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Question Number 201325 by sonukgindia last updated on 04/Dec/23

Commented by mr W last updated on 05/Dec/23

$$doyouhavetherightanswer?$$

Answered by mr W last updated on 04/Dec/23

Commented by mr W last updated on 04/Dec/23

$$R=radiusofbigquartercircle$$$$r=radiusofsmallsemicircle$$$$OB=\frac{R}{2}$$$$AB=\frac{\sqrt{3}R}{2}$$$$BC=\sqrt{{\left(2r\right)}^2-{\left(\frac{\sqrt{3}R}{2}\right)}^2}$$$$CO=\frac{R}{2}-\sqrt{{\left(2r\right)}^2-{\left(\frac{\sqrt{3}R}{2}\right)}^2}$$$$EF=r-\frac{R}{2}+\sqrt{{\left(2r\right)}^2-{\left(\frac{\sqrt{3}R}{2}\right)}^2}=\frac{BC}{2}$$$$2(\frac{R}{2}-r)=\sqrt{{\left(2r\right)}^2-{\left(\frac{\sqrt{3}R}{2}\right)}^2}$$$$4(\frac{R^2}{4}-Rr+r^2)=4r^2-\frac{3R^2}{4}$$$$\Rightarrow r=\frac{7R}{16}$$$$\frac{semi}{quarter}=\frac{\pi r^2}{2}\times \frac{4}{\pi R^2}=2{\left(\frac{r}{R}\right)}^2$$$$=2{\left(\frac{7}{16}\right)}^2=\frac{49}{128}\approx 38.2\mathrm{\%}$$

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