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Learning Resources

Frustums, Cones & Density

Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.

Form 3 Pathway: N/A

First Principles

Objective

Calculate the volume, surface area, and mass/density of cones and frustums.

Concrete Scenario: A grain bucket used in agricultural markets in Eldoret is shaped like an inverted cone with the top sliced off — a shape called a frustum. Knowing its volume allows farmers to calculate the exact mass of maize (density \(\rho = 720\,\text{kg/m}^3\)) the bucket can hold.

Geometric Insight:

  • Cone Volume: Derived by summing circular discs from base to apex: \[V_{\text{cone}} = \frac{1}{3}\pi r^2 h\]
  • Frustum Volume: Formed by cutting a smaller cone of top radius \(r\) from a larger cone of base radius \(R\). Slicing leaves a height \(h\): \[V_{\text{frustum}} = \frac{1}{3}\pi h (R^2 + Rr + r^2)\]
  • Density & Mass Link: Mass equals volume multiplied by density: \(m = \rho \times V\).

Cone vs. Frustum Structure

hTop Radius (r)Base Radius (R)

Key Formulas

\[V_{\text{cone}} = \frac{1}{3}\pi r^2 h\] — Volume of a complete cone.
\[V_{\text{frustum}} = \frac{1}{3}\pi h (R^2 + Rr + r^2)\] — Volume of a frustum with base radius \(R\), top radius \(r\), and height \(h\).
\[\text{Curved Surface Area}_{\text{frustum}} = \pi (R + r) \ell \quad \text{where } \ell = \sqrt{h^2 + (R - r)^2}\] — Slant height and lateral surface area.
\[\rho = \frac{m}{V} \implies m = \rho V\] — Density relationship linking volume to total mass.

Worked Examples

Example 1 (Easy) — Cone Volume:

Problem: Calculate the volume of a cone with radius \(r = 3\,\text{cm}\) and vertical height \(h = 5\,\text{cm}\) (take \(\pi = 3.142\)).

  1. Formula: \[V = \frac{1}{3}\pi r^2 h\]
  2. Substitute values: \[V = \frac{1}{3} \times 3.142 \times 3^2 \times 5 = \frac{1}{3} \times 3.142 \times 9 \times 5\]
  3. Simplify: \[V = 3.142 \times 15 = 47.13\,\text{cm}^3\]
Example 2 (Medium) — Frustum Volume:

Problem: A frustum has bottom radius \(R = 6\,\text{cm}\), top radius \(r = 3\,\text{cm}\), and vertical height \(h = 4\,\text{cm}\). Find its volume in terms of \(\pi\).

  1. Formula: \[V = \frac{1}{3}\pi h (R^2 + Rr + r^2)\]
  2. Evaluate bracket: \[6^2 + (6 \times 3) + 3^2 = 36 + 18 + 9 = 63\]
  3. Calculate volume: \[V = \frac{1}{3}\pi \times 4 \times 63 = 84\pi \approx 263.89\,\text{cm}^3\]
Example 3 (Hard) — Mass & Density of a Frustum:

Problem: A metal frustum has \(R = 8\,\text{cm}\), \(r = 4\,\text{cm}\), and height \(h = 10\,\text{cm}\). If its density is \(\rho = 7.8\,\text{g/cm}^3\), calculate its mass using \(\pi = \frac{22}{7}\).

  1. Bracket evaluation: \[R^2 + Rr + r^2 = 8^2 + (8 \times 4) + 4^2 = 64 + 32 + 16 = 112\]
  2. Calculate volume: \[V = \frac{1}{3} \times \frac{22}{7} \times 10 \times 112 = \frac{1}{3} \times 22 \times 10 \times 16 = \frac{3520}{3} \approx 1173.33\,\text{cm}^3\]
  3. Calculate mass: \[m = \rho \times V = 7.8 \times 1173.33 = 9152\,\text{g} = 9.152\,\text{kg}\]

Common Mistakes

Mistake Using the full height of the original cone instead of the height \(h\) of the frustum in the frustum formula.
Correction \(h\) in the frustum formula refers strictly to the vertical distance between the two parallel circular faces.
Why it feels right Misidentifying the height given in multi-part geometric diagrams.
Mistake Forgetting the \(\frac{1}{3}\) factor when using the frustum volume formula.
Correction Just like a cone or pyramid, the frustum volume formula always includes the \(\frac{1}{3}\) factor.
Why it feels right The bracket term \((R^2 + Rr + r^2)\) is complex, leading students to drop the outer factor.

Real World

Agricultural Grain Storage: Calculating storage capacities of conical and frustum-shaped silos in cereal board depots across Kenya.
Civil Engineering & Roadwork: Determining concrete volume and tonnage required to cast frustum-shaped bridge pier foundations.
Manufacturing: Estimating plastic raw material requirements to manufacture buckets and basins of frustum geometry.

Practice

A rectangular metal block has a mass of 540 g and a volume of 180 cm\u00B3. What is its density in g/cm\u00B3? (Type only the number, e.g., 3)
Review the concepts above.
A frustum of a cone has a height of 10 cm, lower base radius of 8 cm, and upper base radius of 4 cm. What is the volume of the frustum in cm\u00B3? (Use \(\pi = 22/7\) and round to the nearest whole number). (Type only the number, e.g., 1173)
Review the concepts above.
A cone of height 20 cm and base radius 9 cm is cut parallel to its base, leaving a frustum of height 8 cm. What is the volume of the frustum in cm\u00B3? (Use \(\pi = 3.14\) and round to the nearest whole number). (Type only the number, e.g., 1329)
Review the concepts above.
A frustum of a right circular cone has lower radius 8 cm, upper radius 5 cm, and vertical height 12 cm. What is its volume in cm\u00B3? (Use \(\pi = 22/7\) and round to 1 decimal place). (Type only the number, e.g., 1621.7)
Review the concepts above.
A right circular cone has vertical height 12 cm and base radius 5 cm. Calculate its total volume in cm\u00B3 using \(\pi = 3.142\) (round to 1 decimal place). (Type only the number, e.g., 314.2)
Review the concepts above.
A solid consists of a cone of height 10 cm attached on top of a frustum of height 8 cm. The common junction radius is 4 cm and the frustum base radius is 6 cm. Using \(\pi = 3.14\), find total volume in cm\u00B3. (Type only the number, e.g., 804)
Review the concepts above.