Question Number 224176 by fantastic last updated on 23/Aug/25
Commented by fantastic last updated on 23/Aug/25
$${whats}\:{the}\:{green}\:{area}? \\ $$$$ \\ $$$${please}\:{give}\:{me}\:{a}\:{detailed}\:{solution}\: \\ $$
Answered by mr W last updated on 23/Aug/25
Commented by mr W last updated on 23/Aug/25
$${R}=\mathrm{10} \\ $$$${r}=\frac{{R}}{\mathrm{2}} \\ $$$${OA}=\sqrt{\mathrm{2}}{r} \\ $$$${AC}={r} \\ $$$${OC}={R} \\ $$$$\mathrm{cos}\:\alpha=\frac{\left(\sqrt{\mathrm{2}}{r}\right)^{\mathrm{2}} +{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}{rR}}=\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{8}}\:\Rightarrow\mathrm{sin}\:\alpha=\frac{\sqrt{\mathrm{14}}}{\mathrm{8}} \\ $$$$\mathrm{cos}\:\beta=−\frac{\left(\sqrt{\mathrm{2}}{r}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} −{R}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}{r}^{\mathrm{2}} }=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$$\frac{{shaded}\:{area}}{\mathrm{2}}=\frac{\sqrt{\mathrm{2}}{rR}\:\mathrm{sin}\:\alpha}{\mathrm{2}}+\frac{{r}^{\mathrm{2}} \beta}{\mathrm{2}}−\frac{{R}^{\mathrm{2}} \alpha}{\mathrm{2}} \\ $$$${shaded}\:{area}=\left(\frac{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\alpha}{\mathrm{2}}+\frac{\beta}{\mathrm{4}}−\alpha\right){R}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\left(\frac{\sqrt{\mathrm{7}}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}}{\mathrm{4}}−\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{14}}}{\mathrm{8}}\right){R}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\approx\mathrm{14}.\mathrm{638} \\ $$
Commented by fantastic last updated on 24/Aug/25
$${Thanks}\:{sir} \\ $$
Commented by Ghisom_ last updated on 24/Aug/25
$$\mathrm{I}\:\mathrm{get}\:\mathrm{25}\left(\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}−\mathrm{arctan}\:\frac{\mathrm{23}\sqrt{\mathrm{7}}}{\mathrm{67}}\right)\:\mathrm{which}\:\mathrm{has}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{value} \\ $$
Commented by fantastic last updated on 24/Aug/25
Answered by Frix last updated on 06/Sep/25
![c_1 : x^2 +y^2 −100=0 c_2 : x^2 +y^2 −10x−10y+25=0 Rotate by 45° ⇒ x=((√2)/2)(u+v)∧y=((√2)/2)(v−u) c_1 : u^2 +v^2 −100=0 c_2 : u^2 +v^2 −10(√2)v+25=0 c_1 ∩c_2 = (((±((5(√(14)))/4))),(((25(√2))/4)) ) Upper half circles c_1 : v=(√(100−u^2 )) c_2 : v=5(√2)+(√(25−u^2 )) Area = 2∫_0 ^((5(√(14)))/4) (5(√2)+(√(25−u^2 ))−(√(100−u^2 )))du= =[10(√2)u+u(√(25−u^2 ))+25sin^(−1) (u/5) −(u(√(100−u^2 ))+100sin^(−1) (u/(10)))]_0 ^((5(√(14)))/4) = =((25(√7))/2)+25sin^(−1) ((√(14))/4) −100sin^(−1) ((√(14))/8) ≈14.6381259530 [=((25(√7))/2)−25sin^(−1) ((23(√(14)))/(128)) =((25(√7))/2)−25tan^(−1) ((23(√7))/(67))]](https://www.tinkutara.com/question/Q224383.png)
$${c}_{\mathrm{1}} :\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{100}=\mathrm{0} \\ $$$${c}_{\mathrm{2}} :\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{10}{x}−\mathrm{10}{y}+\mathrm{25}=\mathrm{0} \\ $$$$\mathrm{Rotate}\:\mathrm{by}\:\mathrm{45}°\:\Rightarrow\:{x}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left({u}+{v}\right)\wedge{y}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left({v}−{u}\right) \\ $$$${c}_{\mathrm{1}} :\:{u}^{\mathrm{2}} +{v}^{\mathrm{2}} −\mathrm{100}=\mathrm{0} \\ $$$${c}_{\mathrm{2}} :\:{u}^{\mathrm{2}} +{v}^{\mathrm{2}} −\mathrm{10}\sqrt{\mathrm{2}}{v}+\mathrm{25}=\mathrm{0} \\ $$$${c}_{\mathrm{1}} \cap{c}_{\mathrm{2}} =\begin{pmatrix}{\pm\frac{\mathrm{5}\sqrt{\mathrm{14}}}{\mathrm{4}}}\\{\frac{\mathrm{25}\sqrt{\mathrm{2}}}{\mathrm{4}}}\end{pmatrix} \\ $$$$\mathrm{Upper}\:\mathrm{half}\:\mathrm{circles} \\ $$$${c}_{\mathrm{1}} :\:{v}=\sqrt{\mathrm{100}−{u}^{\mathrm{2}} } \\ $$$${c}_{\mathrm{2}} :\:{v}=\mathrm{5}\sqrt{\mathrm{2}}+\sqrt{\mathrm{25}−{u}^{\mathrm{2}} } \\ $$$$\mathrm{Area}\:=\:\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\mathrm{5}\sqrt{\mathrm{14}}}{\mathrm{4}}} {\int}}\left(\mathrm{5}\sqrt{\mathrm{2}}+\sqrt{\mathrm{25}−{u}^{\mathrm{2}} }−\sqrt{\mathrm{100}−{u}^{\mathrm{2}} }\right){du}= \\ $$$$=\left[\mathrm{10}\sqrt{\mathrm{2}}{u}+{u}\sqrt{\mathrm{25}−{u}^{\mathrm{2}} }+\mathrm{25sin}^{−\mathrm{1}} \:\frac{{u}}{\mathrm{5}}\:−\left({u}\sqrt{\mathrm{100}−{u}^{\mathrm{2}} }+\mathrm{100sin}^{−\mathrm{1}} \:\frac{{u}}{\mathrm{10}}\right)\right]_{\mathrm{0}} ^{\frac{\mathrm{5}\sqrt{\mathrm{14}}}{\mathrm{4}}} = \\ $$$$=\frac{\mathrm{25}\sqrt{\mathrm{7}}}{\mathrm{2}}+\mathrm{25sin}^{−\mathrm{1}} \:\frac{\sqrt{\mathrm{14}}}{\mathrm{4}}\:−\mathrm{100sin}^{−\mathrm{1}} \:\frac{\sqrt{\mathrm{14}}}{\mathrm{8}} \\ $$$$\approx\mathrm{14}.\mathrm{6381259530} \\ $$$$ \\ $$$$\left[=\frac{\mathrm{25}\sqrt{\mathrm{7}}}{\mathrm{2}}−\mathrm{25sin}^{−\mathrm{1}} \:\frac{\mathrm{23}\sqrt{\mathrm{14}}}{\mathrm{128}}\:=\frac{\mathrm{25}\sqrt{\mathrm{7}}}{\mathrm{2}}−\mathrm{25tan}^{−\mathrm{1}} \:\frac{\mathrm{23}\sqrt{\mathrm{7}}}{\mathrm{67}}\right] \\ $$