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Cambridge IGCSE Mathematics — 0580 Extended

Topic 5.4: Mensuration — Surface Area and Volume

Cuboid

A cuboid has six rectangular faces in three equal pairs. Neither formula below is on the sheet, but both come straight from that fact.

\[ V = \ell w h \qquad SA = 2(\ell w + \ell h + wh) \]

A cuboid measures \(5 \times 4 \times 3\). Find its volume and surface area.

Isometric cuboid with length 5, width 4 and height 3, hidden edges dashed, showing volume 60 and surface area 94
\(V = 5 \times 4 \times 3 = 60\) and \(SA = 2(20 + 15 + 12) = 94\).

For surface area, find the three distinct face areas, add them, then double. Listing all six faces works too but takes longer.

Prisms: \(V = A\ell\)

A prism is any solid with the same cross-section all the way along its length. Its volume formula is given:

\[ V = A\ell \]

where \(A\) is the area of that uniform cross-section and \(\ell\) is the length. Everything in Topic 5.2 can therefore become a volume.

A prism has a right-angled \(3\)–\(4\)–\(5\) triangular cross-section and length \(6\). Find its volume.

Isometric triangular prism of length 6 with a green shaded 3-4-5 right-angled triangular end face
Cross-section \(A = \tfrac{1}{2} \times 3 \times 4 = 6\), so \(V = A\ell = 6 \times 6 = 36\).

Surface area of a prism is not given: add the two end faces to the rectangles that wrap around the sides. For the prism above, the wrap is a rectangle of length \(6\) and width equal to the triangle's perimeter \(3 + 4 + 5 = 12\), giving \(72\), plus the two triangles \(2 \times 6 = 12\), so \(SA = 84\).

Cylinder

A cylinder is a prism with a circular cross-section. Both of these are on the formula sheet:

\[ V = \pi r^2 h \qquad \text{curved surface area} = 2\pi r h \]

A cylinder has radius \(3\) and height \(8\). Find its volume, curved surface area and total surface area, in terms of \(\pi\).

Isometric cylinder of radius 3 and height 8 beside the curved surface unrolled into a rectangle of width 2 pi r and height h
Unrolled, the curved surface is a rectangle \(2\pi r\) by \(h\), so \(2\pi(3)(8) = 48\pi\). With the two circles, \(SA = 48\pi + 18\pi = 66\pi\), and \(V = \pi(3)^2(8) = 72\pi\).

Total surface area of a closed cylinder is therefore \(2\pi r h + 2\pi r^2\). An open pipe or an open-topped can has fewer circles — count what the question actually describes.

Cone

Given on the sheet:

\[ V = \tfrac{1}{3}\pi r^2 h \qquad \text{curved surface area} = \pi r l \]

Two different lengths appear: \(h\) is the perpendicular height from apex to base centre, and \(l\) is the slant height along the sloping surface. They are linked by Pythagoras: \(l^2 = r^2 + h^2\).

A cone has radius \(3\) and perpendicular height \(4\). Find its volume and total surface area, in terms of \(\pi\).

Isometric cone with radius 3, dashed perpendicular height 4 and blue slant 5, next to a true right-angled 3-4-5 triangle
\(l = \sqrt{3^2 + 4^2} = 5\). Then \(V = \tfrac{1}{3}\pi(9)(4) = 12\pi\) and \(SA = \pi(3)(5) + \pi(3)^2 = 15\pi + 9\pi = 24\pi\).

Method

  1. Decide which length the question gave you, \(h\) or \(l\).
  2. Use \(l^2 = r^2 + h^2\) to get the other one.
  3. Volume needs \(h\); curved surface area needs \(l\).

Pyramid

Given on the sheet:

\[ V = \tfrac{1}{3}Ah \]

\(A\) is the area of the base — square, rectangular, triangular, whatever the question shows — and \(h\) is the perpendicular height from the apex down to the base plane.

A pyramid has a square base of side \(4\) and perpendicular height \(6\). Find its volume.

Isometric square-based pyramid with base 4 by 4 and a dashed perpendicular height of 6 from the apex to the centre of the base
\(A = 4 \times 4 = 16\), so \(V = \tfrac{1}{3} \times 16 \times 6 = 32\).

Surface area of a pyramid is not given: add the base to the triangular faces, each of which uses the slant height of that face, not the pyramid's perpendicular height.

Sphere

Both given on the sheet:

\[ V = \tfrac{4}{3}\pi r^3 \qquad SA = 4\pi r^2 \]

A sphere has radius \(5\). Find its volume and surface area, in terms of \(\pi\).

Sphere of radius 5 shown with its equator drawn as an ellipse, volume 500 pi over 3 and surface area 100 pi
\(V = \tfrac{4}{3}\pi(125) = \dfrac{500\pi}{3}\) and \(SA = 4\pi(25) = 100\pi\).

Paper 2 (non-calculator)

Cube the radius before touching the fraction: \(5^3 = 125\), then \(\tfrac{4}{3} \times 125 = \tfrac{500}{3}\). Leaving \(\dfrac{500\pi}{3}\) is a complete exact answer.

What counts as a prism

The syllabus lists a cylindrical sector as an example of a prism, because it still has one cross-section repeated along its length.

Isometric solid formed by extruding a 60 degree circular sector, with the green sector face marked as the uniform cross-section
The cross-section is a \(60^\circ\) sector, area \(\tfrac{60}{360}\pi r^2\), so \(V = A \times h\) exactly as for any prism.
SolidVolumeSurface area
Cuboid\(\ell w h\) (not given)\(2(\ell w + \ell h + wh)\) (not given)
Prism\(A\ell\) (given)ends \(+\) wrap (not given)
Cylinder\(\pi r^2 h\) (given)curved \(2\pi r h\) (given)
Cone\(\tfrac{1}{3}\pi r^2 h\) (given)curved \(\pi r l\) (given)
Pyramid\(\tfrac{1}{3}Ah\) (given)base \(+\) triangles (not given)
Sphere\(\tfrac{4}{3}\pi r^3\) (given)\(4\pi r^2\) (given)

Try this

A cone has radius \(6\,\text{cm}\) and slant height \(10\,\text{cm}\). Find its volume in terms of \(\pi\).

Show answer
Answer
  1. Volume needs \(h\), so use \(l^2 = r^2 + h^2\)

    \[ h = \sqrt{10^2 - 6^2} = \sqrt{64} = 8 \]
  2. Substitute into \(V = \tfrac{1}{3}\pi r^2 h\)

    \[ V = \tfrac{1}{3}\pi(36)(8) = 96\pi\,\text{cm}^3 \]

Exam Traps

  • In a cone, \(\pi r l\) uses the slant and \(\tfrac{1}{3}\pi r^2 h\) uses the perpendicular height. Swapping them is the single most common loss of marks in this topic.
  • The sheet gives only the curved surface area of a cylinder and cone. If the solid is closed, you must add \(2\pi r^2\) or \(\pi r^2\) yourself.
  • \(V = A\ell\) needs the area of the cross-section, not of the face you happen to be looking at. In a lying-down prism the cross-section is the end, not the long rectangle.

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