b}$$$$\Rightarrow 4ab={BD}^2=2^2=4$$$$\Rightarrow ab=1$$$$A_{shaded}=\frac{\pi }{2}(R^2-a^2-b^2)=\pi ab=\pi$$

Commented by I want to learn more last updated on 16/Aug/20

$$\mathrm{I}\mathrm{appreciate},\mathrm{thanks}\mathrm{sir}.$$

Commented by I want to learn more last updated on 16/Aug/20

$$\mathrm{Sir},\mathrm{answer}\mathrm{in}\mathrm{book}\mathrm{is}4\pi ,\mathrm{That}\mathrm{means}\mathrm{the}\mathrm{textbook}\mathrm{is}\mathrm{a}\mathrm{mistake}.$$

Commented by mr W last updated on 16/Aug/20

$$yes.$$

Commented by I want to learn more last updated on 16/Aug/20

$$\mathrm{Thanks}\mathrm{sir}\mathrm{for}\mathrm{your}\mathrm{time}.\mathrm{I}\mathrm{appreciate}.$$

Commented by mr W last updated on 16/Aug/20

$$justlookatthespecialcase:$$$$BD=R=2$$$$\Rightarrow a=b=\frac{R}{2}=1$$$$A_{shade}=\frac{\pi }{2}(2^2-1^2-1^2)=\pi$$

Commented by I want to learn more last updated on 17/Aug/20

$$\mathrm{Thanks}\mathrm{for}\mathrm{more}\mathrm{clarification}\mathrm{sir}$$

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