What is the difference between area and perimeter in Mathematics?

In Mathematics, specifically in the study of Mensuration, the area refers to the amount of space inside a two-dimensional shape, measured in square units. The perimeter, on the other hand, is the total length of the boundary of the shape. Understanding these concepts is crucial for solving problems related to geometry in the Cambridge IGCSE curriculum.

How do you calculate the perimeter of a rectangle?

To calculate the perimeter of a rectangle, you add the lengths of all four sides. The formula is P = 2(l + w), where 'l' is the length and 'w' is the width. This formula is essential for Cambridge IGCSE Mathematics students to master as it applies to various real-world problems.

What is the formula for finding the area of a triangle?

The area of a triangle can be found using the formula A = 1/2 * base * height. This formula calculates the space inside the triangle by multiplying the base by the height and then dividing by two. Understanding this formula is key for students studying Mensuration in the Cambridge IGCSE Mathematics syllabus.

Why is understanding areas and perimeters important in real life?

Understanding areas and perimeters is important because these concepts are used in various real-life applications, such as architecture, engineering, and land measurement. For Cambridge IGCSE students, mastering these concepts helps in solving practical problems and enhances spatial reasoning skills.

How can I find the area of a circle?

The area of a circle is found using the formula A = πr², where 'r' is the radius of the circle. This formula is crucial for Cambridge IGCSE Mathematics students as it helps in calculating the space enclosed by a circular boundary, a common problem in Mensuration.

Why is "Forgetting to use squared units for area" a common mistake in Areas and Perimeters?

Why it happens: Speed. How to avoid it: Areas are always in square units: cm², m², km².

Why is "Using a slant side as the perpendicular height" a common mistake in Areas and Perimeters?

Why it happens: Confusing slant length with perpendicular distance. How to avoid it: Perpendicular height makes a right angle with the base. Source: 0580/42 — recurring

Why is "Forgetting the $\frac{1}{2}$ in the triangle area formula" a common mistake in Areas and Perimeters?

Why it happens: Speed. How to avoid it: Triangle area: half base times height. ALWAYS the half.

Why is "Double-counting an interior edge in a composite-shape perimeter" a common mistake in Areas and Perimeters?

Why it happens: Adding all edges including internal ones. How to avoid it: Perimeter = ONLY the outer boundary edges. Don't include internal cuts.

Cambridge IGCSE Mathematics (0580)

Areas and Perimeters

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Short Notes - Areas and Perimeters

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Detailed Study Notes

Detailed notes on Mensuration for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Areas and Perimeters of 2D Shapes — Cambridge IGCSE 0580 Maths Extended (2026)

Rectangles, triangles, parallelograms, trapeziums, and composite shapes. Memorise the formulae, watch units, and split composite shapes into shapes you know.

At a glance

Rectangle: area $= l \times w$ . Perimeter $= 2(l + w)$ .
Triangle: area $= \dfrac{1}{2} \times b \times h$ . Use the perpendicular height.
Parallelogram: area $= b \times h$ (perpendicular height, NOT slant).
Trapezium: area $= \dfrac{1}{2}(a + b) \times h$ where $a, b$ are parallel sides.
Composite shapes: split into rectangles + triangles + circles; sum or subtract.
Units: area in $\text{cm}^{2}, \text{m}^{2}$ etc.; perimeter in $\text{cm}, \text{m}$ .

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E6.1 — Calculate perimeter and area of rectangles, triangles, parallelograms and trapeziums.
E6.2 — Calculate area and perimeter of composite shapes.

Rectangle and square

$A = l \times w$ . Perimeter $= 2(l + w)$ .

Rectangle. Length $l$ , width $w$ . $A = l \times w, \quad P = 2(l + w).$

Square. Side $s$ . $A = s^{2}, \quad P = 4s.$

Worked. Rectangle $8\,\text{cm}$ by $5\,\text{cm}$ .

$A = 8 \times 5 = 40\,\text{cm}^{2}$ .
$P = 2(8 + 5) = 26\,\text{cm}$ .

Rectangle: $A = lw$ , $P = 2(l + w)$ .
Square: $A = s^{2}$ , $P = 4s$ .
Area in squared units, perimeter in linear units.

Triangle

$A = \dfrac{1}{2} \times \text{base} \times \text{perpendicular height}$ .

Formula. $A = \dfrac{1}{2} \times b \times h.$

The height $h$ must be perpendicular to the chosen base — measure the right-angle distance from the opposite vertex to the base line.

The height is the right-angle distance from the vertex to the base — never the slant side.

Worked. Triangle base $10\,\text{cm}$ , perpendicular height $6\,\text{cm}$ .

$A = \dfrac{1}{2} \times 10 \times 6 = 30\,\text{cm}^{2}$ .

Perimeter. Add all three sides — measure or use Pythagoras for right-angled triangles.

Heron's formula (for advanced cases). When you know all three sides $a, b, c$ but no height: $s = \dfrac{a + b + c}{2}, \quad A = \sqrt{s(s - a)(s - b)(s - c)}.$

Cambridge rarely needs this at IGCSE — usually a perpendicular height or use of $\dfrac{1}{2}ab\sin C$ (see Trigonometry).

$A = \dfrac{1}{2} b h$ — height perpendicular to base.
Perimeter: sum of three sides.
Right-angled triangle: $A = \dfrac{1}{2} \times \text{the two perpendicular sides}$ .

Parallelogram

$A = b \times h$ where $h$ is the perpendicular distance between the two parallel sides.

Formula. $A = b \times h.$

Same principle as the triangle: $h$ is the perpendicular height, NOT the slant side.

Worked. Parallelogram base $9\,\text{cm}$ , perpendicular height $4\,\text{cm}$ .

$A = 9 \times 4 = 36\,\text{cm}^{2}$ .

Perimeter. Two pairs of equal sides: $P = 2(a + b),$ where $a$ and $b$ are the two distinct side lengths.

Rhombus. All four sides equal. $A = b \times h$ still works. Or use the diagonals: $A = \dfrac{1}{2} d_{1} d_{2}$ .

Parallelogram: $A = bh$ (perpendicular).
Perimeter: $2(a + b)$ .
Rhombus diagonals: $A = \dfrac{1}{2} d_{1} d_{2}$ .

Trapezium

$A = \dfrac{1}{2}(a + b) \times h$ where $a, b$ are the parallel sides.

Formula. $A = \dfrac{1}{2}(a + b) \times h.$

Average the two parallel sides, then multiply by the perpendicular distance between them.

Average the two parallel sides, then multiply by the perpendicular height between them.

Worked. Trapezium with parallel sides $8\,\text{cm}$ and $14\,\text{cm}$ , perpendicular height $5\,\text{cm}$ .

$A = \dfrac{1}{2}(8 + 14) \times 5 = \dfrac{1}{2} \times 22 \times 5 = 55\,\text{cm}^{2}$ .

Perimeter. Sum of all four sides.

Tip. Identify the two PARALLEL sides first — they're the only ones that go into the formula.

$A = \dfrac{1}{2}(a + b) h$ .
$a, b$ = parallel sides; $h$ = perpendicular distance.
Perimeter: sum of all four sides.

Composite shapes

Split the shape into rectangles, triangles, semicircles, etc. Add or subtract their areas.

Strategy.

Look at the diagram and split it into shapes you know (rectangle, triangle, semicircle, …).
Find each piece's area separately.
ADD if the pieces fill the figure together. SUBTRACT if you've drawn a bounding rectangle and need to remove a missing corner.

Worked. L-shaped figure: outer rectangle $12\,\text{cm} \times 8\,\text{cm}$ with a $5\,\text{cm} \times 3\,\text{cm}$ rectangle removed from one corner.

Outer area: $12 \times 8 = 96\,\text{cm}^{2}$ .
Removed: $5 \times 3 = 15\,\text{cm}^{2}$ .
L-shape area: $96 - 15 = 81\,\text{cm}^{2}$ .

Find the bounding rectangle's area, then subtract the missing corner.

Perimeter of composite. Trace round the OUTSIDE of the figure adding every edge. Don't miss the inner corner edges if it's an L-shape — they're outside-facing too.

Tip. When the figure has missing labels, deduce them: opposite sides of the bounding rectangle must add up to the same total.

Split into known shapes.
Add for filled-in regions; subtract for cut-outs.
Perimeter: trace the outside, add every edge.
Deduce missing lengths from the bounding rectangle.

Quick recap

Rectangle: $A = lw$ . Perimeter: $2(l + w)$ .
Triangle: $A = \dfrac{1}{2} bh$ (perpendicular height).
Parallelogram: $A = bh$ .
Trapezium: $A = \dfrac{1}{2}(a + b)h$ .
Composite: split, then add or subtract.
Area in squared units; perimeter in linear units.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Area — region inside the boundary, measured in squared units.
Perimeter — total length of the boundary, measured in linear units.
Perpendicular height — the right-angle distance from a vertex/parallel-side to the chosen base.
Composite shape — a figure made up of two or more standard shapes joined or cut.

How it’s examined

Area and perimeter questions appear on every Paper 4 — typically embedded in a composite or word-problem question worth 3-5 marks. Paper 2 has 2-3 mark single-shape items. Examiner reports flag using slant heights instead of perpendicular heights as the recurring error.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E6.1-6.2); 0580/22 May/Jun 2024 — Q7 (composite area); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Areas and Perimeters

Step-by-step solutions to past-paper-style questions on areas and perimeters, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Areas and Perimeters

Almost every areas and perimeters exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single find / work out / calculate instruction for one area or perimeter — a named shape (rectangle, triangle, trapezium, kite) with its dimensions given, or an inverse version that gives the area and asks for a missing side.

How to approach it

Quote the correct formula first ( $A=\dfrac{1}{2}(a+b)h$ , $A=\dfrac{1}{2}ab\sin C$ , $A=\dfrac{1}{2}d_1 d_2$ ), substitute, then evaluate. For an inverse question, substitute the known area and solve the resulting equation.

Common trap

Using a slant side as the perpendicular height, or dropping the $\frac{1}{2}$ . Examiner reports flag both, plus area answers left without squared units.

Multi-step problem

Recognise it by

Several quantities to combine — two shapes to compare as a ratio, a side recovered by Pythagoras before the area formula, or an unknown width producing a quadratic.

How to approach it

Break the problem into stages: find each area / missing length first, then carry out the comparison, ratio simplification or equation solving as the final step.

Common trap

For a border of uniform width $w$ , writing $(\,l+w)(\,w+w)$ instead of $(\,l+2w)$ — the width is added to both ends of each dimension. Examiner reports flag this repeatedly.

Word problem

Recognise it by

A real-world context — tiling a floor, costing materials, sold-in-boxes packaging — with no formula stated and often a units conversion buried in the wording.

How to approach it

Convert all measurements to the same unit first, translate the context into an area calculation, then handle the practical step (rounding up to whole boxes, multiplying by a unit cost).

Common trap

Rounding the number of boxes / tiles down. Examiner reports note you must buy enough whole units to cover the job, so always round up.

1Area and perimeter of a rectangle
CoreDirect calculation• rectangle
Question
A rectangle is $7$ cm by $4$ cm. Find its area and perimeter.
Step-by-step solution
1. Step 1
  Area = length × width.
  $A = 7 \times 4 = 28\,\text{cm}^{2}$
2. Step 2
  Perimeter = $2(\text{l} + \text{w})$ .
  $P = 2(7 + 4) = 22\,\text{cm}$
Answer
$A = 28\,\text{cm}^{2},\ P = 22\,\text{cm}$
2Area of a triangle
CoreDirect calculation• triangle
Question
Find the area of a triangle with base $9$ cm and height $6$ cm.
Step-by-step solution
1. Step 1
  $A = \frac{1}{2} \times \text{base} \times \text{height}$ .
  $A = \dfrac{1}{2} \times 9 \times 6 = 27\,\text{cm}^{2}$
Answer
$27\,\text{cm}^{2}$
3Area of a trapezium
ExtendedDirect calculation• Adapted from 0580/22 May/Jun 2024 Q7• trapezium
Question
Find the area of a trapezium with parallel sides $5$ cm and $11$ cm, separated by perpendicular distance $4$ cm.
Step-by-step solution
1. Step 1
  $A = \dfrac{1}{2}(a + b)h$ .
  $A = \dfrac{1}{2}(5 + 11)(4) = 32$
Answer
$32\,\text{cm}^{2}$
4Area of a composite shape
ExtendedDirect calculation• composite
Question
An L-shape is formed by removing a $3\,\text{cm} \times 2\,\text{cm}$ rectangle from the corner of a $7\,\text{cm} \times 5\,\text{cm}$ rectangle. Find its area.
Step-by-step solution
1. Step 1
  Larger area minus smaller area.
  $7 \times 5 - 3 \times 2 = 35 - 6 = 29$
Answer
$29\,\text{cm}^{2}$
5Area of a parallelogram
CoreDirect calculation• parallelogram
Question
Find the area of a parallelogram with base $12$ cm and perpendicular height $5$ cm.
Step-by-step solution
1. Step 1
  $A = \text{base} \times \text{perpendicular height}$ .
  $A = 12 \times 5 = 60$
Answer
$60\,\text{cm}^{2}$
6Triangle area with $\frac{1}{2}ab\sin C$
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q15• triangle, trigonometry, sin C
Question
Triangle $ABC$ has $AB = 9\,\text{cm}$ , $AC = 12\,\text{cm}$ and $\angle BAC = 68\degree$ . Find the area of the triangle, correct to 3 significant figures.
Step-by-step solution
1. Step 1
  Use $A = \dfrac{1}{2}ab\sin C$ where $a$ and $b$ are the two sides enclosing the angle $C$ .
2. Step 2
  Substitute.
  $A = \dfrac{1}{2}(9)(12)\sin 68\degree = 54\sin 68\degree$
3. Step 3
  Evaluate.
  $A \approx 54 \times 0.9272 \approx 50.1\,\text{cm}^{2}$
Answer
$A \approx 50.1\,\text{cm}^{2}$
Examiner tip
The examiner report flags candidates who use $\frac{1}{2}\times\text{base}\times\text{height}$ here. There is no perpendicular height — the included angle method is the correct route.
7Ratio of areas
ExtendedMulti-step problem• ratio, area
Question
Triangle $T_1$ has base $8\,\text{cm}$ and height $5\,\text{cm}$ . Triangle $T_2$ has base $12\,\text{cm}$ and height $7\,\text{cm}$ . Find the ratio of the area of $T_1$ to the area of $T_2$ in simplest form.
Step-by-step solution
1. Step 1
  Compute both areas.
  $A_1 = \dfrac{1}{2}(8)(5) = 20,\quad A_2 = \dfrac{1}{2}(12)(7) = 42$
2. Step 2
  Write as a ratio and simplify.
  $20 : 42 = 10 : 21$
Answer
$10 : 21$
8Find a dimension given the area
ExtendedDirect calculation• Adapted from 0580/22 Oct/Nov 2023 Q12• inverse, find dimension
Question
A trapezium has parallel sides of $a\,\text{cm}$ and $13\,\text{cm}$ , perpendicular distance $6\,\text{cm}$ , and area $66\,\text{cm}^{2}$ . Find $a$ .
Step-by-step solution
1. Step 1
  Use $A = \dfrac{1}{2}(a + b)h$ .
  $66 = \dfrac{1}{2}(a + 13)(6)$
2. Step 2
  Simplify.
  $66 = 3(a + 13)$
3. Step 3
  Solve.
  $a + 13 = 22 \implies a = 9\,\text{cm}$
Answer
$a = 9\,\text{cm}$
Examiner tip
The examiner report flags candidates who forget the $\frac{1}{2}$ and obtain $a = -2$ (negative length — impossible). A negative or absurd answer is the cue to recheck the formula.
9Perimeter of a compound shape
ExtendedDirect calculation• compound shape, perimeter
Question
An L-shape is formed from a $10\,\text{cm}\times 6\,\text{cm}$ rectangle with a $4\,\text{cm}\times 3\,\text{cm}$ rectangle removed from one corner. Find the perimeter of the L-shape.
Step-by-step solution
1. Step 1
  Trace the outer boundary. The L-shape has six straight edges: $10, 6, 4, 3, (10-4)=6, (6-3)=3$ .
2. Step 2
  Sum the edges.
  $P = 10 + 6 + 4 + 3 + 6 + 3 = 32\,\text{cm}$
3. Step 3
  Cross-check: the perimeter of any L-shape that fits inside an $a\times b$ bounding rectangle equals $2(a+b)$ , here $2(10+6) = 32$ .
Answer
$P = 32\,\text{cm}$
Examiner tip
The examiner report flags candidates who include the interior cut-out edges. Perimeter follows the OUTER boundary only.
10Area of a kite
ExtendedDirect calculation• kite, diagonals
Question
A kite has diagonals of lengths $14\,\text{cm}$ and $9\,\text{cm}$ . Find its area.
Step-by-step solution
1. Step 1
  The diagonals of a kite (and a rhombus) are perpendicular, so area = $\dfrac{1}{2}d_1 d_2$ .
2. Step 2
  Substitute.
  $A = \dfrac{1}{2}(14)(9) = 63\,\text{cm}^{2}$
Answer
$63\,\text{cm}^{2}$
11Area of a rhombus from a diagonal and side
ExtendedMulti-step problem• rhombus
Question
A rhombus has side $10\,\text{cm}$ and one diagonal of length $16\,\text{cm}$ . Find the area.
Step-by-step solution
1. Step 1
  Diagonals of a rhombus bisect each other at right angles. Half of $d_1 = 8$ .
2. Step 2
  Use Pythagoras on the right triangle with hypotenuse $10$ and one leg $8$ to find half of $d_2$ .
  $\left(\dfrac{d_2}{2}\right)^{2} = 10^{2} - 8^{2} = 36 \implies \dfrac{d_2}{2} = 6$
3. Step 3
  Hence $d_2 = 12$ . Apply the diagonal area formula.
  $A = \dfrac{1}{2}(16)(12) = 96\,\text{cm}^{2}$
Answer
$96\,\text{cm}^{2}$
12Combined area + perimeter problem
ChallengeMulti-step problem• Adapted from 0580/42 May/Jun 2023 Q10• compound shape, multi-step
Question
A path of uniform width $w\,\text{m}$ surrounds a rectangular lawn measuring $15\,\text{m}$ by $9\,\text{m}$ . The total area of the lawn plus the path is $260\,\text{m}^{2}$ . Find $w$ .
Step-by-step solution
1. Step 1
  The outer rectangle has dimensions $(15 + 2w)\times(9 + 2w)$ .
2. Step 2
  Set the outer area equal to $260$ .
  $(15 + 2w)(9 + 2w) = 260$
3. Step 3
  Expand.
  $135 + 30w + 18w + 4w^{2} = 260 \implies 4w^{2} + 48w - 125 = 0$
4. Step 4
  Use the quadratic formula.
  $w = \dfrac{-48 + \sqrt{48^{2} + 16\times 125}}{8} = \dfrac{-48 + \sqrt{4304}}{8}$
5. Step 5
  Evaluate.
  $w \approx \dfrac{-48 + 65.61}{8} \approx 2.20\,\text{m}$
Answer
$w \approx 2.20\,\text{m}$ (3 s.f.)
Examiner tip
The examiner report flags candidates who write $(15 + w)(9 + w)$ , adding only one width. The path adds $w$ to BOTH sides of each dimension, so use $+2w$ .
13Worded mensuration — tiles and area
ChallengeWord problem• worded, real-world
Question
A rectangular floor measures $5.4\,\text{m}$ by $3.6\,\text{m}$ . It is to be tiled using square tiles of side $30\,\text{cm}$ . (a) Find the number of tiles needed. (b) Each tile costs $1.85 and tiles are sold in boxes of $25$ . Find the cost of buying just enough complete boxes.
Step-by-step solution
1. Step 1
  Convert to consistent units. $5.4\,\text{m} = 540\,\text{cm}$ , $3.6\,\text{m} = 360\,\text{cm}$ .
2. Step 2
  Number of tiles along each side: $540/30 = 18$ and $360/30 = 12$ .
3. Step 3
  Total tiles needed.
  $18 \times 12 = 216$
4. Step 4
  Number of boxes: $216/25 = 8.64$ , so round UP to $9$ boxes.
5. Step 5
  Cost.
  $9 \times 25 \times \$1.85 = \$416.25$
Answer
(a) $216$ tiles (b) $416.25
Examiner tip
The examiner report flags candidates who round the number of boxes DOWN to $8$ . You must buy enough WHOLE boxes to cover the floor, so always round up.

Key Formulae — Areas and Perimeters

The formulae you need to memorise for areas and perimeters on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Rectangle
$A = l \times w,\ \ P = 2(l + w)$
$l, w$
length and width
When to use
All rectangle area/perimeter problems.
Triangle
$A = \dfrac{1}{2} \times \text{base} \times \text{height}$
When to use
Area of any triangle when base and perpendicular height are known.
Trapezium
$A = \dfrac{1}{2}(a + b)h$
$a, b$
lengths of the two parallel sides
$h$
perpendicular distance between them
When to use
Area of a trapezium.
Parallelogram
$A = bh$
$b$
base length
$h$
perpendicular height
When to use
Area of any parallelogram.

Key Definitions and Keywords — Areas and Perimeters

Definitions to memorise and the exact keywords mark schemes credit for areas and perimeters answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Perimeter
Examiner keyword
Total length of the boundary of a 2D shape.
Area
Examiner keyword
Amount of 2D space enclosed by a shape, measured in squared units (cm², m², …).
Perpendicular height
Examiner keyword
The shortest distance from the base to the opposite vertex / parallel side. Often NOT one of the given sides.
Composite shape
A shape made by combining simpler shapes (rectangles, triangles). Decompose into pieces or subtract from a larger shape.

Common Mistakes and Misconceptions — Areas and Perimeters

The traps other students keep falling into on areas and perimeters questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Forgetting to use squared units for area
Why it happens
Speed.
How to avoid it
Areas are always in square units: cm², m², km².
✕Using a slant side as the perpendicular height
0580/42 — recurring
Why it happens
Confusing slant length with perpendicular distance.
How to avoid it
Perpendicular height makes a right angle with the base.
✕Forgetting the $\frac{1}{2}$ in the triangle area formula
Why it happens
Speed.
How to avoid it
Triangle area: half base times height. ALWAYS the half.
✕Double-counting an interior edge in a composite-shape perimeter
Why it happens
Adding all edges including internal ones.
How to avoid it
Perimeter = ONLY the outer boundary edges. Don't include internal cuts.

Areas and Perimeters — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Areas and Perimeters

Short Notes - Areas and Perimeters

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Areas and Perimeters

1Area and perimeter of a rectangle

2Area of a triangle

3Area of a trapezium

4Area of a composite shape

5Area of a parallelogram

6Triangle area with 12absin⁡C\frac{1}{2}ab\sin C21​absinC

7Ratio of areas

8Find a dimension given the area

9Perimeter of a compound shape

10Area of a kite

11Area of a rhombus from a diagonal and side

12Combined area + perimeter problem

13Worded mensuration — tiles and area

Rectangle

Triangle

Trapezium

Parallelogram

Perimeter

Area

Perpendicular height

Composite shape

✕Forgetting to use squared units for area

✕Using a slant side as the perpendicular height

✕Forgetting the 12\frac{1}{2}21​ in the triangle area formula

✕Double-counting an interior edge in a composite-shape perimeter

Areas and Perimeters — frequently asked questions

6Triangle area with $\frac{1}{2}ab\sin C$

✕Forgetting the $\frac{1}{2}$ in the triangle area formula