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Question Number 201035 by sonukgindia last updated on 28/Nov/23

Commented by mr W last updated on 28/Nov/23

$$thesidelengthofthesquarecannot$$$$beequaltotheradiusofthesmaller$$$$semi-circles!$$

Commented by mr W last updated on 28/Nov/23

Answered by a.lgnaoui last updated on 29/Nov/23

$$\mathbf{P}\mathrm{ink}\mathrm{triangle}\mathbf{ABC}$$$$\mathbf{B}\mathrm{lue}\mathrm{triangle}\mathbf{BDE}$$$$\mathbf{BD}=2\times \frac{\mathbf{R}}{4}=\frac{\mathbf{R}}{2}\mathbf{BH}=\frac{\mathbf{R}}{4}$$$$\Rightarrow \mathbf{Blue}\mathbf{Area}=\left(\frac{\mathbf{R}}{4}\times \frac{\mathbf{R}}{2}\right)\times \frac{1}{2}=\frac{{\mathbf{R}}^2}{16}\left(1\right)$$$$\mathbf{BC}//\mathbf{OE}{\mathbf{BC}}^2={\mathbf{OH}}^2+{\mathbf{BH}}^2={\mathbf{OB}}^2$$$${\mathbf{R}}^2={\mathbf{OH}}^2+\frac{{\mathbf{R}}^2}{4^2}={\mathbf{AB}}^2+\frac{{\mathbf{R}}^2}{4}$$$$\Rightarrow \mathbf{AB}=\frac{\mathbf{R}\sqrt{3}}{2}\mathbf{AC}=\frac{\mathbf{R}}{2}$$$$\mathbf{Pink}\mathbf{Area}=\frac{1}{2}\left(\mathbf{AB}\times \mathbf{AC}\right)=\frac{{\mathbf{R}}^2\sqrt{3}}{8}$$$$$$$$$$$$\frac{\mathbf{Pink}}{\mathbf{Blue}}=\frac{\frac{{\mathbf{R}}^2\sqrt{3}}{8}}{\frac{{\mathbf{R}}^2}{16}}=\frac{{\mathbf{R}}^2\sqrt{3}}{8}\times \frac{16}{{\mathbf{R}}^2}=2\sqrt{3}\mathrm{X}$$$$(VoirResultatexactdessous:3,1334)$$$$$$

Commented by a.lgnaoui last updated on 28/Nov/23

Commented by mr W last updated on 28/Nov/23

$$wrong!$$$$BD=DF\ne \frac{R}{2}$$

Commented by a.lgnaoui last updated on 28/Nov/23

$$\mathrm{OM}=\mathrm{ON}=\mathrm{R}$$

Commented by mr W last updated on 29/Nov/23

$$thesidelengthofthesquareisnot$$$$equaltotheradiusofsmallersemi$$$$circles!$$

Commented by a.lgnaoui last updated on 29/Nov/23

$$\mathrm{soit}\mathit{\alpha }=\measuredangle \mathrm{CHO}\tan \mathit{\alpha }=\frac{\mathbf{x}/2}{\mathbf{x}}=\frac{1}{2}$$$$\Rightarrow \frac{\mathrm{OC}}{\mathrm{OH}}=\frac{1}{2}=\frac{\frac{\mathbf{R}}{2}}{\mathbf{x}+\mathbf{CD}\cos \mathit{\alpha }}$$$$1=\frac{\mathbf{R}}{\mathbf{x}+\frac{\mathbf{R}}{2}\cos 26,56}\mathit{\alpha }=\left(\tan {}^{-1}\frac{1}{2}\right)=26,56$$$$\mathrm{donc}\frac{2\mathbf{R}}{2\mathbf{x}+\mathbf{R}\cos 26,56}=1$$$$2\mathbf{x}+\mathbf{R}\cos \cos 26,56=2\mathbf{R}$$$$\mathbf{x}=\frac{\mathbf{R}(2-\cos 26,56)}{2}$$$$\mathrm{alors}\mathbf{Blue}\mathbf{Area}=\frac{1}{2}\left(\mathbf{BD}\times \mathbf{x}\right)$$$$=\frac{{\mathbf{x}}^2}{4}=\frac{{\mathbf{R}}^2{(2-\cos 26,56)}^2}{16}$$$$$$$$\mathbf{OH}=\mathbf{x}+\frac{\mathbf{R}}{2}\cos 26,56=\mathbf{R}$$$$\mathbf{BC}=\mathbf{OE}=\mathbf{R}$$$$\mathbf{Pink}\mathbf{Aeea}=\frac{{\mathbf{R}}^2\sqrt{3}}{8}$$$$$$$$\Rightarrow \frac{\mathbf{Pink}\mathbf{Ara}}{\mathbf{Blue}\mathbf{Area}}=\frac{{\mathbf{R}}^2\sqrt{3}}{8}\times \frac{16}{{\mathbf{R}}^2(2-\cos 26,56)}$$$$=\frac{2\sqrt{3}}{2-0,894466}=3,1334$$

Commented by mr W last updated on 29/Nov/23

$$ihavesaid:you\mathit{a}\mathit{s}\mathit{s}\mathit{u}\mathit{m}\mathit{e}\mathit{d}$$$$thatDB=CO.butthisiswrong!!!$$$$thatmeans\tan \mathrm{\angle }CHO\ne \frac{1}{2}.$$$$thereforetherestiswrong.$$

Answered by mr W last updated on 29/Nov/23

Commented by mr W last updated on 29/Nov/23

$$r=radiusofsmallersemi-circles$$$$a=sidelengthofsquare$$$$OA=\frac{a}{2}$$$${AD}^2=h_2^2=OA\times CA=\frac{a}{2}(2r-\frac{a}{2})=ar-\frac{a^2}{4}$$$${AE}^2=h_1^2=CA\times AH=(2r-\frac{a}{2})(2r+\frac{a}{2})=4r^2-\frac{a^2}{4}$$$$ED=h_1-h_2=a$$$$\sqrt{4r^2-\frac{a^2}{4}}-\sqrt{ar-\frac{a^2}{4}}=a$$$$\sqrt{4r^2-\frac{a^2}{4}}-a=\sqrt{ar-\frac{a^2}{4}}$$$$4r^2-\frac{a^2}{4}+a^2-2a\sqrt{4r^2-\frac{a^2}{4}}=ar-\frac{a^2}{4}$$$$a^2-ar+4r^2=2a\sqrt{4r^2-\frac{a^2}{4}}$$$$2a^4-2{ra}^3-7r^2{a}^2-8r^{3}a+16r^4=0$$$$let\lambda =\frac{a}{r}$$$$\Rightarrow 2{\lambda }^4-2{\lambda }^3-7{\lambda }^2-8\lambda +16=0$$$$\Rightarrow \lambda \approx 1.049913358904$$$$thatmeansa\approx 1.049913358904r\ne r$$

Commented by mr W last updated on 29/Nov/23

$$\frac{EG}{ED}=\frac{AB}{AD}=\frac{r-\frac{a}{2}}{\sqrt{ar-\frac{a^2}{4}}}=\frac{2-\lambda }{\sqrt{4\lambda -{\lambda }^2}}$$$$bluearea=\frac{ED\times EG}{2}=\frac{{\lambda }^2(2-\lambda )r^2}{2\sqrt{4\lambda -{\lambda }^2}}$$$${FE}^2+{FB}^2={BA}^2+{AE}^2$$$${FE}^2+r^2={(r-\frac{a}{2})}^2+(4r^2-\frac{a^2}{4})$$$${FE}^2=4r^2-ar$$$$FE=r\sqrt{4-\lambda }$$$$pinkarea=\frac{FB\times FE}{2}=\frac{r^2\sqrt{4-\lambda }}{2}$$$$\frac{pink}{blue}=\frac{\sqrt{4-\lambda }}{2}\times \frac{2\sqrt{4\lambda -{\lambda }^2}}{{\lambda }^2(2-\lambda )}$$$$=\frac{4-\lambda }{(2-\lambda )\lambda \sqrt{\lambda }}\approx 2.886299319$$

Commented by mr W last updated on 29/Nov/23

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