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Question Number 209813 by mr W last updated on 22/Jul/24

	Answered by mahdipoor last updated on 22/Jul/24
	$R= radius red circle r = radius white circle { ((2R=8+2+2r)),((C_1 C_2 =(√(r^2 +(R−(r+2))^2 ))=R−r)) :} ⇒R^2 +2r^2 +4+4r−2Rr−4R=R^2 +r^2 −2Rr ⇒r^2 +4r+4=4R=20+4r ⇒r^2 =16 ⇒r=4⇒R=9⇒S_(red) =(π/2)R^2 −πr^2 =((49π)/2) thank mr.w , i edited.$
	$R = r a d i u s r e d c i r c l e$ $r = r a d i u s w h i t e c i r c l e$ ${\begin{cases} 2 R = 8 + 2 + 2 r \\ C_{1} C_{2} = \sqrt{r^{2} + {(R - (r + 2))}^{2}} = R - r \end{cases}$ $\Rightarrow R^{2} + 2 r^{2} + 4 + 4 r - 2 R r - 4 R = R^{2} + r^{2} - 2 R r$ $\Rightarrow r^{2} + 4 r + 4 = 4 R = 20 + 4 r \Rightarrow r^{2} = 16$ $\Rightarrow r = 4 \Rightarrow R = 9 \Rightarrow S_{r e d} = \frac{π}{2} R^{2} - π r^{2} = \frac{49 π}{2}$ $t h a n k m r . w, i e d i t e d .$
	Commented by mr W last updated on 22/Jul/24

	$g r e a t!$
	Answered by cherokeesay last updated on 22/Jul/24

	Commented by mr W last updated on 22/Jul/24
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