Mensuration: perimeter, area, surface area, volume

Geometry – Mensuration (IGCSE 0580)

Goal: Calculate perimeter, area, surface area and volume of all plane and solid figures required by the Cambridge IGCSE syllabus, using the correct units, showing clear working and rounding to the required number of significant figures.

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0. Units, prefixes & quick conversion table

• Length: mm, cm, m, km (1 m = 100 cm, 1 km = 1000 m, 1 cm = 10 mm)
• Area: mm², cm², m² (square the length unit)
• Volume / Capacity: mm³, cm³, m³, mL (1 cm³ = 1 mL), L (1 L = 1000 mL)

From → To	Factor
mm → cm	÷ 10
cm → m	÷ 100
m → km	÷ 1000
cm³ → mL	1 cm³ = 1 mL
mL → L	÷ 1000
cm³ → L	÷ 1000
mm³ → cm³	÷ 1000
cm³ → m³	÷ 1 000 000

Example: Convert 250 cm³ to litres.
250 cm³ = 250 mL = 0.250 L

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1. Perimeter

When a figure is a combination of simpler shapes, find the perimeter of each part, add the external edges and subtract any interior edges that are counted twice.

Example: Find the perimeter of an “L‑shaped” figure made from two rectangles 8 cm × 3 cm and 5 cm × 3 cm sharing a 3 cm side.

P = (8+3+5+3) + (8+5) = 32 cm

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2. Area of plane figures

Figure	Formula	Key notes
Rectangle	A = l × w	Base × height
Square	A = a²	Side squared
Triangle	A = ½ b h	h ⟂ b
Parallelogram	A = b h	h = perpendicular distance between the parallel sides
Trapezium	A = ½ (a + b) h	a,b = parallel sides
Circle	A = πr²	r = radius
Sector (θ°)	A = (θ⁄360) πr²	Fraction of the whole circle
Regular polygon (n sides, side s)	A = (n s²) ⁄ (4 tan(π⁄n))	Useful for 5‑sided+ shapes

2.1 Compound plane figures

Break the figure into known shapes, find each area, then add (or subtract) them.

Example: An “L‑shaped” figure consists of a 10 cm × 6 cm rectangle with a 4 cm × 6 cm rectangle removed.

A = 10 × 6 – 4 × 6 = 60 – 24 = 36 cm²

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3. Arcs, sectors & segments

• Arc length (portion of the circumference):
  l = (θ⁄360) × 2πr = (θ⁄360) × C where C = 2πr.
• Sector area (already in the table):
  A = (θ⁄360) πr²
• Segment area (area between chord and arc):
  Asegment = Asector – Atriangle where the triangle is formed by the two radii and the chord.

Example: Find the length of a 60° arc of a circle with radius 10 cm.

l = (60⁄360) × 2π × 10 = (1⁄6) × 20π = 10π⁄3 ≈ 10.47 cm

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4. Surface area of solids

For nets, remember that the total surface area equals the sum of the areas of all the faces shown in the net.

Example – Surface area of a right triangular prism (base right‑triangle with legs 4 cm, 3 cm, hypotenuse 5 cm; height h = 10 cm).

Base area B = ½ × 3 × 4 = 6 cm²
Perimeter of base p = 3 + 4 + 5 = 12 cm
SA = 2B + p h = 2×6 + 12×10 = 12 + 120 = 132 cm²

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5. Volume of solids

Solid	Formula	Key parts
Cuboid	V = l w h	l = length, w = width, h = height
Cube	V = a³	a = side length
Cylinder	V = πr²h	r = radius, h = height
Sphere	V = 4⁄3 πr³	r = radius
Cone	V = 1⁄3 πr²h	r = radius, h = vertical height
Right prism (base area B)	V = B h	h = perpendicular height
Right pyramid (base area B)	V = 1⁄3 B h	h = perpendicular height

5.1 Compound solids

Find the volume of each component, then add (or subtract) them. Ensure all dimensions are in the same unit before substituting.

Example: A solid consists of a cylinder (r = 4 cm, h = 10 cm) topped by a hemisphere of the same radius.

V_cylinder = πr²h = π×4²×10 = 160π cm³
V_hemisphere = ½ × (4⁄3 πr³) = ½ × (4⁄3 π×64) = 128⁄3 π cm³
V_total = 160π + 128⁄3 π = (480 + 128)⁄3 π = 608⁄3 π ≈ 637 cm³ (3 sf)

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6. Worked example – Composite figure (area, perimeter, surface area & volume)

Problem: An “L‑shaped” solid is formed by joining a cuboid (10 cm × 6 cm × 4 cm) to a second cuboid (4 cm × 6 cm × 4 cm) along their 6 cm × 4 cm faces. Determine:

1. Surface area of the combined solid.
2. Volume of the combined solid.
3. Perimeter of the base (the shape seen from above).

Solution:

1. Surface area

   Surface area of each cuboid:
           SA₁ = 2(lw + lh + wh) = 2(10·6 + 10·4 + 6·4) = 2(60 + 40 + 24) = 248 cm²
           SA₂ = 2(4·6 + 4·4 + 6·4)   = 2(24 + 16 + 24) = 128 cm²
   
           When joined, the common face (6 × 4 cm) is hidden on both sides:
           Area hidden = 2 × (6·4) = 48 cm²
   
           Total SA = SA₁ + SA₂ – hidden = 248 + 128 – 48 = 328 cm²
           

2. Volume

   V₁ = 10·6·4 = 240 cm³
           V₂ = 4·6·4 = 96 cm³
           V_total = 240 + 96 = 336 cm³
           

3. Base perimeter (viewed from above the L‑shape)

   Perimeter = 10 + 6 + 4 + 6 + 4 = 30 cm
           

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7. Quick revision checklist

• Identify the shape(s) – note any regularity, right‑angle or symmetry.
• Write down the exact formula(s) needed – use the Recall Boxes.
• Convert all measurements to the same unit before substituting.
• For compound figures, split into known parts, calculate each part, then add or subtract.
• Substitute values, keep intermediate results with units, and simplify step‑by‑step.
• Round the final answer to the required number of significant figures and attach the correct unit (cm, cm², cm³, etc.).