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Question-201325




Question Number 201325 by sonukgindia last updated on 04/Dec/23
Commented by mr W last updated on 05/Dec/23
do you have the right answer?
$${do}\:{you}\:{have}\:{the}\:{right}\:{answer}? \\ $$
Answered by mr W last updated on 04/Dec/23
Commented by mr W last updated on 04/Dec/23
R=radius of big quarter circle  r=radius of small semi  circle  OB=(R/2)  AB=(((√3)R)/2)  BC=(√((2r)^2 −((((√3)R)/2))^2 ))  CO=(R/2)−(√((2r)^2 −((((√3)R)/2))^2 ))  EF=r−(R/2)+(√((2r)^2 −((((√3)R)/2))^2 ))=((BC)/2)  2((R/2)−r)=(√((2r)^2 −((((√3)R)/2))^2 ))  4((R^2 /4)−Rr+r^2 )=4r^2 −((3R^2 )/4)  ⇒r=((7R)/(16))  ((semi)/(quarter))=((πr^2 )/2)×(4/(πR^2 ))=2((r/R))^2   =2((7/(16)))^2 =((49)/(128))≈38.2%
$${R}={radius}\:{of}\:{big}\:{quarter}\:{circle} \\ $$$${r}={radius}\:{of}\:{small}\:{semi}\:\:{circle} \\ $$$${OB}=\frac{{R}}{\mathrm{2}} \\ $$$${AB}=\frac{\sqrt{\mathrm{3}}{R}}{\mathrm{2}} \\ $$$${BC}=\sqrt{\left(\mathrm{2}{r}\right)^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{3}}{R}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$${CO}=\frac{{R}}{\mathrm{2}}−\sqrt{\left(\mathrm{2}{r}\right)^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{3}}{R}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$${EF}={r}−\frac{{R}}{\mathrm{2}}+\sqrt{\left(\mathrm{2}{r}\right)^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{3}}{R}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{{BC}}{\mathrm{2}} \\ $$$$\mathrm{2}\left(\frac{{R}}{\mathrm{2}}−{r}\right)=\sqrt{\left(\mathrm{2}{r}\right)^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{3}}{R}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\mathrm{4}\left(\frac{{R}^{\mathrm{2}} }{\mathrm{4}}−{Rr}+{r}^{\mathrm{2}} \right)=\mathrm{4}{r}^{\mathrm{2}} −\frac{\mathrm{3}{R}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{7}{R}}{\mathrm{16}} \\ $$$$\frac{{semi}}{{quarter}}=\frac{\pi{r}^{\mathrm{2}} }{\mathrm{2}}×\frac{\mathrm{4}}{\pi{R}^{\mathrm{2}} }=\mathrm{2}\left(\frac{{r}}{{R}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{2}\left(\frac{\mathrm{7}}{\mathrm{16}}\right)^{\mathrm{2}} =\frac{\mathrm{49}}{\mathrm{128}}\approx\mathrm{38}.\mathrm{2\%} \\ $$

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