Study Companion
Topic 09 | Measurement

Surface area and volume of composite objects

Year 10 core: surface area and volume of composite objects formed from prisms and cylinders, with practical applications in packaging, tanks, and construction.

50-65 min Printable practice Answer key Challenge included
VC2M10M01
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What you will learn

Worked example 0 Real-world example: rainwater tank capacity

A rainwater tank consists of a cylindrical body with radius 0.60.6 m and height 1.51.5 m, sitting on a rectangular concrete slab 1.41.4 m by 1.41.4 m by 0.10.1 m thick. Find (a) the total volume of the assembly, and (b) the exposed surface area to be painted (the curved surface and top of the cylinder only — the slab is unpainted).

  1. Cylinder volume: V=π(0.6)2(1.5)=0.54π≈1.696V = \pi(0.6)^2(1.5) = 0.54\pi \approx 1.696 m3^3 (capacity ≈1696\approx 1696 L).
  2. Slab volume: 1.4×1.4×0.1=0.1961.4 \times 1.4 \times 0.1 = 0.196 m3^3.
  3. Total volume ≈1.696+0.196=1.892\approx 1.696 + 0.196 = 1.892 m3^3.
  4. Curved surface area of cylinder: 2π(0.6)(1.5)=1.8π≈5.6552\pi(0.6)(1.5) = 1.8\pi \approx 5.655 m2^2.
  5. Top circle: π(0.6)2=0.36π≈1.131\pi(0.6)^2 = 0.36\pi \approx 1.131 m2^2.
  6. Area to paint ≈5.655+1.131=6.786\approx 5.655 + 1.131 = 6.786 m2^2.

Key idea: always identify which faces are exposed (visible) and which are hidden where components join.

1. Strategy for composite objects

Every composite-object problem follows the same two-step strategy:

Volume: add (or subtract) the volumes of each component.

Surface area: add the surface areas of each component, then subtract the hidden (shared) faces — those internal areas where the shapes meet.

Composite volume
Vcomposite=V1+V2+⋯(or subtract if a piece is removed)V_{\text{composite}} = V_1 + V_2 + \cdots \quad\text{(or subtract if a piece is removed)}
Composite surface area
SAcomposite=SA1+SA2−2×Ashared\text{SA}_{\text{composite}} = \text{SA}_1 + \text{SA}_2 - 2 \times A_{\text{shared}}

The factor of 22 appears because the shared face is counted once in each component’s full SA.

Worked example 1 Cylinder sitting on a prism

A cylinder of radius 33 cm and height 55 cm sits centred on top of a rectangular prism 10×10×410 \times 10 \times 4 cm. Find the total exposed surface area.

  1. Prism full SA: 2(100)+4(40)=200+160=3602(100) + 4(40) = 200 + 160 = 360 cm2^2.
  2. Cylinder full SA: 2π(9)+2π(3)(5)=18π+30π=48π≈150.802\pi(9) + 2\pi(3)(5) = 18\pi + 30\pi = 48\pi \approx 150.80 cm2^2.
  3. Shared face (circle where cylinder meets prism top): A=π(3)2=9π≈28.27A = \pi(3)^2 = 9\pi \approx 28.27 cm2^2.
  4. Subtract shared face twice (once from prism top, once from cylinder base): exposed SA ≈360+150.80−2(28.27)≈454.26\approx 360 + 150.80 - 2(28.27) \approx 454.26 cm2^2.

2. Volume of composite objects

Worked example 2 Prism with a cylindrical hole

A rectangular block of steel is 20×12×820 \times 12 \times 8 cm. A cylindrical hole of radius 22 cm is drilled all the way through the 2020 cm length. Find the remaining volume.

  1. Block volume: 20×12×8=192020 \times 12 \times 8 = 1920 cm3^3.
  2. Cylinder removed: π(2)2(20)=80π≈251.33\pi(2)^2(20) = 80\pi \approx 251.33 cm3^3.
  3. Remaining volume: 1920−251.33≈1668.671920 - 251.33 \approx 1668.67 cm3^3.
Worked example 3 L-shaped prism

An L-shaped concrete pad can be split into two rectangular prisms: one is 4×2×0.34 \times 2 \times 0.3 m and the other is 3×2×0.33 \times 2 \times 0.3 m (with a 1×21 \times 2 overlap removed). Total volume:

  1. Think of the L as a full 4×3×0.34 \times 3 \times 0.3 rectangle minus a 1×1×0.31 \times 1 \times 0.3 corner block.
  2. Full rectangle: 4×3×0.3=3.64 \times 3 \times 0.3 = 3.6 m3^3.
  3. Corner removed: 1×1×0.3=0.31 \times 1 \times 0.3 = 0.3 m3^3.
  4. Volume =3.6−0.3=3.3= 3.6 - 0.3 = 3.3 m3^3.

3. Surface area of composite objects

Worked example 4 Half-cylinder on a prism (shed roof)

A garden shed has a rectangular base 33 m ×\times 44 m ×\times 22 m high, with a half-cylinder roof of radius 1.51.5 m running along the 44 m length. Find the total exposed surface area.

  1. Prism without its top face: base =12= 12, two long sides =2(4×2)=16= 2(4 \times 2) = 16, two short sides =2(3×2)=12= 2(3 \times 2) = 12. Total =12+16+12=40= 12 + 16 + 12 = 40 m2^2. (The top is covered by the roof.)
  2. Half-cylinder curved surface: 12(2π)(1.5)(4)=6π≈18.85\frac{1}{2}(2\pi)(1.5)(4) = 6\pi \approx 18.85 m2^2.
  3. Two semicircular ends: 2×12π(1.5)2=2.25π≈7.072 \times \frac{1}{2}\pi(1.5)^2 = 2.25\pi \approx 7.07 m2^2.
  4. Total exposed SA ≈40+18.85+7.07=65.92\approx 40 + 18.85 + 7.07 = 65.92 m2^2.

Note: the top rectangle of the prism and the flat face of the half-cylinder cancel each other, so neither appears in the final count.

Worked example 5 Construction: concrete column and base

A square concrete base 0.8×0.8×0.20.8 \times 0.8 \times 0.2 m supports a cylindrical column of radius 0.20.2 m and height 33 m. Find the total volume of concrete.

  1. Base volume: 0.8×0.8×0.2=0.1280.8 \times 0.8 \times 0.2 = 0.128 m3^3.
  2. Column volume: π(0.2)2(3)=0.12π≈0.377\pi(0.2)^2(3) = 0.12\pi \approx 0.377 m3^3.
  3. Total ≈0.128+0.377=0.505\approx 0.128 + 0.377 = 0.505 m3^3.

Key formulas

Rectangular prism volume
V=l×w×hV = l \times w \times h
Cylinder volume
V=πr2hV = \pi r^2 h
Cylinder surface area
SA=2πr2+2πrh\text{SA} = 2\pi r^2 + 2\pi r h
Prism surface area
SA=2Abase+Pbase×L\text{SA} = 2A_{\text{base}} + P_{\text{base}} \times L

Practice

Fluency

Tier 1: basic calculations

    1. A cylinder (r=4r = 4 cm, h=10h = 10 cm) sits on top of a cube of side 1010 cm. Find the total volume.
    2. A rectangular prism 8×6×58 \times 6 \times 5 cm has a cylindrical hole (r=2r = 2 cm) drilled through the 55 cm height. Find the remaining volume.
    3. Two rectangular prisms are joined end-to-end: one is 6×4×36 \times 4 \times 3 cm, the other is 8×4×38 \times 4 \times 3 cm. Find the total volume and the exposed surface area.
    4. A cylinder (r=3r = 3 cm, h=7h = 7 cm) has a hemisphere (r=3r = 3 cm) on top. Find the total volume.
    5. Find the exposed surface area in question 4. (Hemisphere curved SA =2πr2= 2\pi r^2.)
    6. An L-shaped block is formed from a 10×6×410 \times 6 \times 4 cm prism with a 4×3×44 \times 3 \times 4 cm block removed from one corner. Find the volume.
    7. A half-cylinder (r=5r = 5 cm, l=12l = 12 cm) sits on top of a rectangular prism 10×12×610 \times 12 \times 6 cm. Find the total volume.
    8. Find the total surface area for the solid in question 1, given that the cylinder sits centred on the top face of the cube.
Reasoning

Tier 2: mixed practice

    1. A cylindrical water tank (r=0.5r = 0.5 m, h=1.8h = 1.8 m) needs to be insulated on the curved surface and top only. Insulation costs $18 per m2^2. Find the total cost.
    2. A swimming pool has a uniform rectangular cross-section 10×510 \times 5 m. It is 1.21.2 m deep at one end and 2.42.4 m at the other (the floor slopes uniformly). Find the volume of water when the pool is full.
    3. A factory chimney consists of a rectangular base 2×2×12 \times 2 \times 1 m topped by a cylinder of radius 0.40.4 m and height 88 m. Find (a) the total volume, and (b) the total exposed surface area (the chimney is open at the top).
    4. A packing box 30×20×1530 \times 20 \times 15 cm contains a cylindrical can (r=5r = 5 cm, h=15h = 15 cm) standing upright. What percentage of the box volume is wasted space?
    5. Two cylinders are joined: a large one (r=6r = 6 cm, h=10h = 10 cm) with a smaller one (r=3r = 3 cm, h=8h = 8 cm) centred on top. Find the total exposed surface area.
    6. A solid is made by cutting a hemisphere (r=4r = 4 cm) from the top of a cylinder (r=4r = 4 cm, h=10h = 10 cm). Find the remaining volume.
Reasoning

Tier 3: explain and apply

    1. Explain why you must subtract the shared face area twice (not once) when finding the exposed surface area of two joined solids.
    2. A composite tank is a cylinder (r=1r = 1 m, h=2h = 2 m) with a cone on top (same radius, height 0.50.5 m). The cone volume is 13πr2h\frac{1}{3}\pi r^2 h. Find the total capacity in litres and explain why the cone adds relatively little capacity.
    3. A manufacturer needs a container with volume 20002000 cm3^3. Design A is a single cylinder; Design B is a cube with a hemisphere on top. For each, find dimensions that achieve the target volume and compare total surface areas to determine which uses less material.
    4. A rectangular prism a×a×2aa \times a \times 2a has a cylinder of radius a4\frac{a}{4} drilled through its longest dimension. Express the remaining volume as a function of aa.

Challenge

Reasoning

Harder reasoning

    1. A silo consists of a cylinder of radius 33 m and height 1010 m topped by a hemisphere. Find (a) the total volume, and (b) the total external surface area. If grain fills the silo to 34\frac{3}{4} of its capacity, find the volume of grain.
    2. A trophy is made from a rectangular prism base 8×8×28 \times 8 \times 2 cm, with a cylinder (r=2r = 2 cm, h=12h = 12 cm) rising from its centre, and a solid sphere (r=3r = 3 cm) on top of the cylinder. Find the total volume and the total exposed surface area. (Sphere SA =4πr2= 4\pi r^2; sphere V=43πr3V = \frac{4}{3}\pi r^3.)
    3. An underground pipe is a hollow cylinder with outer radius 1515 cm and inner radius 1212 cm, running for 5050 m. Find the volume of material in the pipe wall.
    4. A composite solid is formed by attaching a square-based pyramid (base 6×66 \times 6 cm, slant height 55 cm) to the top of a cube of side 66 cm. Find the total exposed surface area. (Lateral area of a pyramid =12×perimeter×slant height= \frac{1}{2} \times \text{perimeter} \times \text{slant height}.)
Answers

Answer key

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Tier 1

    1. Cube volume =1000= 1000 cm3^3; cylinder volume =π(16)(10)=160π≈502.65= \pi(16)(10) = 160\pi \approx 502.65 cm3^3. Total ≈1502.65\approx 1502.65 cm3^3.
    2. Prism =240= 240 cm3^3; hole =π(4)(5)=20π≈62.83= \pi(4)(5) = 20\pi \approx 62.83 cm3^3. Remaining ≈177.17\approx 177.17 cm3^3.
    3. Volume =72+96=168= 72 + 96 = 168 cm3^3. The joined face (4×3=124 \times 3 = 12 cm2^2) is hidden. SA of each prism: 2(24+18+12)=1082(24+18+12) = 108 and 2(32+24+12)=1362(32+24+12) = 136. Exposed SA =108+136−2(12)=220= 108 + 136 - 2(12) = 220 cm2^2.
    4. Cylinder =π(9)(7)=63π= \pi(9)(7) = 63\pi; hemisphere =23π(27)=18π= \frac{2}{3}\pi(27) = 18\pi. Total =81π≈254.47= 81\pi \approx 254.47 cm3^3.
    5. Cylinder curved SA =2π(3)(7)=42π= 2\pi(3)(7) = 42\pi; cylinder base =9π= 9\pi; hemisphere curved =2π(9)=18π= 2\pi(9) = 18\pi. Total =(42+9+18)π=69π≈216.77= (42 + 9 + 18)\pi = 69\pi \approx 216.77 cm2^2. (Top circle of cylinder is replaced by hemisphere, so not counted.)
    6. Full prism =240= 240; removed =48= 48. Volume =192= 192 cm3^3.
    7. Prism =720= 720 cm3^3; half-cylinder =12π(25)(12)=150π≈471.24= \frac{1}{2}\pi(25)(12) = 150\pi \approx 471.24 cm3^3. Total ≈1191.24\approx 1191.24 cm3^3.
    8. Cube full SA =600= 600; cylinder full SA =2π(16)+2π(4)(10)=32π+80π=112π≈351.86= 2\pi(16) + 2\pi(4)(10) = 32\pi + 80\pi = 112\pi \approx 351.86; shared circle =16π≈50.27= 16\pi \approx 50.27. Exposed SA ≈600+351.86−2(50.27)≈851.32\approx 600 + 351.86 - 2(50.27) \approx 851.32 cm2^2.

Tier 2

    1. Curved SA =2π(0.5)(1.8)=1.8π= 2\pi(0.5)(1.8) = 1.8\pi; top =π(0.25)=0.25π= \pi(0.25) = 0.25\pi. Total area =2.05π≈6.44= 2.05\pi \approx 6.44 m2^2. Cost =6.44×18≈= 6.44 \times 18 \approx $115.92.
    2. Cross-section is a trapezium with parallel sides 1.21.2 and 2.42.4, width 1010. Area =12(1.2+2.4)(10)=18= \frac{1}{2}(1.2 + 2.4)(10) = 18 m2^2. Volume =18×5=90= 18 \times 5 = 90 m3^3.
    3. (a) Base =4= 4 m3^3; column =π(0.16)(8)=1.28π≈4.02= \pi(0.16)(8) = 1.28\pi \approx 4.02 m3^3. Total ≈8.02\approx 8.02 m3^3. (b) Base: bottom =4= 4, four sides =4(2×1)=8= 4(2 \times 1) = 8, top exposed =4−π(0.16)≈3.50= 4 - \pi(0.16) \approx 3.50. Column: curved =2π(0.4)(8)=6.4π≈20.11= 2\pi(0.4)(8) = 6.4\pi \approx 20.11. (Open top, so no top circle.) Total SA ≈4+8+3.50+20.11=35.61\approx 4 + 8 + 3.50 + 20.11 = 35.61 m2^2.
    4. Box =9000= 9000 cm3^3; can =π(25)(15)=375π≈1178.10= \pi(25)(15) = 375\pi \approx 1178.10 cm3^3. Wasted ≈9000−1178.10=7821.90\approx 9000 - 1178.10 = 7821.90 cm3^3. Percentage ≈86.9%\approx 86.9\%.
    5. Large cylinder SA: curved =2π(6)(10)=120π= 2\pi(6)(10) = 120\pi; base =36π= 36\pi; top annulus (ring) =36π−9π=27π= 36\pi - 9\pi = 27\pi. Small cylinder: curved =2π(3)(8)=48π= 2\pi(3)(8) = 48\pi; top =9π= 9\pi. Total exposed =(120+36+27+48+9)π=240π≈753.98= (120 + 36 + 27 + 48 + 9)\pi = 240\pi \approx 753.98 cm2^2.
    6. Cylinder =π(16)(10)=160π= \pi(16)(10) = 160\pi; hemisphere =23π(64)=128π3= \frac{2}{3}\pi(64) = \frac{128\pi}{3}. Remaining =160π−128π3=352π3≈368.61= 160\pi - \frac{128\pi}{3} = \frac{352\pi}{3} \approx 368.61 cm3^3.

Tier 3

    1. Each component’s full SA includes the shared face as part of its own total. When the two solids join, that face is hidden on both solids — it disappears from the outside of solid 1 and from the outside of solid 2. Since it was counted once in each SA calculation, you must subtract it twice to get the correct exposed SA.
    2. Cylinder =π(1)2(2)=2π= \pi(1)^2(2) = 2\pi m3^3; cone =13π(1)2(0.5)=π6= \frac{1}{3}\pi(1)^2(0.5) = \frac{\pi}{6} m3^3. Total =13π6≈6.81= \frac{13\pi}{6} \approx 6.81 m3≈6807^3 \approx 6807 L. The cone adds only π6≈0.52\frac{\pi}{6} \approx 0.52 m3^3 (≈524\approx 524 L), about 7.7%7.7\% of the total, because the cone formula includes the 13\frac{1}{3} factor and the cone height is small.
    3. Design A (cylinder): choose rr and hh with πr2h=2000\pi r^2 h = 2000. For example r=6r = 6 cm gives h=200036π≈17.68h = \frac{2000}{36\pi} \approx 17.68 cm; SA =2π(36)+2π(6)(17.68)≈226.19+666.18≈892.4= 2\pi(36) + 2\pi(6)(17.68) \approx 226.19 + 666.18 \approx 892.4 cm2^2. Design B (cube + hemisphere): cube side ss with hemisphere r=s/2r = s/2; volume =s3+23π(s/2)3=s3(1+π12)≈1.262s3=2000= s^3 + \frac{2}{3}\pi(s/2)^3 = s^3(1 + \frac{\pi}{12}) \approx 1.262 s^3 = 2000, so s≈11.67s \approx 11.67 cm. SA =5s2+2π(s/2)2=5s2+πs22≈6.571s2≈894.7= 5s^2 + 2\pi(s/2)^2 = 5s^2 + \frac{\pi s^2}{2} \approx 6.571 s^2 \approx 894.7 cm2^2. Both designs use roughly similar material; the optimal choice depends on exact dimensions.
    4. V=a×a×2a−π(a4)2(2a)=2a3−πa38=a3(2−π8)V = a \times a \times 2a - \pi\left(\frac{a}{4}\right)^2(2a) = 2a^3 - \frac{\pi a^3}{8} = a^3\left(2 - \frac{\pi}{8}\right).

Challenge

    1. (a) Cylinder =π(9)(10)=90π= \pi(9)(10) = 90\pi; hemisphere =23π(27)=18π= \frac{2}{3}\pi(27) = 18\pi. Total =108π≈339.29= 108\pi \approx 339.29 m3^3. (b) Base circle =9π= 9\pi; curved cylinder =2π(3)(10)=60π= 2\pi(3)(10) = 60\pi; hemisphere curved =2π(9)=18π= 2\pi(9) = 18\pi. Total SA =(9+60+18)π=87π≈273.32= (9 + 60 + 18)\pi = 87\pi \approx 273.32 m2^2. Grain volume =34(108π)=81π≈254.47= \frac{3}{4}(108\pi) = 81\pi \approx 254.47 m3^3.
    2. Base prism =128= 128 cm3^3; cylinder =π(4)(12)=48π= \pi(4)(12) = 48\pi; sphere =43π(27)=36π= \frac{4}{3}\pi(27) = 36\pi. Total volume =128+84π≈391.88= 128 + 84\pi \approx 391.88 cm3^3. SA: prism bottom =64= 64; prism four sides =4(16)=64= 4(16) = 64; prism top annulus =64−4π= 64 - 4\pi; cylinder curved =2π(2)(12)=48π= 2\pi(2)(12) = 48\pi; sphere =4π(9)=36π= 4\pi(9) = 36\pi (but a circle π(4)\pi(4) is hidden where sphere meets cylinder, subtract 2π(4)=8π2\pi(4) = 8\pi). Sphere sits on top so small circle hidden; actually sphere r=3>r=3 > cylinder r=2r=2, so the sphere rests on the cylinder rim; exposed sphere SA =36π−π(4)=32π= 36\pi - \pi(4) = 32\pi. Total SA ≈64+64+(64−4π)+48π+32π≈192+76π≈430.73\approx 64 + 64 + (64 - 4\pi) + 48\pi + 32\pi \approx 192 + 76\pi \approx 430.73 cm2^2.
    3. Outer volume =π(0.15)2(50)=1.125π= \pi(0.15)^2(50) = 1.125\pi m3^3; inner volume =π(0.12)2(50)=0.72π= \pi(0.12)^2(50) = 0.72\pi m3^3. Wall volume =(1.125−0.72)π=0.405π≈1.272= (1.125 - 0.72)\pi = 0.405\pi \approx 1.272 m3^3.
    4. Cube has 5 exposed faces (top replaced by pyramid base): 5×36=1805 \times 36 = 180 cm2^2. Pyramid lateral area =12(24)(5)=60= \frac{1}{2}(24)(5) = 60 cm2^2. Total exposed SA =180+60=240= 180 + 60 = 240 cm2^2.

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