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Frustum of a cone: Class 10 practice

A frustum is a cone with its top cut off parallel to the base. Read the method, follow the three worked examples, then try the six questions.

What you should know first

  • Cone formulas
  • Pythagoras

The idea behind frustum of a cone

A frustum is a cone with its top cut off parallel to the base. Volume = (1/3)πh(R² + r² + Rr); slant height l = √(h² + (R − r)²); curved surface area = π(R + r)l.

Step-by-step method

  1. Write R, r and h.
  2. Find l if needed.
  3. Use the volume or CSA formula.

Worked example 1 (Basic)

A frustum has radii 16 cm and 4 cm and height 9 cm. Find its volume (in terms of π).

  1. V = (1/3)π × 9(256 + 16 + 64)
  2. = 1008π cm³
  3. Answer: 1008π cm³

Worked example 2 (Standard)

A frustum has radii 15 cm and 12 cm and height 12 cm. Find its slant height and curved surface area.

  1. l = √(12² + 3²) = 12.37 cm
  2. CSA = π(15 + 12) × 12.37 = 1049.2 cm²
  3. Answer: l = 12.37 cm, CSA = 1049.2 cm²

Worked example 3 (Challenge)

A frustum has radii 17 cm and 2 cm and height 15 cm. Find its slant height and curved surface area.

  1. l = √(15² + 15²) = 21.21 cm
  2. CSA = π(17 + 2) × 21.21 = 1266.22 cm²
  3. Answer: l = 21.21 cm, CSA = 1266.22 cm²

Common mistakes and misconceptions

  • Using (R + r) in the slant height
  • Forgetting the Rr term

Practice questions

  1. (Basic) A frustum has radii 18 cm and 12 cm and height 11 cm. Find its volume (in terms of π).
  2. (Basic) A frustum has radii 18 cm and 7 cm and height 10 cm. Find its volume (in terms of π).
  3. (Standard) A frustum has radii 12 cm and 8 cm and height 3 cm. Find its slant height and curved surface area.
  4. (Standard) A frustum has radii 18 cm and 6 cm and height 6 cm. Find its slant height and curved surface area.
  5. (Challenge) A frustum has radii 6 cm and 4 cm and height 6 cm. Find its slant height and curved surface area.
  6. (Challenge) A frustum has radii 6 cm and 4 cm and height 4 cm. Find its slant height and curved surface area.

Answer key with full working

  • Q1: V = (1/3)π × 11(324 + 144 + 216) → = 2508π cm³ → Answer: 2508π cm³
  • Q2: V = (1/3)π × 10(324 + 49 + 126) → = 4990/3π cm³ → Answer: 4990/3π cm³
  • Q3: l = √(3² + 4²) = 5 cm → CSA = π(12 + 8) × 5 = 314.16 cm² → Answer: l = 5 cm, CSA = 314.16 cm²
  • Q4: l = √(6² + 12²) = 13.42 cm → CSA = π(18 + 6) × 13.42 = 1011.57 cm² → Answer: l = 13.42 cm, CSA = 1011.57 cm²
  • Q5: l = √(6² + 2²) = 6.32 cm → CSA = π(6 + 4) × 6.32 = 198.69 cm² → Answer: l = 6.32 cm, CSA = 198.69 cm²
  • Q6: l = √(4² + 2²) = 4.47 cm → CSA = π(6 + 4) × 4.47 = 140.5 cm² → Answer: l = 4.47 cm, CSA = 140.5 cm²

Syllabus note

This skill is listed under "Surface Areas and Volumes" for Class 10 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Related practice and notes

Frequently asked questions

Which chapter is frustum of a cone in?

"Surface Areas and Volumes", Class 10 Maths (NCERT/CBSE). Other boards may place it differently.

How were the answers checked?

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

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