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Question Number 67716 by TawaTawa last updated on 30/Aug/19

	Answered by mind is power last updated on 30/Aug/19
	$let c_1 =demi circle of r=2 c2=circle r=1 c_1 ..x^2 +y^2 =4 c_2 =x^2 +(y−h)^2 =1 withe (0,h) center of c_2 c_1 ∩c_2 ⇒y^2 −(y−h)^2 =3 ⇒2yh−h^2 =3⇒y=((3+h^2 )/(2h))=y_0 x^2 =(4−y^2 )=((16h^2 −(3+h^2 )^2 )/(4h^2 ))=((−h^4 +10h^2 −9)/(4h^2 ))=((−(h^2 −9)(h^2 −1))/(4h^2 )) ⇒h^2 ∈[1,9] either c_2 ∩c_1 ={} x=+_− (√((−(h^2 −9)(h^2 −1))/(4h^2 )))=+_− x_0 assum h^2 ∈[1.9] D(+x_0 .y_0 ).C=(−x_0 .y_0 ) DA+AB=∫_(−x_(0 ) ) ^(+x_0 ) (√(1+((dy/dx))^2 ))dx...c_1 (dy/dx)=d((√(4−x^2 )))/dx=((−x)/(√(4−x^2 )))⇒((dy/dx))^2 =(x^2 /(4−x^2 )) DA+AB=∫_(−x_0 ) ^(+x_0 ) (√(4/(4−x^2 )))dx x=2sin(δ)⇒dx=2cos(δ)dδ DA+AB=∫_(arcsin(−((x0)/2))) ^(arcsin(((x0)/2))) (√(4/(4−4sin^2 (δ)))).2cos(δ)dδ) =∫2.((2cos(δ))/(∣2cos(δ)∣))dδ=4∫_0 ^(arcsin((x_0 /2))) dδ=4arcsin((x_0 /2)) same idea too find DC+CB...$
	$l e t c_{1} = d e m i c i r c l e o f r = 2$ $c 2 = c i r c l e r = 1$ $c_{1} . . x^{2} + y^{2} = 4$ $c_{2} = x^{2} + {(y - h)}^{2} = 1$ $w i t h e (0, h) c e n t e r o f c_{2}$ $c_{1} \cap c_{2} \Rightarrow y^{2} - {(y - h)}^{2} = 3$ $\Rightarrow 2 y h - h^{2} = 3 \Rightarrow y = \frac{3 + h^{2}}{2 h} = y_{0}$ $x^{2} = (4 - y^{2}) = \frac{16 h^{2} - {(3 + h^{2})}^{2}}{4 h^{2}} = \frac{- h^{4} + 10 h^{2} - 9}{4 h^{2}} = \frac{- (h^{2} - 9) (h^{2} - 1)}{4 h^{2}}$ $\Rightarrow h^{2} \in [1, 9] e i t h e r c_{2} \cap c_{1} = {}$ $x = \underline{+} \sqrt{\frac{- (h^{2} - 9) (h^{2} - 1)}{4 h^{2}}} = \underline{+} x_{0}$ $a s s u m h^{2} \in [1.9]$ $D (+ x_{0} . y_{0}) . C = (- x_{0} . y_{0})$ $D A + A B = \int_{- x_{0}}^{+ x_{0}} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x . . . c_{1}$ $\frac{d y}{d x} = d (\sqrt{4 - x^{2}}) / d x = \frac{- x}{\sqrt{4 - x^{2}}} \Rightarrow {(\frac{d y}{d x})}^{2} = \frac{x^{2}}{4 - x^{2}}$ $D A + A B = \int_{- x_{0}}^{+ x_{0}} \sqrt{\frac{4}{4 - x^{2}}} d x$ $x = 2 s i n (δ) \Rightarrow d x = 2 c o s (δ) d δ$ $D A + A B = \int_{a r c s i n (- \frac{x 0}{2})}^{a r c s i n (\frac{x 0}{2})} \sqrt{\frac{4}{4 - 4 {s i n}^{2} (δ)}} . 2 c o s (δ) d δ)$ $= \int 2 . \frac{2 c o s (δ)}{∣ 2 c o s (δ) ∣} d δ = 4 \int_{0}^{a r c s i n (\frac{x_{0}}{2})} d δ = 4 a r c s i n (\frac{x_{0}}{2})$ $s a m e i d e a t o o f i n d D C + C B . . .$
	Commented by TawaTawa last updated on 30/Aug/19

	$God bless you sir .$
	Answered by mr W last updated on 30/Aug/19

	Commented by mr W last updated on 30/Aug/19

	$B D = 2 \times 1 \times \sin \frac{α}{2} = 2 \times 2 \times \sin \frac{β}{2}$ $\Rightarrow \sin \frac{α}{2} = 2 \sin \frac{β}{2} . . . (i)$ $E F = 1 + 2 - A C = 1 + 2 - \frac{1}{2} = \frac{5}{2}$ $E F = 1 \times \cos \frac{α}{2} + 2 \times \cos \frac{β}{2} = \frac{5}{2}$ $\Rightarrow \cos \frac{α}{2} = \frac{5}{2} - 2 \cos \frac{β}{2} . . . (i i)$ ${(i)}^{2} + {(i i)}^{2} :$ $\sin^{2} \frac{α}{2} + \cos^{2} \frac{α}{2} = 4 \sin^{2} \frac{β}{2} + \frac{25}{4} - 10 \cos \frac{β}{2} + 4 \cos^{2} \frac{β}{2}$ $1 = 4 + \frac{25}{4} - 10 \cos \frac{β}{2}$ $\Rightarrow \cos \frac{β}{2} = \frac{37}{40}$ $\Rightarrow \sin \frac{β}{2} = \frac{\sqrt{40^{2} - 37^{2}}}{40} = \frac{\sqrt{231}}{40}$ $\Rightarrow \sin β = 2 \times \frac{37}{40} \times \frac{\sqrt{231}}{40} = \frac{37 \sqrt{231}}{800}$ $\Rightarrow β = \sin^{- 1} \frac{37 \sqrt{231}}{800}$ $\sin \frac{α}{2} = 2 \times \frac{\sqrt{231}}{40} = \frac{\sqrt{231}}{20}$ $\cos \frac{α}{2} = \frac{\sqrt{20^{2} - 231}}{20} = \frac{13}{20}$ $\Rightarrow \sin α = 2 \times \frac{\sqrt{231}}{20} \times \frac{13}{20} = \frac{13 \sqrt{231}}{200}$ $\Rightarrow α = \sin^{- 1} \frac{13 \sqrt{231}}{200}$ $p e r i m e t e r o f A B C D = p$ $p = 1 \times α + 2 \times β$ $= \sin^{- 1} \frac{13 \sqrt{231}}{200} + 2 \sin^{- 1} \frac{37 \sqrt{231}}{800}$ $\approx 2.974$ $a r e a o f A B C D = A$ $A = \frac{1^{2}}{2} (α - \sin α) + \frac{2^{2}}{2} (β - \sin β)$ $= \frac{1}{2} (\sin^{- 1} \frac{13 \sqrt{231}}{200} - \frac{13 \sqrt{231}}{200}) + 2 (\sin^{- 1} \frac{37 \sqrt{231}}{800} - \frac{37 \sqrt{231}}{800})$ $\approx 0.367$
	Commented by TawaTawa last updated on 30/Aug/19

	$Wow, God bless you sir .$
	Commented by TawaTawa last updated on 30/Aug/19

	$I appreciate your time sir .$
	Commented by TawaTawa last updated on 31/Aug/19

	$Sir, i checked the workings, i understand the rest but i {don}^{'} t understand$ $how you got equation (i) and (ii) . But i get the rest . Can you let$ $me understand (i) and (ii) . Sorry for disturbing too much . And$ $Thanks for helping always . God will reward you sir .$
	Commented by mr W last updated on 31/Aug/19

	Commented by mr W last updated on 31/Aug/19

	$B D = 2 \times G D$ $G D = E D \times \sin \frac{α}{2} = 1 \times \sin \frac{α}{2}$ $G D = F D \times \sin \frac{β}{2} = 2 \times \sin \frac{β}{2}$ $\Rightarrow \sin \frac{α}{2} = 2 \sin \frac{β}{2}$ $E F = r_{1} + r_{2} - A C = 1 + 2 - \frac{1}{2} = \frac{5}{2}$ $E F = E G + G F = E D \times \cos \frac{α}{2} + F D \times \cos \frac{β}{2}$ $= 1 \times \cos \frac{α}{2} + 2 \times \cos \frac{β}{2} = \frac{5}{2}$ $\Rightarrow \cos \frac{α}{2} = \frac{5}{2} - 2 \cos \frac{β}{2}$
	Commented by TawaTawa last updated on 31/Aug/19

	$I appreciate sir . God bless you more$
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