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Question Number 25718 by ajfour last updated on 13/Dec/17

Commented by mrW1 last updated on 13/Dec/17

$$Case1:\theta =\frac{\pi }{2}$$$$h_i=\sqrt{h^2+R^2}$$$$A_i=\frac{2R\times h_i}{2}=R\sqrt{R^2+h^2}$$$$$$$$Case2:\theta ={\cos }^{-1}[\frac{1}{2}{\left(\frac{h}{R}\right)}^2-1]$$$$...$$

Commented by mrW1 last updated on 13/Dec/17

$$AD=2R\sin \frac{\theta }{2}$$$${AC}^2={CD}^2+{AD}^2$$$$4R^2=h^2+4R^2{\sin }^2\frac{\theta }{2}$$$$4R^2=h^2+2R^2(1-\cos \theta )$$$$2R^2=h^2-2R^2\cos \theta$$$$\Rightarrow \cos \theta =\frac{1}{2}{\left(\frac{h}{R}\right)}^2-1$$$$$$$${AB}^2=h^2+2R^2(1+\cos \theta )$$$$=h^2+2R^2\times \frac{1}{2}{\left(\frac{h}{R}\right)}^2$$$$=2h^2$$$$AB=\sqrt{2}h$$$$h_i=\sqrt{{\left(2R\right)}^2-{\left(\frac{\sqrt{2}h}{2}\right)}^2}=\sqrt{4R^2-\frac{h^2}{2}}$$$$A_i=\frac{1}{2}\times \sqrt{2}h\times \sqrt{4R^2-\frac{h^2}{2}}$$$$=\frac{h}{2}\sqrt{8R^2-h^2}$$

Commented by mrW1 last updated on 13/Dec/17

$$Therearealways2possiblities:$$$$\theta =\frac{\pi }{2}or{\cos }^{-1}[\frac{1}{2}{\left(\frac{h}{R}\right)}^2-1].$$

Commented by ajfour last updated on 13/Dec/17

$$yessir,iunderstand.Ishould$$$$havementionedAB=AC.$$

Commented by ajfour last updated on 13/Dec/17

$$Furtherinwhatvolumeratiodoes$$$$aplane(coincidingwithplaneof$$$$showntriangle)dividethe$$$$cylindricalvolumeinto?$$

Commented by mrW1 last updated on 13/Dec/17

Commented by ajfour last updated on 13/Dec/17

$$\theta =90°if△ABCisisosceles.$$

Commented by ajfour last updated on 14/Dec/17

$$ThankyouSir.$$

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