Mathematics • Year 10 • Unit 1 • Lesson 19
Volume of Composite Solids — Skill Drill
Build fluency with the two composite-volume strategies from Lesson 19: the addition strategy (split into simple solids, add their volumes) and the subtraction strategy (Vcomposite = Venclosing − Vhole). 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 toy consists of a rectangular block 8 cm × 6 cm × 4 cm with a cylinder of radius 2 cm and height 6 cm placed on top. Find the total volume in exact form and to 1 d.p.
Step 1 — Spot the strategy.
Two pieces joined → use the ADDITION strategy.
Reason: the block and cylinder don't overlap — they sit on top of each other. Find each volume and add.
Step 2 — Volume of the block (rectangular prism).
Vblock = 8 × 6 × 4 = 192 cm³
Reason: lwh for a rectangular prism.
Step 3 — Volume of the cylinder.
Vcyl = π(2)²(6) = 24π cm³
Reason: πr²h with r = 2, h = 6.
Step 4 — Add the two volumes.
Vtotal = 192 + 24π cm³
Reason: keep the exact form first — it's easier to mark and easier to round at the end.
Step 5 — Convert to a decimal (1 d.p.).
Vtotal ≈ 192 + 75.4 ≈ 267.4 cm³
Reason: 24π ≈ 75.398. Add and round to 1 d.p. at the very end.
Answer: V = 192 + 24π ≈ 267.4 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 concrete block measures 2 m × 1.5 m × 0.8 m, with a cylindrical hole of diameter 0.4 m drilled all the way through its 0.8 m height. Find the volume of concrete (to 2 d.p.).
Step 1 — Spot the strategy: something has been removed from a larger solid, so use the __________________ strategy.
Step 2 — Volume of the enclosing block:
Vblock = 2 × 1.5 × 0.8 = ______ m³
Step 3 — Convert diameter to radius for the hole:
r = 0.4 ÷ 2 = ______ m
Step 4 — Volume of the cylindrical hole (height = block height = 0.8 m):
Vhole = π × (______)² × 0.8 = ______π m³ ≈ ______ m³
Step 5 — Subtract to get the concrete volume:
Vconcrete = ______ − ______ ≈ ______ m³ (2 d.p.)
3. You do — independent practice
Show your working in the space under each problem. Choose addition or subtraction at the start of each. The first four are foundation (one strategy, integer answers). The middle two are standard (two solids with π). The last two are extension (multi-step in syllabus).
Foundation — pick a strategy
3.1 A solid is made from a cube of side 6 cm with a smaller cube of side 2 cm removed from one corner. Find the remaining volume. 1 mark
3.2 Two stacked rectangular boxes: the bottom box is 10 × 8 × 5 cm and the top box is 4 × 4 × 4 cm. Find the total volume. 1 mark
3.3 An L-shaped prism is made of two rectangular blocks joined: 6 × 4 × 3 cm and 5 × 4 × 3 cm. Find the total volume. 1 mark
3.4 A rectangular prism 10 × 10 × 4 cm has a square hole of side 2 cm cut right through the 4 cm height. Find the remaining volume. 1 mark
Standard — composite with a cylinder
3.5 A composite is a rectangular prism 10 × 8 × 5 cm with a cylinder of radius 3 cm and height 5 cm placed on top. Find the total volume in exact form and to 1 d.p. 2 marks
3.6 A cube of side 10 cm has a cylindrical hole of radius 2 cm drilled all the way through its 10 cm height. Find the remaining volume in exact form and to 1 d.p. 2 marks
Extension — push your thinking
3.7 A storage shed is a rectangular prism 10 m × 6 m × 3 m with a half-cylinder roof (radius 3 m) running along the 10 m length. Find the total volume in exact form and to 1 d.p. 3 marks
3.8 A solid metal rod is a cylinder of radius 2 cm and length 50 cm. A cylindrical hole of radius 1 cm is drilled along its entire length. Find the volume of metal remaining in exact form. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (concrete block 2 × 1.5 × 0.8 m with d = 0.4 m hole)
Step 1: subtraction strategy.
Step 2: Vblock = 2 × 1.5 × 0.8 = 2.4 m³.
Step 3: r = 0.4 ÷ 2 = 0.2 m.
Step 4: Vhole = π × (0.2)² × 0.8 = 0.032π m³ ≈ 0.1005 m³.
Step 5: Vconcrete = 2.4 − 0.1005 ≈ 2.30 m³ (2 d.p.).
3.1 — Cube 6 minus cube 2
V = 6³ − 2³ = 216 − 8 = 208 cm³.
3.2 — Two stacked boxes
V = 10×8×5 + 4×4×4 = 400 + 64 = 464 cm³.
3.3 — L-shaped prism (6×4×3 + 5×4×3)
V = 72 + 60 = 132 cm³.
3.4 — Block 10×10×4 minus 2×2 square hole through
Vblock = 10 × 10 × 4 = 400 cm³.
Vhole = 2 × 2 × 4 = 16 cm³.
V = 400 − 16 = 384 cm³.
3.5 — Block 10×8×5 + cylinder r=3, h=5 on top
Vblock = 10 × 8 × 5 = 400 cm³.
Vcyl = π(3)²(5) = 45π cm³.
Total = 400 + 45π ≈ 541.4 cm³ (1 d.p.).
3.6 — Cube 10 with cylindrical hole r=2 through
Vcube = 10³ = 1000 cm³.
Vhole = π(2)²(10) = 40π cm³.
Remaining = 1000 − 40π ≈ 874.3 cm³ (1 d.p.).
3.7 — Shed 10×6×3 m with half-cylinder roof (r = 3, L = 10)
Vprism = 10 × 6 × 3 = 180 m³.
Vhalf-cyl = ½ × π × 3² × 10 = 45π m³ ≈ 141.4 m³.
Total = 180 + 45π ≈ 321.4 m³ (1 d.p.).
3.8 — Rod r=2, L=50 with r=1 hole drilled lengthwise
Vrod = π(2)²(50) = 200π cm³.
Vhole = π(1)²(50) = 50π cm³.
Metal remaining = 200π − 50π = 150π cm³ (≈ 471.2 cm³).