Question Number 214905 by Spillover last updated on 23/Dec/24
Commented by mr W last updated on 25/Dec/24
$${how}\:{about}\:{following}\:{similar}\:{but} \\ $$$${more}\:{challenging}\:{question}? \\ $$
Commented by mr W last updated on 24/Dec/24
Answered by BaliramKumar last updated on 23/Dec/24
Commented by Spillover last updated on 23/Dec/24
$${great} \\ $$
Answered by Spillover last updated on 23/Dec/24
Answered by dionigi last updated on 24/Dec/24
$$ \\ $$$${set}\:{r}\:{the}\:{radius}\:{of}\:{little}\:{circles} \\ $$$${set}\:{W}\:{the}\:{union}\:{of}\:{little}\:{circles} \\ $$$${W}\:=\:\mathrm{7}\pi{r}^{\mathrm{2}} \\ $$$$ \\ $$$${set}\:{Y}\:{the}\:{great}\:{circle} \\ $$$${Y}\:=\:\mathrm{9}\pi{r}^{\mathrm{2}} \\ $$$$ \\ $$$${set}\:{H}\:{the}\:{hexagon}\:{joining}\: \\ $$$${the}\:{centers}\:{of}\:{the}\:{peripheral}\:{circles} \\ $$$${H}\:=\:\mathrm{6}\sqrt{\mathrm{3}}\:{r}^{\mathrm{2}} \\ $$$$ \\ $$$${set}\:{B}\:{the}\:{area}\:{of}\:{the}\:{hexagon}\: \\ $$$${but}\:{not}\:{the}\:{little}\:{circles} \\ $$$${H}\cap{W}\:=\:\left(\mathrm{1}+\frac{\mathrm{6}}{\mathrm{3}}\right)\:\pi{r}^{\mathrm{2}} \:=\:\mathrm{3}\pi{r}^{\mathrm{2}} \\ $$$${B}\:=\:{H}\:−\:\left({H}\cap{W}\right)\:=\:\left(\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{3}\pi\right){r}^{\mathrm{2}} \\ $$$$ \\ $$$${set}\:{A}\:{the}\:{area}\:{of}\:{great}\:{circle}\: \\ $$$${but}\:{not}\:{the}\:{little}\:{circles}\: \\ $$$${and}\:{not}\:{the}\:{hexagon} \\ $$$${A}\:=\:\mathrm{9}\pi{r}^{\mathrm{2}} −\mathrm{7}\pi{r}^{\mathrm{2}} \:−\left(\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{3}\pi\right){r}^{\mathrm{2}} \\ $$$${A}\:=\:\left(\mathrm{5}\pi\:−\mathrm{6}\sqrt{\mathrm{3}}\right){r}^{\mathrm{2}} \\ $$$$ \\ $$$${expected}\:{area}\:=\:\frac{{A}}{\mathrm{6}}\:=\:\left(\frac{\mathrm{5}\pi}{\mathrm{6}}−\sqrt{\mathrm{3}}\right){r}^{\mathrm{2}} \\ $$
Commented by Spillover last updated on 23/Dec/24
$${great} \\ $$
Answered by Spillover last updated on 23/Dec/24
