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Question Number 50829 by Raj Singh last updated on 21/Dec/18

Commented by Raj Singh last updated on 21/Dec/18

$t h i s i s c l a s s 10 p r o b l e m s o s o l v e$ $b y g e n e r a l m e t h o d$
	Answered by tanmay.chaudhury50@gmail.com last updated on 21/Dec/18
	$area of square=a^2 area of ADCA sector=(1/4)×πa^2 DC=x axis DA=y axis sector ADCA part of circle x^2 +y^2 =a^2 sector DCBD part of circle (x−a)^2 +y^2 =a^2 solving x^2 =(x−a)^2 x^2 =x^2 −2xa+a^2 so 2xa=a^2 x=(a/2) area of big white portion=W ∫_0 ^(a/2) (√(a^2 −(x−a)^2 )) dx+∫_(a/2) ^a (√(a^2 −x^2 )) dx=W so required shaded area=2((1/4)πa^2 −W) formula ∫(√(A^2 −X^2 )) dX=(X/2)(√(A^2 −X^2 )) +(A^2 /2)sin^(−1) ((X/A)) I_1 =∣((x−a)/2)(√(a^2 −(x−a)^2 )) +(a^2 /2)sin^(−1) (((x−a)/a))∣_0 ^(a/2) ={(((−a)/4)×a((√3)/2))+(a^2 /2)sin^(−1) (((−1)/2))−(a^2 /2)sin^(−1) (−1)} =((−(√3))/8)a^2 −(a^2 /2)((π/6))+(a^2 /2)×(π/2) I_2 =∣(x/2)(√(a^2 −x^2 )) +(a^2 /2)sin^(−1) ((x/a))∣_(a/2) ^a =(a/2)×0+(a^2 /2)×(π/2)−(a/4)×((a(√3))/2)−(a^2 /2)×(π/6) =((πa^2 )/4)−((a^2 (√3))/8)−((πa^2 )/(12)) W=2(((πa^2 )/4)−((√3)/8)a^2 −((πa^2 )/(12)))=((πa^2 )/2)−(((√3) a^2 )/4)−((πa^2 )/6) so required area 2[((πa^2 )/4)−((πa^2 )/2)+(((√3) a^2 )/4)+((πa^2 )/6)] =2[((3πa^2 −6πa^2 +3(√3) a^2 +2πa^2 )/(12))] =(1/6)[3(√3) a^2 −πa^2 ] pls check$
	$a r e a o f s q u a r e = a^{2}$ $a r e a o f A D C A s e c t o r = \frac{1}{4} \times π a^{2}$ $D C = x a x i s D A = y a x i s$ $s e c t o r A D C A p a r t o f c i r c l e x^{2} + y^{2} = a^{2}$ $s e c t o r D C B D p a r t o f c i r c l e {(x - a)}^{2} + y^{2} = a^{2}$ $s o l v i n g$ $x^{2} = {(x - a)}^{2}$ $x^{2} = x^{2} - 2 x a + a^{2} s o 2 x a = a^{2} x = \frac{a}{2}$ $a r e a o f b i g w h i t e p o r t i o n = W$ $\int_{0}^{\frac{a}{2}} \sqrt{a^{2} - {(x - a)}^{2}} d x + \int_{\frac{a}{2}}^{a} \sqrt{a^{2} - x^{2}} d x = W$ $s o r e q u i r e d s h a d e d a r e a = 2 (\frac{1}{4} π a^{2} - W)$ $f o r m u l a \int \sqrt{A^{2} - X^{2}} d X = \frac{X}{2} \sqrt{A^{2} - X^{2}} + \frac{A^{2}}{2} {s i n}^{- 1} (\frac{X}{A})$ $I_{1} =∣ \frac{x - a}{2} \sqrt{a^{2} - {(x - a)}^{2}} + \frac{a^{2}}{2} {s i n}^{- 1} (\frac{x - a}{a}) ∣_{0}^{\frac{a}{2}}$ $= {(\frac{- a}{4} \times a \frac{\sqrt{3}}{2}) + \frac{a^{2}}{2} {s i n}^{- 1} (\frac{- 1}{2}) - \frac{a^{2}}{2} {s i n}^{- 1} (- 1)}$ $= \frac{- \sqrt{3}}{8} a^{2} - \frac{a^{2}}{2} (\frac{π}{6}) + \frac{a^{2}}{2} \times \frac{π}{2}$ $I_{2} =∣ \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} {s i n}^{- 1} (\frac{x}{a}) ∣_{\frac{a}{2}}^{a}$ $= \frac{a}{2} \times 0 + \frac{a^{2}}{2} \times \frac{π}{2} - \frac{a}{4} \times \frac{a \sqrt{3}}{2} - \frac{a^{2}}{2} \times \frac{π}{6}$ $= \frac{π a^{2}}{4} - \frac{a^{2} \sqrt{3}}{8} - \frac{π a^{2}}{12}$ $W = 2 (\frac{π a^{2}}{4} - \frac{\sqrt{3}}{8} a^{2} - \frac{π a^{2}}{12}) = \frac{π a^{2}}{2} - \frac{\sqrt{3} a^{2}}{4} - \frac{π a^{2}}{6}$ $s o r e q u i r e d a r e a$ $2 [\frac{π a^{2}}{4} - \frac{π a^{2}}{2} + \frac{\sqrt{3} a^{2}}{4} + \frac{π a^{2}}{6}]$ $= 2 [\frac{3 π a^{2} - 6 π a^{2} + 3 \sqrt{3} a^{2} + 2 π a^{2}}{12}]$ $= \frac{1}{6} [3 \sqrt{3} a^{2} - π a^{2}]$ $p l s c h e c k$
	Answered by mr W last updated on 21/Dec/18

	$(2)$ $R = r a d i u s o f b i g c i r c l e R$ $r = r a d i u s o f s m a l l c i r c l e$ $R B = 2 R - 2 r = 9$ $\Rightarrow R - r = 4.5 \Rightarrow r = R - 4.5$ $A E = O A - O E = R - \sqrt{r^{2} - {(R - r)}^{2}} = R - \sqrt{R (2 r - R)} = 5$ $R - \sqrt{R (R - 9)} = 5$ $R - 5 = \sqrt{R (R - 9)}$ $R^{2} - 10 R + 25 = R^{2} - 9 R$ $- 10 R + 25 = - 9 R$ $\Rightarrow R = 25$ $\Rightarrow r = 20.5$ $A_{c o l o r} = π (R^{2} - r^{2}) = π (R + r) (R - r)$ $= π \times 45.5 \times 4.5 = 643$
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