C6.4; E6.4 - Surface area of solids

C6.4; E6.4 - Surface area of solids

A container needs enough material to cover its faces; leaving its top open saves material even when its side dimensions stay the same. C6.4 and E6.4 develop surface areas of prisms, pyramids, cylinders and cones. You will use nets, count exposed faces and select the height or sloping edge that each area calculation needs.

Nets and exposed faces

How much sheet material makes a closed box, and how much is saved by leaving its top open? Think of unfolding its surface into flat faces: calculate each face once, then add only the faces required.

Surface area and net

A solid's surface area is the total area of the specified surfaces. A net is an arrangement of flat faces that folds to form a solid. An exposed face is one on the outside, rather than hidden at a join or absent at an opening.

A closed cuboid has three pairs of matching rectangular faces. A right prism has two matching end faces and a rectangular side face for each straight edge of its cross-section. Multiplying the cross-section's perimeter by the prism length adds those side rectangles; it does not include the two ends.

[DIAGRAM: asset_name: Faces in two nets; asset_slug: cuboid_pyramid_nets; caption: Add each face once. A pyramid's triangular face height lies in its face.; description: A valid six-rectangle net of a cuboid with dimensions 6 cm, 4 cm and 3 cm is shown beside a square-based pyramid net. The pyramid has a 6 cm square base and four triangular faces each with perpendicular face height 5 cm. Paired cuboid faces share pale fill colours; dashed fold lines and the pyramid face-height right angle are shown.]
Add each face once. A pyramid's triangular face height lies in its face.

There is no supplied total-area formula for a cuboid, prism or pyramid: build the area from its faces. The triangle area formula is supplied on the Papers 1–4 lists. Required formulas are supplied within questions on Papers 5 and 6.

Worked example 1: six rectangular faces

Question. Calculate the total surface area of a closed cuboid measuring 6 cm by 4 cm by 3 cm.

Plan. Find the three different rectangular areas and count each twice.

S=2(6×4)+2(6×3)+2(4×3)three pairs of faces=48+36+24areas in square centimetres=108 cm2\begin{aligned} S&=2(6\times4)+2(6\times3)+2(4\times3) &&\text{three pairs of faces}\\ &=48+36+24 &&\text{areas in square centimetres}\\ &=108\text{ cm}^2 \end{aligned}

Answer. 108 cm2108\text{ cm}^2.

Check. The six net faces have areas 24, 24, 18, 18, 12 and 12 cm212\text{ cm}^2, totalling 108.

The next solid still has matching ends, but each end is triangular, so there are three side rectangles rather than four.

Worked example 2: five faces of a triangular prism

Question. A right triangular prism is 6 cm long. Its end is a right-angled triangle with perpendicular sides 3 cm and 4 cm and third side 5 cm. Calculate its total surface area.

Plan. Add two triangular ends and three rectangles running the full length.

Aend=12×3×4=6 cm2one triangleS=2×6+6(3+4+5)ends plus side rectangles=12+72=84 cm2\begin{aligned} A_{\text{end}}&=\frac12\times3\times4=6\text{ cm}^2 &&\text{one triangle}\\ S&=2\times6+6(3+4+5) &&\text{ends plus side rectangles}\\ &=12+72=84\text{ cm}^2 \end{aligned}

Answer. 84 cm284\text{ cm}^2.

Check. Listing the side areas separately gives 18, 24 and 30; adding the two ends gives 18+24+30+6+6=8418+24+30+6+6=84.

A pyramid has no matching top face. Its triangular side faces meet at a vertex. To calculate a triangle's area, use its perpendicular height within that face, not the perpendicular height of the whole solid.

Worked example 3: a square-based pyramid

Question. A square-based pyramid has base side 6 cm. Each triangular face has perpendicular face height 5 cm. The perpendicular height of the solid is 4 cm. Calculate its total surface area.

Plan. Add the square base and four triangular faces, using face height 5 cm.

Abase=62=36 cm2square baseAface=12×6×5=15 cm2height within a faceS=36+4×15=96 cm2all five faces\begin{aligned} A_{\text{base}}&=6^2=36\text{ cm}^2 &&\text{square base}\\ A_{\text{face}}&=\frac12\times6\times5=15\text{ cm}^2 &&\text{height within a face}\\ S&=36+4\times15=96\text{ cm}^2 &&\text{all five faces} \end{aligned}

Answer. 96 cm296\text{ cm}^2.

Check. The four triangles in the net cover 60 cm260\text{ cm}^2; the square covers 36. The solid height 4 cm is unused because this is an area calculation on the faces.

Removing a face subtracts its area. For example, the cuboid's 6 cm by 4 cm lid covers 24 cm224\text{ cm}^2, so its open-top area is 108−24=84 cm2108-24=84\text{ cm}^2. A thin sheet is treated as having negligible thickness: count the sheet area once unless the question specifically asks for both sides to be painted.

Watch out

Do not calculate volume instead of area. In the pyramid, 12×6×4\frac12\times6\times4 uses the solid height in the wrong plane; each side triangle needs face height 5 cm.

Decide which faces are present before adding their areas.

Cylinder surface

Cutting a cylinder's curved surface along its height and unrolling it gives a rectangle. The rectangle's width is the circumference 2πr2\pi r, and its height is hh, so its area is 2πrh2\pi rh.

Curved surface area of a cylinder

Acurved=2πrhA_{\text{curved}}=2\pi rh

For a closed cylinder, add two circular ends, each of area πr2\pi r^2. For an open-top container, add just its circular base. The curved-area and circle-area formulas are on the supplied lists for Papers 1–4; required formulas are supplied within questions on Papers 5 and 6.

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