Question-137437

Question Number 137437 by I want to learn more last updated on 02/Apr/21

Answered by mr W last updated on 02/Apr/21

((r−2)/(r+2))=((6−r)/(6+r)) r^2 =12 r=2(√3)

$$\frac{{r}−\mathrm{2}}{{r}+\mathrm{2}}=\frac{\mathrm{6}−{r}}{\mathrm{6}+{r}} \\ $$$${r}^{\mathrm{2}} =\mathrm{12} \\ $$$${r}=\mathrm{2}\sqrt{\mathrm{3}} \\ $$

Commented by I want to learn more last updated on 02/Apr/21

Thanks sir. Please what rule did you use.

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{Please}\:\mathrm{what}\:\mathrm{rule}\:\mathrm{did}\:\mathrm{you}\:\mathrm{use}. \\ $$

Commented by I want to learn more last updated on 02/Apr/21

$$\mathrm{I}\:\mathrm{mean}\:\mathrm{how}\:\mathrm{you}\:\mathrm{place}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{sir}. \\ $$

Commented by nadovic last updated on 03/Apr/21

The circles are externally tangential and also have common intersecting tangents, so they are similar. ∴ the ratios of consecutive radii are equal. (r/2) = (6/r) r^2 = 12 r = 2(√3)

$$\mathrm{The}\:\mathrm{circles}\:\mathrm{are}\:\mathrm{externally}\:\mathrm{tangential}\: \\ $$$$\mathrm{and}\:\mathrm{also}\:\mathrm{have}\:\mathrm{common}\:\mathrm{intersecting}\:\: \\ $$$$\mathrm{tangents},\:\mathrm{so}\:\mathrm{they}\:\mathrm{are}\:\mathrm{similar}.\:\therefore\:\mathrm{the} \\ $$$$\mathrm{ratios}\:\mathrm{of}\:{consecutive}\:\mathrm{radii}\:\mathrm{are}\:\mathrm{equal}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{r}}{\mathrm{2}}\:=\:\frac{\mathrm{6}}{{r}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{r}^{\mathrm{2}} \:=\:\mathrm{12} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{r}\:\:\:=\:\mathrm{2}\sqrt{\mathrm{3}} \\ $$

Commented by mr W last updated on 03/Apr/21

Commented by I want to learn more last updated on 03/Apr/21

$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{understand}\:\mathrm{it}\:\mathrm{better}\:\mathrm{now}. \\ $$

Commented by I want to learn more last updated on 03/Apr/21

$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}. \\ $$

Answered by mnjuly1970 last updated on 03/Apr/21

2(√(2r)) +2(√(6r)) =(√((8+2r)^2 −16)) 2(√(2r)) +2(√(6r)) =(√(4r^2 +32r+48)) 2r+6r+2(√(12)) r=r^2 +8r+12 −2(√(12)) r+r^2 +12=0 r=(√(12)) =2(√3)

$$\mathrm{2}\sqrt{\mathrm{2}{r}}\:+\mathrm{2}\sqrt{\mathrm{6}{r}}\:=\sqrt{\left(\mathrm{8}+\mathrm{2}{r}\right)^{\mathrm{2}} −\mathrm{16}} \\ $$$$\mathrm{2}\sqrt{\mathrm{2}{r}}\:+\mathrm{2}\sqrt{\mathrm{6}{r}}\:=\sqrt{\mathrm{4}{r}^{\mathrm{2}} +\mathrm{32}{r}+\mathrm{48}} \\ $$$$\mathrm{2}{r}+\mathrm{6}{r}+\mathrm{2}\sqrt{\mathrm{12}}\:{r}={r}^{\mathrm{2}} +\mathrm{8}{r}+\mathrm{12} \\ $$$$−\mathrm{2}\sqrt{\mathrm{12}}\:{r}+{r}^{\mathrm{2}} +\mathrm{12}=\mathrm{0} \\ $$$$\:\:{r}=\sqrt{\mathrm{12}}\:=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\:\: \\ $$

Commented by I want to learn more last updated on 03/Apr/21

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

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