Mathematics • Year 10 • Unit 1 • Lesson 20
Volume of Pyramids, Cones and Spheres, Skill Drill
Build fluency with the three "tapered solid" volume formulas from Lesson 20: pyramid V = ⅓ × Abase × h, cone V = ⅓πr²h, and sphere V = ⁴⁄₃πr³. Plus the one-third rule: a pyramid (or cone) has exactly one-third the volume of a prism (or cylinder) with the same base and height.
1. I do, fully worked example
Read every step. Each one has a short reason on the right so you can see why, not just what.
Problem. A square-based pyramid has base side 6 cm and perpendicular height 10 cm. Find its volume.
Step 1, Spot the rule.
Pyramid → V = ⅓ × Abase × h.
Reason: every pyramid follows the one-third rule. The base is whatever polygon the pyramid sits on; here it's a square.
Step 2, Find the base area.
Abase = 6 × 6 = 36 cm²
Reason: square of side 6 has area side² = 36.
Step 3, Substitute into the formula (use the perpendicular height, not a slant).
V = ⅓ × 36 × 10
Reason: Abase = 36, h = 10. The lesson warns: always use the perpendicular height.
Step 4, Evaluate.
V = ⅓ × 360 = 120 cm³
Reason: multiply 36 × 10 first, then take a third.
Step 5, Check with the one-third rule.
A prism with the same base would have V = 36 × 10 = 360 cm³; the pyramid is one-third of that = 120 cm³ ✓
Answer: V = 120 cm³.
2. We do, fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 4 marks
Problem. A spherical water balloon has radius 12 cm. Find its volume in litres (to 2 d.p.).
Step 1, Spot the rule: sphere → V = __________________ πr³.
Step 2, Cube the radius:
r³ = 12³ = ______ cm³
Step 3, Substitute into the formula:
V = (4/3) × π × ______ = ______π cm³
Step 4, Convert to a decimal (1 d.p.):
V ≈ ______ cm³
Step 5, Convert cm³ to litres (1 cm³ = 1 mL; 1000 mL = 1 L):
V ≈ ______ mL ÷ 1000 ≈ ______ L (2 d.p.)
3. You do, independent practice
Show your working in the space under each problem. Give answers in exact form (in terms of π where appropriate) and as a decimal to 1 d.p. unless told otherwise. The first four are foundation. The middle two are standard. The last two are extension.
Foundation, single-formula tapered solids
3.1 A cone has radius 3 cm and perpendicular height 8 cm. Find its volume in exact form. 1 mark
3.2 A sphere has radius 6 cm. Find its volume in exact form. 1 mark
3.3 A square-based pyramid has base side 8 cm and perpendicular height 6 cm. Find its volume. 1 mark
3.4 A rectangular-based pyramid has base 5 cm × 4 cm and perpendicular height 6 cm. Find its volume. 1 mark
Standard, diameter and the one-third rule
3.5 A sphere has diameter 10 cm. Find its volume in exact form and to 1 d.p. 2 marks
3.6 A cone and a cylinder have the same radius and the same height. The cylinder has volume 60π cm³. Use the one-third rule to find the cone's volume. 2 marks
Extension, push your thinking
3.7 A cone has radius 4 cm and perpendicular height 9 cm. Find its volume in exact form. Then find what fraction of a sphere of the same radius (r = 4 cm) the cone would fill. 3 marks
3.8 A solid metal ball-bearing has radius 0.5 cm. Find its volume in exact form (as a fraction with π) and to 4 d.p. 2 marks
Section 2, We do (sphere r = 12)
Step 1: V = (4/3) πr³.
Step 2: r³ = 12³ = 1728 cm³.
Step 3: V = (4/3) × π × 1728 = 2304π cm³.
Step 4: V ≈ 7238.2 cm³.
Step 5: V ≈ 7238.2 mL ÷ 1000 ≈ 7.24 L (2 d.p.).
3.1, Cone (r = 3, h = 8)
V = ⅓ × π(3)²(8) = ⅓ × π(9)(8) = 24π cm³ (≈ 75.4 cm³).
3.2, Sphere (r = 6)
V = (4/3)π(6)³ = (4/3)π(216) = 288π cm³ (≈ 904.8 cm³).
3.3, Square pyramid (base 8, h = 6)
Abase = 64 cm². V = ⅓ × 64 × 6 = 128 cm³.
3.4, Rectangular pyramid (5 × 4 × h = 6)
Abase = 20 cm². V = ⅓ × 20 × 6 = 40 cm³.
3.5, Sphere (d = 10)
r = 5. V = (4/3)π(5)³ = (4/3)π(125) = 500π/3 ≈ 523.6 cm³ (1 d.p.).
3.6, Cone vs cylinder (same r, same h, cylinder = 60π)
One-third rule: Vcone = ⅓ × Vcylinder = ⅓ × 60π = 20π cm³.
3.7, Cone (r = 4, h = 9) vs sphere (r = 4)
Vcone = ⅓π(4)²(9) = ⅓π(16)(9) = 48π cm³.
Vsphere = (4/3)π(4)³ = (4/3)π(64) = 256π/3 cm³.
Fraction = (48π) ÷ (256π/3) = 48 × 3 ÷ 256 = 144/256 = 9/16.
So the cone holds 9/16 of the sphere's volume.
3.8, Ball bearing (r = 0.5 cm)
V = (4/3)π(0.5)³ = (4/3)π(0.125) = 0.5π/3 = π/6 cm³ ≈ 0.5236 cm³ (4 d.p.).