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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

$h o w a b o u t f o l l o w i n g s i m i l a r b u t$ $m o r e c h a l l e n g i n g q u e s t i o n ?$
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

	$g r e a t$
	Answered by Spillover last updated on 23/Dec/24

	Answered by dionigi last updated on 24/Dec/24

	$s e t r t h e r a d i u s o f l i t t l e c i r c l e s$ $s e t W t h e u n i o n o f l i t t l e c i r c l e s$ $W = 7 π r^{2}$ $s e t Y t h e g r e a t c i r c l e$ $Y = 9 π r^{2}$ $s e t H t h e h e x a g o n j o i n i n g$ $t h e c e n t e r s o f t h e p e r i p h e r a l c i r c l e s$ $H = 6 \sqrt{3} r^{2}$ $s e t B t h e a r e a o f t h e h e x a g o n$ $b u t n o t t h e l i t t l e c i r c l e s$ $H \cap W = (1 + \frac{6}{3}) π r^{2} = 3 π r^{2}$ $B = H - (H \cap W) = (6 \sqrt{3} - 3 π) r^{2}$ $s e t A t h e a r e a o f g r e a t c i r c l e$ $b u t n o t t h e l i t t l e c i r c l e s$ $a n d n o t t h e h e x a g o n$ $A = 9 π r^{2} - 7 π r^{2} - (6 \sqrt{3} - 3 π) r^{2}$ $A = (5 π - 6 \sqrt{3}) r^{2}$ $e x p e c t e d a r e a = \frac{A}{6} = (\frac{5 π}{6} - \sqrt{3}) r^{2}$
	Commented by Spillover last updated on 23/Dec/24

	$g r e a t$
	Answered by Spillover last updated on 23/Dec/24

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