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Perimeter and area of common and composite shapes: QCE Essential Mathematics Unit 3

Syllabus dot point

“Calculate perimeters and areas of common shapes (rectangles, triangles, parallelograms, trapeziums, circles) and composite shapes, convert units of area, and apply them to practical problems such as flooring, fencing and painting”

QCEEssential MathematicsUnit 3: Measurement, scales and chance7 min read

Quick answer

Perimeter is the distance around a shape; area is the surface inside it. Use the formulas for rectangles, triangles, parallelograms, trapeziums and circles, split composite shapes into simple ones (adding or subtracting areas), convert square units carefully, and round up quantities such as tiles or paint so the job can be finished.

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  1. What this dot point is asking
  2. The answer
  3. Practice questions

What this dot point is asking

You need to find perimeters and areas of common and composite shapes and use them in practical problems: how much fencing, how many tiles, how much paint or turf, and what it costs. These problems appear across the Unit 3 CIA and problem-solving tasks.

The answer

Perimeter and area

  • Perimeter is the distance around the outside of a shape. Units: mm, cm, m, km.
  • Area is the amount of surface a shape covers. Units: mm², cm², m², hectares, km².

Formulas

Shape Area Perimeter
Rectangle A=lwA = lw P=2(l+w)P = 2(l + w)
Triangle A=12bhA = \frac{1}{2}bh Add the three sides
Parallelogram A=bhA = bh Add the four sides
Trapezium A=12(a+b)hA = \frac{1}{2}(a + b)h Add the four sides
Circle A=πr2A = \pi r^2 C=2πr=πdC = 2\pi r = \pi d

The height hh is always perpendicular (at right angles) to the base, not the slanted side. Check your formula sheet for the exact formulas provided in the CIA.

Composite shapes

Split the shape into simple shapes, then:

  • add areas when shapes are joined (an L-shaped room);
  • subtract areas when a shape is cut out (a lawn with a pond).

For perimeter, trace around the outside only; do not include internal lines where shapes join.

Area units

Because area is two-dimensional, conversions are squared: 1 m² = 100 cm × 100 cm = 10 000 cm². A hectare is 10 000 m² (a 100 m by 100 m square).

Worked example

A wall is 4.2 m wide and 2.7 m high, with a window 1.2 m by 1.5 m. One litre of paint covers 12 m², and the wall needs two coats. How many litres are needed?

  1. Wall area = 4.2 × 2.7 = 11.34 m².
  2. Window area = 1.2 × 1.5 = 1.8 m².
  3. Area to paint = 11.34 minus 1.8 = 9.54 m².
  4. Two coats = 2 × 9.54 = 19.08 m².
  5. Paint = 19.08 divided by 12 = 1.59 L, so buy 2 L.
Common traps
Using the diameter in πr2\pi r^2
Halve the diameter first.
Using the slanted side as the height
The height must be perpendicular to the base.
Converting square units as if they were lengths
1 m² is 10 000 cm², not 100 cm².

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marks
A rectangular backyard is 15 m long and 9 m wide. Fencing is needed on three sides (not along the house, which runs along one 15 m side). Find the length of fencing needed and the area of the yard.
Show worked solution →

Fencing = 15 + 9 + 9 = 33 m (the two short sides and the long side opposite the house).

Area = 15 times 9 = 135 m².

Marking guide: 1 mark for identifying the three sides, 1 mark for 33 m, 1 mark for 135 m².

core4 marks
An L-shaped room is made of a 5 m by 4 m rectangle joined to a 3 m by 2 m rectangle. Floor tiles are sold in boxes that cover 1.8 m² each. How many boxes are needed?
Show worked solution →

Area = (5 times 4) + (3 times 2) = 20 + 6 = 26 m².

Boxes = 26 divided by 1.8 = 14.4, so 15 boxes (round up so the whole floor is covered).

Many tilers add about 10% for cuts and breakages: 26 times 1.1 = 28.6 m², which needs 28.6 divided by 1.8 = 15.9, so 16 boxes.

Marking guide: 1 mark for splitting into rectangles, 1 mark for 26 m², 1 mark for dividing by box coverage, 1 mark for rounding up to 15 boxes.

exam6 marks
A circular garden bed of diameter 4 m sits in the middle of a 10 m by 8 m lawn. Turf costs 12 dollars per square metre, and edging for the garden bed costs 9 dollars per metre. Find the total cost of re-turfing the lawn (not the garden bed) and edging the garden bed, and comment on whether your answer is reasonable.
Show worked solution →
Lawn area
Rectangle = 10 times 8 = 80 m². Circle radius = 2 m, so circle area = π×22=12.57\pi \times 2^2 = 12.57 m². Turf area = 80 minus 12.57 = 67.43 m².
Turf cost
67.434 times $12 = $809.20.
Edging
Circumference = π×4=12.57\pi \times 4 = 12.57 m. Cost = 12.57 times $9 = $113.10.

Total = $922.30 (about $922).

Reasonableness. About 70 m² at about $12 is about $840, plus about 13 m of edging at about $9 is about $120, giving about $960. The answer is close, so it is reasonable. In practice, turf is sold in whole rolls, so the real cost may be slightly higher.

Marking guide: 1 mark for the rectangle area, 1 mark for the circle area using the radius, 1 mark for turf area and cost, 1 mark for circumference and edging cost, 1 mark for the total, 1 mark for a reasonableness comment.

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