Mathematics Standard • Year 11 • Module 2 • Lesson 10

Volume of Pyramids, Cones, and Spheres

Build fluency with the one-third rule V = ⅓Ah for any pyramid or cone, V = ⅓πr²h for a cone, and V = ⅔πr³ for a sphere.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete each volume formula.

Any pyramid or cone (general rule): V = ______________________

Cone (specific): V = ______________________

Sphere: V = ______________________ Hemisphere: V = ______________________

Q1.2 True or false: a cone holds exactly the same volume as a cylinder with the same radius and height.

True False

If false, what is the correct relationship? __________________________________

Q1.3 A cone has slant height ℓ = 10 m and base radius r = 6 m. What MUST you use Pythagoras to find before using V = ⅓πr²h? ____________________________

Stuck? Revisit lesson § Key Formulas panel and Card 1 (One-Third Rule).

2. Worked example — cone given vertical height

Follow each line of working. Every step has a reason on the right.

Problem. A cone has base radius r = 5 cm and vertical height h = 12 cm. Find the volume in exact form (in terms of π) and to 2 d.p.

Step 1 — Apply V = ⅓πr²h.

V = ⅓ × π × 25 × 12

Reason: substitute r = 5 (r² = 25) and h = 12 directly. Use vertical height, not slant.

Step 2 — Simplify in exact form.

V = (25 × 12) / 3 × π = 300π / 3 = 100π cm³

Reason: keep π exact. Compute 300 / 3 = 100.

Step 3 — Evaluate decimal.

V = 100π ≈ 314.16 cm³

Reason: evaluate at the final step only. Units are cm³ (cubic) for volume.

Conclusion. V = 100π cm³ ≈ 314.16 cm³.

3. Faded example — sphere from diameter

A spherical water balloon has diameter 18 cm. Find its volume in exact form (in terms of π). Fill in each blank line. 4 marks

Step 1 — Find the radius (always halve d).

r = ____ ÷ 2 = ____ cm

Step 2 — Cube the radius: r³ = ____³ = ________

Step 3 — Apply V = ⅔πr³: V = (4/3) × π × ________

Step 4 — Simplify: V = ____ π cm³

Conclusion. V = ________ π cm³.

Stuck? Revisit lesson § Worked Example 4 — Sphere. The exponent is 3 (cubic), not 2.

4. Graduated practice — pyramids, cones, spheres

Show working. Decimals to 2 d.p. unless told otherwise.

Foundation — single-step substitution (4 questions)

Q	Problem	Answer
4.1 1	Square pyramid: base side 6 m, vertical height 10 m. Find V.
4.2 1	Rectangular pyramid: base 8 cm × 5 cm, vertical height 9 cm. Find V.
4.3 1	Cone: r = 7 cm, vertical height 15 cm. Find V in exact form (in terms of π).
4.4 1	Sphere: r = 6 cm. Find V in exact form.

Standard — typical HSC difficulty (6 questions)

Halve any diameter to find r first. Use vertical (not slant) height in cone formula.

4.5 An ice-cream cone has diameter 6 cm and vertical height 10 cm. Find V in cm³ to 2 d.p. 2 marks

4.6 A sphere has diameter 10 m. Find V to 2 d.p. 2 marks

4.7 A hemisphere has radius 9 cm. Find V in exact form. 2 marks

4.8 A cone has slant height 13 cm and base radius 5 cm. (a) Use Pythagoras to find the vertical height. (b) Find the volume in exact form. 3 marks

4.9 A pyramid has volume 200 cm³ and a square base of side 10 cm. Find its vertical height. 2 marks

4.10 A cone has volume 48π cm³ and base radius 4 cm. Find its vertical height. 2 marks

Extension — reverse and composite (2 questions)

4.11 A sphere has volume 500π/3 cm³. Find its radius. 2 marks

4.12 A solid consists of a cylinder (r = 4 cm, h = 10 cm) with a cone on top (same r, vertical height 6 cm). Find the total volume in exact form. 3 marks

Stuck on 4.11? V = ⅔πr³ ⇒ r³ = (3V)/(4π). Cancel the π first, then take cube root.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 (pyramid b=6) I used base area = 6² = 36 m² (not 6).

For 4.4 (sphere r=6) I cubed the radius (216), not squared it (36).

For 4.6 (sphere d=10) I halved the diameter to r = 5 BEFORE substituting.

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Answers — Do not peek before attempting

Q1.1 — Four formulas

Pyramid/cone general: V = ⅓Ah. Cone: V = ⅓πr²h. Sphere: V = ⅔πr³. Hemisphere: V = ⅙πr³.

Q1.2 — Cone vs cylinder

False. A cone holds exactly one-third the volume of the cylinder with the same radius and height: V_cone = ⅓ V_cyl.

Q1.3 — Cone given slant height

The vertical (perpendicular) height h: h = √(ℓ² − r²). The volume formula uses h, not ℓ.

Q3 — Faded example (sphere d = 18)

Step 1: r = 18 ÷ 2 = 9 cm.
Step 2: r³ = 9³ = 729.
Step 3: V = (4/3) × π × 729.
Step 4: V = (4 × 729) / 3 × π = 2916/3 × π = 972π cm³.

Q4.1 — Square pyramid b = 6, h = 10

A = 36 m². V = ⅓ × 36 × 10 = 360/3 = 120 m³.

Q4.2 — Rectangular pyramid 8 × 5, h = 9

A = 40 cm². V = ⅓ × 40 × 9 = 360/3 = 120 cm³.

Q4.3 — Cone r = 7, h = 15

V = ⅓ × π × 49 × 15 = (49 × 15)/3 × π = 735/3 × π = 245π cm³ (≈ 769.69 cm³).

Q4.4 — Sphere r = 6

r³ = 216. V = (4/3) × π × 216 = 864/3 × π = 288π cm³ (≈ 904.78 cm³).

Q4.5 — Ice-cream cone d = 6, h = 10

r = 3 cm. V = ⅓ × π × 9 × 10 = 30π ≈ 94.25 cm³.

Q4.6 — Sphere d = 10 m

r = 5 m. V = (4/3) × π × 125 = 500π/3 ≈ 523.60 m³.

Q4.7 — Hemisphere r = 9

V = ⅙ × π × 729 = 1458/3 × π = 486π cm³ (≈ 1526.81 cm³).

Q4.8 — Cone ℓ = 13, r = 5 (find h then V)

(a) h = √(13² − 5²) = √(169 − 25) = √144 = 12 cm.
(b) V = ⅓ × π × 25 × 12 = 300π/3 = 100π cm³ (≈ 314.16 cm³).

Q4.9 — Pyramid V = 200, square base b = 10 (find h)

A = 100. 200 = ⅓ × 100 × h = (100h)/3. So 600 = 100h, h = 6 cm.

Q4.10 — Cone V = 48π, r = 4 (find h)

48π = ⅓ × π × 16 × h = (16π h)/3. Divide both sides by π: 48 = (16h)/3. So 144 = 16h, h = 9 cm.

Q4.11 — Sphere V = 500π/3 (find r)

500π/3 = (4/3)πr³. Multiply both sides by 3 and divide by 4π: r³ = (3 × 500π)/(3 × 4π) = 500/4 = 125. So r = 5 cm.

Q4.12 — Cylinder + cone on top (r = 4 throughout)

V_cyl = π(16)(10) = 160π. V_cone = ⅓π(16)(6) = 96π/3 = 32π. Total = 160π + 32π = 192π cm³ (≈ 603.19 cm³).