Question Number 220602 by fantastic last updated on 16/May/25
Answered by A5T last updated on 16/May/25
$$\mathrm{Let}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{and}\:\mathrm{circumference}\:\mathrm{of}\:\mathrm{semicircle}\:, \\ $$$$\mathrm{radius}\:\mathrm{and}\:\mathrm{circumference}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{be}\:\mathrm{r}_{\mathrm{s}} \:\mathrm{and}\:\mathrm{c}_{\mathrm{s},} \: \\ $$$$\mathrm{r}\:\mathrm{and}\:\mathrm{c}\:\mathrm{respectively}. \\ $$$$\mathrm{c}_{\mathrm{s}} =\frac{\mathrm{2}\pi\mathrm{r}_{\mathrm{s}} }{\mathrm{2}}+\mathrm{2r}_{\mathrm{s}} \:\mathrm{and}\:\mathrm{c}=\mathrm{2}\pi\mathrm{r} \\ $$$$\mathrm{c}_{\mathrm{s}} >\mathrm{c}\:\Rightarrow\:\mathrm{r}_{\mathrm{s}} \left(\pi+\mathrm{2}\right)>\mathrm{2}\pi\mathrm{r}\Rightarrow\mathrm{r}<\frac{\mathrm{r}_{\mathrm{s}} \left(\pi+\mathrm{2}\right)}{\mathrm{2}\pi}<\mathrm{0}.\mathrm{82r}_{\mathrm{s}} \\ $$$$\left.\Rightarrow\:\mathrm{b}\right) \\ $$
Commented by fantastic last updated on 16/May/25
$${thanks}\:{sir} \\ $$
Answered by mr W last updated on 16/May/25
$${semi}−{circle}\:{with}\:{radius}\:{R}: \\ $$$${p}_{\mathrm{1}} =\left(\mathrm{2}+\pi\right){R} \\ $$$${circle}\:{with}\:{radius}\:{r}: \\ $$$${p}_{\mathrm{2}} =\mathrm{2}\pi{r} \\ $$$${p}_{\mathrm{1}} ={p}_{\mathrm{2}} \\ $$$$\left(\mathrm{2}+\pi\right){R}=\mathrm{2}\pi{r} \\ $$$$\frac{{r}}{{R}}=\frac{\mathrm{2}+\pi}{\mathrm{2}\pi}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\pi}\approx\mathrm{0}.\mathrm{818} \\ $$$$\left.\Rightarrow{b}\right)\:{is}\:{correct}. \\ $$
Commented by fantastic last updated on 16/May/25
$${thanks}\:{sir} \\ $$$$ \\ $$