Question

The perimeters of the two circular ends of a frustum of a cone are 48 cm and 36 cm. If the height of the frustum is 11 cm, then find its volume and curved surface area. [CBSE 2014]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{Perimeter}\mathrm{of}\mathrm{upper}\mathrm{end},C=48\mathrm{cm}, \\ \mathrm{Perimeter}\mathrm{of}\mathrm{lower}\mathrm{end},c=36\mathrm{cm}\mathrm{and} \\ \mathrm{Height},h=11\mathrm{cm} \\ \mathrm{Let}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{upper}\mathrm{end}\mathrm{be}R\mathrm{and}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{lower}\mathrm{end}\mathrm{be}r. \\ \mathrm{As},C=48\mathrm{cm} \\ \Rightarrow 2\mathrm{\pi }R=48 \\ \Rightarrow R=\frac{48}{2\mathrm{\pi }} \\ \Rightarrow R=\frac{24}{\mathrm{\pi }}\mathrm{cm} \\ \mathrm{Similarly},\mathrm{c}=36\mathrm{cm} \\ \Rightarrow r=\frac{36}{2\mathrm{\pi }} \\ \Rightarrow r=\frac{18}{\mathrm{\pi }}\mathrm{cm} \\ \mathrm{And},l=\sqrt{{\left(R-r\right)}^2+h^2} \\ =\sqrt{{\left(\frac{24}{\mathrm{\pi }}-\frac{18}{\mathrm{\pi }}\right)}^2+{11}^2} \\ =\sqrt{{\left(\frac{6}{\mathrm{\pi }}\right)}^2+{11}^2} \\ =\sqrt{{\left(\frac{6\times 7}{22}\right)}^2+{11}^2} \\ =\sqrt{{\left(\frac{21}{11}\right)}^2+{11}^2} \\ =\sqrt{\frac{441+121\times 121}{121}} \\ =\sqrt{\frac{441+14641}{121}} \\ =\frac{\sqrt{15082}}{11}\mathrm{cm} \\ \mathrm{Now}, \\ \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{frustum}=\frac{1}{3}\mathrm{\pi }h\left(R^2+r^2+Rr\right) \\ =\frac{1}{3}\times \mathrm{\pi }\times 11\times \left[{\left(\frac{24}{\mathrm{\pi }}\right)}^2+{\left(\frac{18}{\mathrm{\pi }}\right)}^2+\left(\frac{24}{\mathrm{\pi }}\right)\times \left(\frac{18}{\mathrm{\pi }}\right)\right] \\ =\frac{11\mathrm{\pi }}{3}\times \left[\frac{576}{{\mathrm{\pi }}^2}+\frac{324}{{\mathrm{\pi }}^2}+\frac{432}{{\mathrm{\pi }}^2}\right] \\ =\frac{11\mathrm{\pi }}{3}\times \frac{1332}{{\mathrm{\pi }}^2} \\ =\frac{11}{3}\times \frac{1332}{\mathrm{\pi }} \\ =\frac{11}{3}\times \frac{1332\times 7}{22} \\ =1554{\mathrm{cm}}^3 \\ \mathrm{Also}, \\ \mathrm{Curved}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{frustum}=\mathrm{\pi }\left(R+r\right)l \\ =\frac{22}{7}\times \left(\frac{24}{\mathrm{\pi }}+\frac{18}{\mathrm{\pi }}\right)\times \frac{\sqrt{15082}}{11} \\ =\frac{22}{7}\times \frac{42}{\mathrm{\pi }}\times \frac{\sqrt{15082}}{11} \\ =\frac{22}{7}\times \frac{42\times 7}{22}\times \frac{\sqrt{15082}}{11} \\ \approx 42\times 11.164436 \\ \approx 468.91{\mathrm{cm}}^2 \end{gathered}$$