Surface area of composite solids | Year 10

01 / LEARN

Understand the idea.

Identify the outside

Surface area measures the exposed skin. List the faces of each component and mark where they touch. A shared face lies inside the assembled object, so neither of its two copies belongs to the outside area.

Adding complete component areas is a useful start, provided both copies of every hidden contact region are then removed. Alternatively, count only exposed faces directly. A face list helps prevent missing a base or counting a join.

A_{o u t s i d e} = A_{1} + A_{2} - 2 A_{c o n t a c t}

Curved and flat surfaces

Curved components have flat boundaries too. A cylinder has one curved surface and two circular ends; a hemisphere has one curved surface and one flat circular face. Matching circular faces become hidden when these solids are joined.

Keep the exposed curved surfaces and any uncovered flat faces. Use consistent lengths, squared units and exact multiples of $π$ unless a decimal is requested. Surface area never measures the space inside the solid.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Joined blocks

A 6 by 4 by 3 cm cuboid joins a 2 by 4 by 3 cm cuboid along their 4 by 3 cm faces. Find the outside area.

Worked solution

1
Calculate both complete cuboid areas.
$\begin{aligned} A_{1} & = 2 (6 \cdot 4 + 6 \cdot 3 + 4 \cdot 3) = 108 {cm}^{2} \\ A_{2} & = 2 (2 \cdot 4 + 2 \cdot 3 + 4 \cdot 3) = 52 {cm}^{2} \end{aligned}$
Each cuboid has three pairs of congruent rectangular faces.
2
Find the shared face area.
$A_{contact} = 4 \cdot 3 = 12 {cm}^{2}$
The stated join is the $4 cm$ by $3 cm$ rectangle.
3
Remove both hidden faces.
$A_{outside} = 108 + 52 - 2 (12) = 136 {cm}^{2}$
Each complete cuboid area counted one copy of the internal join.

Answer

136 cm²

EXAMPLE 02

Domed cylinder

A cylinder of radius 3 cm and height 10 cm has a matching hemispherical top and an exposed flat base. Find its outside area.

Worked solution

1
Find the exposed cylinder curve.
$A_{cylinder curve} = 2 π r h = 2 π (3) (10) = 60 π {cm}^{2}$
The circular top is hidden beneath the dome.
2
Find the hemisphere curve and exposed base.
$\begin{aligned} A_{dome} & = 2 π {(3)}^{2} = 18 π {cm}^{2} \\ A_{base} & = π {(3)}^{2} = 9 π {cm}^{2} \end{aligned}$
The dome's flat circle is hidden, but the cylinder's lower base is exposed.
3
Add only the exposed surfaces.
$A = 60 π + 18 π + 9 π = 87 π {cm}^{2}$
No circular contact face contributes to the outside skin.

Answer

$87 π {cm}^{2}$

03 / YOUR TURN

Try it yourself.

Check your understanding

Two cubes of edge 2 cm share a square face. Find their outside area.

Show answer & reasoning

40 cm²

Each cube has area $6 (2^{2}) = 24 {cm}^{2}$ . Remove the two hidden squares: $24 + 24 - 2 (2^{2}) = 40 {cm}^{2}$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Two cubes of edge 3 cm share a whole face. Find the exposed area.
2 marks
Two cubes of edge 4 cm share a whole face. Find the exposed area.
2 marks
Join 10 by 6 by 4 cm and 2 by 6 by 4 cm cuboids along their 6 by 4 faces. Find exposed area.
2 marks
A cube of edge 6 cm has a square-pyramid roof with triangular slant height 5 cm. Find exposed area; the square join is hidden.
3 marks
A radius 2 cm, height 5 cm cylinder has a matching hemisphere on top and an exposed flat bottom. Find outside area.
2 marks
A cube of edge 5 cm supports a square-pyramid roof of slant height 7 cm. Find exposed area.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Two cubes of edge 3 cm share a whole face. Find the exposed area.
2 marks
$2 \times 54 - 2 \times 9 = 90$ .
90 cm²
Two cubes of edge 4 cm share a whole face. Find the exposed area.
2 marks
The result is an 8 by 4 by 4 cuboid.
$2 (32 + 32 + 16) = 160$ .
160 cm²
Join 10 by 6 by 4 cm and 2 by 6 by 4 cm cuboids along their 6 by 4 faces. Find exposed area.
2 marks
The result is 12 by 6 by 4.
$2 (72 + 48 + 24) = 288$ .
288 cm²
A cube of edge 6 cm has a square-pyramid roof with triangular slant height 5 cm. Find exposed area; the square join is hidden.
3 marks
Five cube faces:180.
Four roof triangles: $4 \times \frac{1}{2} \times 6 \times 5 = 60$ .
$Total = 240$ .
240 cm²
A radius 2 cm, height 5 cm cylinder has a matching hemisphere on top and an exposed flat bottom. Find outside area.
2 marks
Curved cylinder $20 π$ ; hemisphere $8 π$ ; base $4 π$ .
$32 π {cm}^{2}$
A cube of edge 5 cm supports a square-pyramid roof of slant height 7 cm. Find exposed area.
3 marks
Five square faces $= 125$ .
Four roof faces $= 4 \times \frac{1}{2} \times 5 \times 7 = 70$ .
$Total = 195$ .
195 cm²

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum

Surface area of composite solids; volume of composite solids; proportion/scaling of objects; related length/area/volume factors; measurement error; the effect of rounding and repeated approximations; practical design models; evaluating assumptions and suitable precision.

Official curriculum source ↗

In this lesson

Understand the idea.

Identify the outside

Curved and flat surfaces

See the method in action.

Joined blocks

Calculate both complete cuboid areas.

Find the shared face area.

Remove both hidden faces.

Domed cylinder

Find the exposed cylinder curve.

Find the hemisphere curve and exposed base.

Add only the exposed surfaces.

Try it yourself.

Check your understanding

Keep practising