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Question Number 62363 by Tawa1 last updated on 20/Jun/19

Answered by MJS last updated on 20/Jun/19

$$\mathrm{big}\mathrm{circle}(\mathrm{let}R=1)$$$$x=\pm \sqrt{1-y^2}$$$$\mathrm{lines}\mathrm{of}\mathrm{tbe}\mathrm{triangle}$$$$x=\pm \frac{\sqrt{3}}{3}y$$$$\mathrm{looking}\mathrm{for}\mathrm{an}y>0\mathrm{for}\mathrm{which}$$$$$$$$\frac{\sqrt{3}}{3}y-(-\frac{\sqrt{3}}{3}y)=\sqrt{1-y^2}-\frac{\sqrt{3}}{3}y$$$$\sqrt{3}y=\sqrt{1-y^{i2}}$$$$3y^2=1-y^2$$$$y=\frac{1}{2}$$$$\Rightarrow r=\frac{\sqrt{3}}{6}$$$$\frac{\frac{3}{2}{r}^2\pi }{\frac{1}{2}{R}^2\pi }=\frac{1}{4}\Rightarrow 25\mathrm{\%}\mathrm{are}\mathrm{shaded}$$

Commented by Tawa1 last updated on 20/Jun/19

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

Answered by mr W last updated on 20/Jun/19

$$R=radiusofbigsemicircle$$$$r=radiussmallsemicircles$$$$R^2={\left(3r\right)}^2+{\left(\sqrt{3}r\right)}^2$$$$\Rightarrow R^2=12r^2$$$$A_{shade}=3\times \frac{\pi r^2}{2}=\frac{3\pi r^2}{2}$$$$A_{total}=\frac{\pi R^2}{2}$$$$\Rightarrow \frac{A_{shade}}{A_{total}}=\frac{3r^2}{R^2}=\frac{3r^2}{12r^2}=\frac{1}{4}$$

Commented by mr W last updated on 20/Jun/19

Commented by Tawa1 last updated on 20/Jun/19

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

Commented by Tawa1 last updated on 20/Jun/19

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

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