Mathematics • Year 9 • Unit 4 • Lesson 7
Composite Solids
Build fluency with the decompose-and-add (or decompose-and-subtract) approach to composite solids — one step at a time, from a fully worked example through guided practice to independent problems.
1. I do — fully worked example
Read every line. Each step has a short reason on the right so you can see why, not just what.
Problem. A solid consists of a cylinder (radius $3$ cm, height $4$ cm) sitting on top of a cube of side $6$ cm. Find the total volume in cm³, in terms of $\pi$ and as a decimal to 1 d.p.
Step 1 — Decompose into simple shapes.
Two parts: (i) a cube, (ii) a cylinder. Add their volumes.
Reason: the lesson tells us to identify simple parts and add. Joined volumes don't lose anything internally.
Step 2 — Volume of the cube.
$V_{\text{cube}} = 6^3 = 216$ cm³
Reason: a cube is a rectangular prism with all sides equal — $V = s^3$.
Step 3 — Volume of the cylinder.
$V_{\text{cyl}} = \pi r^2 h = \pi \times 3^2 \times 4 = 36\pi$ cm³
Reason: cylinder formula from Lesson 6. Square the radius first, then multiply.
Step 4 — Add the parts.
$V_{\text{total}} = 216 + 36\pi$ cm³
Reason: the cylinder sits ON TOP of the cube. No volume is shared, so we just add.
Step 5 — Decimal answer.
$36\pi \approx 113.1$, so $V \approx 216 + 113.1 = 329.1$ cm³
Answer: $\mathbf{216 + 36\pi}$ cm³ $\mathbf{\approx 329.1}$ cm³.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 5 marks
Problem. A rectangular block measuring $10$ cm × $8$ cm × $6$ cm has a cylindrical hole of radius $2$ cm drilled straight through the $6$ cm dimension. Find the remaining volume in terms of $\pi$.
Step 1 — Decompose: This is a subtractive composite. Volume = volume of block __________ volume of hole.
Step 2 — Volume of the block:
$V_{\text{block}} = 10 \times \_\_\_ \times \_\_\_ = \_\_\_\_$ cm³
Step 3 — Volume of the cylindrical hole. The hole goes through the $6$ cm dimension, so $h_{\text{hole}} = 6$ cm and $r = 2$ cm.
$V_{\text{hole}} = \pi \times \_\_^2 \times \_\_\_ = \_\_\_\_\pi$ cm³
Step 4 — Subtract:
$V_{\text{remaining}} = \_\_\_\_ - \_\_\_\_\pi$ cm³
Step 5 — Decimal estimate (use $\pi \approx 3.14$):
$V_{\text{remaining}} \approx \_\_\_\_\_\_\_\_$ cm³
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation (single decomposition). The middle two are standard (combine ideas). The last two are extension (surface area with hidden faces).
Foundation — simple decomposition
3.1 Two cubes of side $4$ cm are stacked one on top of the other. Find the total volume. 1 mark
3.2 A solid is a rectangular prism ($5$ cm × $4$ cm × $3$ cm) with a smaller rectangular prism ($2$ cm × $4$ cm × $3$ cm) stuck to one of the $4 \times 3$ faces. Find the total volume. 1 mark
3.3 A cube of side $10$ cm has a cube of side $4$ cm cut out of one corner. Find the remaining volume. 1 mark
3.4 A cylinder of radius $5$ cm and height $8$ cm has a cylinder of radius $2$ cm and height $8$ cm removed from its centre (a pipe). Find the remaining volume in terms of $\pi$. 1 mark
Standard — combine two ideas
3.5 A solid is a rectangular prism $8$ cm × $6$ cm × $4$ cm with a half-cylinder ($r = 3$ cm, length $8$ cm) sitting on top along the long edge. Find the total volume in terms of $\pi$. 2 marks
3.6 A "house" solid is a rectangular prism $5$ m × $4$ m × $3$ m with a triangular prism roof on top (the triangle has base $5$ m, perpendicular height $2$ m, and the roof prism is $4$ m long). Find the total volume. 2 marks
Extension — push your thinking
3.7 Two cubes of side $5$ cm are joined face-to-face. Find (a) the total volume, and (b) the total surface area (remember to subtract the hidden faces). 3 marks
3.8 A wooden block $12$ cm × $8$ cm × $5$ cm has a cylindrical hole of radius $1.5$ cm drilled all the way through the $5$ cm dimension. Find the remaining volume, leaving your answer in terms of $\pi$. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (block $10\times 8\times 6$ with hole $r=2$)
Step 1: subtractive — volume of block minus volume of hole.
Step 2: $V_{\text{block}} = 10 \times \mathbf{8} \times \mathbf{6} = \mathbf{480}$ cm³.
Step 3: $V_{\text{hole}} = \pi \times \mathbf{2}^2 \times \mathbf{6} = \mathbf{24}\pi$ cm³.
Step 4: $V_{\text{remaining}} = \mathbf{480} - \mathbf{24}\pi$ cm³.
Step 5: $480 - 24 \times 3.14 \approx 480 - 75.4 \approx \mathbf{404.6}$ cm³.
3.1 — Two cubes side 4 stacked
Each cube: $4^3 = 64$ cm³. Total: $2 \times 64 = \mathbf{128}$ cm³. (No volume is hidden — we just add.)
3.2 — Rectangular prism with smaller prism stuck on
$V_1 = 5 \times 4 \times 3 = 60$ cm³. $V_2 = 2 \times 4 \times 3 = 24$ cm³. Total $= \mathbf{84}$ cm³.
3.3 — Cube with corner cut out
$V_{\text{big}} = 10^3 = 1000$ cm³. $V_{\text{cut}} = 4^3 = 64$ cm³. Remaining $= 1000 - 64 = \mathbf{936}$ cm³.
3.4 — Cylindrical pipe
$V_{\text{outer}} = \pi \times 25 \times 8 = 200\pi$ cm³.
$V_{\text{inner}} = \pi \times 4 \times 8 = 32\pi$ cm³.
Remaining $= 200\pi - 32\pi = \mathbf{168\pi}$ cm³.
3.5 — Prism + half-cylinder on top
$V_{\text{prism}} = 8 \times 6 \times 4 = 192$ cm³.
$V_{\text{half-cyl}} = \tfrac{1}{2} \times \pi \times 3^2 \times 8 = \tfrac{1}{2} \times 72\pi = 36\pi$ cm³.
Total $= \mathbf{192 + 36\pi}$ cm³ $\approx 305.1$ cm³.
3.6 — House solid
$V_{\text{walls}} = 5 \times 4 \times 3 = 60$ m³.
Roof triangle area $= \tfrac{1}{2} \times 5 \times 2 = 5$ m². $V_{\text{roof}} = 5 \times 4 = 20$ m³.
Total $= 60 + 20 = \mathbf{80}$ m³.
3.7 — Two cubes side 5 joined face-to-face
(a) $V = 2 \times 5^3 = \mathbf{250}$ cm³ (volumes simply add).
(b) SA of one cube $= 6 \times 5^2 = 150$ cm². Two separate cubes: $2 \times 150 = 300$ cm². Hidden faces: each cube loses one $5 \times 5 = 25$ cm² face, so subtract $2 \times 25 = 50$ cm². Total SA $= 300 - 50 = \mathbf{250}$ cm².
The lesson's "subtract twice the overlap" rule applied directly.
3.8 — Wooden block with hole
$V_{\text{block}} = 12 \times 8 \times 5 = 480$ cm³.
Hole length = 5 cm (the dimension drilled through). $V_{\text{hole}} = \pi \times (1.5)^2 \times 5 = \pi \times 2.25 \times 5 = 11.25\pi$ cm³.
Remaining $= \mathbf{480 - 11.25\pi}$ cm³ $\approx 444.7$ cm³.