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Question Number 184955 by mnjuly1970 last updated on 14/Jan/23

Answered by HeferH last updated on 14/Jan/23

Commented by HeferH last updated on 14/Jan/23

$$A_{OAB}=\frac{a\cdot 6}{2}=\frac{\pi }{6}\cdot 6\cdot \frac{6}{2}=3\pi \ast (a=arc)$$$$A_{OCA}=3\sqrt{3}$$$$A_{ACB}=3\pi -3\sqrt{3}$$$$A_{A^{\prime }CB}=\frac{3}{4}\cdot (6-\frac{6}{\sqrt{3}})=\frac{9\sqrt{3}-9}{2\sqrt{3}}=\frac{9-3\sqrt{3}}{2}$$$$Blue=\frac{9-3\sqrt{3}}{2}+(3\pi -3\sqrt{3})=\frac{9-3\sqrt{3}+6\pi -6\sqrt{3}}{2}$$$$=\frac{6\pi +9-9\sqrt{3}}{2}$$

Answered by JDamian last updated on 15/Jan/23

$$$$$$a=\mid \overrightarrow{OA}\mid =R\cos 60°=\frac{R}{2}$$$$b=\mid \overrightarrow{OB}\mid =R\cos 30°=\frac{\sqrt{3}}{2}R$$$$\mid \overrightarrow{AB}\mid =b-a=\frac{\sqrt{3}-1}{2}R$$$$S_{blue}={\int }_a^b\sqrt{R^2-x^2}dx-S_{\stackrel{△}{ABC}}$$$$S_{\stackrel{△}{ABC}}=\frac{1}{2}\mid \overrightarrow{AB}\mid \cdot R\cdot \sin 30°=\frac{R^2}{8}(\sqrt{3}-1)$$$$$$$${\int }_a^b\sqrt{R^2-x^2}dx=$$$$=\left|\begin{array}{ccc}x=R\cdot \sin \phi & \to & dx=R\cdot \cos \phi d\phi \\ a=\frac{R}{2}=R\cdot \sin \left(\frac{\pi }{6}\right)& \to & {\phi }_a=\frac{\pi }{6}\\ b=R\frac{\sqrt{3}}{2}=R\cdot \sin \left(\frac{\pi }{3}\right)& \to & {\phi }_b=\frac{\pi }{3}\end{array}\right|=$$$$={\int }_{\pi /6}^{\pi /3}{R}^2{\cos }^2\phi d\phi =\frac{R^2}{2}{[\phi +\frac{\sin 2\phi }{2}]}_{\pi /6}^{\pi /3}=$$$$=\frac{R^2}{2}(\frac{\pi }{3}-\frac{\pi }{6})=\frac{\pi }{12}{R}^2$$$$$$$$S_{blue}=R^2(\frac{\pi }{12}-\frac{\sqrt{3}-1}{8})={\left(\frac{R}{2}\right)}^2(\frac{\pi }{3}-\frac{\sqrt{3}-1}{2})◼$$

Commented by JDamian last updated on 15/Jan/23

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