What are wrong units?

Perimeter is in cm, area in cm^2. Stating the wrong unit costs a mark.

SEAB Original-style practice: A circle has radius 7\ cm. Taking π = 227, find (a) its circumference and (b) its area.

(a) Circumference = 2π r = 2 × 227 × 7 = 44\ cm. (b) Area = π r^2 = 227 × 7^2 = 227 × 49 = 154\ cm^2. What markers reward: the correct formulae (2π r for circumference, π r^2 for area), correct substitution, and the right units (cm for length, cm^2 for area). Using π = 227 with r = 7 keeps the numbers exact.

SingaporeMathsSyllabus dot point

How do we find the perimeter and area of rectangles, triangles, circles and shapes made from them?

Calculate the perimeter and area of rectangles, triangles, parallelograms, trapeziums and circles, and of composite figures

A focused answer to the N(A)-Level Mathematics outcome on perimeter and area. Formulae for rectangles, triangles, parallelograms, trapeziums and circles, plus composite shapes and correct units.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to find the perimeter and area of common plane figures - rectangles, triangles, parallelograms, trapeziums and circles - and of composite shapes built from them. Perimeter is the distance around a shape; area is the space inside it. Knowing the formulae and the correct units is the whole skill.

The answer

Perimeter

Perimeter is the total length around the edge of a shape, found by adding all the side lengths. For a rectangle of length $l$ and width $w$ , the perimeter is $2l + 2w$ . Perimeter is a length, so its unit is cm, m and so on.

Area of straight-sided figures

Area measures the space enclosed, in square units (cm $^2$ , m $^2$ ). The key formulae are:

Rectangle: $\text{area} = l \times w$ .
Triangle: $\text{area} = \dfrac{1}{2} \times \text{base} \times \text{height}$ .
Parallelogram: $\text{area} = \text{base} \times \text{height}$ .
Trapezium: $\text{area} = \dfrac{1}{2}(a + b) \times h$ , where $a$ and $b$ are the parallel sides.

The "height" is always the perpendicular distance, not a slanted side.

Circles

A circle of radius $r$ has:

\text{circumference} = 2\pi r, \qquad \text{area} = \pi r^2

The radius is the distance from the centre to the edge; the diameter is twice the radius. Use the value of $\pi$ the question asks for (often $3.142$ or $\dfrac{22}{7}$ ).

Composite figures

A composite figure is made from simpler shapes. To find its area, either add the areas of the parts, or take a whole shape and subtract the area of any pieces removed. Sketching the split clearly is the key step.

Worked example

A trapezium has parallel sides of $8\ \text{cm}$ and $12\ \text{cm}$ and a perpendicular height of $5\ \text{cm}$ . A semicircle of diameter $8\ \text{cm}$ sits on the shorter parallel side. Find the total area, taking $\pi = 3.142$ .

Step 1: Area of the trapezium

\text{area} = \frac{1}{2}(a + b) \times h = \frac{1}{2}(8 + 12) \times 5 = \frac{1}{2} \times 20 \times 5 = 50\ \text{cm}^2.

Step 2: Radius of the semicircle

The diameter is $8\ \text{cm}$ , so the radius is $4\ \text{cm}$ .

Step 3: Area of the semicircle

A semicircle is half a circle:

\frac{1}{2} \pi r^2 = \frac{1}{2} \times 3.142 \times 4^2 = \frac{1}{2} \times 3.142 \times 16 = 25.1\ \text{cm}^2.

Step 4: Add the parts

50 + 25.1 = 75.1\ \text{cm}^2.

So the total area is about $75.1\ \text{cm}^2$ .

Examples in context

Example 1. Fencing a garden. To fence a rectangular garden $15\ \text{m}$ by $8\ \text{m}$ , you need the perimeter: $2(15) + 2(8) = 46\ \text{m}$ of fencing. To turf it, you need the area: $15 \times 8 = 120\ \text{m}^2$ of turf. The same shape gives a length for the fence and an area for the turf.

Example 2. An athletics track end. The curved end of a running track is a semicircle, so its boundary length is half a circumference, $\dfrac{1}{2} \times 2\pi r = \pi r$ . Recognising parts of circles inside real shapes lets you apply the circle formulae to composite figures.

Try this

Cue. Find the area of a triangle with base $10\ \text{cm}$ and height $6\ \text{cm}$ . Compute $\dfrac{1}{2} \times 10 \times 6 = 30\ \text{cm}^2$ .
Cue. Find the circumference of a circle with radius $5\ \text{cm}$ , taking $\pi = 3.142$ . Compute $2 \times 3.142 \times 5 = 31.42\ \text{cm}$ .
Cue. A square has area $49\ \text{cm}^2$ . Its side is $\sqrt{49} = 7\ \text{cm}$ , so its perimeter is $4 \times 7 = 28\ \text{cm}$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original3 marksA circle has radius

7\ \text{cm}

. Taking

\pi = \dfrac{22}{7}

, find (a) its circumference and (b) its area.

Show worked answer →

(a) Circumference $= 2\pi r = 2 \times \dfrac{22}{7} \times 7 = 44\ \text{cm}$ .

(b) Area $= \pi r^2 = \dfrac{22}{7} \times 7^2 = \dfrac{22}{7} \times 49 = 154\ \text{cm}^2$ .

What markers reward: the correct formulae ( $2\pi r$ for circumference, $\pi r^2$ for area), correct substitution, and the right units (cm for length, cm $^2$ for area). Using $\pi = \dfrac{22}{7}$ with $r = 7$ keeps the numbers exact.

Original4 marksA rectangle measures

10\ \text{cm}

6\ \text{cm}

. A square of side

2\ \text{cm}

is cut from one corner. Find the area of the remaining shape.

Show worked answer →

Area of the whole rectangle: $10 \times 6 = 60\ \text{cm}^2$ .

Area of the square cut out: $2 \times 2 = 4\ \text{cm}^2$ .

Remaining area: $60 - 4 = 56\ \text{cm}^2$ .

What markers reward: finding each area separately and subtracting the removed piece, with correct square-unit labels. This "whole minus part" method is the standard approach for composite shapes.