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Question Number 186868 by Mingma last updated on 11/Feb/23

Commented by Mingma last updated on 11/Feb/23
Shaded area =?
	Answered by HeferH last updated on 11/Feb/23

	Commented by Mingma last updated on 11/Feb/23
	Good, but can you show your work
	Commented by HeferH last updated on 12/Feb/23

	$S o r r y s i r, i w a s t i r e d a t t h e t i m e b u t i h a v e$ $t h e s o l u t i o n i n m i n d (g e t t i n g t h e a n g l e a n d$ $t h e n t h e a r e a o f t h e c i r c u l a r s e c t o r) I m$ $p r o b a b l y g o n n a d o i t t o m o r r o w$
	Commented by HeferH last updated on 13/Feb/23

	Commented by HeferH last updated on 13/Feb/23

	$r : r a d i u s o f t h e i n c i r c l e$ ${A O}^{2} = x^{2} + {(A B - x)}^{2}$ $r = \frac{5 \sqrt{2}}{\sqrt{3}} = \frac{5 \sqrt{6}}{3}$ $A O = \frac{10 \sqrt{6}}{3}$ $\Rightarrow {(\frac{10 \sqrt{6}}{3})}^{2} = x^{2} + {(10 - x)}^{2}$ $x = \frac{15 \pm 5 \sqrt{3}}{3}; x < r \Rightarrow x = \frac{15 - 5 \sqrt{3}}{3}$ $\sin α = \frac{x}{r} = (\frac{15 - 5 \sqrt{3}}{3}) \cdot \frac{3}{5 \sqrt{6}} = \frac{3 - \sqrt{3}}{\sqrt{6}} = \frac{\sqrt{6} - \sqrt{2}}{2}$ $α = \sin^{- 1} (\frac{\sqrt{6} - \sqrt{2}}{2}) \approx 31 ° \Rightarrow$ $β = 14 ° \Rightarrow$ $180 ° - 2 β = 180 ° - 28 ° = 152 °$ $A r e a s h a d e d = π r^{2} - 2 A r e a (△) - A r e a (s e c t o r)$ $A r e a (s e c t o r) = \frac{a r c \cdot r}{2} = \frac{152 ° π}{180 °} {(\frac{5 \sqrt{6}}{3})}^{2} \cdot \frac{1}{2}$ $= \frac{38 π}{9} \cdot \frac{5}{3} = \frac{190 π}{27}$ $2 A r e a (△) = x (x + \sqrt{r^{2} - x^{2}})$ $= \frac{15 - 5 \sqrt{3}}{3} [\frac{15 - 5 \sqrt{3}}{3} + \frac{15 - 5 \sqrt{3}}{3} \sqrt{\frac{25 \cdot 6}{9} - {(\frac{15 - 5 \sqrt{3}}{3})}^{2}}]$ $\approx 11.8$ $\Rightarrow A r e a s h a d e d = π r^{2} - 2 A r e a (△) - A r e a (s e c t o r)$ $= \frac{260 π}{27} - 11.8 \approx 18.452$ $A r e a s h a d e d \approx 18.452$
	Answered by mr W last updated on 12/Feb/23

	Commented by mr W last updated on 12/Feb/23

	$s i d e l e n g t h o f e q u i l a t e r a l t r i a n g l e :$ $s = 10 \sqrt{2}$ $r a d i u s o f c i r c l e :$ $r = \frac{s \tan 30 °}{2} = \frac{5 \sqrt{2}}{\sqrt{3}} = \frac{5 \sqrt{6}}{3}$ $O A = \frac{10 \sqrt{2}}{2} - \frac{5 \sqrt{6}}{3} = \frac{5 (3 \sqrt{2} - \sqrt{6})}{3}$ $\sin α = \frac{O A}{\sqrt{2} \times O B} = \frac{5 (3 \sqrt{2} - \sqrt{6}) \times 3}{\sqrt{2} \times 3 \times 5 \sqrt{6}} = \frac{\sqrt{6} - \sqrt{2}}{2}$ $θ = π - \frac{π}{4} - α = \frac{3 π}{4} - \sin^{- 1} \frac{\sqrt{6} - \sqrt{2}}{2}$ $\sin θ = \sin (\frac{π}{4} + α) = \frac{1}{\sqrt{2}} (\sin α + \cos α)$ $= \frac{1}{\sqrt{2}} (\frac{\sqrt{6} - \sqrt{2}}{2} + \sqrt{\sqrt{3} - 1}) = \frac{\sqrt{3} - 1}{2} + \sqrt{\frac{\sqrt{3} - 1}{2}}$ $s h a d e d a r e a :$ $A_{s h a d e} = 2 (\frac{θ r^{2}}{2} - \frac{r \times O A \sin θ}{2})$ $= θ r^{2} - r \times O A \sin θ$ $= (\frac{3 π}{4} - \sin^{- 1} \frac{\sqrt{6} - \sqrt{2}}{2}) \times {(\frac{5 \sqrt{6}}{3})}^{2} - \frac{5 \sqrt{6}}{3} \times \frac{5 (3 \sqrt{2} - \sqrt{6})}{3} \times (\frac{\sqrt{3} - 1}{2} + \sqrt{\frac{\sqrt{3} - 1}{2}})$ $= \frac{50}{3} (\frac{3 π}{4} - \sin^{- 1} \frac{\sqrt{6} - \sqrt{2}}{2}) - \frac{50 (\sqrt{3} - 1)}{3} \times (\frac{\sqrt{3} - 1}{2} + \sqrt{\frac{\sqrt{3} - 1}{2}})$ $\approx 18.354437$
	Commented by Mingma last updated on 12/Feb/23
	Excellent
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