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Question-14073




Question Number 14073 by tawa tawa last updated on 27/May/17
Commented by ajfour last updated on 27/May/17
Commented by ajfour last updated on 27/May/17
Commented by ajfour last updated on 27/May/17
Q(1).  A=(1/2)(2x)y  y=rsec θ+r    =r(1+sec θ)  x=ycot θ    =r(1+sec θ)cot θ  A=xy=r^2 (1+sec^2 θ)cot θ .
$$\boldsymbol{{Q}}\left(\mathrm{1}\right). \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{x}\right){y} \\ $$$${y}={r}\mathrm{sec}\:\theta+{r} \\ $$$$\:\:={r}\left(\mathrm{1}+\mathrm{sec}\:\theta\right) \\ $$$${x}={y}\mathrm{cot}\:\theta \\ $$$$\:\:={r}\left(\mathrm{1}+\mathrm{sec}\:\theta\right)\mathrm{cot}\:\theta \\ $$$${A}={xy}={r}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{sec}\:^{\mathrm{2}} \theta\right)\mathrm{cot}\:\theta\:. \\ $$
Commented by tawa tawa last updated on 27/May/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by ajfour last updated on 27/May/17
Q(2).  h=rcosec θ+r  h=r(1+cosec θ)  R=htan θ =r(1+cosec θ)tan θ  V=(1/3)πR^2 h  V =(1/3)r^3 (1+cosec θ)^3 tan^2 θ .
$$\boldsymbol{{Q}}\left(\mathrm{2}\right). \\ $$$${h}={r}\mathrm{cosec}\:\theta+{r} \\ $$$${h}={r}\left(\mathrm{1}+\mathrm{cosec}\:\theta\right) \\ $$$${R}={h}\mathrm{tan}\:\theta\:={r}\left(\mathrm{1}+\mathrm{cosec}\:\theta\right)\mathrm{tan}\:\theta \\ $$$${V}=\frac{\mathrm{1}}{\mathrm{3}}\pi{R}^{\mathrm{2}} {h} \\ $$$${V}\:=\frac{\mathrm{1}}{\mathrm{3}}{r}^{\mathrm{3}} \left(\mathrm{1}+\mathrm{cosec}\:\theta\right)^{\mathrm{3}} \mathrm{tan}\:^{\mathrm{2}} \theta\:. \\ $$
Commented by tawa tawa last updated on 27/May/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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