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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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