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Question Number 68305 by TawaTawa last updated on 08/Sep/19

Commented by ajfour last updated on 10/Sep/19

Commented by ajfour last updated on 10/Sep/19

$$Area{S}_{req}=\underset{⌣}{GCE}$$$$(imagedistortedpurposely$$$$sothatareaBCGistobe$$$$enlarged).$$$$considerAtobecenterof$$$$semicircle.$$$$S_1=sectorareaA\stackrel{⌢}{CE}$$$$S_2=area(△ABE)$$$$S_3=area(△BCD)$$$$S_4=sectorareaD\stackrel{⌢}{GC}$$$$S_{req}=S_1-S_2+S_3-S_4$$$$C(\frac{2}{5},\frac{4}{5});E(\frac{1}{2},\frac{1}{2});$$$$G(\frac{2-\sqrt{6}}{2},\frac{\sqrt{6}}{2});B(\frac{2}{7},\frac{5}{7})$$$$S_1=\frac{r^2\theta }{2}withr=\frac{1}{2},\theta ={\cos }^{-1}\frac{4}{5}$$$$\Rightarrow S_1=\frac{1}{8}{\sin }^{-1}\frac{3}{5}$$$$S_2=\frac{1}{2}r(y_B-r)=\frac{1}{4}(\frac{5}{7}-\frac{1}{2})=\frac{3}{56}$$$$S_3=\frac{1}{2}\left|\begin{array}{ccc}1& 0& 1\\ 2/5& 4/5& 1\\ 2/7& 5/7& 1\end{array}\right|$$$$=(\frac{4}{5}-\frac{5}{7})+(\frac{2}{7}-\frac{8}{35})=\frac{28-25+10-8}{35}$$$$=\frac{1}{7}$$$$S_4=\frac{R^2\varphi }{2}withR=1$$$$and\varphi ={\sin }^{-1}\left(y_C\right)-\frac{\pi }{4}$$$$\Rightarrow S_4=\frac{1}{2}{\sin }^{-1}\frac{4}{5}-\frac{\pi }{8}$$$$S_{req}=\frac{1}{8}{\sin }^{-1}\frac{3}{5}-\frac{3}{56}+\frac{1}{7}-\frac{1}{2}{\sin }^{-1}\frac{4}{5}+\frac{\pi }{8}.$$

Commented by TawaTawa last updated on 10/Sep/19

$$\mathrm{Wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by mr W last updated on 08/Sep/19

$$letsidelengthofsquare=1$$$$takeAasoriginandADasx-axis.$$$$eqn.ofbigcircle:$$$${(x-1)}^2+y^2=1$$$$eqn.ofsmallcircle:$$$$x^2+{(y-\frac{1}{2})}^2=\frac{1}{4}$$$$eqn.oflineBD:$$$$y=1-x$$$$$$$$intersectionofbigandsmallcircles:$$$${(x-1)}^2+y^2=1...\left(i\right)$$$$x^2+{(y-\frac{1}{2})}^2=\frac{1}{4}...\left(ii\right)$$$$\left(i\right)-\left(ii\right):$$$$\Rightarrow y=2x$$$${(x-1)}^2+4x^2=1$$$$5x^2-2x+1=1$$$$x(5x-2)=0$$$$\Rightarrow x=0orx=\frac{2}{5}$$$$intersectionofbigcircleandBD:$$$${(x-1)}^2+{(1-x)}^2=1$$$$x^2-2x-\frac{1}{2}=0$$$$\Rightarrow x=1-\sqrt{1+\frac{1}{2}}=\frac{2-\sqrt{6}}{2}$$$$intersectionofsmallcircleandBD:$$$$x^2+{(\frac{1}{2}-x)}^2=\frac{1}{4}$$$$x(2x-1)=0$$$$\Rightarrow x=0orx=\frac{1}{2}$$$$bigcircle:$$$$\Rightarrow y=\sqrt{x(2-x)}$$$$smallcircle:$$$$y=\frac{1}{2}+\sqrt{\frac{1}{4}-x^2}$$$$$$$$shadedarea:$$$$A={\int }_{\frac{2-\sqrt{6}}{2}}^{\frac{2}{5}}[\sqrt{x(2-x)}-1+x]dx+{\int }_{\frac{2}{5}}^{\frac{1}{2}}[\frac{1}{2}+\sqrt{\frac{1}{4}-x^2}-1+x]dx$$$$A={\int }_{\frac{2-\sqrt{6}}{2}}^{\frac{2}{5}}[\sqrt{x(2-x)}-1+x]dx+{\int }_{\frac{2}{5}}^{\frac{1}{2}}[\sqrt{\frac{1}{4}-x^2}-\frac{1}{2}+x]dx$$$$A={[\frac{(x-1)\sqrt{x(2-x)}+{\sin }^{-1}(x-1)}{2}-x+\frac{x^2}{2}]}_{\frac{2-\sqrt{6}}{2}}^{\frac{2}{5}}+{[\frac{{\sin }^{-1}\left(2x\right)}{8}+\frac{x\sqrt{\frac{1}{4}-x^2}}{2}-\frac{x}{2}+\frac{x^2}{2}]}_{\frac{2}{5}}^{\frac{1}{2}}$$$$......$$

Commented by TawaTawa last updated on 08/Sep/19

$$\mathrm{Wow},\mathrm{thanks}\mathrm{for}\mathrm{your}\mathrm{time}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$$$\mathrm{Waiting}\mathrm{for}\mathrm{the}\mathrm{rest}\mathrm{sir}$$

Commented by TawaTawa last updated on 10/Sep/19

$$\mathrm{This}\mathrm{area}\mathrm{is}\mathrm{too}\mathrm{tough}.\mathrm{The}\mathrm{integration}\mathrm{formed}\mathrm{is}\mathrm{complex}.$$

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