Frustum of Cone

When a right circular cone is cut by a plane such that it is parallel to the circular base of the cone, it is divided into two parts – one part with vertex and the other solid part is called the frustum of the cone.

So we can define the frustum of a cone as – If a right circular cone is cut off by a plane parallel to its base, the portion of the cone between the cutting plane and the base of the cone is called the frustum of the cone.

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In the above figure, the portion between the parallel plane and the circular base is called the frustum of the cone.

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From here you can see that a frustum of a cone has two unequal flat circular bases and a curved surface.

Now let us see how to calculate the volume of the frustum of a cone.

The Volume of Frustum of Cone

Some important terms related to the frustum of a cone  will be used in this article which are listed below: 

• Height: The perpendicular distance between the two circular bases is the height of the frustum of a cone. From the above figure, ‘h’ is the height of the frustum of a cone.

• Slant Height: The slant height of a frustum of a right circular cone is the line segment joining the extreme point of two parallel radii, drawn in the same direction, of the two circular bases.

Now let us derive the frustum formula for volume. The frustum of the cone formula will be very useful to find the volume of combined figures.

Formula to Calculate the Volume of Frustum of Cone

We can calculate the volume of frustum of a cone by subtracting the volume of the smaller cone from the volume of the larger cone.

The larger cone is considered as cone 1 and the smaller cone as cone 2.

Let us consider h as the height, l as the slant height, and r  as the radius of the larger cone 1.

And consider h’ as the height, l’ as the slant height, and r’  as the radius of the smaller cone 2.

The height of frustum be H and its slant height is L.

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Now, the volume of the frustum of a cone = Volume of right circular cone 1 – Volume of smaller cone 2.

The volume of right circular larger cone 1 = \[\frac {1}{3} \pi r^2 h\]

Volume of right circular smaller cone 2 = \[\frac {1}{3} \pi r'^2 h'\]

Therefore, the volume of the frustum of a cone V = \[\frac {1}{3} \pi r^2 h - \frac {1}{3} r'^2 h'\]

= \[\frac {1}{3} \pi (r^2 h - r'^2 h')\]  .........equation (1)

Now consider  

ΔOO’D and ΔOPB

∠DOO’ = ∠BOP…( common angles)

As CD ll AB

∠O’DO= ∠PBO ….(corresponding angles)

Therefore, ΔOO’D ~ ΔOPB ( by AA criteria).

Hence, according to similar triangles property, the ratio of corresponding sides must be equal.

We get \[\frac {h'}{h} = \frac {r'}{r}\]

h = \[\frac {h'r}{r'}\]……..equation (2)

Substituting the value of equation (2) in equation (1), we get, 

The volume of the frustum of a cone V=\[\frac {1}{3} \pi r^2 h - r'^2 h'^2\]

V=\[\frac{1} {3} \pi\lgroup r^2\lgroup\frac {rh'} {r'}\rgroup -r'^2 h'^2\rgroup\]

V=\[\frac{1} {3} \pi\lgroup \frac {r^3 -h'}{r'}- \frac {r'^3h'}{r'} \rgroup\]

V=\[\frac{1} {3} \pi h\lgroup \frac {r^3 -r'^3}{r'} \rgroup\]

From the figure we get,

h = H + h’.....................equation (3)

Substituting the value in equation (2), we get

\[\frac {h’} {h} = \frac {r’}{r}\]

\[\frac {h’} {H+h} = \frac {r’}{r}\]

\[h’  = H (\frac {r’}{r-r’})\]

Substituting this in the volume of the frustum of a cone V=\[\frac{1} {3} \pi h'\lgroup \frac {r^3 -r'^3}{r'} \rgroup\], we get

V= \[\frac{1} {3} \pi H\lgroup \frac {r'}{r-r’}  \rgroup \lgroup \frac {r^3-r’^3}{r’} \rgroup\]

V=\[\frac{1} {3} \pi H\lgroup \frac {r^3 -r'^3}{r-r'} \rgroup\]

V=\[\frac{1} {3} \pi H\lgroup \frac {(r-r')(r^2+rr'+r'^2)}{r-r'} \rgroup\]

V= \[\frac{1} {3} \pi H\lgroup  r^2 +rr' +r^{2} \rgroup\]

This is the frustum formula for volume.

Similarly, we can calculate the surface area of the frustum of a cone.

Curved surface area of larger right circular cone 1 = πrl

Curved surface area of smaller right circular cone 2 = πr’l’

Curved surface area of the frustum of cone 

= Curved surface area of larger right circular cone 1 – Curved surface area of smaller right circular cone 2 

= πrl − πr’l’

= π ( rl - r’l’)

From figure, we know  ΔOOD ~ ΔOPB

 Therefore, l’/l = r’/r…...equation (4)

But l’=l-L

Substitute this in equation (4), we get

l-Ll=r'r

l= \[L \lgroup \frac {r} {r-r’} \rgroup\]

The circumference of the base is the length of arc S and S’

 which is given by

S= 2πr \[\frac {1} {2} \times 2\pi rl - \frac {1}{2} \times 2\pi r’ \lgroup \frac {lr’} {r} \rgroup \]

S' = 2πr’

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

A=\[\frac {1} {2} \times 2\pi rl - \frac {1}{2} \times 2\pi r’ \lgroup \frac {lr’} {r} \rgroup \]

A= \[\pi l ( r - \frac {r’^2}{r} )\]

A= \[\pi l ( \frac {r^2 - r’^2}{r} )\]

Substituting the value of l

l= \[L (\frac {r}{r-r’} )\]

Curved surface area of frustum A= \[\pi L\lgroup \frac {r}{r-r'}\rgroup(\frac {r^2-r'^2}{r}) \]

A= \[\pi L (r + r’)\]

This is the frustum formula for curved surface area.

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

Solved Examples

Example 1: If the radii of the circular ends of a conical bucket which is 45 cm high are 28 cm and 7 cm, find the capacity of the bucket (Use π= \[\frac {22}{7}\]).

Solution: Clearly, the bucket forms a frustum of a cone such that the radii of its circular ends are r₁ = 28 cm, r₂ = 7 cm, and height h = 45 cm. 

Let V be the capacity of the bucket.

Then, by frustum of cone formula,

V = Volume of the frustum    

V=\[\frac {1} {3}\pi H (r^2 + r r'+r’^2)\] 

V=\[\frac {1} {3} \times \frac{22}{7} \times 45(28^2 + 28 \times 7 + 7^2)\]

   = 330 x 147 cm³ 

   = 48510 cm³.

Example 2: The slant height of the frustum of a cone is 4 cm, and the perimeter of its circular bases is 18 cm and 6 cm, respectively. Find the curved surface area of the frustum.

Solution: Let r₁ and r₂ be the radii of the circular base of the frustum, l be the slant height, and h be the height.

We have l = 4cm, 

2 π r1 = 18, r1 = \[\frac {9}{\pi}\]

and 

2 π r2 = 6 , r2 = \[\frac {3}{\pi}\] 

By frustum of a cone formula,

Curved surface area = \[\pi (r_{1}+r_{2})l\] \[\pi  (\frac {9} {\pi} + \frac {3}{\pi}) \times 4   \]

=  \[\pi (\frac {9}{\pi} + \frac {3}{\pi}) \times 4\]

=  48 cm²

Quiz Time

1. The slant height of the frustum of a cone is 8 cm and the perimeters of its circular ends are 20 cm and 24 cm, respectively. Find the curved surface area of a cone.

2. The radii of the circular ends of a frustum of height 6 cm are 14 cm and 6 cm, respectively. Find the lateral surface area and surface area of the frustum.