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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...

$${let}\:{c}_{\mathrm{1}} ={demi}\:{circle}\:{of}\:{r}=\mathrm{2} \\ $$$${c}\mathrm{2}={circle}\:{r}=\mathrm{1} \\ $$$${c}_{\mathrm{1}} ..{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4} \\ $$$${c}_{\mathrm{2}} ={x}^{\mathrm{2}} +\left({y}−{h}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${withe}\:\left(\mathrm{0},{h}\right)\:{center}\:{of}\:{c}_{\mathrm{2}} \\ $$$${c}_{\mathrm{1}} \cap{c}_{\mathrm{2}} \Rightarrow{y}^{\mathrm{2}} −\left({y}−{h}\right)^{\mathrm{2}} =\mathrm{3} \\ $$$$\Rightarrow\mathrm{2}{yh}−{h}^{\mathrm{2}} =\mathrm{3}\Rightarrow{y}=\frac{\mathrm{3}+{h}^{\mathrm{2}} }{\mathrm{2}{h}}={y}_{\mathrm{0}} \\ $$$${x}^{\mathrm{2}} =\left(\mathrm{4}−{y}^{\mathrm{2}} \right)=\frac{\mathrm{16}{h}^{\mathrm{2}} −\left(\mathrm{3}+{h}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}{h}^{\mathrm{2}} }=\frac{−{h}^{\mathrm{4}} +\mathrm{10}{h}^{\mathrm{2}} −\mathrm{9}}{\mathrm{4}{h}^{\mathrm{2}} }=\frac{−\left({h}^{\mathrm{2}} −\mathrm{9}\right)\left({h}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{4}{h}^{\mathrm{2}} } \\ $$$$\Rightarrow{h}^{\mathrm{2}} \in\left[\mathrm{1},\mathrm{9}\right]\:\:{either}\:{c}_{\mathrm{2}} \cap{c}_{\mathrm{1}} =\left\{\right\} \\ $$$${x}=\underset{−} {+}\sqrt{\frac{−\left({h}^{\mathrm{2}} −\mathrm{9}\right)\left({h}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{4}{h}^{\mathrm{2}} }}=\underset{−} {+}{x}_{\mathrm{0}} \\ $$$${assum}\:{h}^{\mathrm{2}} \in\left[\mathrm{1}.\mathrm{9}\right] \\ $$$${D}\left(+{x}_{\mathrm{0}} .{y}_{\mathrm{0}} \right).{C}=\left(−{x}_{\mathrm{0}} .{y}_{\mathrm{0}} \right) \\ $$$${DA}+{AB}=\int_{−{x}_{\mathrm{0}\:} } ^{+{x}_{\mathrm{0}} } \sqrt{\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} }{dx}...{c}_{\mathrm{1}} \\ $$$$\frac{{dy}}{{dx}}={d}\left(\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\right)/{dx}=\frac{−{x}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\Rightarrow\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\frac{{x}^{\mathrm{2}} }{\mathrm{4}−{x}^{\mathrm{2}} } \\ $$$${DA}+{AB}=\int_{−{x}_{\mathrm{0}} } ^{+{x}_{\mathrm{0}} } \sqrt{\frac{\mathrm{4}}{\mathrm{4}−{x}^{\mathrm{2}} }}{dx} \\ $$$${x}=\mathrm{2}{sin}\left(\delta\right)\Rightarrow{dx}=\mathrm{2}{cos}\left(\delta\right){d}\delta \\ $$$$\left.{DA}+{AB}=\int_{{arcsin}\left(−\frac{{x}\mathrm{0}}{\mathrm{2}}\right)} ^{{arcsin}\left(\frac{{x}\mathrm{0}}{\mathrm{2}}\right)} \sqrt{\frac{\mathrm{4}}{\mathrm{4}−\mathrm{4}{sin}^{\mathrm{2}} \left(\delta\right)}}.\mathrm{2}{cos}\left(\delta\right){d}\delta\right) \\ $$$$=\int\mathrm{2}.\frac{\mathrm{2}{cos}\left(\delta\right)}{\mid\mathrm{2}{cos}\left(\delta\right)\mid}{d}\delta=\mathrm{4}\int_{\mathrm{0}} ^{{arcsin}\left(\frac{{x}_{\mathrm{0}} }{\mathrm{2}}\right)} {d}\delta=\mathrm{4}{arcsin}\left(\frac{{x}_{\mathrm{0}} }{\mathrm{2}}\right) \\ $$$${same}\:{idea}\:\:{too}\:{find}\:{DC}+{CB}... \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by TawaTawa last updated on 30/Aug/19

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\: \\ $$

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

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

BD=2×1×sin (α/2)=2×2×sin (β/2) ⇒sin (α/2)=2 sin (β/2) ...(i) EF=1+2−AC=1+2−(1/2)=(5/2) EF=1×cos (α/2)+2×cos (β/2)=(5/2) ⇒cos (α/2)=(5/2)−2 cos (β/2) ...(ii) (i)^2 +(ii)^2 : sin^2 (α/2)+cos^2 (α/2)=4 sin^2 (β/2)+((25)/4)−10 cos (β/2)+4 cos^2 (β/2) 1=4+((25)/4)−10 cos (β/2) ⇒cos (β/2)=((37)/(40)) ⇒sin (β/2)=((√(40^2 −37^2 ))/(40))=((√(231))/(40)) ⇒sin β=2×((37)/(40))×((√(231))/(40))=((37(√(231)))/(800)) ⇒β=sin^(−1) ((37(√(231)))/(800)) sin (α/2)=2×((√(231))/(40))=((√(231))/(20)) cos (α/2)=((√(20^2 −231))/(20))=((13)/(20)) ⇒sin α=2×((√(231))/(20))×((13)/(20))=((13(√(231)))/(200)) ⇒α=sin^(−1) ((13(√(231)))/(200)) perimeter of ABCD=p p=1×α+2×β =sin^(−1) ((13(√(231)))/(200))+2 sin^(−1) ((37(√(231)))/(800)) ≈2.974 area of ABCD=A A=(1^2 /2)(α−sin α)+(2^2 /2)(β−sin β) =(1/2)(sin^(−1) ((13(√(231)))/(200))−((13(√(231)))/(200)))+2(sin^(−1) ((37(√(231)))/(800))−((37(√(231)))/(800))) ≈0.367

$${BD}=\mathrm{2}×\mathrm{1}×\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}=\mathrm{2}×\mathrm{2}×\mathrm{sin}\:\frac{\beta}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}=\mathrm{2}\:\mathrm{sin}\:\frac{\beta}{\mathrm{2}}\:\:\:\:...\left({i}\right) \\ $$$${EF}=\mathrm{1}+\mathrm{2}−{AC}=\mathrm{1}+\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${EF}=\mathrm{1}×\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}+\mathrm{2}×\mathrm{cos}\:\frac{\beta}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}}−\mathrm{2}\:\mathrm{cos}\:\frac{\beta}{\mathrm{2}}\:\:\:\:...\left({ii}\right) \\ $$$$\left({i}\right)^{\mathrm{2}} +\left({ii}\right)^{\mathrm{2}} : \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\frac{\alpha}{\mathrm{2}}+\mathrm{cos}^{\mathrm{2}} \:\frac{\alpha}{\mathrm{2}}=\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\frac{\beta}{\mathrm{2}}+\frac{\mathrm{25}}{\mathrm{4}}−\mathrm{10}\:\mathrm{cos}\:\frac{\beta}{\mathrm{2}}+\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\beta}{\mathrm{2}} \\ $$$$\mathrm{1}=\mathrm{4}+\frac{\mathrm{25}}{\mathrm{4}}−\mathrm{10}\:\mathrm{cos}\:\frac{\beta}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{cos}\:\frac{\beta}{\mathrm{2}}=\frac{\mathrm{37}}{\mathrm{40}} \\ $$$$\Rightarrow\mathrm{sin}\:\frac{\beta}{\mathrm{2}}=\frac{\sqrt{\mathrm{40}^{\mathrm{2}} −\mathrm{37}^{\mathrm{2}} }}{\mathrm{40}}=\frac{\sqrt{\mathrm{231}}}{\mathrm{40}} \\ $$$$\Rightarrow\mathrm{sin}\:\beta=\mathrm{2}×\frac{\mathrm{37}}{\mathrm{40}}×\frac{\sqrt{\mathrm{231}}}{\mathrm{40}}=\frac{\mathrm{37}\sqrt{\mathrm{231}}}{\mathrm{800}} \\ $$$$\Rightarrow\beta=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{37}\sqrt{\mathrm{231}}}{\mathrm{800}} \\ $$$$\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}=\mathrm{2}×\frac{\sqrt{\mathrm{231}}}{\mathrm{40}}=\frac{\sqrt{\mathrm{231}}}{\mathrm{20}} \\ $$$$\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}=\frac{\sqrt{\mathrm{20}^{\mathrm{2}} −\mathrm{231}}}{\mathrm{20}}=\frac{\mathrm{13}}{\mathrm{20}} \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=\mathrm{2}×\frac{\sqrt{\mathrm{231}}}{\mathrm{20}}×\frac{\mathrm{13}}{\mathrm{20}}=\frac{\mathrm{13}\sqrt{\mathrm{231}}}{\mathrm{200}} \\ $$$$\Rightarrow\alpha=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{13}\sqrt{\mathrm{231}}}{\mathrm{200}} \\ $$$$ \\ $$$${perimeter}\:{of}\:{ABCD}={p} \\ $$$${p}=\mathrm{1}×\alpha+\mathrm{2}×\beta \\ $$$$=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{13}\sqrt{\mathrm{231}}}{\mathrm{200}}+\mathrm{2}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{37}\sqrt{\mathrm{231}}}{\mathrm{800}} \\ $$$$\approx\mathrm{2}.\mathrm{974} \\ $$$$ \\ $$$${area}\:{of}\:{ABCD}={A} \\ $$$${A}=\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{2}}\left(\alpha−\mathrm{sin}\:\alpha\right)+\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{2}}\left(\beta−\mathrm{sin}\:\beta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{13}\sqrt{\mathrm{231}}}{\mathrm{200}}−\frac{\mathrm{13}\sqrt{\mathrm{231}}}{\mathrm{200}}\right)+\mathrm{2}\left(\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{37}\sqrt{\mathrm{231}}}{\mathrm{800}}−\frac{\mathrm{37}\sqrt{\mathrm{231}}}{\mathrm{800}}\right) \\ $$$$\approx\mathrm{0}.\mathrm{367} \\ $$

Commented by TawaTawa last updated on 30/Aug/19

Wow, God bless you sir.

$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by TawaTawa last updated on 30/Aug/19

I appreciate your time sir.

$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}. \\ $$

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.

$$\mathrm{Sir},\:\mathrm{i}\:\mathrm{checked}\:\mathrm{the}\:\mathrm{workings},\:\mathrm{i}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{rest}\:\mathrm{but}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand} \\ $$$$\mathrm{how}\:\mathrm{you}\:\mathrm{got}\:\mathrm{equation}\:\left(\mathrm{i}\right)\:\mathrm{and}\:\left(\mathrm{ii}\right).\:\mathrm{But}\:\mathrm{i}\:\mathrm{get}\:\mathrm{the}\:\mathrm{rest}.\:\:\mathrm{Can}\:\mathrm{you}\:\mathrm{let}\: \\ $$$$\mathrm{me}\:\mathrm{understand}\:\left(\mathrm{i}\right)\:\mathrm{and}\:\left(\mathrm{ii}\right).\:\:\mathrm{Sorry}\:\mathrm{for}\:\mathrm{disturbing}\:\mathrm{too}\:\mathrm{much}.\:\mathrm{And} \\ $$$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{helping}\:\mathrm{always}.\:\mathrm{God}\:\mathrm{will}\:\mathrm{reward}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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

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

BD=2×GD GD=ED×sin (α/2)=1×sin (α/2) GD=FD×sin (β/2)=2×sin (β/2) ⇒sin (α/2)=2 sin (β/2) EF=r_1 +r_2 −AC=1+2−(1/2)=(5/2) EF=EG+GF=ED×cos (α/2)+FD×cos (β/2) =1×cos (α/2)+2×cos (β/2)=(5/2) ⇒cos (α/2)=(5/2)−2 cos (β/2)

$${BD}=\mathrm{2}×{GD} \\ $$$${GD}={ED}×\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}=\mathrm{1}×\mathrm{sin}\:\frac{\alpha}{\mathrm{2}} \\ $$$${GD}={FD}×\mathrm{sin}\:\frac{\beta}{\mathrm{2}}=\mathrm{2}×\mathrm{sin}\:\frac{\beta}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}=\mathrm{2}\:\mathrm{sin}\:\frac{\beta}{\mathrm{2}} \\ $$$${EF}={r}_{\mathrm{1}} +{r}_{\mathrm{2}} −{AC}=\mathrm{1}+\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${EF}={EG}+{GF}={ED}×\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}+{FD}×\mathrm{cos}\:\frac{\beta}{\mathrm{2}} \\ $$$$=\mathrm{1}×\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}+\mathrm{2}×\mathrm{cos}\:\frac{\beta}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}}−\mathrm{2}\:\mathrm{cos}\:\frac{\beta}{\mathrm{2}} \\ $$

Commented by TawaTawa last updated on 31/Aug/19

I appreciate sir. God bless you more

$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{more} \\ $$

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