Frustum is a Latin word which means ‘piece cut off’. When a solid (generally a cone or a pyramid) is cut in such a manner that base of the solid and the plane cutting the solid are parallel to each other, part of solid which remains between the parallel cutting plane and the base is known as frustum of that solid. 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.

The volume and the surface area of frustum of cone can be calculated as follows:

In the cone given above, the frustum can be considered as the difference of two right circular cones.

Let the larger cone which has a height equal to \(h\) units, slant height as \(l\) units and radius as \(r\) units be named as cone 1 and the smaller right circular cone be named as cone 2 whose height is given as \(h’\) units, radius as \(r’\) units and the slant height as \(l’\) units.

The height of frustum is \(H\) units and its slant height is \(L\) units.

The volume of right circular cone 1 = \(\frac{1}{3}~ πr^2 h\)

Similarly, the volume of right circular cone 2 = \(\frac{1}{3}~ πr’^2 h’\)

Therefore, the volume of the frustum of cone can be given as:

Volume of frustum of cone \(V\) = \(\frac{1}{3}~ πr^2~ h~-~\frac{1}{3}~ πr’^2 h’\)

Volume of frustum \(V\) = \(\frac{1}{3}~ π(r^2 ~h~-~r’^2 h’)\) —(1)

From the figure given above, in the \(∆OO’D\) and \(∆OPB\);

\(∠DOO’\) = \(∠BOP\) (*Common Angle*)

As \(\overleftrightarrow{CD}\)||\(\overleftrightarrow{AB}\)(Plane dividing the cone is parallel to the base)

\(⇒~∠O’DO\) = \(∠PBO\) (*Corresponding Angles*)

Thus, \(∆OO’D\)~\(∆OPD\)(By *AA criterion of similarity*)

Hence, according to condition for similar triangles the ratio of corresponding sides must be equal:

\(⇒~\frac{h’}{h}\) = \(\frac{r’}{r}\) —(2)

Substituting this value in equation (1),

\(V\) = \(\frac{1}{3}~ π\left(r^2 \left(\frac{rh’}{r’}\right)~-~r’^2 h’\right)\)

\(⇒~V\) = \(\frac{1}{3}~ πh’~ \left(\frac{r^3~-~r’^3}{r’}\right)\) —(3)

As it can be seen from fig 3,

\(h\) = \(H~+~h’\)

Substituting this value in equation (2),

\(\frac{h’}{H~+~h’}\) = \(\frac{r’}{r}\)

\(⇒~\frac{h’}{H}\) = \(\frac{r’}{r~-~r’}\)

\(⇒~h’\) = \(H\left(\frac{r’}{r~-~r’}\right)\)

Substituting this value of \(h’\) in equation (3), we get

\(V\) = \(\frac{1}{3}~ π(H)~\left({r’}{r~-~r’}\right)\left({r^3~-~r’^3}{r’}\right)\)

\(⇒~V\) = \(\frac{1}{3}~ πH~(r^2~+~r’r~+~r’^2)\)

This gives the required volume of frustum of cone.

Similarly, the surface area will also be the difference of the surface areas of both the cones.

Curved surface area of right circular cone 1 = \(πrl\)

Curved surface area of right circular cone 2 = \(πr’l’\)

Curved surface area of the frustum of cone = \(πrl~-~πr’l’\)

Also, from fig. 3 ,in the \(∆OAB\) and \(∆OCD\);

\(∠AOB\) = \(∠COD\) (Common Angle)

As \(\overleftrightarrow{CD}\) || \(\overleftrightarrow{AB}\) (Plane dividing the cone is parallel to the base)

\(⇒~∠OCD\) = \(∠OAB\) (Corresponding Angles)

Thus, \(∆OAB\)~\(∆OCD\) (By AA criterion of similarity)

Lateral surface area of frustum of the cone is the difference of the areas of sector of circles (\(s\) and \(s’\)) with radii \(r\) and \(r’\) and common central angle \(θ\) as shown in the figure.

From fig. 3 as the \(∆OO’D\) ~ \(∆OPB\) therefore:

\(⇒~\frac{l’}{l}\) = \(\frac{r’}{r}\) —(4)

Also from the figure given above: \(l’\) = \(l~-~L\)

Substituting this value of \(l’\) in equations (4),

\(\frac{l~-~L}{l}\) = \(\frac{r’}{r}\)

\(⇒~l\) = \(L~\left({r}{r~-~r’}\right)\) —(5)

The circumference of the base is the length of arc \(S\) and \(S’\), which is given by:

\(S’\) = \(2πr’\)

\(S\) = \(2πr\)

From the figure,

Curved surface area of frustum, \(A\) = \(\frac{1}{2}~(Sl~-~S’ l’)\)

\(⇒~A\) = \(\frac{1}{2}~×~2πrl~-~\frac{1}{2}~×~2πr’\left({lr’}{r}\right)\)

\(⇒~A\) = \(πl~\left(r~-~\frac{r’^2}{r}\right)\)

\(⇒~A\) = \(πl~\left(\frac{r^2~-~r’^2}{r}\right)\)

From equation (5), substituting the value of \(l\),

\(⇒~A\) = \(πL~\left({r}{r~-~r’}\right)\left(\frac{r^2~-~r’^2}{r}\right)\)

\(⇒~A\) = \(πL~(r~+~r’)\)

Thus, the curved surface area of frustum of cone, \(A\) = \(πL~(r~+~r’)\)

Where, \(L\) = \(√{H^2~+~(r~-~r’)^2}\)

The total surface area is given as the sum of curved surface area and the area of the base. Thus the total surface area of the frustum is:

*Total surface area of frustum of cone* = \(πL~(r~+~r’)~+~πr^2~+~πr’^2\)

Thus, the problems based on frustum of solids and cones can be easily solved. Learning is an unending journey. All you need to learn is just a click away. Please visit our website www.byjus.com and download BYJU’s-The Learning App to achieve success.