Question-51526

Question Number 51526 by ajfour last updated on 27/Dec/18

Commented by ajfour last updated on 27/Dec/18

$${For}\:{the}\:{area}\:{in}\:{black}\:{and}\:{yellow} \\ $$$${to}\:{be}\:{the}\:{same},\:{find}\:{r}\:{in}\:{terms}\:{of}\:{R}. \\ $$$$ \\ $$

Answered by mr W last updated on 28/Dec/18

cos θ=((R−r)/(R+r))=((1−(r/R))/(1+(r/R)))=((1−λ)/(1+λ)) sin θ=((√((1+λ)^2 −(1−λ)^2 ))/(1+λ))=((2(√λ))/(1+λ)) A_(black) =(((R+r)(R+r) sin θ)/2)−((R^2 θ)/2)−(r^2 /2)(π−θ) A_(black) =(((R+r)^2 sin θ)/2)−(((R^2 −r^2 )θ)/2)−((πr^2 )/2)=πr^2 (R+r)^2 sin θ−(R^2 −r^2 )θ=3πr^2 ⇒2(1+λ)(√λ)−(1−λ^2 ) cos^(−1) ((1−λ)/(1+λ))=3πλ^2 ⇒λ=0.0886 ⇒r=0.0886R

$$\mathrm{cos}\:\theta=\frac{{R}−{r}}{{R}+{r}}=\frac{\mathrm{1}−\frac{{r}}{{R}}}{\mathrm{1}+\frac{{r}}{{R}}}=\frac{\mathrm{1}−\lambda}{\mathrm{1}+\lambda} \\ $$$$\mathrm{sin}\:\theta=\frac{\sqrt{\left(\mathrm{1}+\lambda\right)^{\mathrm{2}} −\left(\mathrm{1}−\lambda\right)^{\mathrm{2}} }}{\mathrm{1}+\lambda}=\frac{\mathrm{2}\sqrt{\lambda}}{\mathrm{1}+\lambda} \\ $$$$ \\ $$$${A}_{{black}} =\frac{\left({R}+{r}\right)\left({R}+{r}\right)\:\mathrm{sin}\:\theta}{\mathrm{2}}−\frac{{R}^{\mathrm{2}} \theta}{\mathrm{2}}−\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\pi−\theta\right) \\ $$$${A}_{{black}} =\frac{\left({R}+{r}\right)^{\mathrm{2}} \:\mathrm{sin}\:\theta}{\mathrm{2}}−\frac{\left({R}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)\theta}{\mathrm{2}}−\frac{\pi{r}^{\mathrm{2}} }{\mathrm{2}}=\pi{r}^{\mathrm{2}} \\ $$$$\left({R}+{r}\right)^{\mathrm{2}} \:\mathrm{sin}\:\theta−\left({R}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)\theta=\mathrm{3}\pi{r}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}\left(\mathrm{1}+\lambda\right)\sqrt{\lambda}−\left(\mathrm{1}−\lambda^{\mathrm{2}} \right)\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}−\lambda}{\mathrm{1}+\lambda}=\mathrm{3}\pi\lambda^{\mathrm{2}} \\ $$$$\Rightarrow\lambda=\mathrm{0}.\mathrm{0886} \\ $$$$\Rightarrow{r}=\mathrm{0}.\mathrm{0886}{R} \\ $$

Commented by mr W last updated on 28/Dec/18