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Question Number 47006 by Tawa1 last updated on 03/Nov/18

Commented by Tawa1 last updated on 05/Nov/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by ajfour last updated on 03/Nov/18

$$letA(0,0)$$$$eq.ofarcwithcenterA$$$$x^2+y^2=1$$$$whenx=\frac{1}{2},y=\frac{\sqrt{3}}{2}$$$$wheny=\frac{1}{2},x=\frac{\sqrt{3}}{2}$$$$reqdarea=4{\int }_{1/2}^{\sqrt{3}/2}(\sqrt{1-x^2}-\frac{1}{2})dx$$$$=4{[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x-\frac{x}{2}]}_{1/2}^{\sqrt{3}/2}$$$$=4(\frac{\pi }{12}-\frac{\sqrt{3}-1}{4})=\frac{\pi }{3}+1-\sqrt{3}.$$

Commented by gunawan last updated on 05/Nov/18

$$\mathrm{Area}=1^2-4(\frac{1}{2}\times \left(1^2\right)\sin 60°-\frac{30°}{360°}\times \pi \times \left(1^2\right)$$$$\mathrm{Area}=1^2-4(\frac{\sqrt{3}}{4}-\frac{1}{12}\pi )$$$$=1-\sqrt{3}+\frac{\pi }{3}$$

Commented by Tawa1 last updated on 05/Nov/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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