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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?

$${how}\:{about}\:{following}\:{similar}\:{but} \\ $$$${more}\:{challenging}\:{question}? \\ $$

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

$${great} \\ $$

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 = 7πr^2     set Y the great circle  Y = 9πr^2     set H the hexagon joining   the centers of the peripheral circles  H = 6(√3) r^2     set B the area of the hexagon   but not the little circles  H∩W = (1+(6/3)) πr^2  = 3πr^2   B = H − (H∩W) = (6(√3)−3π)r^2     set A the area of great circle   but not the little circles   and not the hexagon  A = 9πr^2 −7πr^2  −(6(√3)−3π)r^2   A = (5π −6(√3))r^2     expected area = (A/6) = (((5π)/6)−(√3))r^2

$$ \\ $$$${set}\:{r}\:{the}\:{radius}\:{of}\:{little}\:{circles} \\ $$$${set}\:{W}\:{the}\:{union}\:{of}\:{little}\:{circles} \\ $$$${W}\:=\:\mathrm{7}\pi{r}^{\mathrm{2}} \\ $$$$ \\ $$$${set}\:{Y}\:{the}\:{great}\:{circle} \\ $$$${Y}\:=\:\mathrm{9}\pi{r}^{\mathrm{2}} \\ $$$$ \\ $$$${set}\:{H}\:{the}\:{hexagon}\:{joining}\: \\ $$$${the}\:{centers}\:{of}\:{the}\:{peripheral}\:{circles} \\ $$$${H}\:=\:\mathrm{6}\sqrt{\mathrm{3}}\:{r}^{\mathrm{2}} \\ $$$$ \\ $$$${set}\:{B}\:{the}\:{area}\:{of}\:{the}\:{hexagon}\: \\ $$$${but}\:{not}\:{the}\:{little}\:{circles} \\ $$$${H}\cap{W}\:=\:\left(\mathrm{1}+\frac{\mathrm{6}}{\mathrm{3}}\right)\:\pi{r}^{\mathrm{2}} \:=\:\mathrm{3}\pi{r}^{\mathrm{2}} \\ $$$${B}\:=\:{H}\:−\:\left({H}\cap{W}\right)\:=\:\left(\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{3}\pi\right){r}^{\mathrm{2}} \\ $$$$ \\ $$$${set}\:{A}\:{the}\:{area}\:{of}\:{great}\:{circle}\: \\ $$$${but}\:{not}\:{the}\:{little}\:{circles}\: \\ $$$${and}\:{not}\:{the}\:{hexagon} \\ $$$${A}\:=\:\mathrm{9}\pi{r}^{\mathrm{2}} −\mathrm{7}\pi{r}^{\mathrm{2}} \:−\left(\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{3}\pi\right){r}^{\mathrm{2}} \\ $$$${A}\:=\:\left(\mathrm{5}\pi\:−\mathrm{6}\sqrt{\mathrm{3}}\right){r}^{\mathrm{2}} \\ $$$$ \\ $$$${expected}\:{area}\:=\:\frac{{A}}{\mathrm{6}}\:=\:\left(\frac{\mathrm{5}\pi}{\mathrm{6}}−\sqrt{\mathrm{3}}\right){r}^{\mathrm{2}} \\ $$

Commented by Spillover last updated on 23/Dec/24

great

$${great} \\ $$

Answered by Spillover last updated on 23/Dec/24

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