Question Number 25718 by ajfour last updated on 13/Dec/17
Commented by mrW1 last updated on 13/Dec/17
$${Case}\:\mathrm{1}:\:\theta=\frac{\pi}{\mathrm{2}} \\ $$$${h}_{{i}} =\sqrt{{h}^{\mathrm{2}} +{R}^{\mathrm{2}} } \\ $$$${A}_{{i}} =\frac{\mathrm{2}{R}×{h}_{{i}} }{\mathrm{2}}={R}\sqrt{{R}^{\mathrm{2}} +{h}^{\mathrm{2}} } \\ $$$$ \\ $$$${Case}\:\mathrm{2}:\:\theta=\mathrm{cos}^{−\mathrm{1}} \left[\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{h}}{{R}}\right)^{\mathrm{2}} −\mathrm{1}\right] \\ $$$$… \\ $$
Commented by mrW1 last updated on 13/Dec/17
$${AD}=\mathrm{2}{R}\mathrm{sin}\:\frac{\theta}{\mathrm{2}} \\ $$$${AC}^{\mathrm{2}} ={CD}^{\mathrm{2}} +{AD}^{\mathrm{2}} \\ $$$$\mathrm{4}{R}^{\mathrm{2}} ={h}^{\mathrm{2}} +\mathrm{4}{R}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}} \\ $$$$\mathrm{4}{R}^{\mathrm{2}} ={h}^{\mathrm{2}} +\mathrm{2}{R}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$\mathrm{2}{R}^{\mathrm{2}} ={h}^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} \mathrm{cos}\:\theta \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{h}}{{R}}\right)^{\mathrm{2}} −\mathrm{1} \\ $$$$ \\ $$$${AB}^{\mathrm{2}} ={h}^{\mathrm{2}} +\mathrm{2}{R}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\theta\right) \\ $$$$={h}^{\mathrm{2}} +\mathrm{2}{R}^{\mathrm{2}} ×\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{h}}{{R}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{2}{h}^{\mathrm{2}} \\ $$$${AB}=\sqrt{\mathrm{2}}{h} \\ $$$${h}_{{i}} =\sqrt{\left(\mathrm{2}{R}\right)^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{2}}{h}}{\mathrm{2}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{4}{R}^{\mathrm{2}} −\frac{{h}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$${A}_{{i}} =\frac{\mathrm{1}}{\mathrm{2}}×\sqrt{\mathrm{2}}{h}×\sqrt{\mathrm{4}{R}^{\mathrm{2}} −\frac{{h}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$=\frac{{h}}{\mathrm{2}}\sqrt{\mathrm{8}{R}^{\mathrm{2}} −{h}^{\mathrm{2}} } \\ $$
Commented by mrW1 last updated on 13/Dec/17
$${There}\:{are}\:{always}\:\mathrm{2}\:{possiblities}: \\ $$$$\theta=\frac{\pi}{\mathrm{2}}\:{or}\:\mathrm{cos}^{−\mathrm{1}} \left[\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{h}}{{R}}\right)^{\mathrm{2}} −\mathrm{1}\right]. \\ $$
Commented by ajfour last updated on 13/Dec/17
$${yes}\:{sir},\:{i}\:{understand}.\:{I}\:{should} \\ $$$${have}\:{mentioned}\:{AB}={AC}\:. \\ $$
Commented by ajfour last updated on 13/Dec/17
$${Further}\:{in}\:{what}\:{volume}\:{ratio}\:{does} \\ $$$${a}\:{plane}\:\left({coinciding}\:{with}\:{plane}\:{of}\right. \\ $$$$\left.{shown}\:{triangle}\right)\:{divide}\:{the} \\ $$$${cylindrical}\:{volume}\:{into}? \\ $$
Commented by mrW1 last updated on 13/Dec/17
Commented by ajfour last updated on 13/Dec/17
$$\theta=\mathrm{90}°\:{if}\:\bigtriangleup{ABC}\:{is}\:{isosceles}. \\ $$
Commented by ajfour last updated on 14/Dec/17
$${Thank}\:{you}\:{Sir}. \\ $$