Mathematics • Year 9 • Unit 4 • Lesson 7
Composite Solids in the Real World
Use the decompose-and-add (or subtract) method in everyday contexts: a garden shed, a concrete pipe, a swimming pool with a built-in spa, a hollowed-out lighthouse model, and the painting of a barn. Then explain your method in your own words.
1. Word problems
Each problem uses the decompose-and-add (or subtract) strategy from Lesson 7. Show your working — a single final answer with no working only earns half marks.
1.1 — Garden shed. A garden shed is shaped like a "house": a rectangular prism base measuring $3$ m × $2$ m × $2$ m, with a triangular prism roof on top (the triangle has base $3$ m, perpendicular height $1$ m, and the roof prism is $2$ m long).
(a) Find the volume of the walls (rectangular prism).
(b) Find the volume of the roof (triangular prism).
(c) Find the total internal volume of the shed in m³. 3 marks
1.2 — Concrete pipe. A concrete drainage pipe is a cylinder $2$ m long with outer radius $30$ cm and inner radius $25$ cm (so the wall thickness is $5$ cm).
(a) Find the volume of the outer cylinder (treating it as solid).
(b) Find the volume of the hollow inside.
(c) Subtract to find the volume of concrete used, in cm³ and in m³. 4 marks
1.3 — Pool with built-in spa. A rectangular swimming pool is $8$ m × $4$ m × $1.5$ m deep. At one end, a small cylindrical spa (radius $1$ m, depth $0.8$ m) is built into the pool floor (i.e. it extends down a further $0.8$ m below the pool floor).
(a) Find the volume of water in the rectangular pool when full.
(b) Find the extra volume added by the cylindrical spa cavity.
(c) Total capacity of pool + spa, in litres ($1$ m³ $= 1000$ L)? 4 marks
1.4 — Lighthouse model. A model lighthouse for a school project is a cylinder ($r = 4$ cm, $h = 20$ cm) with a smaller cylinder ($r = 3$ cm, $h = 18$ cm) hollowed out from the inside, leaving solid walls and a solid base 2 cm thick.
(a) Find the outer volume.
(b) Find the hollowed-out interior volume.
(c) How much plaster (in cm³) was needed to make the model? Leave your answer in terms of $\pi$. 3 marks
1.5 — Painting a barn. A simple barn is a rectangular prism $10$ m × $6$ m × $4$ m, with a triangular prism roof (triangle base $6$ m, perpendicular height $2$ m, roof length $10$ m). The owner wants to paint just the four external rectangular walls and the two triangular gable ends — not the floor, not the sloping roof, not the doors.
(a) Find the area of the four walls.
(b) Find the area of the two triangular gable ends.
(c) If $1$ litre of paint covers $10$ m², how many litres are needed (round UP to a whole litre)? 3 marks
2. Explain your thinking
This question is about communication, not just answers. Use full sentences. 4 marks
2.1 Two identical cubes of side $5$ cm are joined face-to-face. A classmate says "the total surface area is just $2 \times 6 \times 25 = 300$ cm² because there are still 12 faces in total." In your own words, explain (i) what is wrong with their reasoning, (ii) which key idea from Lesson 7 they have ignored, and (iii) the correct surface area. Refer to "hidden faces" somewhere in your explanation.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Garden shed
(a) $V_{\text{walls}} = 3 \times 2 \times 2 = \mathbf{12}$ m³.
(b) Triangle area $= \tfrac{1}{2} \times 3 \times 1 = 1.5$ m². $V_{\text{roof}} = 1.5 \times 2 = \mathbf{3}$ m³.
(c) Total $= 12 + 3 = \mathbf{15}$ m³.
This is the "house = rectangular + triangular prism" example from the lesson, made concrete.
1.2 — Concrete pipe
Convert: $r_{\text{out}} = 0.30$ m, $r_{\text{in}} = 0.25$ m, $h = 2$ m.
(a) $V_{\text{outer}} = \pi (0.30)^2 \times 2 = 0.18\pi$ m³ $\approx 0.565$ m³.
(b) $V_{\text{inner}} = \pi (0.25)^2 \times 2 = 0.125\pi$ m³ $\approx 0.393$ m³.
(c) Concrete $= 0.18\pi - 0.125\pi = \mathbf{0.055\pi}$ m³ $\approx \mathbf{0.173}$ m³ ($= 173\,000$ cm³).
Real-world note: storm-water pipes are designed exactly this way — outer cylinder minus inner cylinder gives the concrete required.
1.3 — Pool with spa
(a) $V_{\text{pool}} = 8 \times 4 \times 1.5 = \mathbf{48}$ m³.
(b) $V_{\text{spa}} = \pi \times 1^2 \times 0.8 = 0.8\pi \approx \mathbf{2.51}$ m³.
(c) Total volume $\approx 48 + 2.51 = 50.51$ m³ $\approx \mathbf{50\,510}$ L.
1.4 — Lighthouse model
(a) $V_{\text{outer}} = \pi \times 4^2 \times 20 = 320\pi$ cm³.
(b) $V_{\text{hollow}} = \pi \times 3^2 \times 18 = 162\pi$ cm³.
(c) Plaster $= 320\pi - 162\pi = \mathbf{158\pi}$ cm³ $\approx 496.4$ cm³.
1.5 — Painting the barn
(a) Two long walls $10 \times 4 = 40$ m² each $= 80$ m². Two short walls $6 \times 4 = 24$ m² each $= 48$ m². Total walls $= \mathbf{128}$ m².
(b) Two gable triangles $\tfrac{1}{2} \times 6 \times 2 = 6$ m² each $= \mathbf{12}$ m².
(c) Total area $= 128 + 12 = 140$ m². Paint $= 140 \div 10 = 14$ L. Round up: $\mathbf{14}$ L (already whole).
This shows the decomposition idea applied to surface area, not just volume.
2.1 — Explain your thinking (sample response)
My classmate has counted all 12 faces as if the cubes were still separate, but when the cubes are joined face-to-face two faces become hidden: the right face of cube 1 and the left face of cube 2 touch each other and are no longer external surfaces. The key idea from Lesson 7 they have ignored is "subtract twice the overlap area" when finding the surface area of joined solids. The correct surface area is $300 - 2 \times 25 = \mathbf{250}$ cm². Their answer of $300$ cm² counts surfaces that aren't actually visible from outside the new solid.
Marking: 1 mark for naming the hidden faces; 1 for stating the "subtract twice" rule; 1 for the correct $250$ cm²; 1 for clear sentences.