Question-19964

Question Number 19964 by ajfour last updated on 19/Aug/17

Commented by ajfour last updated on 19/Aug/17

$${Inscribed}\:{in}\:{a}\:{right}\:{circular}\:{cone} \\ $$$${is}\:{a}\:{sphere}\:{whose}\:{surface}\:{area}\:{is} \\ $$$${equal}\:{to}\:{the}\:{area}\:{of}\:{the}\:{base}\:{if}\:{the} \\ $$$${cone}.\:{In}\:{what}\:{ratio}\:{is}\:{the}\:{lateral} \\ $$$${side}\:{of}\:{the}\:{cone}\:{divided}\:{by}\:{the}\:{line} \\ $$$${of}\:{tangency}\:{of}\:{the}\:{sphere}\:{and}\:{cone}. \\ $$

Answered by ajfour last updated on 20/Aug/17

4πr^2 =πR^2 ⇒ R=2r BT=OB=R=2r h=((√(AT^( 2) +r^2 ))+r)^2 =(AT+BT)^2 −OB^2 ⇒ AT^( 2) +r^2 +r^2 +2r(√(AT^( 2) +r^2 )) =AT^2 +4r^2 +4r(AT)−4r^2 ⇒ (√(AT^( 2) +r^2 )) =−r+2(AT) ⇒AT^( 2) +r^2 =r^2 +4(AT)^2 −4r(AT) ⇒ 3AT=4r or AT=((4r)/3) Hence ((AT)/(BT)) = ((4r/3)/(2r))=(2/3) .

$$\mathrm{4}\pi{r}^{\mathrm{2}} =\pi{R}^{\mathrm{2}} \:\:\:\:\Rightarrow\:\:\:{R}=\mathrm{2}{r} \\ $$$${BT}={OB}={R}=\mathrm{2}{r} \\ $$$${h}=\left(\sqrt{{AT}^{\:\:\mathrm{2}} +{r}^{\mathrm{2}} }+{r}\right)^{\mathrm{2}} =\left({AT}+{BT}\right)^{\mathrm{2}} −{OB}^{\mathrm{2}} \: \\ $$$$\Rightarrow\:{AT}^{\:\:\mathrm{2}} +{r}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{r}\sqrt{{AT}^{\:\:\mathrm{2}} +{r}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:={AT}^{\mathrm{2}} +\mathrm{4}{r}^{\mathrm{2}} +\mathrm{4}{r}\left({AT}\right)−\mathrm{4}{r}^{\mathrm{2}} \\ $$$$\Rightarrow\:\sqrt{{AT}^{\:\:\mathrm{2}} +{r}^{\mathrm{2}} }\:=−{r}+\mathrm{2}\left({AT}\right) \\ $$$$\Rightarrow{AT}^{\:\:\mathrm{2}} +{r}^{\mathrm{2}} ={r}^{\mathrm{2}} +\mathrm{4}\left({AT}\right)^{\mathrm{2}} −\mathrm{4}{r}\left({AT}\right) \\ $$$$\Rightarrow\:\mathrm{3}{AT}=\mathrm{4}{r}\:\:{or}\:\:\:{AT}=\frac{\mathrm{4}{r}}{\mathrm{3}} \\ $$$${Hence}\:\:\:\frac{{AT}}{{BT}}\:=\:\frac{\mathrm{4}{r}/\mathrm{3}}{\mathrm{2}{r}}=\frac{\mathrm{2}}{\mathrm{3}}\:\:. \\ $$

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