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Question-68305

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Question Number 68305 by TawaTawa last updated on 08/Sep/19
Commented by ajfour last updated on 10/Sep/19
Commented by ajfour last updated on 10/Sep/19
Area S_(req) =GC_(⌣) E^(⌢) (image distorted purposely so that area BCG is to be enlarged). consider A to be center of semicircle. S_1 =sector area ACE^(⌢) S_2 = area(△ABE) S_3 = area(△BCD) S_4 = sector area DGC^(⌢) S_(req) =S_1 −S_2 +S_3 −S_4 C((2/5),(4/5)) ; E((1/2),(1/2)) ; G(((2−(√6))/2), ((√6)/2)) ; B((2/7),(5/7)) S_1 =((r^2 θ)/2) with r=(1/2) , θ=cos^(−1) (4/5) ⇒ S_1 =(1/8)sin^(−1) (3/5) S_2 =(1/2)r(y_B −r)=(1/4)((5/7)−(1/2))=(3/(56)) S_3 =(1/2) determinant ((1,0,1),((2/5),(4/5),1),((2/7),(5/7),1)) =((4/5)−(5/7))+((2/7)−(8/(35)))=((28−25+10−8)/(35)) =(1/7) S_4 =((R^2 φ)/2) with R=1 and φ=sin^(−1) (y_C )−(π/4) ⇒ S_4 =(1/2)sin^(−1) (4/5)−(π/8) S_(req) =(1/8)sin^(−1) (3/5)−(3/(56))+(1/7)−(1/2)sin^(−1) (4/5)+(π/8) .
$${Area}\:\:{S}_{{req}} =\underset{\smile} {{GC}E}\:\:\:\: \\ $$$$\left({image}\:{distorted}\:{purposely}\right. \\ $$$${so}\:{that}\:{area}\:{BCG}\:{is}\:{to}\:{be} \\ $$$$\left.\:{enlarged}\right). \\ $$$${consider}\:{A}\:{to}\:{be}\:{center}\:{of} \\ $$$${semicircle}. \\ $$$${S}_{\mathrm{1}} ={sector}\:{area}\:{A}\overset{\frown} {{CE}} \\ $$$${S}_{\mathrm{2}} =\:{area}\left(\bigtriangleup{ABE}\right) \\ $$$${S}_{\mathrm{3}} =\:{area}\left(\bigtriangleup{BCD}\right) \\ $$$${S}_{\mathrm{4}} =\:{sector}\:{area}\:{D}\overset{\frown} {{GC}} \\ $$$${S}_{{req}} ={S}_{\mathrm{1}} −{S}_{\mathrm{2}} +{S}_{\mathrm{3}} −{S}_{\mathrm{4}} \\ $$$${C}\left(\frac{\mathrm{2}}{\mathrm{5}},\frac{\mathrm{4}}{\mathrm{5}}\right)\:\:\:;\:{E}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)\:; \\ $$$${G}\left(\frac{\mathrm{2}−\sqrt{\mathrm{6}}}{\mathrm{2}},\:\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\right)\:\:;\:\:{B}\left(\frac{\mathrm{2}}{\mathrm{7}},\frac{\mathrm{5}}{\mathrm{7}}\right) \\ $$$${S}_{\mathrm{1}} =\frac{{r}^{\mathrm{2}} \theta}{\mathrm{2}}\:\:\:{with}\:{r}=\frac{\mathrm{1}}{\mathrm{2}}\:,\:\theta=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\Rightarrow\:\:\:\:{S}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{8}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{5}} \\ $$$${S}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{r}\left({y}_{{B}} −{r}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{5}}{\mathrm{7}}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{3}}{\mathrm{56}} \\ $$$${S}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\begin{vmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{2}/\mathrm{5}}&{\mathrm{4}/\mathrm{5}}&{\mathrm{1}}\\{\mathrm{2}/\mathrm{7}}&{\mathrm{5}/\mathrm{7}}&{\mathrm{1}}\end{vmatrix} \\ $$$$\:\:\:=\left(\frac{\mathrm{4}}{\mathrm{5}}−\frac{\mathrm{5}}{\mathrm{7}}\right)+\left(\frac{\mathrm{2}}{\mathrm{7}}−\frac{\mathrm{8}}{\mathrm{35}}\right)=\frac{\mathrm{28}−\mathrm{25}+\mathrm{10}−\mathrm{8}}{\mathrm{35}} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{7}} \\ $$$${S}_{\mathrm{4}} =\frac{{R}^{\mathrm{2}} \phi}{\mathrm{2}}\:\:{with}\:{R}=\mathrm{1}\: \\ $$$${and}\:\phi=\mathrm{sin}^{−\mathrm{1}} \left({y}_{{C}} \right)−\frac{\pi}{\mathrm{4}} \\ $$$$\:\:\:\Rightarrow\:\:{S}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{5}}−\frac{\pi}{\mathrm{8}} \\ $$$${S}_{{req}} =\frac{\mathrm{1}}{\mathrm{8}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{5}}−\frac{\mathrm{3}}{\mathrm{56}}+\frac{\mathrm{1}}{\mathrm{7}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{5}}+\frac{\pi}{\mathrm{8}}\:. \\ $$
Commented by TawaTawa last updated on 10/Sep/19
Wow, God bless you sir.
$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by mr W last updated on 08/Sep/19
let side length of square=1 take A as origin and AD as x−axis. eqn. of big circle: (x−1)^2 +y^2 =1 eqn. of small circle: x^2 +(y−(1/2))^2 =(1/4) eqn. of line BD: y=1−x intersection of big and small circles: (x−1)^2 +y^2 =1 ...(i) x^2 +(y−(1/2))^2 =(1/4) ...(ii) (i)−(ii): ⇒y=2x (x−1)^2 +4x^2 =1 5x^2 −2x+1=1 x(5x−2)=0 ⇒x=0 or x=(2/5) intersection of big circle and BD: (x−1)^2 +(1−x)^2 =1 x^2 −2x−(1/2)=0 ⇒x=1−(√(1+(1/2)))=((2−(√6))/2) intersection of small circle and BD: x^2 +((1/2)−x)^2 =(1/4) x(2x−1)=0 ⇒x=0 or x=(1/2) big circle: ⇒y=(√(x(2−x))) small circle: y=(1/2)+(√((1/4)−x^2 )) shaded area: A=∫_((2−(√6))/2) ^(2/5) [(√(x(2−x)))−1+x]dx+∫_(2/5) ^(1/2) [(1/2)+(√((1/4)−x^2 ))−1+x]dx A=∫_((2−(√6))/2) ^(2/5) [(√(x(2−x)))−1+x]dx+∫_(2/5) ^(1/2) [(√((1/4)−x^2 ))−(1/2)+x]dx A=[(((x−1)(√(x(2−x)))+sin^(−1) (x−1))/2)−x+(x^2 /2)]_((2−(√6))/2) ^(2/5) +[((sin^(−1) (2x))/8)+((x(√((1/4)−x^2 )))/2)−(x/2)+(x^2 /2)]_(2/5) ^(1/2) ......
$${let}\:{side}\:{length}\:{of}\:{square}=\mathrm{1} \\ $$$${take}\:{A}\:{as}\:{origin}\:{and}\:{AD}\:{as}\:{x}−{axis}. \\ $$$${eqn}.\:{of}\:{big}\:{circle}: \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1} \\ $$$${eqn}.\:{of}\:{small}\:{circle}: \\ $$$${x}^{\mathrm{2}} +\left({y}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${eqn}.\:{of}\:{line}\:{BD}: \\ $$$${y}=\mathrm{1}−{x} \\ $$$$ \\ $$$${intersection}\:{of}\:{big}\:{and}\:{small}\:{circles}: \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}\:\:\:\:…\left({i}\right) \\ $$$${x}^{\mathrm{2}} +\left({y}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\Rightarrow{y}=\mathrm{2}{x} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}=\mathrm{1} \\ $$$${x}\left(\mathrm{5}{x}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{0}\:{or}\:{x}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$${intersection}\:{of}\:{big}\:{circle}\:{and}\:{BD}: \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{1}−{x}\right)^{\mathrm{2}} =\mathrm{1}\:\:\: \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{1}−\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}}=\frac{\mathrm{2}−\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$$${intersection}\:{of}\:{small}\:{circle}\:{and}\:{BD}: \\ $$$${x}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}−{x}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\: \\ $$$${x}\left(\mathrm{2}{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{0}\:{or}\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${big}\:{circle}: \\ $$$$\Rightarrow{y}=\sqrt{{x}\left(\mathrm{2}−{x}\right)} \\ $$$${small}\:{circle}: \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−{x}^{\mathrm{2}} } \\ $$$$ \\ $$$${shaded}\:{area}: \\ $$$${A}=\int_{\frac{\mathrm{2}−\sqrt{\mathrm{6}}}{\mathrm{2}}} ^{\frac{\mathrm{2}}{\mathrm{5}}} \left[\sqrt{{x}\left(\mathrm{2}−{x}\right)}−\mathrm{1}+{x}\right]{dx}+\int_{\frac{\mathrm{2}}{\mathrm{5}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left[\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−{x}^{\mathrm{2}} }−\mathrm{1}+{x}\right]{dx} \\ $$$${A}=\int_{\frac{\mathrm{2}−\sqrt{\mathrm{6}}}{\mathrm{2}}} ^{\frac{\mathrm{2}}{\mathrm{5}}} \left[\sqrt{{x}\left(\mathrm{2}−{x}\right)}−\mathrm{1}+{x}\right]{dx}+\int_{\frac{\mathrm{2}}{\mathrm{5}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left[\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}+{x}\right]{dx} \\ $$$${A}=\left[\frac{\left({x}−\mathrm{1}\right)\sqrt{{x}\left(\mathrm{2}−{x}\right)}+\mathrm{sin}^{−\mathrm{1}} \left({x}−\mathrm{1}\right)}{\mathrm{2}}−{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\frac{\mathrm{2}−\sqrt{\mathrm{6}}}{\mathrm{2}}} ^{\frac{\mathrm{2}}{\mathrm{5}}} +\left[\frac{\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)}{\mathrm{8}}+\frac{{x}\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−{x}^{\mathrm{2}} }}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\frac{\mathrm{2}}{\mathrm{5}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$…… \\ $$
Commented by TawaTawa last updated on 08/Sep/19
Wow, thanks for your time sir. God bless you. Waiting for the rest sir
$$\mathrm{Wow},\:\mathrm{thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}.\:\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$$$\mathrm{Waiting}\:\mathrm{for}\:\mathrm{the}\:\mathrm{rest}\:\mathrm{sir} \\ $$
Commented by TawaTawa last updated on 10/Sep/19
This area is too tough. The integration formed is complex.
$$\mathrm{This}\:\mathrm{area}\:\mathrm{is}\:\mathrm{too}\:\mathrm{tough}.\:\mathrm{The}\:\mathrm{integration}\:\mathrm{formed}\:\mathrm{is}\:\mathrm{complex}. \\ $$

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