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Stacking Slices
Stacking Slices
Imagine slicing a Toblerone lengthwise - every slice has the exact same triangular shape! A prism is any 3D shape where you can slice it and get identical cross-sections all the way through. To find its volume, just find the area of one slice and multiply by how many slices (the length).
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Exam Tips
- A cylinder is just a circular prism - same formula principle applies!
- If the cross-section is compound (L-shape, T-shape), split it into rectangles first
- Check units carefully - if given in different units, convert first
Practice Questions
Q1 Find the volume of a cuboid: length 5 cm, width 4 cm, height 3 cm [1 mark]
[1] 5 × 4 × 3 = 60 cm³
Q2 A triangular prism has cross-section area 15 cm² and length 6 cm. Find the volume. [1 mark]
[1] 15 × 6 = 90 cm³
Q3 Find the volume of a cylinder with radius 4 cm and height 6 cm (1 dp) [2 marks]
[1] V = πr²h = π × 4² × 6 = π × 96
[1] V = 301.6 cm³
Q4 A triangular prism: base 4 cm, height 3 cm, length 8 cm. Find volume. [3 marks]
[1] Area of triangle = ½ × 4 × 3 = 6 cm²
[1] Volume = 6 × 8
[1] V = 48 cm³
Q5 Calculate the volume of a cuboid with length 10 cm, width 6 cm, and height 4 cm. [2 marks]
[1] V = l × w × h = 10 × 6 × 4
[1] = 240 cm³
Examiner says: Volume is ALWAYS in cubic units (cm³, m³, mm³) - never forget the ³!
Q6 Find the volume of a cylinder with radius 5 cm and height 8 cm. Give your answer to 1 decimal place. [3 marks]
[1] V = πr²h
[1] V = π × 5² × 8 = π × 25 × 8 = 200π
[1] = 628.3 cm³ (1 dp)
Q7 A triangular prism has a triangle cross-section with base 8 cm and height 5 cm. The prism is 12 cm long. Find its volume. [3 marks]
[1] Cross-sectional area = ½ × 8 × 5 = 20 cm²
[1] Volume = area × length = 20 × 12
[1] = 240 cm³
Q8 A water tank is a cuboid measuring 2 m by 1.5 m by 0.8 m. Find its volume in: (a) m³, (b) litres (1 m³ = 1000 litres). [4 marks]
[1] (a) V = 2 × 1.5 × 0.8
[1] = 2.4 m³
[1] (b) Volume in litres = 2.4 × 1000
[1] = 2400 litres
Q9 A cylindrical can has diameter 10 cm and height 15 cm. Find its volume to the nearest cm³. [4 marks]
[1] Radius = 10 ÷ 2 = 5 cm
[1] V = πr²h = π × 5² × 15
[1] = π × 25 × 15 = 375π
[1] = 1178 cm³ (nearest cm³)
Examiner says: When given diameter, HALVE IT to get radius before using V = πr²h!
Q10 A cuboid has volume 360 cm³. Its length is 10 cm and width is 6 cm. Find the height. [4 marks]
[1] V = l × w × h, so 360 = 10 × 6 × h
[1] 360 = 60h
[1] h = 360 ÷ 60
[1] h = 6 cm
Q11 Find the volume of a cylinder with radius 6 cm and height 10 cm. Leave your answer in terms of π. [3 marks]
[1] V = πr²h
[1] V = π × 6² × 10 = π × 36 × 10
[1] = 360π cm³
Q12 A prism has a trapezoidal cross-section with parallel sides 7 cm and 11 cm, and height 4 cm. The prism is 15 cm long. Find its volume. [5 marks]
[2] Cross-sectional area = ½ × (7 + 11) × 4 = ½ × 18 × 4 = 36 cm²
[2] Volume = area × length = 36 × 15
[1] = 540 cm³
Q13 A cylinder has volume 500 cm³ and height 8 cm. Find its radius. Give your answer to 2 decimal places. [5 marks]
[1] V = πr²h, so 500 = πr² × 8
[1] πr² = 500 ÷ 8 = 62.5
[1] r² = 62.5 ÷ π = 19.894...
[1] r = √19.894
[1] r = 4.46 cm (2 dp)
Q14 An L-shaped prism has a cross-section made of two rectangles: one 10 cm × 3 cm and one 4 cm × 5 cm. The prism is 20 cm long. Find its volume. [6 marks]
[2] Rectangle 1 area = 10 × 3 = 30 cm²
[1] Rectangle 2 area = 4 × 5 = 20 cm²
[1] Total cross-sectional area = 30 + 20 = 50 cm²
[1] Volume = 50 × 20
[1] = 1000 cm³
Q15 The radius of a cylinder is increased by 10% and the height decreased by 10%. Find the percentage change in volume. [6 marks]
[1] Let original r = r, h = h. V = πr²h
[1] New radius = 1.1r, new height = 0.9h
[2] New V = π(1.1r)²(0.9h) = π × 1.21r² × 0.9h = 1.089πr²h
[1] Change = (1.089πr²h - πr²h)/(πr²h) × 100
[1] = 0.089 × 100 = 8.9% increase
Examiner says: Radius is squared, so 10% increase → 1.1² = 1.21 factor!
Q16 A metal cuboid measures 50 cm × 30 cm × 20 cm. The metal has density 7.8 g/cm³. Find the mass of the cuboid in kg. [6 marks]
[1] Volume = 50 × 30 × 20 = 30,000 cm³
[2] Mass = density × volume = 7.8 × 30,000
[1] = 234,000 g
[1] Convert to kg: 234,000 ÷ 1000
[1] = 234 kg
Q17 A cylindrical water tank has diameter 1.2 m and height 2 m. Water flows in at 50 litres per minute. How long does it take to fill the tank? (1 m³ = 1000 litres) [7 marks]
[1] Radius = 1.2 ÷ 2 = 0.6 m
[2] V = πr²h = π × 0.6² × 2 = 0.72π ≈ 2.262 m³
[1] Capacity in litres = 2.262 × 1000 = 2262 litres
[2] Time = 2262 ÷ 50 = 45.24 minutes
[1] = 45 minutes 14 seconds (or 45.2 minutes)
Q18 A cuboid has length (x + 3) cm, width x cm, and height (x - 1) cm. Its volume is 60 cm³. Form and solve an equation to find x. [7 marks]
[1] V = (x+3) × x × (x-1) = 60
[2] x(x+3)(x-1) = 60, expand: x(x² + 3x - x - 3) = x(x² + 2x - 3)
[2] x³ + 2x² - 3x = 60, so x³ + 2x² - 3x - 60 = 0
[1] By trial: x = 3 works (27 + 18 - 9 - 60 = -24, try x=4: 64+32-12-60=24, try x=3.5...)
[1] By systematic trial or calculator: x ≈ 3.39 cm (or exact: solve cubic)
Examiner says: Cubic equations are tricky - use calculator graph or systematic trial!
Q19 A hollow cylindrical pipe has outer radius 5 cm, inner radius 4 cm, and length 50 cm. Find the volume of metal needed to make the pipe. Give your answer to the nearest cm³. [8 marks]
[2] Outer cylinder volume = π × 5² × 50 = 1250π cm³
[2] Inner cylinder volume (hollow) = π × 4² × 50 = 800π cm³
[2] Volume of metal = 1250π - 800π = 450π
[2] = 1414 cm³ (nearest cm³)
Examiner says: Hollow shapes: subtract the inner volume from the outer volume!
Q20 A cuboid box has a square base of side x cm and height h cm. Its volume must be 1000 cm³. Express h in terms of x, and find the value of x that minimises the surface area. [8 marks]
[1] V = x²h = 1000, so h = 1000/x²
[2] Surface area S = 2x² + 4xh = 2x² + 4x(1000/x²) = 2x² + 4000/x
[2] For minimum, dS/dx = 0: 4x - 4000/x² = 0
[2] 4x = 4000/x², x³ = 1000, x = ∛1000
[1] x = 10 cm
Examiner says: This is calculus optimization - A-level territory! The cube minimizes surface area for given volume.
Q21 Two similar cylinders have volumes in the ratio 8:27. The smaller cylinder has radius 4 cm. Find: (a) the radius of the larger cylinder, (b) the ratio of their surface areas. [9 marks]
[2] (a) Volume ratio = 8:27, so linear scale factor = ∛(8/27) = 2/3 or ∛(27/8) = 3/2
[2] Since larger volume, SF = 3/2
[1] Larger radius = 4 × (3/2) = 6 cm
[2] (b) Surface area ratio = (linear SF)² = (3/2)² = 9/4
[2] Ratio = 4:9
Examiner says: For similar shapes: length ratio = ∛(volume ratio), area ratio = (length ratio)²!
Q22 A cylindrical bucket has top radius 15 cm, bottom radius 10 cm, and height 20 cm (frustum of a cone). Estimate its volume by treating it as a prism with average radius. Give answer to nearest 10 cm³. [9 marks]
[2] Average radius = (15 + 10) ÷ 2 = 12.5 cm
[2] Average cross-sectional area = π × 12.5² = 156.25π cm²
[2] Volume ≈ area × height = 156.25π × 20
[2] = 3125π ≈ 9817.48
[1] ≈ 9820 cm³ (nearest 10)
Examiner says: This is an approximation method. Exact frustum formula is more complex!
Q23 A prism has an equilateral triangle cross-section with side s cm. The prism length is 3s cm. Find an expression for the volume in terms of s. [Area of equilateral triangle = (s²√3)/4] [10 marks]
[3] Cross-sectional area = (s²√3)/4
[2] Length = 3s
[3] Volume = area × length = (s²√3)/4 × 3s
[2] V = (3s³√3)/4 or 0.75s³√3
Examiner says: Equilateral triangle area uses √3 - don't try to simplify it to a decimal!
Q24 A semi-cylindrical tunnel has diameter 4 m and length 50 m. Find: (a) the volume of air inside the tunnel, (b) the area of material needed for the curved surface (not the ends). Give answers to 1 dp. [10 marks]
[1] Radius = 4 ÷ 2 = 2 m
[2] (a) Full cylinder V = πr²h = π × 2² × 50 = 200π m³
[2] Semi-cylinder V = ½ × 200π = 100π = 314.2 m³ (1 dp)
[2] (b) Circumference of full circle = 2πr = 4π m
[1] Semi-circle arc = ½ × 4π = 2π m
[2] Curved surface area = arc × length = 2π × 50 = 100π = 314.2 m² (1 dp)
Q25 A regular hexagonal prism has side length 6 cm and height 10 cm. Find its volume. [A regular hexagon of side s has area (3s²√3)/2] [11 marks]
[3] Cross-sectional area = (3 × 6² × √3)/2
[2] = (3 × 36 × √3)/2 = (108√3)/2 = 54√3 cm²
[2] Volume = area × height = 54√3 × 10
[2] = 540√3
[2] ≈ 935.3 cm³
Examiner says: Regular hexagon = 6 equilateral triangles arranged around a center!
Q26 A solid is formed by rotating the region bounded by y = x², y = 0, x = 0, and x = 2 about the x-axis. Find its volume using V = ∫π y² dx. Give answer to 1 dp. [11 marks]
[2] y = x², so y² = x⁴
[2] V = ∫₀² π x⁴ dx
[3] = π [x⁵/5]₀²
[2] = π × (32/5 - 0) = 32π/5
[2] = 20.1 cubic units (1 dp)
Examiner says: This is A-level calculus - volumes of revolution using integration!
Q27 A swimming pool is 25 m long and 10 m wide. The depth increases uniformly from 1 m at the shallow end to 3 m at the deep end. Find the volume of water needed to fill it. [12 marks]
[3] This is a prism with trapezoidal cross-section (viewed from the side)
[2] Cross-section trapezoid: parallel sides 1 m and 3 m, width 25 m
[2] Area = ½ × (1 + 3) × 25 = ½ × 4 × 25 = 50 m²
[2] Alternative: Average depth = (1+3)/2 = 2 m, Volume = 25 × 10 × 2 = 500 m³
[2] Volume = area × width = 50 × 10
[1] = 500 m³
Q28 Prove that if all dimensions of a cuboid are doubled, the volume is multiplied by 8. [12 marks]
[2] Let original cuboid have dimensions l, w, h
[2] Original volume V₁ = l × w × h
[2] After doubling: new dimensions are 2l, 2w, 2h
[2] New volume V₂ = (2l) × (2w) × (2h)
[2] V₂ = 2 × 2 × 2 × l × w × h = 8lwh
[2] V₂ = 8V₁ ✓
Examiner says: This applies to ALL 3D shapes - if linear scale factor = 2, volume scale factor = 2³ = 8!
Q29 A cylindrical tank with radius 2 m is being filled with water at 0.5 m³/min. At what rate is the water level rising? (Give answer in cm/min to 2 dp) [13 marks]
[2] V = πr²h, where r = 2 m (constant), h varies with time
[2] V = π × 2² × h = 4πh
[3] dV/dt = 4π × dh/dt (differentiating with respect to time)
[2] Given dV/dt = 0.5 m³/min
[2] 0.5 = 4π × dh/dt
[1] dh/dt = 0.5/(4π) = 0.0398 m/min
[1] = 3.98 cm/min (2 dp)
Examiner says: This is A-level calculus - related rates of change!
Q30 A cylindrical can must hold 500 ml (500 cm³). Find the radius and height that minimize the surface area (to save material). Give answers to 2 dp. [14 marks]
[2] V = πr²h = 500, so h = 500/(πr²)
[2] Surface area S = 2πr² + 2πrh (top + bottom + curved)
[2] S = 2πr² + 2πr × 500/(πr²) = 2πr² + 1000/r
[3] For minimum, dS/dr = 0: 4πr - 1000/r² = 0
[2] 4πr = 1000/r², so 4πr³ = 1000, r³ = 1000/(4π) = 79.577
[1] r = ∛79.577 = 4.30 cm (2 dp)
[1] h = 500/(π × 4.30²) = 8.60 cm (2 dp)
[1] (Note: h = 2r for minimum surface area!)
Examiner says: For a cylinder with fixed volume, minimum surface area when h = 2r!
Now try it yourself
Quiz · Question 1 of 20
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This topic in real past papers
Every real exam question we've found on volume of cuboids & prisms, with a full worked answer.