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Curved Surface Area of Frustum

Class 10Surface Areas and Volumes

The curved surface area (CSA) of a frustum is the area of its lateral (slanted) surface, excluding the two circular faces. This is the area of the material that forms the "wall" of the frustum.


Knowing the CSA is essential for practical problems like calculating the amount of cloth needed for a lampshade, the metal sheet for a bucket's side, or the paint needed for a flower pot's outer surface.


This topic is part of Chapter 13 (Surface Areas and Volumes) in NCERT Class 10 and is frequently tested alongside volume and total surface area of frustum.

What is Curved Surface Area of Frustum?

Definition: The curved surface area (CSA) of a frustum is the area of the lateral surface that connects the two parallel circular faces.


It does NOT include:

  • The area of the larger circular base (πR²)
  • The area of the smaller circular top (πr²)

Components needed:

  • R = radius of the larger base
  • r = radius of the smaller top
  • l = slant height (not perpendicular height)

Total Surface Area (TSA) includes both circular faces:

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

Curved Surface Area of Frustum Formula

Curved Surface Area of Frustum:

CSA = π(R + r)l


Where:

  • R = radius of the larger base
  • r = radius of the smaller top
  • l = slant height of the frustum

Slant Height Formula:

l = √[h² + (R − r)²]


Where:

  • h = perpendicular height (vertical distance between the two faces)

Total Surface Area:

TSA = π(R + r)l + πR² + πr²

Derivation and Proof

Derivation from cone CSAs:

  1. Consider a full cone with slant height L and base radius R. Its CSA = πRL.
  2. A smaller cone with slant height (L − l) and radius r is cut from the top.
  3. By similar triangles: r/R = (L − l)/L, giving L = Rl/(R − r).
  4. CSA of full cone = πR × Rl/(R − r) = πR²l/(R − r).
  5. Slant height of small cone = L − l = Rl/(R − r) − l = rl/(R − r).
  6. CSA of small cone = πr × rl/(R − r) = πr²l/(R − r).
  7. CSA of frustum = CSA of full cone − CSA of small cone.
  8. = πl/(R − r) × (R² − r²)
  9. = πl/(R − r) × (R + r)(R − r)
  10. = π(R + r)l

The derivation uses the difference of squares factorisation: R² − r² = (R + r)(R − r). The (R − r) cancels, leaving the elegant formula π(R + r)l.

Types and Properties

Comparison of surface area formulas:

SolidCSA FormulaTSA Formula
ConeπRlπRl + πR²
Cylinder2πRh2πRh + 2πR²
Frustumπ(R + r)lπ(R + r)l + πR² + πr²

Verification for r = 0 (frustum becomes cone):

  • CSA = π(R + 0)l = πRl. Matches cone CSA formula.

When to use CSA vs TSA:

  • CSA: Painting the outer wall, wrapping the curved part, cloth for a lampshade.
  • TSA: When both flat faces are also covered — e.g., total metal sheet for a closed frustum container.
  • TSA without one base: For open containers like buckets: π(R + r)l + πr² (bottom only).

Methods

Steps to find CSA of a frustum:

  1. Identify R, r: Larger and smaller radii. Convert from diameters if needed.
  2. Find slant height l:
    • If l is given directly, use it.
    • If h (perpendicular height) and radii are given: l = √[h² + (R − r)²].
  3. Apply CSA = π(R + r)l.
  4. For TSA: Add πR² + πr² to the CSA.

Common mistakes:

  • Using h instead of l in the CSA formula. CSA uses slant height l, not perpendicular height h.
  • Confusing CSA with TSA (forgetting to add or accidentally adding the circular face areas).
  • Using (R − r) instead of (R + r) in the formula.
  • When finding l, using (R + r)² instead of (R − r)² under the square root.

Solved Examples

Example 1: CSA with Given Slant Height

Problem: Find the CSA of a frustum with R = 14 cm, r = 7 cm, and l = 13 cm. (Use π = 22/7)


Solution:

Given:

  • R = 14, r = 7, l = 13

Using CSA = π(R + r)l:

  • = (22/7)(14 + 7)(13)
  • = (22/7)(21)(13)
  • = (22/7) × 273
  • = 22 × 39
  • = 858 cm²

Answer: CSA = 858 cm²

Example 2: Finding Slant Height First

Problem: A frustum has R = 20 cm, r = 8 cm, and h = 16 cm. Find its CSA. (Use π = 3.14)


Solution:

Step 1: Find slant height

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

Step 2: Find CSA

  • CSA = π(R + r)l
  • = 3.14(20 + 8)(20)
  • = 3.14 × 28 × 20
  • = 3.14 × 560
  • = 1758.4 cm²

Answer: CSA = 1758.4 cm²

Example 3: Total Surface Area

Problem: Find the TSA of a frustum with R = 10 cm, r = 4 cm, and l = 15 cm. (Use π = 3.14)


Solution:

Given:

  • R = 10, r = 4, l = 15

CSA:

  • = π(R + r)l = 3.14(14)(15) = 3.14 × 210 = 659.4 cm²

Area of larger base:

  • = πR² = 3.14 × 100 = 314 cm²

Area of smaller top:

  • = πr² = 3.14 × 16 = 50.24 cm²

TSA = CSA + πR² + πr²:

  • = 659.4 + 314 + 50.24
  • = 1023.64 cm²

Answer: TSA = 1023.64 cm²

Example 4: Bucket Metal Sheet

Problem: A bucket (open at top) has R = 21 cm, r = 7 cm, and h = 24 cm. Find the area of metal sheet needed to make it. (Use π = 22/7)


Solution:

Step 1: Find slant height

  • l = √[24² + (21 − 7)²] = √[576 + 196] = √772
  • = √(4 × 193) = 2√193 ≈ 27.78 cm

Step 2: Area needed = CSA + bottom circle (open at top)

  • CSA = π(R + r)l = (22/7)(28)(27.78) = (22/7) × 777.84 = 2444.64 cm²
  • Bottom = πr² = (22/7)(49) = 154 cm²
  • Total = 2444.64 + 154 = 2598.64 cm²

Answer: Metal sheet needed ≈ 2598.64 cm²

Example 5: Lampshade Cloth

Problem: A lampshade is in the shape of a frustum with top radius 5 cm, bottom radius 15 cm, and slant height 20 cm. Find the area of cloth needed. (Use π = 3.14)


Solution:

Given:

  • R = 15, r = 5, l = 20
  • A lampshade is open at both top and bottom, so only CSA is needed.

Using CSA = π(R + r)l:

  • = 3.14(15 + 5)(20)
  • = 3.14 × 20 × 20
  • = 3.14 × 400
  • = 1256 cm²

Answer: Cloth needed = 1256 cm²

Example 6: CSA in Terms of π

Problem: Find the CSA of a frustum with R = 9 cm, r = 5 cm, and h = 12 cm. Leave in terms of π.


Solution:

Step 1: Slant height

  • l = √[12² + (9 − 5)²] = √[144 + 16] = √160 = 4√10 cm

Step 2: CSA

  • = π(9 + 5)(4√10)
  • = π × 14 × 4√10
  • = 56√10 π cm²

Answer: CSA = 56√10 π cm² (≈ 556.2 cm²)

Example 7: Finding Radius Given CSA

Problem: A frustum has CSA = 1320 cm², R = 14 cm, and l = 10 cm. Find r. (Use π = 22/7)


Solution:

Using CSA = π(R + r)l:

  • 1320 = (22/7)(14 + r)(10)
  • 1320 = (220/7)(14 + r)
  • 14 + r = 1320 × 7/220
  • 14 + r = 9240/220
  • 14 + r = 42
  • r = 28 cm

Answer: r = 28 cm

Note: Here r > R, so we re-label: the smaller radius is 14 cm and the larger is 28 cm. The formula gives the same result regardless of which is called R or r.

Example 8: Comparing CSA and Base Areas

Problem: A frustum has R = 7 cm, r = 3.5 cm, and l = 5 cm. Find the CSA and compare it with the sum of the areas of the two circular faces. (Use π = 22/7)


Solution:

CSA:

  • = (22/7)(7 + 3.5)(5) = (22/7)(10.5)(5) = (22/7) × 52.5 = 165 cm²

Area of larger base:

  • = (22/7)(49) = 154 cm²

Area of smaller top:

  • = (22/7)(12.25) = 38.5 cm²

Sum of circular face areas = 154 + 38.5 = 192.5 cm²

Comparison: CSA (165 cm²) < Sum of base areas (192.5 cm²).

Answer: CSA = 165 cm². It is less than the combined circular face areas (192.5 cm²).

Real-World Applications

Manufacturing:

  • Calculating sheet metal for bucket production, funnel making, and tapered container fabrication.

Textile Industry:

  • Finding the amount of fabric for lampshades, skirts (circular), and tent sections.

Painting and Coating:

  • Determining the area to be painted on frustum-shaped objects like flower pots and pillars.

Architecture:

  • Surface area calculations for dome structures and tapered building elements.

Key Points to Remember

  • CSA of frustum = π(R + r)l, where l is the slant height.
  • Slant height: l = √[h² + (R − r)²].
  • TSA = CSA + πR² + πr².
  • CSA uses slant height (l), NOT perpendicular height (h).
  • For an open container (like a bucket): area = CSA + πr² (bottom only).
  • When r = 0, CSA = πRl (reduces to cone CSA).
  • The formula uses (R + r), not (R − r).
  • In the slant height formula, use (R − r)² under the root, not (R + r)².
  • Always check whether the problem asks for CSA, TSA, or area of specific surfaces.
  • This topic is combined with volume questions in CBSE exams for 4–5 mark problems.

Practice Problems

  1. Find the CSA of a frustum with R = 21 cm, r = 14 cm, and l = 10 cm. (Use π = 22/7)
  2. A frustum has R = 10 cm, r = 6 cm, and h = 8 cm. Find its slant height and CSA.
  3. Find the TSA of a frustum with R = 7 cm, r = 3.5 cm, and l = 7 cm. (Use π = 22/7)
  4. A bucket has top radius 15 cm, bottom radius 10 cm, and depth 12 cm. Find the area of tin needed to make it (open at top).
  5. A tent is in the shape of a frustum with R = 8 m, r = 4 m, and slant height = 6 m. Find the area of canvas used.
  6. A frustum has CSA = 2200 cm², l = 10 cm, and R = 42 cm. Find r. (Use π = 22/7)

Frequently Asked Questions

Q1. What is the CSA of a frustum?

The CSA (curved surface area) of a frustum is the area of its lateral (slanted) surface, excluding the two circular faces. Formula: CSA = π(R + r)l, where R and r are the radii and l is the slant height.

Q2. What is the difference between CSA and TSA of a frustum?

CSA includes only the curved lateral surface: π(R + r)l. TSA includes the curved surface plus both circular faces: π(R + r)l + πR² + πr².

Q3. How do I find the slant height?

Use l = √[h² + (R − r)²], where h is the perpendicular height and R, r are the two radii. This comes from the Pythagorean theorem applied to the cross-section.

Q4. Why does the CSA formula use slant height, not perpendicular height?

Because the curved surface is slanted. The CSA is the area of the surface that wraps around the frustum at an angle. Slant height measures the actual length of this surface, while perpendicular height is just the vertical distance.

Q5. What is the area of metal needed for an open bucket?

For a bucket (open at top, closed at bottom): Area = CSA + area of bottom circle = π(R + r)l + πr², where r is the bottom radius.

Q6. How do I remember the CSA formula?

Think of it as similar to the cone CSA (πRl), but with R replaced by (R + r) because the frustum has two circular edges. CSA = π(R + r)l.

Q7. What happens when the frustum becomes a cone (r = 0)?

CSA = π(R + 0)l = πRl, which is the standard CSA formula for a cone.

Q8. Can h and l ever be equal?

Only when R = r (the frustum becomes a cylinder). In that case, l = √[h² + 0] = h. For a proper frustum (R ≠ r), l > h always.

Q9. Is this combined with volume in board exams?

Yes. A typical 5-mark question may ask: 'Find the volume, CSA, and TSA of a frustum with given dimensions.' Knowing all three formulas and the slant height formula is essential.

Q10. How is the frustum CSA formula derived?

By subtracting the CSA of the removed small cone from the CSA of the full cone. Using similar triangles and the difference of squares identity R² − r² = (R+r)(R−r), the result simplifies to π(R+r)l.

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