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Question-25718




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

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