Question-183393

Question Number 183393 by cherokeesay last updated on 25/Dec/22

Answered by mr W last updated on 25/Dec/22

Commented by mr W last updated on 25/Dec/22

(x/2)=(2/(2+2)) ⇒x=1 R=((1+2+2)/2)=(5/2) cos θ=(((R−r)^2 +(R−1)^2 −(r+1)^2 )/(2(R−r)(R−1))) cos θ=(((r+2)^2 −(R−1)^2 −(R−r)^2 )/(2(R−r)(R−1))) (r+2)^2 −(R−1)^2 −(R−r)^2 =(R−r)^2 +(R−1)^2 −(r+1)^2 9(r−(1/4))−(9/4)=7((3/4)−r)+(9/4) ⇒r=(3/4) S=((π×2^2 )/4)=π S_1 =πr^2 S_2 =((π×1^2 )/2)=(π/2) (S/(S_1 +S_2 ))=(π/(πr^2 +(π/2)))=(2/(1+2r^2 ))=((16)/(17)) ✓

$$\frac{{x}}{\mathrm{2}}=\frac{\mathrm{2}}{\mathrm{2}+\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{1} \\ $$$${R}=\frac{\mathrm{1}+\mathrm{2}+\mathrm{2}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\theta=\frac{\left({R}−{r}\right)^{\mathrm{2}} +\left({R}−\mathrm{1}\right)^{\mathrm{2}} −\left({r}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}\left({R}−{r}\right)\left({R}−\mathrm{1}\right)} \\ $$$$\mathrm{cos}\:\theta=\frac{\left({r}+\mathrm{2}\right)^{\mathrm{2}} −\left({R}−\mathrm{1}\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} }{\mathrm{2}\left({R}−{r}\right)\left({R}−\mathrm{1}\right)} \\ $$$$\left({r}+\mathrm{2}\right)^{\mathrm{2}} −\left({R}−\mathrm{1}\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} =\left({R}−{r}\right)^{\mathrm{2}} +\left({R}−\mathrm{1}\right)^{\mathrm{2}} −\left({r}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{9}\left({r}−\frac{\mathrm{1}}{\mathrm{4}}\right)−\frac{\mathrm{9}}{\mathrm{4}}=\mathrm{7}\left(\frac{\mathrm{3}}{\mathrm{4}}−{r}\right)+\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${S}=\frac{\pi×\mathrm{2}^{\mathrm{2}} }{\mathrm{4}}=\pi \\ $$$${S}_{\mathrm{1}} =\pi{r}^{\mathrm{2}} \\ $$$${S}_{\mathrm{2}} =\frac{\pi×\mathrm{1}^{\mathrm{2}} }{\mathrm{2}}=\frac{\pi}{\mathrm{2}} \\ $$$$\frac{{S}}{{S}_{\mathrm{1}} +{S}_{\mathrm{2}} }=\frac{\pi}{\pi{r}^{\mathrm{2}} +\frac{\pi}{\mathrm{2}}}=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{2}{r}^{\mathrm{2}} }=\frac{\mathrm{16}}{\mathrm{17}}\:\checkmark \\ $$

Commented by mr W last updated on 25/Dec/22