Volume of Frustum of Cone – Meaning, Formula & Examples

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A frustum is a Latin word that means ‘piece cut off’. When a solid (generally a cone or a pyramid) is cut in such a manner that the base of the solid and the plane cutting the solid are parallel to each other, part of the solid which remains between the parallel cutting plane and the base is known as a frustum of that solid. Some common examples of a frustum of cone are the shade of a table lamp, bucket, glass tumbler, etc.

The volume of frustum is basically the space occupied by the frustum and is measured in cubic units like $m^{3}$, $cm^{3}$, $ft^{3}$, $in^{3}$, etc.

Let’s learn how to find the volume of frustum of a cone and its methods and formulas.

Frustum of Cone – A 3D Solid Shape

The frustum of cone is the part of the cone without a vertex when the cone is divided into two parts with a plane that is parallel to the base of the cone.

Properties of a Frustum of Cone

The following properties of a frustum of cone will help you to identify a it easily. 

• The frustum of a cone doesn’t contain the vertex of the corresponding cone but contains the base of the cone.
• The frustum of a cone is determined by its height and two radii (corresponding to two bases).
• The height of the frustum of a cone is the perpendicular distance between the centers of the two bases of the frustum.
• If the cone is a right circular cone, then the frustums formed from it also would be right-circular.

Volume of Frustum of Cone

The volume of frustum of cone can be calculated using its height and the areas of its bases. 

Let us consider a frustum of cone of height $h$ and radii of the two bases as $r$ and $\text{R}$. 

The volume of frustum of cone is calculated using the formula $\text{V} = \frac {\pi}{3}h(\text{R}^{2} + \text{R}r + r^{2})$.

Where $h$ is the height of the frustum of the cone(perpendicular distance between two bases)

$r$ is the radius of the smaller base

$R$ is the radius of the larger base

Derivation of Formula for Volume of Frustum of Cone

There are two methods of deriving the formula for the volume of the frustum of a cone. You can use any of these methods to derive the formula.

Method 1 to Derive the Volume of Frustum of Cone Formula

In this method, we’ll use the formula for the volume of a general frustum to derive the volume of frustum of cone formula. If $s_{1}$ and $s_{2}$ are the surface areas of the two bases of a frustum,  and $h$ is the height of the frustum, then the volume of frustum of cone is given by the formula 

$\text{V} = \frac {h}{3}\left(s_{1} + s_{2} + \sqrt{s_{1}s_{2}} \right)$ ————————– (1)

If $r$ and $\text{R}$ are the radii of the two bases of the frustum of a cone, then $s_{1} = \pi r^{2}$ and $s_{2} = \pi \text{R}^{2}$ 

Substituting the values of $s_{1}$ and $s_{2}$ in (1), we get

$\text{V} = \frac {h}{3}\left(\pi r^{2} + \pi \text{R}^{2} + \sqrt{\pi r^{2} \times \pi \text{R}^{2}} \right)$

$=>\text{V} = \frac {h}{3}\left(\pi r^{2} + \pi \text{R}^{2} + \sqrt{(\pi r \text{R})^{2}} \right)$

$=>\text{V} = \frac {h}{3}\left(\pi r^{2} + \pi \text{R}^{2} + \pi r \text{R} \right)$

$=>\text{V} = \frac {\pi h}{3}\left( r^{2} + \text{R}^{2} + r\text{R} \right)$

where $h$ is the height of frustum(perpendicular distance between the two bases of a frustum)

$r$ is the radius of a smaller base of the frustum

$\text{R}$ is the radius of a larger base of the frustum

Method 2 to Derive the Volume of Frustum of Cone Formula

In this method, we’ll use the formula for the volume of a cone to derive the formula for the volume of frustum of a cone. The volume of a cone of height $h$ and radius of base $r$ is given by $V = \frac {1}{3}\pi r^{2} h$.

Now consider a cone of height $\text{H} + h$, where $h$ is the height of a frustum and radius of the base $\text{R}$, then the volume of a cone is given by the formula $\frac {1}{3}\pi \text{R}^{2} \left(\text{H} + h \right)$.

Note: Volume of small cone that is cut by a plane is $\frac {1}{3}\pi r^{2} h$.

Now, we have the volume of frustum of the cone, $\text{V} = \text{The volume of the full cone} – \text{The volume of the cone that is cut}$, which means $\text{V} = \frac {1}{3}\pi \text{R}^{2} \left(\text{H} + h \right) – \frac {1}{3}\pi r^{2} h$. ——————————– (1)

The triangles $\text{OBC}$ and $\text{PQC}$ are similar (by AA property of similarity) and thus,

$\frac {\text{H} + h}{h} = \frac {\text{R}}{r}$ —————————— (2)

$\text{H} + h = \frac {\text{R}h}{r}$ ——————————— (3)

Substituting (3) this in (1), we get

$\text{V} = \pi \text{R}^{2} \times \frac{\text{R}h}{3r} – \frac{\pi r^{2} h}{3}$

$=>\text{V} = \frac{\pi h \left(\text{R}^{3} – r^{3} \right)}{3r}$

Now consider (2)

$\frac {H + h}{h} = \frac {\text{R}}{r} => \frac{\text{H}}{h} + 1 = \frac{\text{R}}{r} => \frac{\text{H}}{h} = \frac{\text{R}}{r} – 1 => \frac{\text{H}}{h} = \frac{\text{R} – r}{r}$

Taking reciprocal of both sides

$\frac{h}{\text{H}} = \frac{r}{\text{R} – r} =>h = \frac{\text{H}r}{\text{R} – r}$

Substituting this value of $h$ in the formula, we get

$\text{V} = \frac{\pi}{3} \times \frac{\text{H}r}{\text{R} – r} \times \frac{\text{R}^{3} – r^{3}}{r}$

Now, using the algebraic identity $a^{3} – b^{3} = \left(a – b \right) \left(a^{2} + ab + b^{2} \right)$ to replace $\left(\text{R}^{3} – r^{3} \right)$, we get

$\text{V} = \frac{\pi}{3} \times \frac{\text{H}r}{\text{R} – r} \times \frac{\left(\text{R} – r \right) \left(R^{2} + \text{R}r + r^{2} \right)}{r}$

$=> \text{V} = \frac{\pi H}{3}\left(\text{R}^{2} + \text{R}r + r^{2} \right)$

where $H$ is the height of frustum(perpendicular distance between the two bases of a frustum)

$r$ is the radius of a smaller base of the frustum

$\text{R}$ is the radius of a larger base of the frustum

Examples

Ex 1: If the radii of the circular ends of a frustum that is $45 cm$ high are $28 cm$ and $7 cm$, find the volume of the frustum.

Height of a frustum $h = 45 cm$.

Radius of lower base $r = 7 cm$.

Radius of upper base $\text{R} = 28 cm$.

Volume of a frustum = $\frac {\pi h}{3}\left( r^{2} + \text{R}^{2} + r\text{R} \right) = \frac {\frac{22}{7} \times 45}{3}\left( 7^{2} + 28^{2} + 7 \times 28 \right)$

$ = \frac {\frac{990}{7}}{3}\left(49 + 784 + 196 \right) = \frac {\frac{990}{7}}{3}\times 1029 = 48510 cm^{3}$.

Ex 2: An open plastic drum of height $63 cm$ with radii of lower and upper ends as $15 cm$ and $25 cm$ respectively is filled with milk. Find the cost of milk which can completely fill the bucket at ₹ $45$ per litre.  

Height of the container (frustum) $h = 63 cm$.

Radius of lower base $r = 15 cm$.

Radius of upper base $\text{R} = 25 cm$.

Volume of a frustum = $\frac {\pi h}{3}\left( r^{2} + \text{R}^{2} + r\text{R} \right) = \frac {\frac{22}{7}\times 63}{3}\left( 15^{2} + 25^{2} + 15 \times 25 \right)$

$ = \frac {198}{3}\left( 225 + 625 + 375 \right) = \frac {198}{3} \times 1225 = 80850 cm^{3} = 80.85 \text{L}$.

Rate of milk = ₹ $45$ per litre

Therefore, the cost of milk = $80.85 \times 45 = $ ₹ $3638.25$.

Conclusion

A frustum of a cone is obtained by slicing a cone in between by a plane parallel to its base. The volume of frustum is the space occupied by the frustum between the two radii of different measures. The formula to find the volume of frustum of cone is given by $V =\frac {\pi h}{3}\left( r^{2} + \text{R}^{2} + r\text{R} \right)$.

Practice Problems

1. If the radii of the circular ends of a frustum that is $60 cm$ high are $20 cm$ and $10 cm$, find the volume of the frustum.
2. The circumference of the two ends of a frustum of a cone is $66 cm$ and $132 cm$ respectively and its height is $25 cm$, find the volume of the frustum.
3. A bucket in the form of a frustum of a cone has end diameters of $1 ft$ and $1.5 ft$ and a height of $2.5 ft$, find the volume of water that can be filled in the bucket.

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