Mathematics • Year 9 • Unit 4 • Lesson 7
Composite Solids — Mixed Challenge
Pull together every idea from Lesson 7: additive and subtractive composites, hidden faces, holes drilled through solids, and the careful tracking of which surfaces count. You'll choose the right tool, spot someone else's mistake, and tackle an open-ended design challenge.
1. Mixed problems — choose the right approach
Each question uses a different combination of ideas from Lesson 7. Decide whether to add, subtract, or both before you start writing. Show your working. 3 marks each
1.1 A solid is a cylinder ($r = 4$ cm, $h = 5$ cm) with a hemisphere ($r = 4$ cm) on top. Find the total volume in terms of $\pi$. (Hemisphere volume $= \tfrac{2}{3}\pi r^3$.)
1.2 A cube of side $6$ cm has a cylindrical hole of radius $1$ cm drilled all the way through one face (so the hole length equals the cube side). Find the remaining volume in terms of $\pi$ and as a decimal to 1 d.p.
1.3 Three identical cubes of side $4$ cm are stacked in an L-shape: two side-by-side on the bottom, one cube sitting on top of the left one. Find the total surface area, accounting for hidden faces.
1.4 A rectangular prism $10$ cm × $6$ cm × $4$ cm has a half-cylinder of radius $3$ cm and length $10$ cm scooped out of its top (along the long edge). Find the remaining volume.
1.5 Two cylinders of radius $5$ cm are joined end-to-end. One has height $8$ cm, the other has height $12$ cm. Find the total surface area in terms of $\pi$, remembering that the two circular faces where they join are hidden.
1.6 A solid is a rectangular prism $12$ cm × $8$ cm × $4$ cm with a rectangular hole $4$ cm × $4$ cm cut all the way through the $4$ cm dimension. Find (a) the volume of the remaining solid, and (b) the total surface area of the remaining solid (including the inner surface of the hole — remember the hole is open at both ends).
2. Find the mistake
Another student has tried to find the total surface area of three cubes of side $4$ cm joined in a straight line (face-to-face-to-face, forming a row). Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — SA of three cubes side 4 in a row:
Line 1: SA of one cube $= 6 \times 4^2 = 96$ cm²
Line 2: SA of three separate cubes $= 3 \times 96 = 288$ cm²
Line 3: Joining hides $2$ faces total, area $= 2 \times 16 = 32$ cm²
Line 4: Total SA $= 288 - 32 = 256$ cm²
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? Count the joins! Three cubes in a row means TWO joins, not one. Each join hides $2$ faces.3. Open-ended challenge — design a $1$-litre container
This question has many valid answers. 4 marks
3.1 Design a composite container made from at least two different solids (e.g. cylinder + half-sphere, two stacked prisms, prism with a cylinder removed, etc.) that holds exactly $1000$ cm³ ($\pm 50$ cm³).
For your design:
(i) Sketch it (label all dimensions clearly).
(ii) State the formula(s) you used.
(iii) Show the calculation that your design totals close to $1000$ cm³.
(iv) Suggest a real-world product that has a similar shape.
Bonus: If you can also state the approximate surface area of your container (correctly accounting for any hidden faces), claim a bonus mark.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Cylinder + hemisphere
$V_{\text{cyl}} = \pi \times 16 \times 5 = 80\pi$ cm³.
$V_{\text{hemi}} = \tfrac{2}{3}\pi \times 64 = \tfrac{128}{3}\pi$ cm³.
Total $= 80\pi + \tfrac{128}{3}\pi = \dfrac{240\pi + 128\pi}{3} = \mathbf{\dfrac{368\pi}{3}}$ cm³ $\approx 385.4$ cm³.
1.2 — Cube with cylindrical hole
$V_{\text{cube}} = 6^3 = 216$ cm³.
$V_{\text{hole}} = \pi \times 1^2 \times 6 = 6\pi$ cm³.
Remaining $= \mathbf{216 - 6\pi}$ cm³ $\approx 197.2$ cm³.
1.3 — Three cubes in an L-shape
SA of each cube $= 6 \times 16 = 96$ cm². Three separate cubes: $3 \times 96 = 288$ cm².
Number of joins: 2 (bottom-bottom and bottom-top of the left cube). Each join hides $2$ faces of area $16$ cm².
Hidden $= 2 \times 2 \times 16 = 64$ cm².
Total SA $= 288 - 64 = \mathbf{224}$ cm².
1.4 — Prism with half-cylinder scoop
$V_{\text{prism}} = 10 \times 6 \times 4 = 240$ cm³.
$V_{\text{scoop}} = \tfrac{1}{2} \times \pi \times 3^2 \times 10 = 45\pi$ cm³.
Remaining $= 240 - 45\pi \approx \mathbf{98.6}$ cm³.
1.5 — Two stacked cylinders
Treat as one combined cylinder of radius $5$ cm and total height $8 + 12 = 20$ cm.
Curved SA: $2\pi r h = 2\pi \times 5 \times 20 = 200\pi$ cm².
Two end circles: $2 \times \pi \times 25 = 50\pi$ cm².
Total SA $= \mathbf{250\pi}$ cm² $\approx 785.4$ cm².
The two circles where the cylinders join are hidden (one from each cylinder), so they don't appear in the answer — they have already been subtracted by treating the join as one continuous cylinder.
1.6 — Prism with square hole through it
(a) $V_{\text{prism}} = 12 \times 8 \times 4 = 384$ cm³. $V_{\text{hole}} = 4 \times 4 \times 4 = 64$ cm³. Remaining $= \mathbf{320}$ cm³.
(b) Original SA of prism $= 2(12 \times 8) + 2(12 \times 4) + 2(8 \times 4) = 192 + 96 + 64 = 352$ cm².
Subtract two $4 \times 4 = 16$ cm² openings (one each side): $352 - 32 = 320$ cm².
Add the four inner walls of the square hole: each is $4$ cm × $4$ cm (depth) $= 16$ cm², so $4 \times 16 = 64$ cm².
Total SA $= 320 + 64 = \mathbf{384}$ cm².
2 — Find the mistake
(a) The mistake is on Line 3.
(b) Three cubes in a row have two joins (cube 1↔cube 2, cube 2↔cube 3), not one. Each join hides $2$ faces, so $4$ faces are hidden in total, not $2$.
(c) Corrected working:
SA of one cube $= 6 \times 4^2 = 96$ cm²
SA of three separate cubes $= 3 \times 96 = 288$ cm²
Joining hides $4$ faces total (two joins × two faces each), area $= 4 \times 16 = 64$ cm²
Total SA $= 288 - 64 = \mathbf{224}$ cm².
This is exactly the answer to question 1.3 — same shape!
3 — Open-ended challenge (sample solution)
Design: cylinder + half-sphere "drink bottle".
Sketch: A cylinder of radius $r = 4$ cm and height $h = 17$ cm, with a half-sphere of the same radius $4$ cm sitting on top (like a smoothie bottle).
Formulas: Cylinder $V = \pi r^2 h$. Half-sphere $V = \tfrac{1}{2} \times \tfrac{4}{3}\pi r^3 = \tfrac{2}{3}\pi r^3$.
Calculation:
$V_{\text{cyl}} = \pi \times 16 \times 17 = 272\pi \approx 854.5$ cm³.
$V_{\text{half-sphere}} = \tfrac{2}{3}\pi \times 64 \approx 134.0$ cm³.
Total $\approx 988.5$ cm³ — within $\pm 50$ cm³ of $1000$. ✓
Real-world product: A cylindrical smoothie bottle or shaker with a domed cap.
Marking: 1 for a sketch with labelled dimensions; 1 for correct formulas; 1 for a calculation that lands in $\pm 50$ cm³ of $1000$; 1 for a real-world product analogy. Bonus mark available for a correct surface-area calculation.