Mathematics • Year 10 • Unit 1 • Lesson 19

Composite Volumes in the Real World

Apply addition and subtraction strategies to real Australian situations — backyard sheds with peaked roofs, grain silos with conical tops, drilled engineering blocks and trophies. Choose the right strategy each time and then explain your method in your own words.

Apply · Real-World Maths

1. Word problems

Each problem is a composite solid from real Australian engineering or design. Decide whether to add simpler volumes or subtract a removed piece, then show every step. Use exact form first and round only at the end.

1.1 — Backyard shed with triangular roof. A garden shed has a rectangular prism base 5 m × 4 m × 2.5 m (height of the walls), plus a triangular-prism roof on top. The roof's triangular cross-section has base 5 m (along the front wall) and perpendicular height 1.5 m; the roof is 4 m long (same as the building depth).

(a) Calculate the volume of the rectangular base.
(b) Calculate the volume of the triangular roof.
(c) Calculate the total internal volume of the shed. 3 marks

Stuck? Addition strategy: V_shed = V_{base prism} + V_{roof triangular prism}.

1.2 — Trophy on a base. A sports trophy consists of a rectangular wooden base measuring 12 cm × 8 cm × 3 cm, with a metal cylinder of radius 3 cm and height 10 cm fixed to the top.

(a) Find the volume of the wooden base.
(b) Find the volume of the metal cylinder in exact form.
(c) Find the total volume of the trophy to 1 d.p. 3 marks

Stuck? Addition: V_total = 12×8×3 + π(3)²(10).

1.3 — Drilled engineering block. A rectangular block of wood measures 20 cm × 15 cm × 10 cm. A cylindrical hole of diameter 6 cm is drilled through the block along its 10 cm height for a pole.

(a) Find the volume of the original block.
(b) Find the volume of the cylindrical hole.
(c) Find the volume of wood remaining (to 1 d.p.). 3 marks

Stuck? Subtraction: V_wood = lwh − πr²h. Halve the diameter for r first.

1.4 — L-shaped concrete slab. An L-shaped backyard concrete slab is 10 cm thick. The L-shape consists of one rectangle 6 m × 4 m joined to a second rectangle 3 m × 2 m. Concrete costs $185 per m³.

(a) Calculate the area of the L-shaped slab.
(b) Calculate the volume of concrete needed.
(c) Calculate the cost of the concrete. 3 marks

Stuck? Addition for area: 6×4 + 3×2. Then ×0.10 m to get m³, then ×$185.

1.5 — Grain silo with conical roof. A grain silo on a rural property consists of a cylindrical section of diameter 4 m and height 8 m, with a conical roof of the same diameter and perpendicular height 2 m sitting on top. (Use V_cone = ⅓πr²h.)

(a) Find the volume of the cylindrical section.
(b) Find the volume of the conical roof.
(c) Find the total capacity of the silo in kilolitres to 1 d.p. 4 marks

Stuck? r = 2 m. V_cyl = π(2)²(8); V_cone = ⅓ × π(2)²(2). 1 m³ = 1 kL.

2. Explain your thinking

This question is about communication, not just numbers. Use full sentences. 4 marks

2.1 A classmate looks at the drilled engineering block from question 1.3 and says: "I'll just split the remaining wood into 4 rectangular blocks around the hole and add them up — that way I don't have to use π at all." Using the lesson's two strategies, explain (i) why subtraction is the more efficient strategy for this problem, (ii) what would actually go wrong if your classmate tried the "4 blocks" approach (think about the shape of the leftover wood once a round hole is drilled), and (iii) when the addition strategy would be the better choice.

Stuck? The corners of the block around a round hole are not rectangular — they have curved edges from the cylinder. Addition is better when the parts are simple shapes that genuinely fit together without overlap or curved boundaries.

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Answers — Do not peek before attempting

1.1 — Shed with triangular roof

(a) V_base = 5 × 4 × 2.5 = 50 m³.
(b) V_roof = ½ × 5 × 1.5 × 4 = 15 m³.
(c) Total = 50 + 15 = 65 m³.

1.2 — Trophy (block + cylinder)

(a) V_base = 12 × 8 × 3 = 288 cm³.
(b) V_cyl = π(3)²(10) = 90π cm³.
(c) Total = 288 + 90π ≈ 570.7 cm³ (1 d.p.).

1.3 — Drilled wood block

(a) V_block = 20 × 15 × 10 = 3000 cm³.
(b) r = 3 cm. V_hole = π(3)²(10) = 90π cm³ (≈ 282.7 cm³).
(c) Remaining = 3000 − 90π ≈ 2717.3 cm³ (1 d.p.).

1.4 — L-shaped concrete slab

(a) Area = 6 × 4 + 3 × 2 = 24 + 6 = 30 m².
(b) V = 30 × 0.10 = 3 m³.
(c) Cost = 3 × $185 = $555.

1.5 — Grain silo (cylinder + cone, d = 4 m)

(a) r = 2 m. V_cyl = π(2)²(8) = 32π m³.
(b) V_cone = ⅓ × π(2)²(2) = 8π/3 m³ = (8π/3) m³.
(c) Total = 32π + 8π/3 = (96π + 8π)/3 = 104π/3 ≈ 108.9 m³ = 108.9 kL (1 d.p.).

2.1 — Explain your thinking (sample response)

(i) Subtraction is more efficient because the block and the hole are both single simple solids whose volumes you can compute in one calculation each — V_block = lwh and V_hole = πr²h. One subtraction gives the answer, and using π is unavoidable because the hole really is circular.
(ii) The "4 blocks around the hole" approach would not work: once a round hole is drilled, the leftover wood does not have flat sides next to the hole — each "block" would have a curved cut-out where the cylinder used to be. Those curved-edge pieces are not rectangular prisms, so lwh doesn't apply and the calculation breaks.
(iii) The addition strategy is better when the composite is genuinely two or more simple shapes joined at a flat face — for example, the trophy in 1.2 (block + cylinder stacked) or the shed in 1.1 (rectangular prism + triangular prism roof). In those cases each part is a clean simple solid and no curved leftover appears.

Marking: 1 for naming subtraction's efficiency, 1 for the curved-edge problem of the "4 blocks" idea, 1 for a clear example where addition is appropriate, 1 for full-sentence communication.