Perimeter and area of squares and rectangles

Learning outcome

Calculate perimeters and areas, recover unknown dimensions, and count tiles using consistent units.

Concepts and assumptions

A rectangle has four right angles and equal opposite sides; a square also has four equal sides. Treat exercise dimensions as ideal exact lengths.

Perimeter measures the complete boundary; area measures the enclosed surface. Adding a rectangle’s four sides gives length + width + length + width, hence perimeter = 2(length + width).

Imagine rows of unit squares: length determines squares per row and width determines rows. Thus area = length × width, extending proportionately to fractional dimensions.

For square side s, perimeter = 4s and area = s². Perimeters use linear units; areas use squared units. Convert all dimensions to one unit before calculating.

Divide rectangular area by one side to find the other. Half the perimeter gives length + width. Perimeter alone does not determine rectangular area: side lengths still matter.

For tiling, assume no gaps or overlaps and no wastage unless specified. An area quotient gives a whole-tile count only when the arrangement fits.

Worked examples

Example 1 — Boundary versus coverage. A rectangular garden is 26 m long and 18 m wide. Find its perimeter and area.

Perimeter = 26 + 18 + 26 + 18 = 88 m. Area = 26 × 18 = 468 m². One complete fence without a gate follows the 88 m boundary; surface covering concerns 468 m².

Example 2 — Recovering dimensions. A rectangle has perimeter 74 m. Its length exceeds its width by 5 m. Find both dimensions and area.

Let width be w metres; length is w + 5. Then 2(w + w + 5) = 74. Divide by 2: 2w + 5 = 37. Thus 2w = 32, giving width 16 m and length 21 m. Area = 21 × 16 = 336 m². Check: 2(21 + 16) = 74 m.

Example 3 — Exact tiling. A rectangular floor is 6 m by 4.5 m. Square tiles have side 30 cm. Find the tile count with aligned rows and no gaps.

Floor dimensions = 600 cm by 450 cm. Tiles along the length = 600/30 = 20. Rows = 450/30 = 15. Both are whole numbers, so no cutting is needed. Total = 20 × 15 = 300 tiles.

Common mistakes

Using area for fencing; using perimeter for flooring; forgetting to halve a perimeter before recovering sides; mixing centimetres with metres; assuming an area ratio proves tiles fit.

Practice questions

1. A square has perimeter 52 cm. Find its side and area.
2. A rectangle has area 216 m² and width 12 m. Find its length and perimeter.
3. A rectangle’s length:width ratio is 5:2 and its area is 360 m². Find its dimensions and perimeter.
4. A wire forms a 24 cm by 12 cm rectangle. It is reshaped into a square without loss. Find the increase in enclosed area.

Worked solutions

1. Side = 52/4 = 13 cm. Area = 13 × 13 = 169 cm².
2. Length = 216/12 = 18 m. Perimeter = 2(18 + 12) = 60 m.
3. Let dimensions be 5x and 2x metres. Then 10x² = 360, so x² = 36 and x = 6 because lengths are positive. Dimensions are 30 m and 12 m; perimeter = 2(30 + 12) = 84 m.
4. Wire length = 2(24 + 12) = 72 cm. Square side = 72/4 = 18 cm. New area = 18² = 324 cm²; original area = 24 × 12 = 288 cm². Increase = 324 − 288 = 36 cm².