Volume of Frustum of Cone

Class 10Surface Areas and Volumes

A frustum of a cone is the solid obtained when a cone is cut by a plane parallel to its base, and the smaller cone above the cut is removed. The remaining solid, shaped like a bucket, is the frustum.

Frustum shapes appear in everyday objects: buckets, lampshades, flower pots, drinking glasses (tapered), and cooling towers. Calculating the volume of a frustum is essential for determining capacity and material requirements.

This topic is part of Chapter 13 (Surface Areas and Volumes) in NCERT Class 10 and carries significant marks in CBSE board exams.

What is Volume of Frustum of Cone?

Definition: A frustum of a cone is the portion of a cone between two parallel planes cutting the cone — typically the base and a plane parallel to it.

Components:

R = radius of the larger base (bottom)
r = radius of the smaller top (upper face)
h = perpendicular height (vertical distance between the two parallel faces)
l = slant height = √[h² + (R − r)²]

Properties:

Both circular faces are parallel to each other.
If r = 0, the frustum becomes a complete cone.
If R = r, the frustum becomes a cylinder.

Volume of Frustum of Cone Formula

Volume of Frustum:

V = (πh/3)(R² + r² + Rr)

Where:

R = radius of the larger base
r = radius of the smaller top
h = perpendicular height
π = 22/7 or 3.14159...

Slant height:

l = √[h² + (R − r)²]

Related formulas:

CSA of frustum = π(R + r)l
TSA of frustum = π(R + r)l + πR² + πr²

Derivation and Proof

Derivation from cone volumes:

Consider a full cone with height H and base radius R.
A smaller cone of height (H − h) and radius r is cut from the top.
By similar triangles: r/R = (H − h)/H, so H = Rh/(R − r).
Volume of full cone = (1/3)πR²H = (1/3)πR² × Rh/(R − r) = πR³h / [3(R − r)].
Height of small cone = H − h = Rh/(R − r) − h = rh/(R − r).
Volume of small cone = (1/3)πr² × rh/(R − r) = πr³h / [3(R − r)].
Volume of frustum = Volume of full cone − Volume of small cone.
= πh / [3(R − r)] × (R³ − r³)
Using the identity: R³ − r³ = (R − r)(R² + r² + Rr)
= πh/3 × (R² + r² + Rr)
= (πh/3)(R² + r² + Rr)

Types and Properties

Special cases of the frustum:

Condition	Shape	Volume Formula
r = 0	Full cone	(1/3)πR²h
R = r	Cylinder	πR²h
R > r > 0	Frustum	(πh/3)(R² + r² + Rr)

Verification for r = 0:

V = (πh/3)(R² + 0 + 0) = (1/3)πR²h. Matches cone formula.

Verification for R = r:

V = (πh/3)(R² + R² + R²) = (πh/3)(3R²) = πR²h. Matches cylinder formula.

Methods

Steps to find the volume of a frustum:

Identify R, r, and h: Read the problem carefully. R is always the larger radius.
If diameters are given: Convert to radii (R = D/2).
Substitute into the formula: V = (πh/3)(R² + r² + Rr).
Compute each term: Calculate R², r², and Rr separately.
Add and multiply: Find the sum, multiply by πh/3.
Use π = 22/7 or 3.14 as specified.

For conversion problems (e.g., melting and recasting):

Find volume of original solid.
Set equal to volume of new shape (volume is conserved).
Solve for the unknown dimension.

Common mistakes:

Using (R − r) instead of R and r separately in the formula.
Confusing h (perpendicular height) with l (slant height).
Forgetting the (1/3) factor.
Using the cone formula (1/3)πR²h instead of the frustum formula.

Solved Examples

Example 1: Basic Volume Calculation

Problem: Find the volume of a frustum with R = 14 cm, r = 7 cm, and h = 6 cm. (Use π = 22/7)

Solution:

Given:

R = 14, r = 7, h = 6

Using V = (πh/3)(R² + r² + Rr):

R² = 196, r² = 49, Rr = 98
R² + r² + Rr = 196 + 49 + 98 = 343
V = (22/7)(6/3)(343)
V = (22/7)(2)(343)
V = (22/7) × 686
V = 22 × 98
V = 2156 cm³

Answer: Volume = 2156 cm³

Example 2: Bucket Volume

Problem: A bucket is in the shape of a frustum with top radius 28 cm, bottom radius 21 cm, and height 30 cm. Find its capacity in litres. (Use π = 22/7)

Solution:

Given:

R = 28 cm, r = 21 cm, h = 30 cm

Using the formula:

R² = 784, r² = 441, Rr = 588
Sum = 784 + 441 + 588 = 1813
V = (22/7)(30/3)(1813)
V = (22/7)(10)(1813)
V = (22 × 10 × 1813)/7
V = 399,060/7
V = 57,008.57 cm³

Converting to litres: 1 litre = 1000 cm³

V = 57,008.57/1000 = 57.01 litres

Answer: Capacity ≈ 57.01 litres

Example 3: Frustum from Given Diameters

Problem: A frustum-shaped vessel has top diameter 40 cm, bottom diameter 20 cm, and depth 21 cm. Find its volume. (Use π = 22/7)

Solution:

Given:

R = 40/2 = 20 cm, r = 20/2 = 10 cm, h = 21 cm

Using the formula:

R² = 400, r² = 100, Rr = 200
Sum = 400 + 100 + 200 = 700
V = (22/7)(21/3)(700)
V = (22/7)(7)(700)
V = 22 × 700
V = 15,400 cm³

Answer: Volume = 15,400 cm³

Example 4: Finding Height Given Volume

Problem: A frustum has R = 10 cm, r = 5 cm, and volume = 2200/7 cm³. Find its height. (Use π = 22/7)

Solution:

Given:

R = 10, r = 5, V = 2200/7

Using V = (πh/3)(R² + r² + Rr):

R² + r² + Rr = 100 + 25 + 50 = 175
2200/7 = (22/7)(h/3)(175)
2200/7 = (22 × 175 × h)/(7 × 3)
2200/7 = (3850h)/21
2200 × 21 = 7 × 3850h
46200 = 26950h
h = 46200/26950
h = 12/7 cm

Answer: Height = 12/7 cm (≈ 1.71 cm)

Example 5: Volume in Terms of π

Problem: Find the volume of a frustum with R = 6 cm, r = 3 cm, h = 4 cm. Leave the answer in terms of π.

Solution:

Given:

R = 6, r = 3, h = 4

Using the formula:

R² + r² + Rr = 36 + 9 + 18 = 63
V = (π × 4/3)(63)
V = (4π/3)(63)
V = 84π cm³

Answer: Volume = 84π cm³

Example 6: Melting and Recasting

Problem: A frustum of metal has R = 15 cm, r = 10 cm, h = 24 cm. It is melted and recast into a cylinder of radius 12 cm. Find the height of the cylinder. (Use π = 22/7)

Solution:

Volume of frustum:

R² + r² + Rr = 225 + 100 + 150 = 475
V = (π × 24/3)(475) = 8π × 475 = 3800π cm³

Volume of cylinder = π × 12² × H = 144πH

Setting equal (volume conserved):

144πH = 3800π
H = 3800/144 = 475/18 ≈ 26.39 cm

Answer: Height of cylinder = 475/18 cm ≈ 26.39 cm

Example 7: Finding Slant Height and Volume

Problem: A frustum has R = 20 cm, r = 8 cm, and h = 16 cm. Find (a) the slant height and (b) the volume. (Use π = 3.14)

Solution:

(a) Slant height:

l = √[h² + (R − r)²]
= √[16² + (20 − 8)²]
= √[256 + 144]
= √400 = 20 cm

(b) Volume:

R² + r² + Rr = 400 + 64 + 160 = 624
V = (3.14 × 16/3)(624)
V = (3.14 × 5.333)(624)
V = 3.14 × 3328
V = 10,449.92 cm³

Answer: (a) l = 20 cm, (b) V ≈ 10,449.92 cm³

Example 8: Frustum with Equal Radii Check

Problem: A solid has R = 7 cm, r = 7 cm, and h = 10 cm. What shape is it, and what is its volume? (Use π = 22/7)

Solution:

Since R = r = 7: The solid is a cylinder, not a frustum.

Using the frustum formula as verification:

V = (π × 10/3)(49 + 49 + 49)
= (π × 10/3)(147)
= 490π
= 490 × 22/7 = 1540 cm³

Using cylinder formula:

V = πr²h = (22/7)(49)(10) = 1540 cm³ ✔

Answer: The solid is a cylinder. Volume = 1540 cm³

Real-World Applications

Daily Life:

Calculating the capacity of buckets, tubs, and tapered containers.
Measuring the volume of flower pots and lampshades.

Engineering:

Volume of cooling towers (hyperboloid frustum shape).
Hopper design in grain storage and material handling.

Construction:

Volume of concrete needed for tapered column bases and pillar foundations.

Manufacturing:

Metal casting: determining the amount of material needed for frustum-shaped components.

Key Points to Remember

Volume of frustum = (πh/3)(R² + r² + Rr).
R = larger radius (base), r = smaller radius (top), h = perpendicular height.
Slant height: l = √[h² + (R − r)²].
Do NOT confuse h (height) with l (slant height).
When r = 0, the formula reduces to the cone formula: (1/3)πR²h.
When R = r, the formula reduces to the cylinder formula: πR²h.
The formula contains R² + r² + Rr (three terms), not (R + r)².
For melting/recasting problems, set volumes equal (volume is conserved).
Always convert diameters to radii before substituting.
This is a high-scoring topic in CBSE Class 10 board exams (3–5 marks).

Practice Problems

Find the volume of a frustum with R = 21 cm, r = 14 cm, and h = 15 cm. (Use π = 22/7)
A bucket (frustum-shaped) has top diameter 30 cm, bottom diameter 20 cm, and depth 14 cm. Find its capacity.
A frustum has R = 12 cm, r = 8 cm, and slant height 13 cm. Find its height and volume.
A glass has the shape of a frustum with top radius 5 cm, bottom radius 3 cm, and height 12 cm. Find how much water it can hold. (Use π = 3.14)
A frustum is melted and recast into a sphere of radius 7 cm. If the frustum had R = 14 cm and r = 7 cm, find its height. (Use π = 22/7)
Find the volume and total surface area of a frustum with R = 10 cm, r = 4 cm, h = 8 cm.

Frequently Asked Questions

Q1. What is a frustum?

A frustum is the portion of a cone between two parallel planes cutting it. When you cut a cone parallel to its base and remove the top part, the remaining solid is a frustum. It has two circular faces of different radii.

Q2. What is the volume formula for a frustum?

V = (πh/3)(R² + r² + Rr), where R and r are the radii of the two circular faces and h is the perpendicular height.

Q3. How is the frustum formula derived?

It is derived by subtracting the volume of the small cone (cut off) from the full cone. Using similar triangles and the identity R³ − r³ = (R − r)(R² + r² + Rr), the result simplifies to (πh/3)(R² + r² + Rr).

Q4. What is the difference between height and slant height of a frustum?

Height (h) is the perpendicular distance between the two parallel faces. Slant height (l) is the distance along the lateral surface. They are related by l = √[h² + (R − r)²].

Q5. What happens when r = 0 in the frustum formula?

When r = 0, the frustum becomes a full cone. V = (πh/3)(R² + 0 + 0) = (1/3)πR²h, which is the standard cone volume formula.

Q6. Is a bucket a frustum?

Yes. A typical bucket is shaped like a frustum of a cone — it has a wider top and a narrower bottom, both circular and parallel.

Q7. Why is the formula R² + r² + Rr and not (R + r)²?

Because the derivation involves R³ − r³ which factors as (R − r)(R² + Rr + r²). Note: (R + r)² = R² + 2Rr + r², which has 2Rr instead of Rr. They are different expressions.

Q8. How do I convert the volume to litres?

If the volume is in cm³, divide by 1000 to convert to litres (1 litre = 1000 cm³). If in m³, multiply by 1000 (1 m³ = 1000 litres).

Q9. Is this topic important for CBSE exams?

Yes. Frustum problems appear regularly in board exams under Surface Areas and Volumes. Questions may ask for volume, capacity in litres, or melting/recasting problems. This carries 3–5 marks.

Q10. Can I use the frustum formula for a cylinder?

Yes. When R = r, the frustum formula gives (πh/3)(R² + R² + R²) = πR²h, which is the cylinder volume. The frustum formula is a generalisation.

Volume of Frustum of Cone

What is Volume of Frustum of Cone?

Volume of Frustum of Cone Formula

Derivation and Proof

Types and Properties

Methods

Solved Examples

Example 1: Basic Volume Calculation

Example 2: Bucket Volume

Example 3: Frustum from Given Diameters

Example 4: Finding Height Given Volume

Example 5: Volume in Terms of π

Example 6: Melting and Recasting

Example 7: Finding Slant Height and Volume

Example 8: Frustum with Equal Radii Check

Real-World Applications

Key Points to Remember

Practice Problems

Frequently Asked Questions

Q1. What is a frustum?

Q2. What is the volume formula for a frustum?

Q3. How is the frustum formula derived?

Q4. What is the difference between height and slant height of a frustum?

Q5. What happens when r = 0 in the frustum formula?

Q6. Is a bucket a frustum?

Q7. Why is the formula R² + r² + Rr and not (R + r)²?

Q8. How do I convert the volume to litres?

Q9. Is this topic important for CBSE exams?

Q10. Can I use the frustum formula for a cylinder?

Related Topics

More Maths Topics

Quick Links