Frustum Examples and Practice Problems

What the Heck Is a Frustum?

A frustum is what you get when you chop the top off a cone or pyramid with a plane parallel to the base. That's it. No fancy geometry speak needed.

Think of a traffic cone. Or a lampshade. Or that cup you drink from. All frustums.

You deal with two circular or polygonal bases (one big, one small) and a curved or flat lateral surface connecting them.

Frustum vs. Cone

A cone has one vertex and one base. A frustum has two parallel bases and no vertex. The frustum is the cone's middle section after the top gets sliced off.

The Formulas You Actually Need

Don't memorize everything. Memorize these three and you're set.

Volume Formula

V = (1/3)πh(R² + Rr + r²)

Where:

For a pyramidal frustum (square or rectangular base), replace the circular terms with areas:

V = (h/3)(A₁ + A₂ + √(A₁ × A₂))

Lateral Surface Area

For a right circular frustum:

SL = πs(R + r)

Where s = slant height = √((R - r)² + h²)

Total Surface Area

STA = SL + πR² + πr²

Example 1: Volume of a Conical Frustum

Problem: A frustum has a height of 8 cm. The larger radius is 5 cm, the smaller radius is 3 cm. Find the volume.

Step 1: Plug into the volume formula.

V = (1/3) × π × 8 × (5² + 5×3 + 3²)

Step 2: Calculate inside the parentheses.

5² = 25

5×3 = 15

3² = 9

25 + 15 + 9 = 49

Step 3: Finish the math.

V = (1/3) × π × 8 × 49

V = (8/3) × 49 × π

V = 392π/3

V ≈ 410.5 cm³

Example 2: Surface Area of a Frustum

Problem: Find the total surface area of a frustum with h = 12 cm, R = 6 cm, r = 4 cm.

Step 1: Find the slant height first.

s = √((6 - 4)² + 12²)

s = √(4 + 144)

s = √148

s ≈ 12.17 cm

Step 2: Calculate lateral surface area.

SL = π × 12.17 × (6 + 4)

SL = π × 12.17 × 10

SL ≈ 382.3 cm²

Step 3: Add the base areas.

Base areas = π(6²) + π(4²) = 36π + 16π = 52π ≈ 163.4 cm²

Step 4: Total surface area.

STA = 382.3 + 163.4 = 545.7 cm²

Practice Problems

Try these before checking the answers below. No cheating.

Problem 1

A bucket shaped like a frustum has a height of 20 cm. The top radius is 15 cm, bottom radius is 10 cm. What's the volume in liters?

Problem 2

Find the lateral surface area of a frustum where h = 7 cm, R = 8 cm, r = 5 cm.

Problem 3

A frustum has volume 500 cm³, height 10 cm, and bottom radius 6 cm. Find the top radius.

Problem 4

A pyramidal frustum has square bases. Bottom base is 8 cm × 8 cm, top base is 4 cm × 4 cm, height is 6 cm. Find the volume.

Answers

Answer 1

V = (1/3) × π × 20 × (15² + 15×10 + 10²)

V = (1/3) × π × 20 × (225 + 150 + 100)

V = (1/3) × π × 20 × 475

V ≈ 9,948 cm³ ≈ 9.95 liters

Answer 2

s = √((8 - 5)² + 7²) = √(9 + 49) = √58 ≈ 7.62 cm

SL = π × 7.62 × (8 + 5) ≈ 311.1 cm²

Answer 3

500 = (1/3) × π × 10 × (36 + 6r + r²)

150/π = 36 + 6r + r²

r² + 6r + 36 - 150/π = 0

Solving: r ≈ 3.5 cm

Answer 4

A₁ = 64 cm², A₂ = 16 cm²

V = (6/3)(64 + 16 + √(64 × 16))

V = 2(80 + 32)

V = 224 cm³

Common Mistakes That Cost You Points

Tools and Methods Compared

Method Best For Speed Accuracy
Direct formula Simple problems with given values Fast High
Break into shapes Complex frustums or when dimensions missing Medium High
Integration Theoretical understanding Slow Very high
Calculator/Spreadsheet Multiple calculations, checking work Very fast Depends on input

Real-World Frustums

You see these shapes constantly and probably never noticed:

Quick Reference Cheat Sheet

Final Warning

Students lose marks on frustum problems for two reasons: wrong formula selection and algebraic errors. Double-check both.

When in doubt, draw it. A quick sketch showing R, r, h, and s will catch most mistakes before they happen.