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Question Number 120133 by prakash jain last updated on 29/Oct/20

Commented by nimnim last updated on 29/Oct/20

$$Letthesmallestdiameterbexandthesecondbey$$$$thenlargestdiameter=(x+y)$$$$xy=2^2=4$$$${Area}_{shaded}=\frac{\pi }{2}{\left(\frac{x+y}{2}\right)}^2-\frac{\pi }{2}{\left(\frac{x}{2}\right)}^2-\frac{\pi }{2}{\left(\frac{y}{2}\right)}^2$$$$=\frac{\pi }{8}(x^2+2xy+y^2-x^2-y^2)$$$$=\frac{\pi }{8}\left(2xy\right)=\frac{\pi }{8}\left(8\right)=\pi$$

Commented by prakash jain last updated on 29/Oct/20

$$3\mathrm{semi}-\mathrm{circles}\mathrm{and}\mathrm{length}\mathrm{of}\mathrm{line}\mathrm{are}$$$$\mathrm{given}.\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{shaded}\mathrm{region}$$

Commented by nimnim last updated on 29/Oct/20

$$Ithinkits\pi$$

Commented by prakash jain last updated on 29/Oct/20

$$\mathrm{I}\mathrm{think}\mathrm{answer}\mathrm{is}\pi .\mathrm{But}\mathrm{need}\mathrm{to}\mathrm{prove}$$$$\mathrm{that}\mathrm{area}\mathrm{stays}\mathrm{no}\mathrm{matter}\mathrm{where}\mathrm{the}$$$$\mathrm{line}\mathrm{is}\mathrm{drawn}$$

Commented by prakash jain last updated on 29/Oct/20

$$\mathrm{Thanks}.$$

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