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Frustum of a 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. The portion between the base and the cutting plane is called the frustum. This topic is covered in CBSE Class 10 Mathematics, Chapter 13 (Surface Areas and Volumes).



The word "frustum" comes from the Latin word meaning "morsel cut off." A frustum has two circular faces (of different radii), a curved surface, and a slant height.



Frustums appear in everyday objects: buckets, lampshades, flowerpots, drinking glasses (truncated cone shape), and funnels. The formulas for volume, curved surface area (CSA), and total surface area (TSA) of a frustum are derived from cone formulas.

What is Frustum of a Cone - Formulas, Derivation & Solved Examples?

Definition: A frustum of a cone is the portion of a cone that lies between its base and a plane cutting the cone parallel to the base.


Key elements of a frustum:

  • R = radius of the larger circular face (base of the original cone)
  • r = radius of the smaller circular face (cut face)
  • h = height of the frustum (perpendicular distance between the two circular faces)
  • l = slant height of the frustum

Relationship between slant height, height, and radii:

l = sqrt[h^2 + (R - r)^2]


How a frustum is formed:

  1. Take a right circular cone with base radius R and height H.
  2. Cut it with a plane parallel to the base at height h₁ from the base (so h₁ < H).
  3. Remove the smaller cone from the top.
  4. The remaining solid is the frustum with larger radius R, smaller radius r, and height h₁.

Frustum of a Cone Formula

Volume of Frustum:

V = (1/3)πh(R^2 + r^2 + Rr)


Curved Surface Area (CSA):

CSA = π(R + r)l


Total Surface Area (TSA):

TSA = π(R + r)l + πR^2 + πr^2


Slant Height:

l = sqrt[h^2 + (R - r)^2]


Where:

  • R = radius of larger base
  • r = radius of smaller base
  • h = height (perpendicular distance between the two bases)
  • l = slant height

Derivation and Proof

Derivation of Volume Formula:

A frustum is a large cone minus a small cone cut from the top.

  1. Let the large cone have height H and base radius R.
  2. Let the small cone (removed) have height H - h and base radius r.
  3. By similar triangles: r/R = (H - h)/H, which gives H = Rh/(R - r).
  4. Volume of large cone = (1/3)πR^2H = (1/3)πR^2 × Rh/(R - r) = (1/3)πR^3h/(R - r).
  5. Volume of small cone = (1/3)πr^2(H - h) = (1/3)πr^2 × rh/(R - r) = (1/3)πr^3h/(R - r).
  6. Volume of frustum = (1/3)πh/(R - r) × (R^3 - r^3).
  7. Using factorisation: R^3 - r^3 = (R - r)(R^2 + r^2 + Rr).
  8. Volume = (1/3)πh × (R^2 + r^2 + Rr). Hence derived.

Derivation of CSA Formula:

  1. CSA of large cone = πRL₁ (where L₁ = slant height of large cone).
  2. CSA of small cone = πrL₂ (where L₂ = slant height of small cone).
  3. CSA of frustum = πRL₁ - πrL₂.
  4. By similar triangles: L₂/L₁ = r/R, so L₂ = rL₁/R.
  5. CSA = πRL₁ - πr(rL₁/R) = πL₁(R^2 - r^2)/R.
  6. Also, l = L₁ - L₂ = L₁(1 - r/R) = L₁(R - r)/R, so L₁ = lR/(R - r).
  7. CSA = π × lR/(R - r) × (R^2 - r^2)/R = πl(R + r)(R - r)/(R - r) = π(R + r)l.

Derivation of Slant Height:

  • Drop a perpendicular from the edge of the smaller circle to the larger base.
  • This forms a right triangle with legs h (height) and (R - r) (difference of radii).
  • By Pythagoras: l^2 = h^2 + (R - r)^2.

Types and Properties

Problems on frustum of a cone in Class 10 include:

Type 1: Volume of a Frustum

  • Given R, r, and h, find the volume using V = (1/3)πh(R^2 + r^2 + Rr).

Type 2: CSA and TSA

  • Find slant height first using l = sqrt[h^2 + (R - r)^2], then compute CSA and TSA.

Type 3: Real-World Objects (Bucket, Flowerpot, Lampshade)

  • Given dimensions of a bucket-shaped object, find volume (capacity) or surface area (material needed).

Type 4: Metal Conversion Problems

  • A frustum-shaped container is melted and recast into another shape (cylinder, sphere, etc.). Find the dimensions of the new shape using volume conservation.

Type 5: Finding Missing Dimensions

  • Given volume and some dimensions, find the missing radius or height.

Type 6: Cost Problems

  • Find the cost of painting the outer surface (TSA) or capacity (volume) at a given rate.

Methods

Step-by-step method for frustum problems:

  1. Identify the shape — confirm it is a frustum (two parallel circular faces of different radii).
  2. List the given values — R (larger radius), r (smaller radius), h (height), l (slant height).
  3. Find slant height if needed — l = sqrt[h^2 + (R - r)^2].
  4. Apply the correct formula:
    • Volume: (1/3)πh(R^2 + r^2 + Rr)
    • CSA: π(R + r)l
    • TSA: π(R + r)l + πR^2 + πr^2
  5. Substitute and calculate — use π = 22/7 or 3.14 as given in the problem.

Common Mistakes to Avoid:

  • Confusing R and r — R is always the LARGER radius.
  • Using l (slant height) where h (vertical height) is needed, or vice versa.
  • Forgetting the (R - r) in the slant height formula (not R + r).
  • In TSA, forgetting to add the areas of both circular faces (πR^2 + πr^2).
  • In bucket problems, the open top means TSA = CSA + πR^2 (only the bottom circle, not both).

Solved Examples

Example 1: Finding Volume of a Frustum

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


Solution:

Given:

  • R = 14 cm, r = 7 cm, h = 6 cm

Using V = (1/3)πh(R^2 + r^2 + Rr):

  • R^2 = 196, r^2 = 49, Rr = 98
  • R^2 + r^2 + Rr = 196 + 49 + 98 = 343
  • V = (1/3) × (22/7) × 6 × 343
  • = (1/3) × (22/7) × 2058
  • = (1/3) × 22 × 294
  • = (1/3) × 6468
  • = 2156 cm^3

Answer: The volume is 2156 cm^3.

Example 2: Finding Slant Height, CSA, and TSA

Problem: A frustum has radii 10 cm and 4 cm, and height 8 cm. Find the slant height, CSA, and TSA. (Use π = 3.14)


Solution:

Given:

  • R = 10 cm, r = 4 cm, h = 8 cm

Slant height:

  • l = sqrt[h^2 + (R - r)^2] = sqrt[64 + 36] = sqrt(100) = 10 cm

CSA:

  • = π(R + r)l = 3.14 × (10 + 4) × 10 = 3.14 × 140 = 439.6 cm^2

TSA:

  • = CSA + πR^2 + πr^2
  • = 439.6 + 3.14 × 100 + 3.14 × 16
  • = 439.6 + 314 + 50.24
  • = 803.84 cm^2

Answer: Slant height = 10 cm, CSA = 439.6 cm^2, TSA = 803.84 cm^2.

Example 3: Bucket Problem (Volume/Capacity)

Problem: A bucket is in the shape of a frustum with diameters 28 cm (top) and 14 cm (bottom), and depth 30 cm. Find the capacity of the bucket in litres. (Use π = 22/7)


Solution:

Given:

  • R = 28/2 = 14 cm (top radius)
  • r = 14/2 = 7 cm (bottom radius)
  • h = 30 cm

Volume:

  • V = (1/3)πh(R^2 + r^2 + Rr)
  • = (1/3) × (22/7) × 30 × (196 + 49 + 98)
  • = (1/3) × (22/7) × 30 × 343
  • = (1/3) × 22 × 30 × 49
  • = (1/3) × 32340
  • = 10780 cm^3

Converting to litres (1 litre = 1000 cm^3):

  • = 10780/1000 = 10.78 litres

Answer: The capacity is 10.78 litres.

Example 4: Metal Sheet Required for a Bucket

Problem: A bucket shaped as a frustum has top diameter 30 cm, bottom diameter 20 cm, and depth 24 cm. Find the area of metal sheet needed to make it (open at the top). (Use π = 3.14)


Solution:

Given:

  • R = 15 cm, r = 10 cm, h = 24 cm

Slant height:

  • l = sqrt[24^2 + (15 - 10)^2] = sqrt[576 + 25] = sqrt(601) ≈ 24.52 cm

Metal sheet = CSA + bottom circle (open top):

  • CSA = π(R + r)l = 3.14 × 25 × 24.52 = 1924.82 cm^2
  • Bottom = πr^2 = 3.14 × 100 = 314 cm^2
  • Total = 1924.82 + 314 = 2238.82 cm^2

Answer: Metal sheet needed = 2238.82 cm^2.

Example 5: Frustum Melted into a Cylinder

Problem: A frustum-shaped solid with radii 21 cm and 7 cm, and height 18 cm is melted and recast into a cylinder of radius 7 cm. Find the height of the cylinder. (Use π = 22/7)


Solution:

Given:

  • Frustum: R = 21 cm, r = 7 cm, h = 18 cm
  • Cylinder: radius = 7 cm, height = H (to find)

Volume of frustum:

  • = (1/3) × (22/7) × 18 × (441 + 49 + 147)
  • = (1/3) × (22/7) × 18 × 637
  • = (1/3) × 22 × 18 × 91
  • = (1/3) × 36036 = 12012 cm^3

Volume of cylinder = Volume of frustum:

  • πr^2H = 12012
  • (22/7) × 49 × H = 12012
  • 154H = 12012
  • H = 78 cm

Answer: The height of the cylinder is 78 cm.

Example 6: Flowerpot Problem

Problem: A flowerpot is in the form of a frustum with top diameter 40 cm, bottom diameter 20 cm, and height 30 cm. Find the volume of soil it can hold. (Use π = 3.14)


Solution:

Given:

  • R = 20 cm, r = 10 cm, h = 30 cm

Volume:

  • V = (1/3) × 3.14 × 30 × (400 + 100 + 200)
  • = (1/3) × 3.14 × 30 × 700
  • = (1/3) × 65940
  • = 21980 cm^3

Answer: The flowerpot can hold 21,980 cm^3 of soil.

Example 7: Cost of Painting a Frustum

Problem: A solid frustum has radii 8 cm and 3 cm with height 12 cm. Find the cost of painting its total surface at Rs. 5 per cm^2. (Use π = 22/7)


Solution:

Given:

  • R = 8 cm, r = 3 cm, h = 12 cm

Slant height:

  • l = sqrt[144 + 25] = sqrt(169) = 13 cm

TSA:

  • CSA = (22/7) × (8 + 3) × 13 = (22/7) × 143 = 449.43 cm^2
  • Top circle = (22/7) × 9 = 28.29 cm^2
  • Bottom circle = (22/7) × 64 = 201.14 cm^2
  • TSA = 449.43 + 28.29 + 201.14 = 678.86 cm^2

Cost:

  • = 678.86 × 5 = Rs. 3394.30

Answer: The cost of painting is Rs. 3394.30.

Example 8: Finding Missing Radius

Problem: The volume of a frustum is 1232 cm^3. The larger radius is 14 cm and the height is 6 cm. Find the smaller radius. (Use π = 22/7)


Solution:

Given:

  • V = 1232 cm^3, R = 14 cm, h = 6 cm

Using V = (1/3)πh(R^2 + r^2 + Rr):

  • 1232 = (1/3) × (22/7) × 6 × (196 + r^2 + 14r)
  • 1232 = (44/7) × (196 + r^2 + 14r)
  • 196 + r^2 + 14r = 1232 × 7/44 = 196
  • r^2 + 14r + 196 = 196
  • r^2 + 14r = 0
  • r(r + 14) = 0
  • r = 0 (not valid for a frustum with two radii) or check the calculation.

Rechecking:

  • If r = 0, the frustum becomes a complete cone. Volume = (1/3) × (22/7) × 6 × 196 = (1/3) × 22 × 6 × 28 = 1232 cm^3. Confirmed.

Answer: The smaller radius is 0 cm — meaning the solid is actually a cone, not a frustum.

Real-World Applications

Household Items:

  • Buckets, flower pots, lamp shades, and drinking glasses are often frustum-shaped. Calculating capacity and material required uses frustum formulas.

Civil Engineering:

  • Foundations of buildings and dams often have frustum-shaped cross-sections for structural stability.

Manufacturing:

  • Metal casting frequently involves frustum-shaped moulds. Knowing the volume helps determine the amount of raw material needed.

Agriculture:

  • Water tanks, grain hoppers, and feed containers are often frustum-shaped for ease of emptying from the bottom.

Architecture:

  • Cooling towers of power plants are hyperboloid structures that approximate frustums in cross-section.

Key Points to Remember

  • A frustum is formed by cutting a cone with a plane parallel to its base and removing the top cone.
  • Volume = (1/3)πh(R^2 + r^2 + Rr).
  • CSA = π(R + r)l, where l = slant height.
  • TSA = CSA + πR^2 + πr^2 (both circular faces).
  • Slant height: l = sqrt[h^2 + (R - r)^2].
  • R is the LARGER radius, r is the SMALLER radius.
  • For open containers (buckets), surface area = CSA + one circular base (πR^2 or πr^2, depending on which end is open).
  • Volume is conserved when a frustum is melted and recast into another shape.
  • When r = 0, the frustum becomes a complete cone. All formulas reduce to cone formulas.
  • When R = r, the frustum becomes a cylinder. Volume = πR^2h, CSA = 2πRh.

Practice Problems

  1. Find the volume and TSA of a frustum with R = 21 cm, r = 14 cm, and h = 10 cm. (Use π = 22/7)
  2. A bucket has top diameter 36 cm, bottom diameter 24 cm, and depth 14 cm. Find its capacity in litres.
  3. A frustum-shaped glass has R = 5 cm, r = 3 cm, and height 10 cm. How much liquid can it hold?
  4. A frustum with R = 20 cm, r = 8 cm, h = 16 cm is melted and recast into a sphere. Find the sphere's radius.
  5. Find the slant height of a frustum with R = 12 cm, r = 5 cm, and h = 24 cm.
  6. An industrial hopper is frustum-shaped with R = 1.5 m, r = 0.5 m, h = 2 m. Find the sheet metal area needed (no top).

Frequently Asked Questions

Q1. What is a frustum of a cone?

A frustum is the solid remaining when a cone is cut by a plane parallel to its base and the smaller cone above the cut is removed. It has two circular faces of different radii.

Q2. What is the volume formula for a frustum?

V = (1/3)πh(R^2 + r^2 + Rr), where R is the larger radius, r is the smaller radius, and h is the height.

Q3. How is the slant height calculated?

l = sqrt[h^2 + (R - r)^2]. It is found using the Pythagoras Theorem on the right triangle formed by h, (R - r), and l.

Q4. What is the CSA of a frustum?

CSA = π(R + r)l, where l is the slant height. This is the area of the curved surface only, excluding the two circular faces.

Q5. How is the frustum formula derived?

Volume of frustum = Volume of large cone - Volume of small cone. Using similar triangles and the factorisation R^3 - r^3 = (R - r)(R^2 + Rr + r^2), we arrive at V = (1/3)πh(R^2 + r^2 + Rr).

Q6. What is the TSA of a frustum?

TSA = CSA + πR^2 + πr^2 = π(R + r)l + πR^2 + πr^2. It includes the curved surface and both circular faces.

Q7. How do frustum formulas relate to cone and cylinder formulas?

When r = 0, the frustum becomes a cone: V = (1/3)πR^2h, CSA = πRl. When R = r, it becomes a cylinder: V = πR^2h, CSA = 2πRh.

Q8. What common objects are frustum-shaped?

Buckets, drinking glasses, flower pots, lamp shades, funnels, and industrial hoppers are commonly frustum-shaped.

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