Mathematics • Year 10 • Unit 1 • Lesson 17
Surface Area of Cylinders, Cones and Spheres — Skill Drill
Build fluency with the three curved-surface formulas from Lesson 17: cylinder TSA = 2πrh + 2πr², cone TSA = πrl + πr² with slant height l = √(r² + h²), and sphere SA = 4πr². One step at a time, from a fully worked example through guided practice to independent problems.
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 closed cylindrical tin has radius 7 cm and height 12 cm. Find its total surface area in exact form and as a decimal (1 d.p.).
Step 1 — Spot the rule.
Closed cylinder → TSA = 2πrh + 2πr².
Reason: a closed cylinder has a curved side (2πrh, the rectangle when unrolled) plus two circular bases (2 × πr²).
Step 2 — Substitute r = 7, h = 12 into the curved part.
Curved SA = 2π(7)(12) = 168π cm²
Reason: 2 × 7 × 12 = 168. Keep π in the answer so it's exact.
Step 3 — Substitute r = 7 into the two bases.
2 bases = 2π(7)² = 2π(49) = 98π cm²
Reason: each base is a circle of area πr² = π × 49. Two of them.
Step 4 — Add the curved part and the bases.
TSA = 168π + 98π = 266π cm²
Reason: both terms are in π so the coefficients add directly.
Step 5 — Convert to a decimal (1 d.p.).
266π ≈ 266 × 3.14159 ≈ 835.7 cm²
Reason: rounding only at the final step keeps the answer accurate.
Answer: TSA = 266π ≈ 835.7 cm² (1 d.p.).
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 cone has radius 5 cm and perpendicular height 12 cm. Find the total surface area.
Step 1 — Spot the rule: a cone has one circular base (πr²) and a curved surface (πrl). Before substituting, we need the __________________ height.
Step 2 — Use Pythagoras to find the slant height l:
l = √(r² + h²) = √(5² + 12²) = √(25 + ______) = √______ = ______ cm
Step 3 — Substitute into the curved surface area:
CSA = πrl = π(5)(______) = ______π cm²
Step 4 — Substitute into the base:
Base = πr² = π(5)² = ______π cm²
Step 5 — Add and convert to a decimal (1 d.p.):
TSA = ______π + ______π = ______π ≈ ______ cm²
3. You do — independent practice
Show your working in the space under each problem. Give answers in exact form (in terms of π) and as a decimal to 1 d.p. unless told otherwise. The first four are foundation (one formula, one substitution). The middle two are standard (slant height or open cylinder). The last two are extension (multi-step in syllabus).
Foundation — single-formula curved solids
3.1 Find the curved surface area of a cylinder with radius 5 cm and height 8 cm. 1 mark
3.2 Find the surface area of a sphere with radius 3 cm. 1 mark
3.3 Find the surface area of a sphere with radius 6 cm. 1 mark
3.4 Find the total surface area of a closed cylinder with radius 4 cm and height 10 cm. 1 mark
Standard — slant height and open cylinder
3.5 A cone has radius 6 cm and perpendicular height 8 cm. (a) Find the slant height. (b) Find the total surface area. 2 marks
3.6 An open-top cylinder (no lid, but with a base) has radius 4 cm and height 9 cm. Find the total surface area. 2 marks
Extension — push your thinking
3.7 A cylindrical can has diameter 7 cm and height 11 cm. The label wraps once around the curved side with no overlap. Find the area of the label. 3 marks
3.8 A hemisphere (half-sphere, including its flat circular base) has radius 5 cm. Find its total surface area. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (cone r = 5, h = 12)
Step 1: slant height.
Step 2: l = √(25 + 144) = √169 = 13 cm.
Step 3: CSA = π(5)(13) = 65π cm².
Step 4: Base = π(5)² = 25π cm².
Step 5: TSA = 65π + 25π = 90π ≈ 282.7 cm².
3.1 — Cylinder CSA (r = 5, h = 8)
CSA = 2πrh = 2π(5)(8) = 80π ≈ 251.3 cm².
3.2 — Sphere SA (r = 3)
SA = 4πr² = 4π(9) = 36π ≈ 113.1 cm².
3.3 — Sphere SA (r = 6)
SA = 4π(6)² = 4π(36) = 144π ≈ 452.4 cm².
3.4 — Closed cylinder TSA (r = 4, h = 10)
TSA = 2πrh + 2πr² = 2π(4)(10) + 2π(16) = 80π + 32π = 112π ≈ 351.9 cm².
3.5 — Cone (r = 6, h = 8)
(a) l = √(6² + 8²) = √(36 + 64) = √100 = 10 cm.
(b) TSA = πrl + πr² = π(6)(10) + π(36) = 60π + 36π = 96π ≈ 301.6 cm².
3.6 — Open-top cylinder (r = 4, h = 9)
CSA = 2π(4)(9) = 72π. One base only = π(4)² = 16π.
TSA = 72π + 16π = 88π ≈ 276.5 cm².
An open-top cylinder has the curved side + 1 base (not 2).
3.7 — Can label (d = 7, h = 11)
r = 7 ÷ 2 = 3.5 cm.
Label area = curved surface only = 2πrh = 2π(3.5)(11) = 77π ≈ 241.9 cm².
The label is the rectangle you get when you unroll the curved side. Width = height = 11 cm; length = circumference = 2π(3.5) = 7π cm.
3.8 — Hemisphere with base (r = 5)
Curved half = ½ × 4πr² = 2π(25) = 50π cm².
One base = π(25) = 25π cm².
Total = 50π + 25π = 75π ≈ 235.6 cm².