2D Mensuration

Grade 9 · Geometry · Cambridge IGCSE 0580 · Age 13–14

Welcome to 2D Mensuration!

Mensuration is the branch of mathematics concerned with measuring geometric figures — their perimeters, areas, arc lengths and sector areas. 2D mensuration is a core topic in Cambridge IGCSE 0580 and appears in nearly every exam paper. You need to know the formulas, apply them correctly, and handle composite shapes and unit conversions.

Key idea: Area measures the space inside a shape (units²) · Perimeter measures the distance around a shape (units)

Rectangles & Triangles

Basic area and perimeter formulas

Parallelogram & Trapezium

Quadrilateral area formulas

Circles & Semicircles

πr² and 2πr

Arc Length & Sector

Angle fraction of a circle

Segment Area

Sector minus triangle

Composite Shapes

Add and subtract regions

Unit Conversions

cm² ↔ m² and more

Interactive Visualiser

Adjust sliders and see live results

1. Rectangle

A rectangle has four right angles. Its area is the product of its length and width; its perimeter is twice the sum of length and width.

Area = l × w Perimeter = 2(l + w)

Worked example: A rectangle has length 9 cm and width 4 cm.
Area = 9 × 4 = 36 cm²
Perimeter = 2(9 + 4) = 2 × 13 = 26 cm

A square is a special rectangle where l = w, so Area = l² and Perimeter = 4l.

2. Triangle

The area of any triangle is half base times perpendicular height. The perpendicular height must be at right angles to the chosen base — it is not necessarily a side of the triangle.

Area = ½ × b × h

Derivation: A triangle is exactly half of a rectangle with the same base and height. Draw a rectangle around any triangle — the triangle always occupies exactly half the area.

Example 1: Base = 10 cm, height = 6 cm → Area = ½ × 10 × 6 = 30 cm²
Example 2: Base = 7 cm, height = 4 cm → Area = ½ × 7 × 4 = 14 cm²

Perimeter of triangle: Simply add all three sides.
For a right-angled triangle with legs 3 cm and 4 cm: hypotenuse = √(9+16) = 5 cm → Perimeter = 3 + 4 + 5 = 12 cm

On exams, the height is always the perpendicular height, NOT the slant side. Look for the right-angle symbol in diagrams.

3. Parallelogram

A parallelogram has two pairs of parallel sides. Its area equals base times perpendicular height (not the slant height). This is because a parallelogram can be rearranged into a rectangle of the same base and height.

Area = b × h Perimeter = 2(a + b)

Derivation: Cut the triangle off the left end of a parallelogram and move it to the right — you get a rectangle with the same base b and same perpendicular height h.

Example: Parallelogram with base 8 cm, perpendicular height 5 cm, and slant side 6 cm.
Area = 8 × 5 = 40 cm²
Perimeter = 2(8 + 6) = 28 cm

The perimeter uses the slant side (the actual length of the side), not the height. Do not confuse them.

4. Trapezium

A trapezium (US: trapezoid) has exactly one pair of parallel sides, called the parallel sides or bases. The perpendicular distance between them is the height.

Area = ½(a + b) × h

Derivation: A trapezium can be split into two triangles with bases a and b and the same height h.
Area = ½ah + ½bh = ½(a + b)h

Example: Parallel sides 5 cm and 9 cm, height 4 cm.
Area = ½(5 + 9) × 4 = ½ × 14 × 4 = 28 cm²

Remember: add the two parallel sides together first, then multiply by ½ and by the perpendicular height.

5. Circle

For a circle of radius r: the circumference (perimeter) uses the diameter d = 2r. The area formula involves π ≈ 3.14159…

Area = πr² Circumference = 2πr = πd

Example 1: Radius = 5 cm.
Area = π × 5² = 25π ≈ 78.54 cm²
Circumference = 2π × 5 = 10π ≈ 31.42 cm

Example 2 — finding radius: Area = 50.27 cm². Find r.
πr² = 50.27 → r² = 50.27/π ≈ 16 → r = 4 cm

Leave answers in terms of π unless asked for a decimal. e.g. "25π cm²" is more exact than "78.54 cm²".

6. Semicircle

A semicircle is exactly half a circle. Its area is half the circle area. Its perimeter includes the curved part (half the circumference) PLUS the diameter.

Area = ½πr² Perimeter = πr + 2r = r(π + 2)

Example: Semicircle with radius 6 cm.
Area = ½ × π × 36 = 18π ≈ 56.55 cm²
Perimeter = π × 6 + 2 × 6 = 6π + 12 ≈ 30.85 cm

Students often forget to add the diameter when finding the perimeter of a semicircle. The diameter is a straight edge that forms part of the boundary.

7. Arc Length

An arc is a portion of the circumference of a circle. A sector with angle θ° at the centre represents a fraction θ/360 of the full circle.

Arc length = (θ / 360) × 2πr

Derivation: Full circumference = 2πr. The arc is θ/360 of the full circle.
Arc = (θ/360) × 2πr

Example 1: θ = 90°, r = 8 cm.
Arc = (90/360) × 2π × 8 = ¼ × 16π = 4π ≈ 12.57 cm

Example 2: θ = 120°, r = 6 cm.
Arc = (120/360) × 2π × 6 = ⅓ × 12π = 4π ≈ 12.57 cm

The perimeter of a sector = arc length + 2r (the two radii that form the straight sides).

8. Sector Area

A sector is a "pizza slice" of a circle — the region bounded by two radii and an arc. Its area is the fraction θ/360 of the full circle area.

Sector area = (θ / 360) × πr²

Example 1: θ = 60°, r = 9 cm.
Sector area = (60/360) × π × 81 = ⅙ × 81π = 13.5π ≈ 42.41 cm²

Example 2: θ = 270°, r = 4 cm.
Sector area = (270/360) × π × 16 = ¾ × 16π = 12π ≈ 37.70 cm²

A sector with θ = 180° is a semicircle. Check: (180/360) × πr² = ½πr². ✓

9. Segment Area

A segment is the region between a chord and its arc (like a "slice of orange" shape). To find its area, subtract the triangle from the sector.

Segment area = Sector area − Triangle area = (θ/360)πr² − ½r²sin θ

Derivation of triangle area: The triangle formed by two radii of length r and the angle θ between them has area = ½ × r × r × sin θ = ½r²sin θ (from the sine area formula).

Example: r = 10 cm, θ = 60°.
Sector = (60/360) × π × 100 = 100π/6 ≈ 52.36 cm²
Triangle = ½ × 100 × sin 60° = 50 × (√3/2) = 25√3 ≈ 43.30 cm²
Segment = 52.36 − 43.30 ≈ 9.06 cm²

Make sure your calculator is in degree mode. sin 60° = √3/2 ≈ 0.866, sin 90° = 1, sin 45° = √2/2 ≈ 0.707.

10. Composite Shapes

A composite shape is made by combining or subtracting simpler shapes. Break it into parts you recognise, calculate each, then add or subtract.

Composite area = sum of individual areas (or subtract cut-out regions)

Example 1 — L-shape: An L-shape can be split into two rectangles.
If the total bounding box is 10 × 8 cm and a 4 × 5 cm rectangle is removed from one corner:
Area = (10 × 8) − (4 × 5) = 80 − 20 = 60 cm²

Example 2 — Rectangle with semicircle on top:
Rectangle: 8 cm × 5 cm → 40 cm²
Semicircle on top (diameter 8 cm, so r = 4 cm): ½ × π × 16 = 8π ≈ 25.13 cm²
Total area = 40 + 8π ≈ 65.13 cm²

For perimeters of composite shapes, only count edges that form the outer boundary. Shared internal edges are NOT counted.

11. Unit Conversions for Area

Since area is two-dimensional, converting units involves squaring the linear conversion factor. This trips up many students — be careful!

1 m = 100 cm ⟹ 1 m² = 100² cm² = 10 000 cm²

Conversion	Linear factor	Area factor
cm ↔ mm	1 cm = 10 mm	1 cm² = 100 mm²
m ↔ cm	1 m = 100 cm	1 m² = 10 000 cm²
km ↔ m	1 km = 1000 m	1 km² = 1 000 000 m²
m ↔ mm	1 m = 1000 mm	1 m² = 1 000 000 mm²
hectare	—	1 ha = 10 000 m²

Example 1: Convert 3.5 m² to cm².
3.5 × 10 000 = 35 000 cm²

Example 2: Convert 45 000 cm² to m².
45 000 ÷ 10 000 = 4.5 m²

Example 3: Convert 850 mm² to cm².
850 ÷ 100 = 8.5 cm²

Memory trick: "To go from larger unit² to smaller unit², MULTIPLY. To go from smaller unit² to larger unit², DIVIDE."

Quick Reference — All Formulas

Shape	Area	Perimeter / Key formula
Rectangle	l × w	2(l + w)
Square	l²	4l
Triangle	½bh	a + b + c
Parallelogram	bh	2(a + b)
Trapezium	½(a+b)h	a + b + c + d
Circle	πr²	2πr
Semicircle	½πr²	πr + 2r
Sector	(θ/360)πr²	Arc + 2r = (θ/360)2πr + 2r
Segment	(θ/360)πr² − ½r²sinθ	Arc + chord

Example 1 — Area of a Composite Shape (Rectangle + Triangle)

A shape consists of a rectangle 12 cm by 7 cm, with a triangle of base 12 cm and height 5 cm sitting on top. Find the total area.

Step 1 — Rectangle: Area = l × w = 12 × 7 = 84 cm²

Step 2 — Triangle: Area = ½ × b × h = ½ × 12 × 5 = 30 cm²

Step 3 — Total: 84 + 30 = 114 cm²

Check: The triangle sits on the full width of the rectangle, so no overlap. Addition is correct.

Example 2 — Perimeter of a Sector

A sector has radius 9 cm and angle 80°. Find its perimeter. Give your answer correct to 3 significant figures.

Step 1 — Arc length: Arc = (θ/360) × 2πr = (80/360) × 2π × 9

Step 2 — Calculate: = (2/9) × 18π = 4π ≈ 12.566 cm

Step 3 — Add the two radii: Perimeter = Arc + 2r = 4π + 18 ≈ 12.566 + 18 = 30.566…

Answer: Perimeter = 30.6 cm (3 s.f.)

Example 3 — Sector Area

Find the area of a sector with radius 12 cm and angle 135°. Give the answer in terms of π.

Step 1 — Write the formula: Sector area = (θ/360) × πr²

Step 2 — Substitute: = (135/360) × π × 144

Step 3 — Simplify the fraction: 135/360 = 3/8

Step 4 — Calculate: = (3/8) × 144π = 54π

Answer: 54π cm² ≈ 169.6 cm²

Example 4 — Segment Area

A chord divides a circle of radius 8 cm into a segment with angle 90° at the centre. Find the area of the minor segment. Give your answer to 3 s.f.

Step 1 — Sector area: (90/360) × π × 64 = ¼ × 64π = 16π ≈ 50.265 cm²

Step 2 — Triangle area: ½r²sinθ = ½ × 64 × sin 90° = 32 × 1 = 32 cm²

Step 3 — Segment area: 50.265 − 32 = 18.265…

Answer: 18.3 cm² (3 s.f.)

Example 5 — Trapezium inside a Rectangle (Shaded Region)

A rectangle is 15 cm by 10 cm. A trapezium with parallel sides 5 cm and 9 cm, and height 6 cm, is cut out from the centre. Find the remaining shaded area.

Step 1 — Rectangle area: 15 × 10 = 150 cm²

Step 2 — Trapezium area: ½(5 + 9) × 6 = ½ × 14 × 6 = 42 cm²

Step 3 — Shaded area: 150 − 42 = 108 cm²

Example 6 — Unit Conversion in a Problem

A circular pond has radius 3.5 m. Find its area in cm².

Step 1 — Area in m²: π × 3.5² = 12.25π ≈ 38.485 m²

Step 2 — Convert to cm²: 1 m² = 10 000 cm², so 38.485 × 10 000 = 384 845 cm²

Alternatively, convert first: r = 3.5 m = 350 cm → Area = π × 350² = 122 500π ≈ 384 845 cm² ✓

Interactive Shape Visualiser

Choose a shape, adjust the sliders, and see the shape drawn with labelled dimensions. Area and perimeter update instantly.

Shape:

Select a shape and adjust the sliders.

Exercise 1 — Rectangles & Triangles

Give all answers to 2 decimal places where necessary. Area in cm², perimeter in cm.

1. A rectangle has length 8 cm and width 5 cm. Find the area (cm²).

2. A rectangle has length 11 cm and width 4 cm. Find the perimeter (cm).

3. A square has side 7 cm. Find the area (cm²).

4. A square has perimeter 36 cm. Find its area (cm²).

5. A triangle has base 10 cm and height 6 cm. Find the area (cm²).

6. A right-angled triangle has legs 6 cm and 8 cm. Find the area (cm²).

Exercise 2 — Parallelogram & Trapezium

Give all answers to 2 decimal places where necessary.

1. Parallelogram: base 9 cm, perpendicular height 4 cm. Find the area (cm²).

2. Parallelogram: base 13 cm, perpendicular height 7 cm. Find the area (cm²).

3. Trapezium: parallel sides 6 cm and 10 cm, height 5 cm. Find the area (cm²).

4. Trapezium: parallel sides 4 cm and 14 cm, height 8 cm. Find the area (cm²).

5. A trapezium has area 45 cm² and height 5 cm. One parallel side is 7 cm. Find the other parallel side (cm).

6. A parallelogram has area 56 cm² and base 8 cm. Find the perpendicular height (cm).

Exercise 3 — Circles & Semicircles

Give answers to 2 decimal places. Use π ≈ 3.14159.

1. Circle with radius 5 cm. Find the area (cm²). Round to 2 d.p.

2. Circle with radius 7 cm. Find the circumference (cm). Round to 2 d.p.

3. Circle with diameter 12 cm. Find the area (cm²). Round to 2 d.p.

4. Circle with circumference 31.42 cm. Find the radius (cm). Round to 2 d.p.

5. Semicircle with radius 4 cm. Find the area (cm²). Round to 2 d.p.

6. Semicircle with radius 6 cm. Find the full perimeter (curved part + diameter) in cm. Round to 2 d.p.

Exercise 4 — Arc Length & Sector Area

Give answers to 2 decimal places.

1. Arc length: θ = 90°, r = 8 cm. Find the arc length (cm). Round to 2 d.p.

2. Arc length: θ = 60°, r = 12 cm. Find the arc length (cm). Round to 2 d.p.

3. Sector area: θ = 120°, r = 9 cm. Find the area (cm²). Round to 2 d.p.

4. Sector area: θ = 45°, r = 10 cm. Find the area (cm²). Round to 2 d.p.

5. Sector perimeter: θ = 72°, r = 5 cm. Find the perimeter (arc + 2 radii) in cm. Round to 2 d.p.

6. A sector has area 30π cm² and radius 6 cm. Find θ (degrees).

Exercise 5 — Composite Shapes & Unit Conversions

Read each question carefully. Give answers to 2 d.p. where necessary.

1. A shape is a rectangle (8 cm × 5 cm) with a semicircle of diameter 8 cm on top. Find the total area (cm²). Round to 2 d.p.

2. A rectangle (10 cm × 6 cm) has a circle of radius 2 cm cut from its centre. Find the remaining area (cm²). Round to 2 d.p.

3. Convert 2.5 m² to cm².

4. Convert 75 000 cm² to m².

5. Convert 450 mm² to cm².

6. An L-shape is a 12 cm × 10 cm rectangle with a 5 cm × 4 cm rectangle removed from one corner. Find the area (cm²).

Practice — 20 Questions

Mixed practice covering all 2D mensuration skills. Give answers to 2 d.p. where needed. Use π ≈ 3.14159.

1. Rectangle: l = 9 cm, w = 6 cm. Area (cm²)?

2. Rectangle: l = 7 cm, w = 3 cm. Perimeter (cm)?

3. Triangle: base 8 cm, height 9 cm. Area (cm²)?

4. Parallelogram: base 11 cm, perp. height 6 cm. Area (cm²)?

5. Trapezium: parallel sides 5 cm and 11 cm, height 4 cm. Area (cm²)?

6. Circle: radius 3 cm. Area (cm²)? Round to 2 d.p.

7. Circle: radius 10 cm. Circumference (cm)? Round to 2 d.p.

8. Semicircle: radius 5 cm. Area (cm²)? Round to 2 d.p.

9. Semicircle: radius 8 cm. Full perimeter (cm)? Round to 2 d.p.

10. Arc length: θ = 180°, r = 7 cm. Arc length (cm)? Round to 2 d.p.

11. Sector area: θ = 90°, r = 6 cm. Area (cm²)? Round to 2 d.p.

12. Arc length: θ = 30°, r = 12 cm. Arc length (cm)? Round to 2 d.p.

13. Sector area: θ = 270°, r = 4 cm. Area (cm²)? Round to 2 d.p.

14. Segment: r = 6 cm, θ = 90°. Area (cm²)? Round to 2 d.p. [Sector − Triangle = (90/360)π×36 − ½×36×sin90°]

15. Composite: rectangle 6×4 cm with triangle (base 6, height 3) on top. Total area (cm²)?

16. Convert 4 m² to cm².

17. Convert 85 000 cm² to m².

18. Convert 300 mm² to cm².

19. Square with area 144 cm². Find the perimeter (cm).

20. Circle: diameter 14 cm. Area (cm²)? Round to 2 d.p.

Challenge — 8 Questions

Harder multi-step problems. Give answers to 2 decimal places. Think carefully before starting each one.

1. A sector has perimeter 30 cm and radius 8 cm. Find the arc length (cm), then find θ to the nearest degree. Enter θ.

2. A window is a rectangle 80 cm wide and 120 cm tall, topped with a semicircle of diameter 80 cm. Find the total area in cm². Round to 2 d.p.

3. A running track consists of two straight sections each 80 m long and two semicircles each of diameter 50 m. Find the total perimeter of the track in m. Round to 2 d.p.

4. The area of a sector is 60 cm² and the angle is 150°. Find the radius (cm). Round to 2 d.p. [r² = Area × 360 / (θ × π)]

5. A segment of a circle has r = 12 cm and θ = 60°. Find the segment area (cm²). Round to 2 d.p. [Sector − ½r²sin60°]

6. A floor 6 m × 4 m is tiled. Each tile is 20 cm × 20 cm. How many tiles are needed? (Convert all to same units first.)

7. A circle is inscribed in a square of side 10 cm (the circle just touches all four sides). Find the area of the four corner regions (square minus circle) in cm². Round to 2 d.p.

8. Two concentric circles have radii 5 cm and 9 cm. Find the area of the annulus (ring) between them in cm². Round to 2 d.p.