Question

the radius and the height of a cone are in the ratio 4:3, the volume of cone is 269 1/2. find its curve surface area

Answer 1

Dear student
$$\begin{gathered} Letradius=4x \\ andheight=3x \\ Volumeofcone=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h} \\ \mathrm{Also},\mathrm{Volume}\mathrm{of}\mathrm{cone}=269\frac{1}{2}\mathrm{unit}\mathrm{cube} \\ \frac{539}{2}=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h} \\ \frac{539}{2}=\frac{1}{3}\times \frac{22}{7}\times {\left(4\mathrm{x}\right)}^2\times 3\mathrm{x} \\ \frac{539\times 21}{2\times 22}=48{\mathrm{x}}^3 \\ \frac{343}{64}={\mathrm{x}}^3 \\ \mathrm{So},\mathrm{x}=\frac{7}{4}\mathrm{unit} \\ \mathrm{So},\mathrm{Radius}=4\mathrm{x}=4\times \frac{7}{4}=7\mathrm{unit} \\ \mathrm{height}=3\mathrm{x}=3\times \frac{7}{4}=\frac{21}{4}\mathrm{unit}. \\ \mathrm{Now},\mathrm{slant}\mathrm{heigth}\left(\mathrm{l}\right)=\sqrt{{\mathrm{h}}^2+{\mathrm{r}}^2}=\sqrt{{\left(\frac{21}{4}\right)}^2+7^2} \\ =\sqrt{\frac{441}{16}+49} \\ =\frac{35}{4}\mathrm{unit}. \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{curve}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{cone}=\mathrm{\pi rl} \\ =\frac{22}{7}\times 7\times \frac{35}{4} \\ =\frac{385}{2}\mathrm{unit}\mathrm{cube}. \end{gathered}$$
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