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Question Number 188078 by Rupesh123 last updated on 25/Feb/23

	Answered by a.lgnaoui last updated on 25/Feb/23

	$f r o m G r a p h e : △ A B C E q u i l a t e r a l e$ $b i g c i r c l e (r a d i u s O F = \sqrt{4 + \frac{{E F}^{2}}{4}})$ $\sin \frac{π}{6} = \frac{1}{2} = \frac{2}{O F} \Rightarrow (O F = 4; E F = 4 \sqrt{3 .})$ $A r e a (b i g {c i r c l e C}_{B} = 16 π)$ $L e n g t h o f △ A B C = 8 \sqrt{3} \Rightarrow A = 96 \sqrt{3}$ $A r e a (A_{1} + A_{2} + A_{3}) = 96 \sqrt{3} - 16 π$ $A r e a (C_{B} - C_{S}); C_{S} = S m a l c i r c l e =$ $16 π - 4 π = 12 π$ $A_{1}^{'} + A_{2} + A_{3}^{'} = [A r e a (s m a l t r i a g l e) - 4 π]$ $= 4 \sqrt{3} \times (4 \sqrt{3} \times \frac{\sqrt{3}}{2}) - 4 π = 24 \sqrt{3} - 4 π$ $T o t a l A r e a = (A_{1} + A_{2} + A_{3}) + (A_{1}^{'} + A_{2}^{'} + A_{3}^{'})$ $(96 \sqrt{3} - 16 π) + (24 \sqrt{3} - 4 π) = 120 \sqrt{3} - 20 π$ $S h a d e d A r e a = 145$
	Commented by Rupesh123 last updated on 25/Feb/23
	Please check your solution well!
	Commented by a.lgnaoui last updated on 25/Feb/23

	$t h a n k y o u .$
	Commented by mr W last updated on 25/Feb/23

	$c l e a r l y w r o n g!$ $t h e s h a d e d a r e a c a n n o t b e m o r e t h a n$ $t h e b i g e q u i l a t e r a l t r i a n g l e!$
	Commented by a.lgnaoui last updated on 25/Feb/23

	$A r e a o f b i g t r i a n g l e$ $= (8 \sqrt{3}) \times \frac{8 \sqrt{3} \times \sqrt{3}}{2} = 32 \times 3 \sqrt{3} = 166, 267$ $166, 267 > 145 (l o g i c)$ $t h e s h a d e d A r e a = (B l e u)$
	Commented by mr W last updated on 25/Feb/23

	$y o u h a v e p a p e r a n d r u l e r, w h y n o t$ $d r a w s u c h a n e q u i l a t e r a l a n d m e a s u r e$ $i t s b a s e a n d h e i g h t a n d t h e n c a l c u l a t e$ $i t s a r e a ? {y o u}^{'} l l s e e i t s a r e a i s a b o u t$ $83 . w h e n i t i s l o g i c f o r y o u t h a t t h e$ $s h a d e d a r e a i s 145, t h e n i m u s t s a y$ $y o u h a v e n o l o g i c .$
	Answered by mr W last updated on 25/Feb/23

	$r a d i u s o f s m a l l c i r c l e r = 2$ $r a d i u s o f b i g c i r c l e R = 4$ $s i d e l e n g t h s m a l l e q u i l a t e r a l s = 4 \sqrt{3}$ $s i d e l e n g t h b i g e q u i l a t e r a l S = 8 \sqrt{3}$ $s h a d e d a r e a = (b i g e q u i l a t e r a l - b i g c i r c l e) + (s m a l l e q u i l a t e r a l - s m a l l c i r c l e)$ $= 5 \times (s m a l l e q u i l a t e r a l - s m a l l c i r c l e)$ $= 5 \times (\frac{\sqrt{3} s^{2}}{4} - π r^{2})$ $= 5 \times (\frac{\sqrt{3} \times 16 \times 3}{4} - π \times 4)$ $= 60 \sqrt{3} - 20 π \approx 41.092$
	Commented by Rupesh123 last updated on 25/Feb/23
	Excellent, sir!
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