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