Australian Curriculum v9 / ACiQ Year 8 Mathematics - Unit 3 - Composite perimeter and area

Composite perimeter and area

Calculate perimeter and area of composite shapes by decomposing them and accounting for shared or missing boundaries.

Updated 2026-07-26 - 13 min read

Composite perimeter and area is taught here as a connected set of decisions, not a list of facts. Work through the prerequisite recall, explicit models, carefully faded examples, misconception repairs and transfer task before using the target in Check, Practice, Review or Rapid Revision.

This note is designed to work with the guided lessons, curated practice, flashcards, Tutor context, Review and Rapid Revision for the same canonical target. The same three evidence checks are used throughout, so feedback can route a learner back to the precise idea that needs repair.

Decompose composite areas

Split a composite shape into non-overlapping familiar shapes, calculate each area and add or subtract exactly once. This relationship must be selected from the quantities and conditions in the problem, then checked against the context.

A dependable reasoning routine

  1. Name the unknowns, units and constraints before calculating.
  2. Choose the relationship represented by decompose composite areas and state why it applies.
  3. Keep exact values for as long as possible, show substitutions and preserve units through each step.
  4. Check the result by substitution, estimation, an alternative representation or the original context.

Repair: A valid decomposition covers the region once; overlaps must be removed or avoided.

The repair matters because the shortcut may appear to work in one familiar example while failing when the method, representation, scale, constraint or accuracy requirement changes. Use the routine above to make the reasoning visible enough for another learner to verify.

Example 1.1

An L-shape is a 10 m by 8 m rectangle with a 4 m by 3 m corner removed. Find its area.

Step 1 - identify the governing idea: Split a composite shape into non-overlapping familiar shapes, calculate each area and add or subtract exactly once.

Step 2 - apply it to this evidence: The full area is 80 m² and the removed corner is 12 m².

Result: 68 m²

The relationship is visible in the working: The full area is 80 m² and the removed corner is 12 m². Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 92 m² — It does not agree with the required relationship: The full area is 80 m² and the removed corner is 12 m².
  • 56 m² — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 80 m² — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 1.2

A shape consists of non-overlapping rectangles 6 cm by 4 cm and 3 cm by 2 cm. Find total area.

Step 1 - identify the governing idea: Split a composite shape into non-overlapping familiar shapes, calculate each area and add or subtract exactly once.

Step 2 - apply it to this evidence: The areas are 24 cm² and 6 cm², which add to 30 cm².

Result: 30 cm²

The relationship is visible in the working: The areas are 24 cm² and 6 cm², which add to 30 cm². Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 15 cm² — It does not agree with the required relationship: The areas are 24 cm² and 6 cm², which add to 30 cm².
  • 28 cm² — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 60 cm² — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 1.3

A 12 m by 9 m rectangle contains a 5 m by 4 m rectangular courtyard. Find the remaining area.

Step 1 - identify the governing idea: Split a composite shape into non-overlapping familiar shapes, calculate each area and add or subtract exactly once.

Step 2 - apply it to this evidence: The outer area 108 m² minus courtyard area 20 m² equals 88 m².

Result: 88 m²

The relationship is visible in the working: The outer area 108 m² minus courtyard area 20 m² equals 88 m². Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 128 m² — It does not agree with the required relationship: The outer area 108 m² minus courtyard area 20 m² equals 88 m².
  • 68 m² — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 108 m² — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Trace composite perimeters

Perimeter is the total length of the outside boundary; shared internal edges are not counted. This relationship must be selected from the quantities and conditions in the problem, then checked against the context.

A dependable reasoning routine

  1. Name the unknowns, units and constraints before calculating.
  2. Choose the relationship represented by trace composite perimeters and state why it applies.
  3. Keep exact values for as long as possible, show substitutions and preserve units through each step.
  4. Check the result by substitution, estimation, an alternative representation or the original context.

Repair: Trace only the external boundary and infer missing lengths from aligned parallel sides where needed.

The repair matters because the shortcut may appear to work in one familiar example while failing when the method, representation, scale, constraint or accuracy requirement changes. Use the routine above to make the reasoning visible enough for another learner to verify.

Example 2.1

A 10 cm by 6 cm rectangle has a 3 cm by 2 cm corner cut from one corner. What is the perimeter?

Step 1 - identify the governing idea: Perimeter is the total length of the outside boundary; shared internal edges are not counted.

Step 2 - apply it to this evidence: The two removed outer lengths are replaced by equal notch lengths, so the perimeter remains 2(10 + 6) = 32 cm.

Result: 32 cm

The relationship is visible in the working: The two removed outer lengths are replaced by equal notch lengths, so the perimeter remains 2(10 + 6) = 32 cm. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 26 cm — It does not agree with the required relationship: The two removed outer lengths are replaced by equal notch lengths, so the perimeter remains 2(10 + 6) = 32 cm.
  • 38 cm — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 60 cm — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 2.2

Two 5 cm squares share one full side. What is the perimeter of the joined shape?

Step 1 - identify the governing idea: Perimeter is the total length of the outside boundary; shared internal edges are not counted.

Step 2 - apply it to this evidence: Together they have 40 cm before joining; the shared 5 cm edge is counted twice and removed twice.

Result: 30 cm

The relationship is visible in the working: Together they have 40 cm before joining; the shared 5 cm edge is counted twice and removed twice. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 40 cm — It does not agree with the required relationship: Together they have 40 cm before joining; the shared 5 cm edge is counted twice and removed twice.
  • 35 cm — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 20 cm — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 2.3

A rectilinear shape has horizontal rightward lengths totalling 14 m. What is the total of its leftward horizontal boundary lengths?

Step 1 - identify the governing idea: Perimeter is the total length of the outside boundary; shared internal edges are not counted.

Step 2 - apply it to this evidence: A closed boundary returns to its starting horizontal position, so rightward and leftward totals match.

Result: 14 m

The relationship is visible in the working: A closed boundary returns to its starting horizontal position, so rightward and leftward totals match. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 7 m — It does not agree with the required relationship: A closed boundary returns to its starting horizontal position, so rightward and leftward totals match.
  • 28 m — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • It cannot be known — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Choose and justify an efficient composite-area strategy

Choose addition, subtraction or rearrangement according to the shape, and verify the result using an alternative decomposition or bounds. This relationship must be selected from the quantities and conditions in the problem, then checked against the context.

A dependable reasoning routine

  1. Name the unknowns, units and constraints before calculating.
  2. Choose the relationship represented by choose and justify an efficient composite-area strategy and state why it applies.
  3. Keep exact values for as long as possible, show substitutions and preserve units through each step.
  4. Check the result by substitution, estimation, an alternative representation or the original context.

Repair: Different non-overlapping decompositions can be equivalent if they cover exactly the same region.

The repair matters because the shortcut may appear to work in one familiar example while failing when the method, representation, scale, constraint or accuracy requirement changes. Use the routine above to make the reasoning visible enough for another learner to verify.

Example 3.1

A frame is an 8 cm by 6 cm rectangle with a centred 6 cm by 4 cm opening. Find its area.

Step 1 - identify the governing idea: Choose addition, subtraction or rearrangement according to the shape, and verify the result using an alternative decomposition or bounds.

Step 2 - apply it to this evidence: Subtracting inner from outer gives 48 − 24 = 24 cm².

Result: 24 cm²

The relationship is visible in the working: Subtracting inner from outer gives 48 − 24 = 24 cm². Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 72 cm² — It does not agree with the required relationship: Subtracting inner from outer gives 48 − 24 = 24 cm².
  • 20 cm² — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 48 cm² — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 3.2

A 9 m by 7 m rectangle has a 3 m by 2 m extension attached without overlap. Find total area.

Step 1 - identify the governing idea: Choose addition, subtraction or rearrangement according to the shape, and verify the result using an alternative decomposition or bounds.

Step 2 - apply it to this evidence: The areas 63 m² and 6 m² add because the regions do not overlap.

Result: 69 m²

The relationship is visible in the working: The areas 63 m² and 6 m² add because the regions do not overlap. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 57 m² — It does not agree with the required relationship: The areas 63 m² and 6 m² add because the regions do not overlap.
  • 126 m² — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 63 m² — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 3.3

A composite floor fits inside a 12 m by 10 m rectangle and has a 2 m by 3 m notch removed. Which bound must its area satisfy?

Step 1 - identify the governing idea: Choose addition, subtraction or rearrangement according to the shape, and verify the result using an alternative decomposition or bounds.

Step 2 - apply it to this evidence: The exact area is 120 − 6 = 114 m², below the outer rectangle and above zero.

Result: 114 m²

The relationship is visible in the working: The exact area is 120 − 6 = 114 m², below the outer rectangle and above zero. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 126 m² — It does not agree with the required relationship: The exact area is 120 − 6 = 114 m², below the outer rectangle and above zero.
  • 120 m² — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 6 m² — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Retrieval check

Try these without looking back at the examples.

  1. A shape consists of non-overlapping rectangles 6 cm by 4 cm and 3 cm by 2 cm. Find total area.
  2. Two 5 cm squares share one full side. What is the perimeter of the joined shape?
  3. A 9 m by 7 m rectangle has a 3 m by 2 m extension attached without overlap. Find total area.

Answers

  1. 30 cm² — The areas are 24 cm² and 6 cm², which add to 30 cm².
  2. 30 cm — Together they have 40 cm before joining; the shared 5 cm edge is counted twice and removed twice.
  3. 69 m² — The areas 63 m² and 6 m² add because the regions do not overlap.

Transfer task

Find an unfamiliar example from school, daily life, a credible news source or another subject. Explain which of the three evidence checks applies. Complete the task, then audit your own response: identify the evidence used, the relationship applied, one plausible misconception and the final reasonableness check. If a peer could not reproduce your reasoning, add the missing step.

Sources