In the following figure, if ((S_1 +S_2 )/( (√S_3 ))) = (6/( (√π))) find the area of the circular crown


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Question Number 176741 by HeferH last updated on 26/Sep/22

$I n t h e f o l l o w i n g f i g u r e, i f \frac{S_{1} + S_{2}}{\sqrt{S_{3}}} = \frac{6}{\sqrt{π}}$ $f i n d t h e a r e a o f t h e c i r c u l a r c r o w n$
Commented by HeferH last updated on 26/Sep/22

	Answered by HeferH last updated on 26/Sep/22

	$A r e a o f t h e c r o w n : π (R^{2} - r^{2})$ $\frac{S_{1} + S_{2}}{r \sqrt{π}} = \frac{6}{\sqrt{π}}$ $S_{1} + S_{2} = 6 r$ $L e t a b e t h e r a d i u s o f S_{2}$ $r a d i u s o f S_{1} = r a d i u s o f S_{3}$ $a^{2} = (R + r) (R - r)$ $a^{2} = R^{2} - r^{2}$ ${S e g}_{1} + {S e g}_{2} + T r i a n g l e = \frac{R^{2} π}{2}$ $\frac{π r^{2} - 2 S_{1}}{2} + \frac{π a^{2} - 2 S_{2}}{2} + \frac{2 r \cdot 2 a}{2} = \frac{R^{2} π}{2}$ $π r^{2} + π a^{2} - 2 S_{1} - 2 S_{2} + 4 a r = π R^{2}$ $π r^{2} + π a^{2} - π R^{2} = 2 (S_{1} + S_{2}) - 4 a r$ $π (r^{2} - R^{2} + a^{2}) = 2 (6 r) - 4 a r$ $π (r^{2} - R^{2} + R^{2} - r^{2}) = 2 (6 r) - 4 a r$ $0 = 12 r - 4 a r$ $4 a r = 12 r$ $a = 3$ $a^{2} = 9$ $A r e a o f t h e c r o w n : π (R^{2} - r^{2}) = 9 π$
	Commented by Tawa11 last updated on 03/Oct/22

	$Great sir$
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