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Question Number 38247 by ajfour last updated on 23/Jun/18

Commented by ajfour last updated on 23/Jun/18

$E a c h o f t h e t h r e e c o l o u r e d a r e a s$ $a r e e q u a l; f i n d \frac{a}{R}, a n d \frac{b}{R} .$
Commented by behi83417@gmail.com last updated on 23/Jun/18

$s i r A j f o u r! p l e a s e c h e c k m y c o m m e n t$ $You can't use 'macro parameter character #' in math mode$
	Answered by behi83417@gmail.com last updated on 25/Jun/18

	$e q . o f c i r c l e : x^{2} + y^{2} = R^{2}$ $z^{2} = R^{2} - b^{2} \Rightarrow z = \sqrt{R^{2} - b^{2}}$ $S_{z} = \underset{- b}{\int^{0}} [(\sqrt{R^{2} - x^{2}} - (\sqrt{R^{2} - b^{2}})] d x =$ $Missing \left or extra \right$ $= [\frac{R^{2}}{2} {s i n}^{- 1} \frac{b}{R} - \frac{3 b}{2} \sqrt{R^{2} - b^{2}}]$ $S_{p i n k} = \frac{π R^{2}}{4} + b^{2} + 2 b \sqrt{R^{2} - b^{2}} + 2 S_{z}$ $S_{b l u e} = \frac{3 π R^{2}}{4} - b^{2} - 2 b \sqrt{R^{2} - b^{2}} - 2 S_{z}$ $\Rightarrow \frac{π R^{2}}{2} - 2 b^{2} - 4 b \sqrt{R^{2} - b^{2}} = 4 S_{z}$ $\frac{π R^{2}}{2} - 2 b^{2} - 4 b \sqrt{R^{2} - b^{2}} = 2 R^{2} {s i n}^{- 1} \frac{b}{R} - 6 b \sqrt{R^{2} - b^{2}}$ $\frac{π R^{2}}{2} - 2 b^{2} + 2 b \sqrt{R^{2} - b^{2}} - 2 R^{2} {s i n}^{- 1} \frac{b}{R} = 0$ $π - 4 \frac{b^{2}}{R^{2}} + 4 \frac{b}{R} \sqrt{1 - \frac{b^{2}}{R^{2}}} - 4 {s i n}^{- 1} \frac{b}{R} = 0$ $\overset{\frac{b}{R} = m}{\Rightarrow} m^{2} - m \sqrt{1 - m^{2}} + {s i n}^{- 1} m - \frac{π}{4} = 0$ $\Rightarrow m = \frac{b}{R} = 0.71$
	Commented by behi83417@gmail.com last updated on 25/Jun/18

	$d e a r m a s t e r m r W 3! t h a n k y o u f o r$ $c o r r e c t i o n . n o w i t i s f i x e d .$
	Commented by ajfour last updated on 25/Jun/18

	$I n t e g r a t i o n i s n o t n e e d e d, i$ $b e l i e v e, i, l l c h e c k w i t h m y$ $a n s w e r S i r .$
	Commented by MrW3 last updated on 25/Jun/18

	$T o {B e h i}^{'} s f a t h e r :$ $p l s r e c h e c k, y o u r f i n a l e q n .$ $m^{2} + m \sqrt{1 - m^{2}} + {c o s}^{- 1} m - \frac{3 π}{4} = 0$ $h a s n o s o l u t i o n .$
	Commented by MrW3 last updated on 25/Jun/18

	$t h e f i n a l e q n . s h o u l d b e :$ $\Rightarrow m^{2} + m \sqrt{1 - m^{2}} + {s i n}^{- 1} m - \frac{π}{4} = 0$
	Answered by MrW3 last updated on 25/Jun/18

	$a^{2} = π R^{2}$ $\Rightarrow \frac{a}{R} = \sqrt{π} \approx 1.7725$ $l e t \sin α = \frac{b}{R} = λ$ $A_{r e d} = b^{2} + b R \cos α + \frac{R^{2}}{2} (2 α + \frac{π}{2}) = \frac{π R^{2}}{2}$ $b^{2} + b R \cos α + (α - \frac{π}{4}) R^{2} = 0$ $l e t λ = \frac{b}{R} \Rightarrow α = \sin^{- 1} λ, \cos α = \sqrt{1 - λ^{2}}$ $\Rightarrow λ^{2} + λ \sqrt{1 - λ^{2}} + \sin^{- 1} λ = \frac{π}{4}$ $\Rightarrow λ = \frac{b}{R} \approx 0.3412$
	Commented by behi83417@gmail.com last updated on 25/Jun/18

	$d e a r A j f o u r! p l e a s e r e c h e c k m y n e w$ $e d i t i o n f o r y o u r q u e s t i o n . t h a n k s .$
	Commented by ajfour last updated on 25/Jun/18

	$T h a n k y o u b o t h, S i r s .$
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