Question Number 177578 by BaliramKumar last updated on 07/Oct/22 | ||
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Commented by BaliramKumar last updated on 07/Oct/22 | ||
area of green colour | ||
Commented by som(math1967) last updated on 07/Oct/22 | ||
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$$\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:−\mathrm{1}\:? \\ $$ | ||
Commented by BaliramKumar last updated on 07/Oct/22 | ||
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$${yes}\:{sir} \\ $$ | ||
Commented by som(math1967) last updated on 07/Oct/22 | ||
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$$\:{Ar}\:.{of}\:\:{circle}=\mathrm{2}\pi \\ $$$${let}\:{R}={rad}\:{of}\:{quadrant} \\ $$$${OB}={R}−\sqrt{\mathrm{2}} \\ $$$${OABC}\:{square}\:\therefore{OB}=\sqrt{\mathrm{2}}×\sqrt{\mathrm{2}}=\mathrm{2} \\ $$$$\therefore{R}=\mathrm{2}+\sqrt{\mathrm{2}} \\ $$$$\:{ar}.\:{of}\:{quadrant}=\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${ar}.{of}\:{quadrant}\:{form}\:{with}\:{OABC} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{4}}×\pi×\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\pi}{\mathrm{2}} \\ $$$$\:{A}_{{green}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{2}\pi−\left(\mathrm{2}−\frac{\pi}{\mathrm{2}}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{4}}\pi×\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{2}}\right)−\frac{\mathrm{3}\pi}{\mathrm{2}}−\mathrm{2}\right] \\ $$$$=\frac{\:\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:−\mathrm{1}\:{sq}\:{unit} \\ $$$$ \\ $$ | ||
Commented by som(math1967) last updated on 07/Oct/22 | ||
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Commented by Tawa11 last updated on 07/Oct/22 | ||
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$$\mathrm{Great}\:\mathrm{sir} \\ $$ | ||
Answered by cherokeesay last updated on 07/Oct/22 | ||
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Commented by BaliramKumar last updated on 07/Oct/22 | ||
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$${great}\:{sir} \\ $$ | ||