Question Number 196304 by cortano12 last updated on 22/Aug/23
Answered by a.lgnaoui last updated on 24/Aug/23
![S=shaded Area S1=Arc(AMC) S2=Arc(OBD) S=S(ABCD)−S1+S2 •Calcul de S(ABCD) OBsin 30=OAsin 45⇒ OA=5(√2) S(ABCD)=AB×AC { ((AB=OBcos 30−OAcos 45)=5((√3) −1))),((AC=2×AM=2×OAsin 45=10)) :} ⇒S(ABCD)=50((√3) −1) •calcul de S1 S1=Area[Arc(OAC)]−[area(OAC)] =π((OA^2 )/4)−OA^2 cos 45sin 45 =((25π)/2)−((50)/2)=((25π−50)/2) •calcul de S2 S2=Area[arc(OBD)]−Area(OBD) =π((OB^2 )/6)−OB^2 cos 30sin 30 =((50π)/3)−((50(√3))/2)= ((100π−150(√3))/6) alors Shaded Area=50((√3) −1)−((25π)/2)+25 + ((100π−150(√3))/6) =(((50)/3)−((25)/2))π+(50−25)(√3) −25 =((25π)/6)+25(√3) −25 Shaded Area= 31,39 cm^2](https://www.tinkutara.com/question/Q196372.png)
$$\:\boldsymbol{\mathrm{S}}=\mathrm{shaded}\:\mathrm{Area} \\ $$$$\boldsymbol{\mathrm{S}}\mathrm{1}=\mathrm{Arc}\left(\mathrm{AMC}\right)\:\:\:\boldsymbol{\mathrm{S}}\mathrm{2}=\mathrm{Arc}\left(\mathrm{OBD}\right) \\ $$$$\:\boldsymbol{\mathrm{S}}=\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{ABCD}}\right)−\boldsymbol{\mathrm{S}}\mathrm{1}+\boldsymbol{\mathrm{S}}\mathrm{2} \\ $$$$\bullet\boldsymbol{\mathrm{Calcul}}\:\boldsymbol{\mathrm{de}}\:\boldsymbol{\mathrm{S}}\left(\mathrm{ABCD}\right) \\ $$$$\:\:\mathrm{OBsin}\:\mathrm{30}=\mathrm{OAsin}\:\mathrm{45}\Rightarrow\:\:\boldsymbol{\mathrm{OA}}=\mathrm{5}\sqrt{\mathrm{2}}\: \\ $$$$\:\:\boldsymbol{\mathrm{S}}\left(\mathrm{ABCD}\right)=\mathrm{AB}×\mathrm{AC} \\ $$$$\begin{cases}{\left.\mathrm{AB}=\mathrm{OBcos}\:\mathrm{30}−\mathrm{OAcos}\:\mathrm{45}\right)=\mathrm{5}\left(\sqrt{\mathrm{3}}\:−\mathrm{1}\right)}\\{\mathrm{AC}=\mathrm{2}×\mathrm{AM}=\mathrm{2}×\mathrm{OAsin}\:\:\mathrm{45}=\mathrm{10}}\end{cases} \\ $$$$\Rightarrow\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{ABCD}}\right)=\mathrm{50}\left(\sqrt{\mathrm{3}}\:−\mathrm{1}\right) \\ $$$$ \\ $$$$\bullet\mathrm{calcul}\:\mathrm{de}\:\boldsymbol{\mathrm{S}}\mathrm{1} \\ $$$$\:\boldsymbol{\mathrm{S}}\mathrm{1}=\mathrm{Area}\left[\mathrm{Arc}\left(\mathrm{OAC}\right)\right]−\left[\mathrm{area}\left(\mathrm{OAC}\right)\right] \\ $$$$\:\:\:=\pi\frac{\mathrm{OA}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{OA}^{\mathrm{2}} \mathrm{cos}\:\mathrm{45sin}\:\mathrm{45} \\ $$$$\:\:=\frac{\mathrm{25}\pi}{\mathrm{2}}−\frac{\mathrm{50}}{\mathrm{2}}=\frac{\mathrm{25}\pi−\mathrm{50}}{\mathrm{2}} \\ $$$$ \\ $$$$\bullet\mathrm{calcul}\:\mathrm{de}\:\boldsymbol{\mathrm{S}}\mathrm{2} \\ $$$$\:\boldsymbol{\mathrm{S}}\mathrm{2}=\mathrm{Area}\left[\mathrm{arc}\left(\mathrm{OBD}\right)\right]−\mathrm{Area}\left(\mathrm{OBD}\right) \\ $$$$\:\:\:=\pi\frac{\mathrm{OB}^{\mathrm{2}} }{\mathrm{6}}−\mathrm{OB}^{\mathrm{2}} \mathrm{cos}\:\mathrm{30sin}\:\mathrm{30} \\ $$$$\:\:\:=\frac{\mathrm{50}\pi}{\mathrm{3}}−\frac{\mathrm{50}\sqrt{\mathrm{3}}}{\mathrm{2}}=\:\:\frac{\mathrm{100}\pi−\mathrm{150}\sqrt{\mathrm{3}}}{\mathrm{6}} \\ $$$$ \\ $$$$\mathrm{alors}\:\: \\ $$$$\:\:\mathrm{Shaded}\:\mathrm{Area}=\mathrm{50}\left(\sqrt{\mathrm{3}}\:−\mathrm{1}\right)−\frac{\mathrm{25}\pi}{\mathrm{2}}+\mathrm{25} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\frac{\mathrm{100}\pi−\mathrm{150}\sqrt{\mathrm{3}}}{\mathrm{6}} \\ $$$$=\left(\frac{\mathrm{50}}{\mathrm{3}}−\frac{\mathrm{25}}{\mathrm{2}}\right)\pi+\left(\mathrm{50}−\mathrm{25}\right)\sqrt{\mathrm{3}}\:−\mathrm{25} \\ $$$$ \\ $$$$=\frac{\mathrm{25}\pi}{\mathrm{6}}+\mathrm{25}\sqrt{\mathrm{3}}\:−\mathrm{25} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{Shaded}}\:\boldsymbol{\mathrm{Area}}=\:\mathrm{31},\mathrm{39}\:\mathrm{cm}^{\mathrm{2}} \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 24/Aug/23
$$\mathrm{look}\:\mathrm{at}\:\mathrm{rectification}\:\:\left(\mathrm{error}\:\mathrm{calcul}\right) \\ $$
Commented by a.lgnaoui last updated on 23/Aug/23
Commented by cortano12 last updated on 24/Aug/23
$$\mathrm{no}\:\mathrm{sir} \\ $$
Commented by cortano12 last updated on 24/Aug/23
$$\mathrm{correct}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 23/Aug/23
![METHOD I r sin α=R sin β ⇒r=((R sin β)/(sin α)) AB=R cos β−r cos α A_(shade) =β(R^2 −r^2 )+(R cos β−r cos α)R sin β−(α−β)r^2 A_(shade) =R^2 [β(1−((sin^2 β)/(sin^2 α)))+sin β (cos β−((sin β)/(tan α)))−(α−β)(( sin^2 β)/(sin^2 α))] A_(shade) =10^2 ×((6((√3)−1)+π)/(24))≈31.391](https://www.tinkutara.com/question/Q196378.png)
$${METHOD}\:{I} \\ $$$${r}\:\mathrm{sin}\:\alpha={R}\:\mathrm{sin}\:\beta \\ $$$$\Rightarrow{r}=\frac{{R}\:\mathrm{sin}\:\beta}{\mathrm{sin}\:\alpha} \\ $$$${AB}={R}\:\mathrm{cos}\:\beta−{r}\:\mathrm{cos}\:\alpha \\ $$$${A}_{{shade}} =\beta\left({R}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)+\left({R}\:\mathrm{cos}\:\beta−{r}\:\mathrm{cos}\:\alpha\right){R}\:\mathrm{sin}\:\beta−\left(\alpha−\beta\right){r}^{\mathrm{2}} \\ $$$${A}_{{shade}} ={R}^{\mathrm{2}} \left[\beta\left(\mathrm{1}−\frac{\mathrm{sin}^{\mathrm{2}} \:\beta}{\mathrm{sin}^{\mathrm{2}} \:\alpha}\right)+\mathrm{sin}\:\beta\:\left(\mathrm{cos}\:\beta−\frac{\mathrm{sin}\:\beta}{\mathrm{tan}\:\alpha}\right)−\left(\alpha−\beta\right)\frac{\:\mathrm{sin}^{\mathrm{2}} \:\beta}{\mathrm{sin}^{\mathrm{2}} \:\alpha}\right] \\ $$$${A}_{{shade}} =\mathrm{10}^{\mathrm{2}} ×\frac{\mathrm{6}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)+\pi}{\mathrm{24}}\approx\mathrm{31}.\mathrm{391} \\ $$
Commented by cortano12 last updated on 24/Aug/23
$$\mathrm{nice}\:\mathrm{solution} \\ $$
Answered by mr W last updated on 23/Aug/23
![METHOD II r=((R sin β)/(sin α)) A_(shade) =2∫_0 ^(R sin β) ((√(R^2 −x^2 ))−(√(r^2 −x^2 )))dx =[R^2 sin^(−1) (x/R)−r^2 sin^(−1) (x/r)+x(√(R^2 −x^2 ))−x(√(r^2 −x^2 ))]_0 ^(R sin β) =R^2 [β−α((sin^2 β)/(sin^2 α))+sin β cos β−((sin^2 β)/(tan α))] =10^2 ×[((6((√3)−1)+π)/(24))]≈31.391](https://www.tinkutara.com/question/Q196379.png)
$${METHOD}\:{II} \\ $$$${r}=\frac{{R}\:\mathrm{sin}\:\beta}{\mathrm{sin}\:\alpha} \\ $$$${A}_{{shade}} =\mathrm{2}\int_{\mathrm{0}} ^{{R}\:\mathrm{sin}\:\beta} \left(\sqrt{{R}^{\mathrm{2}} −{x}^{\mathrm{2}} }−\sqrt{{r}^{\mathrm{2}} −{x}^{\mathrm{2}} }\right){dx} \\ $$$$\:\:=\left[{R}^{\mathrm{2}} \mathrm{sin}^{−\mathrm{1}} \frac{{x}}{{R}}−{r}^{\mathrm{2}} \mathrm{sin}^{−\mathrm{1}} \frac{{x}}{{r}}+{x}\sqrt{{R}^{\mathrm{2}} −{x}^{\mathrm{2}} }−{x}\sqrt{{r}^{\mathrm{2}} −{x}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{{R}\:\mathrm{sin}\:\beta} \\ $$$$\:\:={R}^{\mathrm{2}} \left[\beta−\alpha\frac{\mathrm{sin}^{\mathrm{2}} \:\beta}{\mathrm{sin}^{\mathrm{2}} \:\alpha}+\mathrm{sin}\:\beta\:\mathrm{cos}\:\beta−\frac{\mathrm{sin}^{\mathrm{2}} \:\beta}{\mathrm{tan}\:\alpha}\right] \\ $$$$\:\:=\mathrm{10}^{\mathrm{2}} ×\left[\frac{\mathrm{6}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)+\pi}{\mathrm{24}}\right]\approx\mathrm{31}.\mathrm{391} \\ $$
Commented by mr W last updated on 23/Aug/23
