Perimeter, Area and 2-D Shapes

This lesson covers DfE content statement L2.16 — calculating the perimeter and area of common 2-D shapes, including composite shapes.

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Perimeter

The perimeter is the total distance around the outside of a shape. Add up all the sides.

Worked Example — Fencing a Garden

Scenario: You need to fence a rectangular garden that measures 12 m by 8.5 m. How much fencing do you need?

Perimeter = 2 × (length + width) = 2 × (12 + 8.5) = 2 × 20.5 = 41 m

Worked Example — Irregular Shape

Scenario: An L-shaped room has the following measurements (in metres):

    4m
  ┌─────┐
  │     │ 3m
  │  ┌──┘
  │  │
  │  │ 5m
  │  │
  └──┘
  2m

Missing sides: bottom horizontal = 4 − 2 = 2m (wait — we need to work out which measurements are given).

Let us say the overall shape has these outer measurements: 6m wide at the bottom, 8m tall on the left, with a rectangular notch cut from the top-right corner that is 4m wide and 3m deep.

Perimeter = 6 + 8 + 4 + 3 + (6 − 4) + (8 − 3) = 6 + 8 + 4 + 3 + 2 + 5 = 28 m

Exam Tip: For composite shapes, first work out any missing side lengths. The perimeter is always the OUTSIDE edge — do not include internal lines.

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Area of Standard 2-D Shapes

Shape	Formula	Key facts
Rectangle	A = length × width	All angles 90°
Square	A = side²	Special rectangle
Triangle	A = 1/2 × base × height	Height must be PERPENDICULAR to base
Parallelogram	A = base × perpendicular height	Not the slant height
Trapezium	A = 1/2 × (a + b) × h	a and b are the parallel sides
Circle	A = π × r²	r = radius; use π = 3.14159...

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Rectangle

Worked Example — Painting a Wall

Scenario: A bedroom wall is 4.2 m wide and 2.6 m high. A tin of paint covers 12 m². How many tins are needed?

Area = 4.2 × 2.6 = 10.92 m²

One tin covers 12 m², so 1 tin is enough (with 1.08 m² to spare).

Worked Example — Flooring

Scenario: A room measures 5.5 m by 4 m. Laminate flooring costs £18.50 per square metre. What is the cost?

Area = 5.5 × 4 = 22 m² Cost = 22 × £18.50 = £407.00

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Triangle

The height of a triangle is the perpendicular distance from the base to the opposite vertex — NOT the length of a slanted side.

Worked Example — Decorative Panel

Scenario: A triangular sign has a base of 80 cm and a height of 50 cm. What is its area?

Area = 1/2 × 80 × 50 = 2,000 cm² (or 0.2 m²)

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Parallelogram

Worked Example — Paving Stones

Scenario: Each parallelogram-shaped paving stone has a base of 30 cm and a perpendicular height of 18 cm. How many are needed to cover 5.4 m²?

Area of one stone = 30 × 18 = 540 cm² = 0.054 m² Number needed = 5.4 ÷ 0.054 = 100 stones

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Trapezium

A trapezium has one pair of parallel sides (labelled a and b). The formula averages these parallel sides and multiplies by the height.

graph TD
    subgraph Trapezium
    A["a (top parallel side)"]
    B["b (bottom parallel side)"]
    H["h (perpendicular height)"]
    end
    A --- H --- B

Worked Example — Cross-Section of a Channel

Scenario: An irrigation channel has a trapezoidal cross-section. The top width is 2.4 m, the bottom width is 1.2 m, and the depth (height) is 0.8 m. What is the area of the cross-section?

Area = 1/2 × (2.4 + 1.2) × 0.8 = 1/2 × 3.6 × 0.8 = 1.44 m²

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Circle

Area = π × r² (where r is the radius) Circumference = π × d or 2 × π × r (where d is the diameter)

Remember: the diameter is twice the radius → d = 2r.

Worked Example — Circular Patio

Scenario: You want to build a circular patio with a diameter of 6 m. How much area will it cover?

Radius = 6 ÷ 2 = 3 m Area = π × 3² = π × 9 = 28.27 m² (to 2 d.p.)

Worked Example — Circular Pond Edge

Scenario: A circular pond has a radius of 2.5 m. You want to put edging stones around it. What length of edging do you need?

Circumference = 2 × π × 2.5 = 15.71 m (to 2 d.p.)

Exam Tip: Always check whether the question gives you the radius or the diameter. If it gives the diameter, halve it to get the radius before using the area formula. This is one of the most common errors in geometry questions.

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Composite (Compound) Shapes

Real-world shapes are often made up of several standard shapes combined. Break the shape into rectangles, triangles, circles, etc., find each area, then add (or subtract) them.

Worked Example — L-Shaped Room