Composite figures, paths and shaded regions

Learning outcome

Calculate composite regions, distinguish inner and outer paths, and correct for overlapping areas.

Concepts and assumptions

Use ideal dimensions, consistent units, uniform-width paths and square-cornered borders.

Add non-overlapping areas; subtract cut-outs. If added pieces overlap, subtract their intersection once to correct double counting. Outer perimeter excludes shared internal edges.

An inner border of width w occupies both opposite sides: central dimensions decrease by 2w. Subtract central area from original area. Require 2w smaller than both original dimensions.

An outer border adds 2w to both dimensions. Subtract original area from enlarged area. Equal-width inner and outer borders need not have equal areas.

Circular calculations use approximate pi = 22/7 without further rounding.

Worked examples

Example 1 — Joined regions. Attach a semicircle outside a 20 m by 14 m rectangle, using one full 14 m side as its diameter. Find area and outer perimeter. Use pi = 22/7; no further rounding.

Radius = 14/2 = 7 m. Area = 20 × 14 + (22/7) × 7²/2 = 280 + 77 = 357 m². Exclude the shared diameter. Outer perimeter = 20 + 20 + 14 + (22/7) × 7 = 76 m.

Example 2 — Two path designs. Compare separate 2 m inner and outer borders for a 30 m by 22 m rectangular garden.

Garden area = 30 × 22 = 660 m². Inner central dimensions = (30 − 4) by (22 − 4) = 26 m by 18 m. Inner border = 660 − 26 × 18 = 192 m². Outer dimensions = (30 + 4) by (22 + 4) = 34 m by 26 m. Outer border = 34 × 26 − 660 = 224 m².

Example 3 — Crossing paths. A 40 m by 30 m park has perpendicular central paths: one 2 m wide spanning its full length, the other 3 m wide spanning its full width. Find unpaved area.

Path areas = 40 × 2 = 80 m² and 30 × 3 = 90 m². Intersection = 2 × 3 = 6 m². Total path area = 80 + 90 − 6 = 164 m². Unpaved area = 40 × 30 − 164 = 1,036 m².

Common mistakes

Counting shared edges; changing dimensions by only one border width; confusing inner and outer paths; subtracting overlapping paths twice.

Practice questions

1. An 18 cm-wide, 12 cm-high rectangle loses an 8 cm-wide, 5 cm-high rectangle from its top-right corner. Find the remaining area and perimeter.
2. Find the area of a 1.5 m inner border in a 20 m by 16 m garden.
3. Find the area of a 1.5 m outer border around a 24 m by 18 m garden.
4. A circle centred in a 28 cm square touches all four sides. Find the shaded area outside the circle but inside the square. Use pi = 22/7; no further rounding.

Worked solutions

1. Area = 18 × 12 − 8 × 5 = 176 cm². Remaining outer sides: 12 − 5 = 7 cm and 18 − 8 = 10 cm. Perimeter = 18 + 7 + 8 + 5 + 10 + 12 = 60 cm.
2. Central dimensions = 20 − 3 = 17 m and 16 − 3 = 13 m. Border area = 20 × 16 − 17 × 13 = 99 m².
3. Outer dimensions = 24 + 3 = 27 m and 18 + 3 = 21 m. Border area = 27 × 21 − 24 × 18 = 135 m².
4. Using pi = 22/7 without further rounding, radius = 28/2 = 14 cm. Shaded area = 28² − (22/7) × 14² = 784 − 616 = 168 cm².