MathematicsExam code: 0580

Volume and Surface Area

Lengths, Areas and Volumes

PractiseFlashcards

Prisms and cylinders

  • A prism is any solid with the same cross-section all the way along, and its volume is that cross-sectional area times its length
  • On this specification "prism" is defined broadly, so a cylinder is a prism with a circular cross-section, and even a cylindrical sector counts
  • So a triangular prism whose cross-section has area 24 cm² and whose length is 15 cm has volume 24 × 15 = 360 cm³
  • A cuboid is a prism with a rectangular cross-section, and its volume is length × width × height
A cuboid with its hidden edges dashed and its three dimensions named: length along the front bottom edge, width along the bottom edge running back, and height up the side
  • For a cylinder the cross-section is a circle, which is why V = πr²h
A cylinder with the radius r marked across the top face from its centre and the height h marked up the side, the hidden part of the base drawn as a dashed curve
Source: Volume by Save My Exams

Worked example

A volume in m³, cm³ and litres

A water tank is a cuboid 1.5 m long, 0.8 m wide and 0.6 m deep. Find its volume in m³ and in cm³, and the number of litres it holds.

Solution:

  • Volume = 1.5 × 0.8 × 0.6 = 0.72 m³
  • One metre is 100 cm, so 1 m³ is 100 × 100 × 100 = 1 000 000 cm³
  • Volume = 0.72 × 1 000 000 = 720 000 cm³
  • One litre is 1000 cm³, so divide by 1000
  • Capacity = 720 000 ÷ 1000 = 720 litres
  • Check by changing the lengths to centimetres first: 150 × 80 × 60 = 720 000 cm³
  • The length factor of 100 is cubed because three lengths are multiplied, so multiplying 0.72 by 100 would be wrong

Pyramids, cones and spheres

  • A pyramid has volume ⅓ × base area × perpendicular height, and the height must run straight down from the apex to the base
A square-based pyramid with a red arrow running straight down from the apex to the centre of the base, marked h, which is the perpendicular height rather than the length of a sloping edge
Source: Volume by Save My Exams
  • A cone is a pyramid on a circular base, so its volume is ⅓πr²h, again using the perpendicular height
  • The sloping edge of a cone is longer than its perpendicular height, and Pythagoras converts between them since the radius, the height and the sloping edge form a right-angled triangle
A cone with the perpendicular height h drawn from the apex down to the centre of the base, a small square marking the right angle where it meets the radius r
Source: Volume by Save My Exams
  • A sphere has volume 4/3 πr³, and a hemisphere is half of that
A sphere with its radius r drawn from the centre out to the surface, the equator shown as a curve with its hidden half dashed
Source: Volume by Save My Exams

Worked example

A cone from its sloping edge

A cone has base radius 5 cm and sloping edge 13 cm. Find its volume in terms of π.

A cone standing on its circular base. The sloping edge from the rim up to the apex is labelled 13 cm and the base radius is labelled 5 cm. A dashed vertical line runs from the centre of the base to the apex and a dashed radius runs out to the rim, meeting at a marked right angle; the vertical height carries no label. The figure is marked NOT TO SCALE.

Solution:

  • The volume formula needs the perpendicular height, but the sloping edge is given
  • The radius, the height and the sloping edge form a right-angled triangle, with the sloping edge as the hypotenuse
  • By Pythagoras, h² = 13² − 5² = 169 − 25 = 144
  • So h = 12 cm
  • Volume = ⅓ × π × 5² × 12 = ⅓ × π × 25 × 12
  • Volume = 100π cm³

Worked example

Finding a radius from a volume

A cylinder has radius 4 cm and height 18 cm. A sphere has the same volume as the cylinder. Find the radius of the sphere.

Two solids side by side: on the left an upright cylinder with a dashed radius of 4 cm across its top and a height of 18 cm, and on the right a sphere with its equator drawn and a dashed radius labelled r cm. The figure is marked NOT TO SCALE.

Solution:

  • Volume of the cylinder = π × 4² × 18 = π × 16 × 18 = 288π cm³
  • Call the sphere's radius r and set its volume equal to that: 4/3 πr³ = 288π
  • Divide both sides by π: 4/3 r³ = 288
  • Multiply by 3 and divide by 4: r³ = 288 × 3 ÷ 4 = 216
  • Take the cube root: r = ∛216 = 6 cm
  • Check: 4/3 × π × 6³ = 4/3 × 216π = 288π
  • Keeping π as a symbol lets it cancel, so no calculator is needed

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