Frustum of a Cone: Volume, Surface Areas, and Practical Applications for CBSE Class 10

Take an ordinary right circular cone—such as an ice-cream cone or a party hat—and slice straight through it with a sharp blade parallel to its base. Remove the small cone at the top. What geometric shape remains at the bottom?

The remaining solid is a familiar, everyday practical structure: a drinking water glass, a metallic bucket, a Turkish fez cap, or a coffee mug. In geometry, this sliced cone is called the Frustum of a Cone (derived from the Latin word frustum, meaning a "piece cut off").

In CBSE Class 10 Mathematics, Chapter 12 (Surface Areas and Volumes), mastering the formulas for slant height, curved surface area, total surface area, and volume of a frustum of a cone completes the solid mensuration syllabus.

What You Will Learn

• Geometric definition and anatomy of a Frustum of a Cone
• Two unequal circular base radii ($r_1$ and $r_2$), vertical height ($h$), and slant height ($l$)
• Derivation and formula for Slant Height: $l = \sqrt{h^2 + (r_1 - r_2)^2}$
• Formula for the Curved Surface Area (CSA): $\pi (r_1 + r_2) l$
• Formula for the Total Surface Area (TSA): $\pi (r_1 + r_2) l + \pi r_1^2 + \pi r_2^2$
• Formula for the Volume: $\frac{1}{3}\pi h (r_1^2 + r_2^2 + r_1 r_2)$
• Step-by-step solved CBSE board examination problems (the drinking glass and metallic milk bucket)
• Presentation guidelines and arithmetic verification

1. Anatomy of a Frustum of a Cone

                                      /                                     /  \  <-- Removed Small Cone
                                    /____                                   (  r2  ) <-- Smaller Top Base (Radius r2)
                                  /                                  Slant  /          \  Vertical Height h
                         Height /                                           (      r1      ) <-- Larger Bottom Base (Radius r1)

A frustum of a cone is bounded by:

1. A larger circular base of radius $r_1$.
2. A smaller circular base of radius $r_2$ ($r_1 > r_2$).
3. A vertical perpendicular height $h$ connecting the centers of the two circular bases.
4. An inclined curved lateral surface with slant height $l$.

2. Derivation of the Slant Height ($l$)

Draw a perpendicular from the edge of the smaller top base to the larger bottom base:

• A right-angled triangle is formed with vertical side $h$ and horizontal base $(r_1 - r_2)$.
• By the Pythagoras Theorem:

Slant Height Formula

$\mathbf{l = \sqrt{h^2 + (r_1 - r_2)^2}}$

(Notice how this generalizes the standard cone formula: when $r_2 = 0$, it collapses back to $l = \sqrt{h^2 + r_1^2}$!)

3. Surface Area and Volume Formulas of a Frustum

1. Curved Surface Area (CSA):

$\mathbf{\text{CSA of Frustum} = \pi (r_1 + r_2) l}$

2. Total Surface Area (TSA):

The total surface area includes the curved lateral surface plus both circular flat bases: $\mathbf{\text{TSA of Frustum} = \pi (r_1 + r_2) l + \pi r_1^2 + \pi r_2^2}$

3. Open Bucket Surface Area (One Circular End Open):

For a practical bucket or drinking glass that is closed at the bottom (radius $r_2$) and open at the top (radius $r_1$): $\mathbf{\text{Surface Area of Open Bucket} = \pi (r_1 + r_2) l + \pi r_2^2}$

4. Volume of a Frustum of a Cone:

$\mathbf{\text{Volume} = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)}$

(Notice that if $r_2 = 0$, this formula becomes $\frac{1}{3}\pi r_1^2 h$, the exact volume of a standard cone!)

4. Solved CBSE Board Examination Problems

Solved Example 1: Capacity of a Drinking Glass (NCERT Classic)

Problem: A drinking glass is in the shape of a frustum of a cone of height $14\text{ cm}$. The diameters of its two circular ends are $4\text{ cm}$ and $2\text{ cm}$. Find the capacity of the glass. (Use $\pi = 22/7$).

Solution:

1. Analyze Dimensions:
   • Height $h = 14\text{ cm}$.
   • Top radius $r_1 = \frac{4}{2} = \mathbf{2\text{ cm}}$.
   • Bottom radius $r_2 = \frac{2}{2} = \mathbf{1\text{ cm}}$.
2. Apply Volume Formula: $\text{Capacity (Volume)} = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$
3. Substitute Values: $\text{Volume} = \frac{1}{3} \times \frac{22}{7} \times 14 \times (2^2 + 1^2 + 2 \times 1)$ $\text{Volume} = \frac{1}{3} \times 22 \times 2 \times (4 + 1 + 2)$ $\text{Volume} = \frac{44}{3} \times 7 = \frac{308}{3} = \mathbf{102\frac{2}{3}\text{ cm}^3 = 102.67\text{ cm}^3}$
4. Therefore, <u>the capacity of the drinking glass is $102\frac{2}{3}\text{ cm}^3$ (or $102.67\text{ cm}^3$)</u>.

Solved Example 2: Slant Height and CSA of a Frustum

Problem: The slant height of a frustum of a cone is $4\text{ cm}$ and the perimeters (circumferences) of its circular ends are $18\text{ cm}$ and $6\text{ cm}$. Find the curved surface area of the frustum.

Solution:

1. Analyze Given Perimeters:
   • Circumference of top base: $2\pi r_1 = 18\text{ cm} \implies \pi r_1 = 9\text{ cm}$.
   • Circumference of bottom base: $2\pi r_2 = 6\text{ cm} \implies \pi r_2 = 3\text{ cm}$.
   • Slant height $l = 4\text{ cm}$.
2. Apply the CSA Formula: $\text{CSA} = \pi (r_1 + r_2) l = (\pi r_1 + \pi r_2) l$
3. Substitute Directly (No need to calculate individual radii!): $\text{CSA} = (9 + 3) \times 4 = 12 \times 4 = \mathbf{48\text{ cm}^2}$
4. Therefore, <u>the curved surface area of the frustum is $48\text{ cm}^2$</u>.

Solved Example 3: The Metallic Milk Bucket (5-Mark Classic)

Problem: A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height $16\text{ cm}$ with radii of its lower and upper ends as $8\text{ cm}$ and $20\text{ cm}$, respectively. Find the cost of the milk which can completely fill the container, at the rate of ₹$20\text{ per litre}$. Also find the cost of metal sheet used, if it costs ₹$8\text{ per } 100\text{ cm}^2$. (Use $\pi = 3.14$).

Solution:

1. List Given Dimensions:

   • Upper radius $r_1 = 20\text{ cm}$, Lower radius $r_2 = 8\text{ cm}$, Height $h = 16\text{ cm}$.

2. Part A: Calculate Volume of the Bucket: $V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$ $V = \frac{1}{3} \times 3.14 \times 16 \times [20^2 + 8^2 + (20 \times 8)]$ $V = \frac{50.24}{3} \times [400 + 64 + 160] = \frac{50.24}{3} \times 624 = 50.24 \times 208 = \mathbf{10449.92\text{ cm}^3}$

   • Convert to litres ($1\text{ litre} = 1000\text{ cm}^3$): $\text{Volume in litres} = \frac{10449.92}{1000} = \mathbf{10.45\text{ litres}}$
   • Cost of milk at ₹$20/\text{litre}$: $\text{Cost of milk} = 10.45 \times 20 = \mathbf{₹\,209}$

3. Part B: Calculate Area and Cost of Metal Sheet Used:

   • Slant height $l = \sqrt{h^2 + (r_1 - r_2)^2} = \sqrt{16^2 + (20 - 8)^2} = \sqrt{256 + 144} = \sqrt{400} = \mathbf{20\text{ cm}}$.
   • Metal sheet area (CSA $+$ closed bottom base): $\text{Area} = \pi (r_1 + r_2) l + \pi r_2^2 = \pi [(20 + 8) \times 20 + 8^2] = \pi [28 \times 20 + 64] = \pi [560 + 64] = 3.14 \times 624 = \mathbf{1959.36\text{ cm}^2}$
   • Cost of metal sheet at ₹$8$ per $100\text{ cm}^2$: $\text{Cost of metal sheet} = \frac{1959.36}{100} \times 8 = 19.5936 \times 8 = \mathbf{₹\,156.75}$

Conclusion: <u>The cost of the milk is ₹$209$, and the cost of the metal sheet used is ₹$156.75$</u>.

5. Summary and Examination Tips

Exam Tip: In the bucket problem, always ensure you add the smaller bottom base ($\pi r_2^2$) to the CSA, NOT the larger top base! A bucket is open at the top and closed at the bottom.

Common Mistake: In the volume formula, writing $(r_1 + r_2)^2$ instead of $(r_1^2 + r_2^2 + r_1 r_2)$. Notice that $(r_1 + r_2)^2 = r_1^2 + r_2^2 + 2r_1 r_2$, which has an extra $r_1 r_2$! The correct term has no coefficient of 2.