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Question Number 146960 by mathdanisur last updated on 16/Jul/21

	Answered by TheHoneyCat last updated on 16/Jul/21

	$let A be the fuction that gives the area of a surface$ $let Δ = A B O$ $let x be the hatched area$ $let D be the intersection between [AO] and the circle C (O, O R)$ $let y be the portion A O D of the circle$ $A (y) = π B C \frac{180 - 90 - \overset{Λ}{A}}{360}$ $A (A B O) = \sqrt{{B C}^{2} - {(\frac{B C}{2})}^{2}} = B C \frac{\sqrt{3}}{2}$ $A (x) = A (A B O) - A (y)$ $A (x) = B C {(\frac{\sqrt{3}}{2} - π \frac{180 - 90 - \overset{Λ}{A}}{360})}_{◼}$ $I l e t y o u p l o t t h a t o n y o u r c a l c u l a t o r i f y o u w a n t a v a l u e$ $b y t h e w a y : y o u g a v e o n e i n f o r m a t i o n w i t h o u t n e c e s s i t y,$ $i n d e e d : \overset{Λ}{A} = 30 ° c a n b e d e d u c e d f r o m t h e f a c t t h a t A O = 2 B O$ $; -)$
	Commented by mathdanisur last updated on 16/Jul/21

	$t h a n k s S e r b u t a n s w e r ?$
	Answered by ajfour last updated on 16/Jul/21

	$A D = O D = O B = 6$ $A B = 6 \sqrt{3}$ $A r e a △ A B O = 18 \sqrt{3}$ $A r e a o f s e c t o r \overset{⌢}{B O A} = \frac{1}{2} r^{2} θ$ $\frac{1}{2} \times 36 \times \frac{π}{3} = 6 π$ $h a t c h e d s e c t i o n a r e a$ $= 18 \sqrt{3} - 6 π s q . u n i t s$
	Commented by mathdanisur last updated on 17/Jul/21

	$t h a n k y o u S e r$
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