Question Number 220320 by Tawa11 last updated on 11/May/25
Commented by fantastic last updated on 11/May/25
$${is}\:{o}_{\mathrm{1}\:} {amd}\:{o}_{\mathrm{2}} \:{are}\:{the}\:{centers}\:{of}\:{the}\:{big}\:{and}\:{small}\:{coicles}\:{respectivly}??? \\ $$
Commented by mr W last updated on 11/May/25
$${the}\:{diagram}\:{is}\:{terrible}!\: \\ $$$${a}\:{good}\:{diagram}\:{is}\:{the}\:{base}\:{for}\:{a}\:{good} \\ $$$${solution}. \\ $$
Commented by fantastic last updated on 11/May/25
$${yeah}\:{i}\:{cam}\:{not}\:{understand} \\ $$
Answered by mr W last updated on 11/May/25
Commented by mr W last updated on 11/May/25
![AB:BC:CD=3:7:2 ⇒AB=3k, BC=7k, CD=2k O_1 A=O_1 D=R=((3k+7k+2k)/2)=6k O_1 B=R−AB=6k−3k=3k O_1 C=R−CD=6k−2k=4k O_1 O_2 =R−r=6k−r 3kr^2 +4kr^2 =7k[(6k−r)^2 +3k×4k] ⇒r=4k ⇒ratio of radii =(R/r)=((6k)/(4k))=3:2](https://www.tinkutara.com/question/Q220329.png)
$${AB}:{BC}:{CD}=\mathrm{3}:\mathrm{7}:\mathrm{2} \\ $$$$\Rightarrow{AB}=\mathrm{3}{k},\:{BC}=\mathrm{7}{k},\:{CD}=\mathrm{2}{k} \\ $$$${O}_{\mathrm{1}} {A}={O}_{\mathrm{1}} {D}={R}=\frac{\mathrm{3}{k}+\mathrm{7}{k}+\mathrm{2}{k}}{\mathrm{2}}=\mathrm{6}{k} \\ $$$${O}_{\mathrm{1}} {B}={R}−{AB}=\mathrm{6}{k}−\mathrm{3}{k}=\mathrm{3}{k} \\ $$$${O}_{\mathrm{1}} {C}={R}−{CD}=\mathrm{6}{k}−\mathrm{2}{k}=\mathrm{4}{k} \\ $$$${O}_{\mathrm{1}} {O}_{\mathrm{2}} ={R}−{r}=\mathrm{6}{k}−{r} \\ $$$$\mathrm{3}{kr}^{\mathrm{2}} +\mathrm{4}{kr}^{\mathrm{2}} =\mathrm{7}{k}\left[\left(\mathrm{6}{k}−{r}\right)^{\mathrm{2}} +\mathrm{3}{k}×\mathrm{4}{k}\right] \\ $$$$\Rightarrow{r}=\mathrm{4}{k} \\ $$$$\Rightarrow{ratio}\:{of}\:{radii}\:=\frac{{R}}{{r}}=\frac{\mathrm{6}{k}}{\mathrm{4}{k}}=\mathrm{3}:\mathrm{2} \\ $$
Commented by Tawa11 last updated on 11/May/25
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$$$\mathrm{I}\:\mathrm{can}\:\mathrm{see}\:\mathrm{clearly}\:\mathrm{from}\:\mathrm{your}\:\mathrm{diagram}. \\ $$