Question-208021

Question Number 208021 by efronzo1 last updated on 02/Jun/24

Answered by A5T last updated on 02/Jun/24

((CD)/(CD+r))=(r/(r+AB))⇒((r+AB)/(r+CD))=(r/(CD)) ((r+AB)/(8+r))=((r+CD)/(3+r))⇒((r+AB)/(r+CD))=((8+r)/(3+r))=(r/(CD)) ⇒CD=((r(3+r))/(8+r)) ((AB)/(AB+r))=(r/(CD+r))=AB(CD+r)=r(AB)+r^2 AB=(r^2 /(CD))=((r(8+r))/(3+r)) (r+AB)^2 +(r+CD)^2 =(11+2r)^2 ⇒(r+AB)^2 =(11+2r)^2 −r^2 (1+((3+r)/(8+r)))^2 AB=(√((11+2r)^2 −r^2 (1+((3+r)/(8+r)))^2 ))−r=((r(8+r))/(3+r)) ⇒r=12⇒Diameter=24

$$\frac{{CD}}{{CD}+{r}}=\frac{{r}}{{r}+{AB}}\Rightarrow\frac{{r}+{AB}}{{r}+{CD}}=\frac{{r}}{{CD}} \\ $$$$\frac{{r}+{AB}}{\mathrm{8}+{r}}=\frac{{r}+{CD}}{\mathrm{3}+{r}}\Rightarrow\frac{{r}+{AB}}{{r}+{CD}}=\frac{\mathrm{8}+{r}}{\mathrm{3}+{r}}=\frac{{r}}{{CD}} \\ $$$$\Rightarrow{CD}=\frac{{r}\left(\mathrm{3}+{r}\right)}{\mathrm{8}+{r}} \\ $$$$\frac{{AB}}{{AB}+{r}}=\frac{{r}}{{CD}+{r}}={AB}\left({CD}+{r}\right)={r}\left({AB}\right)+{r}^{\mathrm{2}} \\ $$$${AB}=\frac{{r}^{\mathrm{2}} }{{CD}}=\frac{{r}\left(\mathrm{8}+{r}\right)}{\mathrm{3}+{r}} \\ $$$$\left({r}+{AB}\right)^{\mathrm{2}} +\left({r}+{CD}\right)^{\mathrm{2}} =\left(\mathrm{11}+\mathrm{2}{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left({r}+{AB}\right)^{\mathrm{2}} =\left(\mathrm{11}+\mathrm{2}{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{3}+{r}}{\mathrm{8}+{r}}\right)^{\mathrm{2}} \\ $$$${AB}=\sqrt{\left(\mathrm{11}+\mathrm{2}{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{3}+{r}}{\mathrm{8}+{r}}\right)^{\mathrm{2}} }−{r}=\frac{{r}\left(\mathrm{8}+{r}\right)}{\mathrm{3}+{r}} \\ $$$$\Rightarrow{r}=\mathrm{12}\Rightarrow{Diameter}=\mathrm{24} \\ $$

Answered by A5T last updated on 02/Jun/24

$${AB}=\sqrt{\mathrm{8}\left(\mathrm{8}+\mathrm{2}{r}\right)}=\frac{{r}\left(\mathrm{8}+{r}\right)}{\mathrm{3}+{r}}\Rightarrow{r}=\mathrm{12} \\ $$

Answered by A5T last updated on 02/Jun/24

(r/(r+AB))=((3+r)/(11+2r))⇒11r+2r^2 =r(3+r)+3AB+rAB AB=((r(8+r))/(3+r)) (8+r)^2 =r^2 +((r^2 (8+r)^2 )/((3+r)^2 ))⇒r=12

$$\frac{{r}}{{r}+{AB}}=\frac{\mathrm{3}+{r}}{\mathrm{11}+\mathrm{2}{r}}\Rightarrow\mathrm{11}{r}+\mathrm{2}{r}^{\mathrm{2}} ={r}\left(\mathrm{3}+{r}\right)+\mathrm{3}{AB}+{rAB} \\ $$$${AB}=\frac{{r}\left(\mathrm{8}+{r}\right)}{\mathrm{3}+{r}} \\ $$$$\left(\mathrm{8}+{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} +\frac{{r}^{\mathrm{2}} \left(\mathrm{8}+{r}\right)^{\mathrm{2}} }{\left(\mathrm{3}+{r}\right)^{\mathrm{2}} }\Rightarrow{r}=\mathrm{12} \\ $$

Answered by efronzo1 last updated on 02/Jun/24

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