Year 9 Mathematics | Victorian Curriculum 2.0
Surface area and volume of prisms and cylinders
Topic 08 | Measurement | Practice

What you will learn

  • calculate surface area of right prisms by unfolding their nets,
  • derive and apply the surface area formula for cylinders,
  • calculate the volume of cylinders using V=πr2hV = \pi r^2 hV=πr2h,
  • find the surface area and volume of composite solids built from prisms and cylinders.
Why surface area and volume together?

Surface area tells you how much material covers the outside (paint, sheet metal, wrapping). Volume tells you how much space is inside (capacity, contents). Many real problems need both: a manufacturer needs volume to know how much a can holds and surface area to know how much aluminium to buy. Mastering both at once trains you to choose the right calculation for each context.

Where you'll see this
  • Packaging design: minimising material (surface area) while maximising contents (volume).
  • Construction: concrete volume for a cylindrical column; paint area for its surface.
  • Cooking: tin can dimensions affect heating time (surface area) and serving size (volume).
  • Engineering: pipe dimensions require both inner volume (flow capacity) and outer surface area (insulation).
Worked example 0 Real-world example: painting a water tank

A cylindrical water tank has radius 1.21.21.2 m and height 222 m. You need to paint the outside (curved surface and top lid only — the base sits on a concrete pad).

  1. Curved surface area: 2πrh=2π(1.2)(2)=4.8π≈15.082\pi r h = 2\pi(1.2)(2) = 4.8\pi \approx 15.082πrh=2π(1.2)(2)=4.8π≈15.08 m2^22.
  2. Top circle: πr2=π(1.2)2=1.44π≈4.52\pi r^2 = \pi(1.2)^2 = 1.44\pi \approx 4.52πr2=π(1.2)2=1.44π≈4.52 m2^22.
  3. Total area to paint: 15.08+4.52≈19.6015.08 + 4.52 \approx 19.6015.08+4.52≈19.60 m2^22.
  4. At $12 per m2^22, paint cost ≈\approx≈ $235.20.

Key idea: real problems often require only part of the total surface area.

1. Surface area of right prisms (using nets)

The surface area (SA) of any right prism is the sum of the areas of all its faces. Unfolding the prism into a net makes every face visible.

Surface area of a right prism
SA=2×Abase+lateral area\text{SA} = 2 \times A_{\text{base}} + \text{lateral area}SA=2×Abase​+lateral area

where the lateral area equals the perimeter of the base times the length of the prism: Pbase×LP_{\text{base}} \times LPbase​×L.

Worked example 1 Rectangular prism (cuboid)

Find the surface area of a box 888 cm by 555 cm by 333 cm.

  1. Two faces of 8×5=408 \times 5 = 408×5=40 cm2^22.
  2. Two faces of 8×3=248 \times 3 = 248×3=24 cm2^22.
  3. Two faces of 5×3=155 \times 3 = 155×3=15 cm2^22.
  4. SA =2(40+24+15)=2×79=158= 2(40 + 24 + 15) = 2 \times 79 = 158=2(40+24+15)=2×79=158 cm2^22.
Worked example 2 Triangular prism

A triangular prism has a right-angled triangular cross-section with legs 666 cm and 888 cm (hypotenuse 101010 cm) and length 121212 cm.

  1. Base area (triangle): 12×6×8=24\tfrac{1}{2} \times 6 \times 8 = 2421​×6×8=24 cm2^22.
  2. Two triangular ends: 2×24=482 \times 24 = 482×24=48 cm2^22.
  3. Three rectangular faces: (6+8+10)×12=288(6 + 8 + 10) \times 12 = 288(6+8+10)×12=288 cm2^22.
  4. SA =48+288=336= 48 + 288 = 336=48+288=336 cm2^22.

2. Surface area of cylinders

When you “unroll” a cylinder, its curved surface becomes a rectangle whose width is the circumference 2πr2\pi r2πr and whose height is hhh.

rtop2πrhcurved surfacerbase
Cylinder net: the curved surface unrolls into a rectangle of width 2 pi r and height h, plus two circular ends.
Surface area of a cylinder
SA=2πr2+2πrh\text{SA} = 2\pi r^2 + 2\pi r hSA=2πr2+2πrh
  • 2πr22\pi r^22πr2: two circular ends.
  • 2πrh2\pi r h2πrh: curved (lateral) surface.
Worked example 3 Full cylinder surface area

Find the total surface area of a cylinder with radius 555 cm and height 121212 cm. Use π≈3.14\pi \approx 3.14π≈3.14.

  1. Two circles: 2π(5)2=50π≈157.02\pi(5)^2 = 50\pi \approx 157.02π(5)2=50π≈157.0 cm2^22.
  2. Curved surface: 2π(5)(12)=120π≈376.82\pi(5)(12) = 120\pi \approx 376.82π(5)(12)=120π≈376.8 cm2^22.
  3. SA ≈157.0+376.8=533.8\approx 157.0 + 376.8 = 533.8≈157.0+376.8=533.8 cm2^22.

3. Volume of cylinders

Volume of a cylinder
V=πr2hV = \pi r^2 hV=πr2h

This follows the same logic as any prism: base area (πr2\pi r^2πr2) times height (hhh).

Worked example 4 Volume of a cylinder

A cylindrical tin has diameter 101010 cm and height 151515 cm. Find its volume and capacity in mL.

  1. Radius =5= 5=5 cm.
  2. V=π(5)2(15)=375π≈1178.1V = \pi(5)^2(15) = 375\pi \approx 1178.1V=π(5)2(15)=375π≈1178.1 cm3^33.
  3. Since 111 cm3=1^3 = 13=1 mL, capacity ≈1178\approx 1178≈1178 mL ≈1.18\approx 1.18≈1.18 L.
Worked example 5 Finding a missing dimension

A cylinder has volume 500500500 cm3^33 and radius 444 cm. Find its height to one decimal place.

  1. V=πr2hV = \pi r^2 hV=πr2h, so h=Vπr2h = \dfrac{V}{\pi r^2}h=πr2V​.
  2. h=500π×16=50050.27≈9.9h = \dfrac{500}{\pi \times 16} = \dfrac{500}{50.27} \approx 9.9h=π×16500​=50.27500​≈9.9 cm.

4. Composite solids

A composite solid is formed by combining two or more simple solids. To find its surface area, add the exposed areas (subtract any faces hidden where the solids join). To find its volume, add the individual volumes.

Worked example 6 Cylinder on top of a rectangular prism

A silo consists of a rectangular base 444 m by 444 m by 333 m high, topped with a half-cylinder of radius 222 m and length 444 m. Find the total volume.

  1. Rectangular prism volume: 4×4×3=484 \times 4 \times 3 = 484×4×3=48 m3^33.
  2. Half-cylinder volume: 12π(2)2(4)=8π≈25.13\tfrac{1}{2} \pi (2)^2 (4) = 8\pi \approx 25.1321​π(2)2(4)=8π≈25.13 m3^33.
  3. Total volume ≈48+25.13=73.13\approx 48 + 25.13 = 73.13≈48+25.13=73.13 m3^33.

Key formulas

Prism surface area
SAprism=2Abase+Pbase×L\text{SA}_{\text{prism}} = 2A_{\text{base}} + P_{\text{base}} \times LSAprism​=2Abase​+Pbase​×L
Cylinder surface area
SAcyl=2πr2+2πrh\text{SA}_{\text{cyl}} = 2\pi r^2 + 2\pi r hSAcyl​=2πr2+2πrh
Cylinder volume
Vcyl=πr2hV_{\text{cyl}} = \pi r^2 hVcyl​=πr2h

Practice

Fluency

Tier 1: basic calculations

    1. Find the surface area of a cuboid 10×6×410 \times 6 \times 410×6×4 cm.
    2. Find the surface area of a cube of side 777 cm.
    3. A triangular prism has an equilateral triangle base of side 555 cm (height ≈4.33\approx 4.33≈4.33 cm) and length 101010 cm. Find its surface area.
    4. Find the total surface area of a cylinder with r=3r = 3r=3 cm and h=10h = 10h=10 cm. Give your answer in terms of π\piπ and as a decimal.
    5. Find the volume of a cylinder with r=6r = 6r=6 cm and h=20h = 20h=20 cm. Leave your answer in terms of π\piπ.
    6. A cylinder has r=4r = 4r=4 cm and h=8h = 8h=8 cm. Find (a) the curved surface area, (b) the total surface area, (c) the volume.
    7. Convert a cylinder volume of 250025002500 cm3^33 to litres.
    8. Find the surface area of a cylinder with diameter 141414 cm and height 999 cm.
Reasoning

Tier 2: mixed practice

    1. A cylindrical can has volume 100010001000 cm3^33 and height 121212 cm. Find its radius to one decimal place.
    2. Two cylinders have the same volume. Cylinder A has r=5r = 5r=5 cm. Cylinder B has r=10r = 10r=10 cm. If A has height 202020 cm, find the height of B.
    3. A closed cylinder uses 600π600\pi600π cm2^22 of sheet metal. If r=10r = 10r=10 cm, find hhh.
    4. A rectangular prism 20×10×820 \times 10 \times 820×10×8 cm has a cylindrical hole of radius 333 cm drilled through its length. Find the remaining volume.
    5. Which has the greater surface area: a cube of side 101010 cm or a cylinder with r=5r = 5r=5 cm and h=10h = 10h=10 cm? Justify.
    6. A prism has a cross-section that is a right trapezium with parallel sides 666 cm and 101010 cm and height 444 cm. Its length is 151515 cm. Find the surface area. (The non-parallel sides are 444 cm and ≈5.66\approx 5.66≈5.66 cm.)
Reasoning

Tier 3: explain and apply

    1. Explain why the formula SA=2πr2+2πrh\text{SA} = 2\pi r^2 + 2\pi r hSA=2πr2+2πrh has two separate terms. What does each represent physically?
    2. A manufacturer wants to double the volume of a cylindrical can without changing the radius. By what factor must the height change?
    3. If you double the radius of a cylinder but keep the height the same, by what factor does the volume increase? Explain why.
    4. A composite solid is a cylinder (r=3r = 3r=3 cm, h=10h = 10h=10 cm) with a hemisphere (r=3r = 3r=3 cm) on top. Find the total surface area. (Hemisphere curved SA =2πr2= 2\pi r^2=2πr2.)
    5. A water pipe is 505050 m long with an inner radius of 555 cm. Find the volume of water it can hold, in litres.

Challenge

Reasoning

Harder reasoning

    1. A cylinder and a cube have the same volume. The cylinder has r=5r = 5r=5 cm and h=10h = 10h=10 cm. Find the side length of the cube to one decimal place.
    2. A closed cylinder has total surface area 200π200\pi200π cm2^22. Express hhh in terms of rrr, and find the radius that maximises the volume. (Hint: substitute into V=πr2hV = \pi r^2 hV=πr2h and look for the maximum.)
    3. A composite solid is a rectangular prism 10×8×610 \times 8 \times 610×8×6 cm with a half-cylinder (radius 444 cm, length 101010 cm) sitting on its top face. Find (a) the total volume, and (b) the total exposed surface area.
    4. A cylindrical tank of radius 0.50.50.5 m and height 1.21.21.2 m is lying on its side. When it is half full, what volume of water does it contain? Give your answer in litres.
Year 9 Mathematics study companion | Practice