Question

A bucket is in the form of a frustum of a cone with a capacity of \[12308.8\] cm$^3$ of water. The radii of the top and bottom circular ends are \[20\]cm and \[12\]cm respectively. Find the height of the bucket and the area of the metal sheet used in its making.

Answer 1

Hint: In geometry, a frustum is a portion of a solid (normally a cone or pyramid) that lies between one or two parallel planes cutting it. A right frustum is a parallel truncation of a right pyramid or right cone.
If all the edges are forced to be identical, a frustum becomes a uniform prism.
A frustum's axis is that of the original cone or pyramid. A frustum is circular if it has circular bases; it is right if the axis is perpendicular to both bases and oblique otherwise.
The height of a frustum is the perpendicular distance between the planes of the two bases.
\[{\text{Volume of Frustum of cone}} = \dfrac{1}{3}\pi h({r_1}^2 + {r_2}^2 + {r_1}{r_2})\]
\[{\text{Surface area of a frustum cone }} = {\text{ }}\pi l\left( {{r_1} + {r_2}} \right)\]
Where \[{{\text{r}}_1}\] and \[{{\text{r}}_2}\] are bigger and smaller radius respectively

Complete answer:

Given:
The bucket is in the form of the frustum of a cone.
Capacity of Frustum of a cone = \[12308.8c{m^3}\]
Bigger radius \[({r_1}) = 20cm\]
Smaller radius \[({r_2}) = 12cm\]
Now on putting the value of \[{{\text{r}}_1}\]and \[{{\text{r}}_2}\]and volume in the formula
We get
\[{\text{Volume of Frustum of cone}} = \dfrac{1}{3}\pi h({r_1}^2 + {r_2}^2 + {r_1}{r_2})\]
\[12308.8 = \dfrac{\pi }{3}h(20 \times 20 + 12 \times 12 + 20 \times 12)\]
\[12308.8 \times 3 = \pi h(400 + 144 + 240)\]
\[{\text{123}}0{\text{8}}.{\text{8}} \times {\text{3 }} = {\text{ }}\dfrac{{22}}{7}h\left( {{\text{784}}} \right)\]
\[h = \dfrac{{12308.8 \times 3 \times 7}}{{22 \times 784}}\]
\[h = \dfrac{{12308.0 \times 3}}{{22 \times 112}}\]
\[h{\text{ = }}\dfrac{{{\text{6,154}}{\text{.4}} \times {\text{3}}}}{{11 \times 12}}\]
\[h = \dfrac{{18,463.2}}{{1232}} = 14.99\]
\[h = 15\] (approximately)
Height of Frustum of cone \[h = 15cm\]
\[{\text{Slant height }}\left( l \right){\text{ of a frustum cone }} = \sqrt {{h^2} + {{\left( {{r^1} - {r^2}} \right)}^2}} \]
Now on putting the value of h and both radius in the formula we get
\[l = \sqrt {{{15}^2} + {{\left( {{\text{2}}0{\text{ }} - {\text{ 12}}} \right)}^2}} = \sqrt {225 + {{\left( {\text{8}} \right)}^2}} \]
\[l = \sqrt {225 + 64} = \sqrt {289} = 17\]
\[l = 17cm\]
\[{\text{Surface area of a frustum cone }} = {\text{ }}\pi l\left( {{r_1} + {r_2}} \right)\]
\[ = {\text{ }}\pi {\text{ }} \times {\text{ 17 }}\left( {{\text{2}}0 + {\text{12}}} \right)\]
\[ = \dfrac{{22}}{7} \times 17\left( {32} \right)\]
\[ = \dfrac{{11968}}{7} = 1709.7\]
\[ = {\text{ 17}}0{\text{9}}.{\text{7}}c{m^2}\]

Note: To visualize a frustum properly, consider an ice-cream cone which is completely filled with ice-cream. When the cone is cut in a manner as shown in the figure, the section left between the base and parallel plane is the frustum of a cone.