Lesson 6: Mensuration | O Level Mathematics

Lesson 6 of 10

60% complete

1. Perimeter and Area of 2D Shapes

Perimeter is the total length of the boundary of a shape (measured in cm, m, etc.). Area is the amount of space enclosed within a shape (measured in cm², m², etc.).

Shape	Perimeter	Area	Diagram Notes
Rectangle	P = 2(l + w)	A = l × w	l = length, w = width
Square	P = 4s	A = s²	s = side length
Triangle	P = a + b + c	A = ½ × base × height	Height must be perpendicular to base
Parallelogram	P = 2(a + b)	A = base × height	Height is perpendicular distance between parallel sides
Trapezium	P = sum of all sides	A = ½(a + b) × h	a, b = parallel sides; h = perpendicular height
Circle	C = 2πr = πd	A = πr²	r = radius, d = diameter

⚠ Height is always PERPENDICULAR! For triangles and parallelograms, the height is the perpendicular distance — never the slant side. Students commonly use the slant side as height and lose all marks.

📐 Worked Example 1 — Composite Shape

A shape consists of a rectangle 10 cm × 6 cm with a semicircle of diameter 6 cm attached to one of the shorter ends. Find the perimeter and area of the composite shape. (Give answers in terms of π where appropriate.)

Perimeter: Two long sides (10 cm each) + one short side (6 cm) + semicircle arc.
Semicircle arc = ½ × π × d = ½ × π × 6 = 3π cm
P = 10 + 10 + 6 + 3π = (26 + 3π) cm ≈ 35.4 cm

Area: Rectangle + Semicircle
Rectangle area = 10 × 6 = 60 cm²
Semicircle radius = 3. Area = ½ × π × 3² = 4.5π cm²
Total area = (60 + 4.5π) cm² ≈ 74.1 cm²

📐 Worked Example 2 — Area by Subtraction

A square of side 12 cm has a circle of radius 4 cm cut from its centre. Find the shaded area remaining.

Area of square = 12² = 144 cm²

Area of circle = π × 4² = 16π cm²

Shaded area = 144 − 16π ≈ 144 − 50.27 = 93.7 cm²

2. Arc Length and Sector Area

Sector: A "pizza slice" of a circle — the region bounded by two radii and an arc. The angle at the centre is called the sector angle θ (in degrees).

Arc Length

l = (θ/360) × 2πr

θ in degrees, r = radius

= fraction of full circumference

Sector Area

A = (θ/360) × πr²

θ in degrees, r = radius

= fraction of full circle area

Perimeter of a Sector = arc length + 2 radii = (θ/360)×2πr + 2r
Do NOT forget to add the two radii when finding the perimeter of a sector!

📐 Worked Example 3 — Arc and Sector

A sector has radius 8 cm and sector angle 135°. Find (a) the arc length (b) the sector area (c) the perimeter of the sector. Give answers to 3 significant figures.

(a) Arc length:
l = (135/360) × 2π × 8 = (3/8) × 16π = 6π ≈ 18.8 cm

(b) Sector area:
A = (135/360) × π × 64 = (3/8) × 64π = 24π ≈ 75.4 cm²

(c) Perimeter:
P = 6π + 2(8) = 6π + 16 ≈ 18.85 + 16 = 34.8 cm

📐 Worked Example 4 — Finding the Angle from Arc Length

A sector of radius 12 cm has an arc length of 10π cm. Find the sector angle.

Use l = (θ/360) × 2πr. Substitute l = 10π, r = 12:
10π = (θ/360) × 24π

Divide both sides by π then solve for θ:
10 = (θ/360) × 24
θ = (10 × 360)/24 = 150°

3. Volume of 3D Shapes

Volume is the amount of three-dimensional space enclosed by a shape. Measured in cm³, m³, litres (1 litre = 1000 cm³).

Shape	Volume Formula	Variables
Cuboid	V = l × w × h	l = length, w = width, h = height
Cube	V = s³	s = side length
Prism (any cross-section)	V = A × l	A = area of cross-section, l = length
Cylinder	V = πr²h	r = radius of base, h = height
Pyramid	V = ⅓ × base area × height	Height is perpendicular height from apex to base
Cone	V = ⅓πr²h	r = base radius, h = perpendicular height
Sphere	V = (4/3)πr³	r = radius
Hemisphere	V = (2/3)πr³	Half of sphere volume

Key Prism Rule: Any solid with a uniform cross-section is a prism. Volume = cross-sectional area × length. This includes cylinders (circular cross-section), triangular prisms, L-shaped prisms, etc.

📐 Worked Example 5 — Triangular Prism

A triangular prism has a right-angled triangular cross-section with legs 5 cm and 12 cm. The prism is 20 cm long. Find the volume.

Area of triangular cross-section:
A = ½ × 5 × 12 = 30 cm²

Volume = cross-section area × length:
V = 30 × 20 = 600 cm³

📐 Worked Example 6 — Composite Solid

A solid consists of a cylinder of radius 4 cm and height 10 cm with a hemisphere of radius 4 cm on top. Find the total volume. (Give answer in terms of π.)

Cylinder volume = πr²h = π × 16 × 10 = 160π cm³

Hemisphere volume = (2/3)πr³ = (2/3)π × 64 = 128π/3 cm³

Total = 160π + 128π/3 = 480π/3 + 128π/3 = 608π/3 cm³ ≈ 636 cm³

4. Surface Area of 3D Shapes

Surface area is the total area of all faces of a 3D shape. Think of it as the area of the net that would fold up to make the shape.

Shape	Surface Area Formula	Notes
Cuboid	SA = 2(lw + lh + wh)	6 rectangular faces in 3 pairs
Cube	SA = 6s²	6 identical square faces
Cylinder	SA = 2πr² + 2πrh	2 circular ends + curved surface (rectangle when unrolled)
Cone	SA = πr² + πrl	Circular base + curved surface. l = slant height
Sphere	SA = 4πr²	No flat faces — entirely curved
Hemisphere	SA = 2πr² + πr²= 3πr²	Curved surface (2πr²) + flat circular base (πr²)
Pyramid (square base)	SA = base² + 4 × (½ × base × slant height)	Square base + 4 triangular faces. Slant height ≠ perpendicular height.

⚠ Slant Height vs Perpendicular Height:
For cones and pyramids, the slant height (l) is the distance from the apex to the midpoint of a base edge along the surface — NOT the vertical height (h). Use Pythagoras to find l: l² = r² + h² (cone) or l² = (½base)² + h² (pyramid).

📐 Worked Example 7 — Cone Surface Area

A cone has base radius 5 cm and perpendicular height 12 cm. Find (a) the slant height (b) the total surface area. Give answer to 3 s.f.

(a) Slant height:
l = √(r² + h²) = √(25 + 144) = √169 = 13 cm

(b) Total surface area:
SA = πr² + πrl = π(5)² + π(5)(13)
= 25π + 65π = 90π ≈ 283 cm²

📐 Worked Example 8 — Open Cylinder (no lid)

A cylindrical container is open at the top. It has radius 6 cm and height 15 cm. Find the total surface area of material needed to make it.

Open cylinder = curved surface + ONE circular base (no top):
Curved surface = 2πrh = 2π(6)(15) = 180π cm²

Base = πr² = π(36) = 36π cm²

Total = 180π + 36π = 216π ≈ 679 cm²

5. Units, Conversions, and Rates

Area and Volume Conversions

Area Conversions

1 cm² = 100 mm²
1 m² = 10,000 cm²
1 km² = 1,000,000 m²
1 hectare = 10,000 m²

Rule: square the linear conversion factor.
1 m = 100 cm → 1 m² = 100² = 10,000 cm²

Volume Conversions

1 cm³ = 1000 mm³
1 m³ = 1,000,000 cm³
1 litre = 1000 cm³
1 m³ = 1000 litres
1 ml = 1 cm³

Rule: cube the linear conversion factor.
1 m = 100 cm → 1 m³ = 100³ = 1,000,000 cm³

📐 Worked Example 9 — Volume and Capacity

A rectangular fish tank is 80 cm long, 40 cm wide and 50 cm deep. It is filled to ¾ of its capacity. Find (a) the full capacity in litres (b) the volume of water in the tank.

Full volume = 80 × 40 × 50 = 160,000 cm³

(a) Full capacity = 160,000 ÷ 1000 = 160 litres

(b) Volume of water = ¾ × 160,000 = 120,000 cm³ = 120 litres

Similar Shapes — Length, Area, Volume Ratios

For similar 3D shapes with linear scale factor k:
Ratio of corresponding lengths = k : 1
Ratio of corresponding surface areas = k² : 1
Ratio of corresponding volumes = k³ : 1

Important: If you know the ratio of volumes = p:q, then the ratio of lengths = ∛p : ∛q

📐 Worked Example 10 — Similar Solids

Two similar cylinders have heights in the ratio 2:5. The surface area of the smaller cylinder is 48π cm². Find the surface area of the larger cylinder. Find the ratio of their volumes.

Linear scale factor k = 5/2

Surface area ratio = k² = (5/2)² = 25/4
Surface area of larger = 48π × (25/4) = 300π cm²

Volume ratio = k³ = (5/2)³ = 125/8
Volume ratio = 8 : 125 (smaller : larger)

6. Problem Solving with Mensuration

Finding Dimensions from Given Area or Volume

📐 Worked Example 11 — Finding Height from Volume

A cone has volume 150π cm³ and base radius 5√2 cm. Find the perpendicular height.

V = ⅓πr²h. Substitute V = 150π, r = 5√2:
150π = ⅓ × π × (5√2)² × h
150π = ⅓ × π × 50 × h

Divide both sides by π then solve:
150 = (50/3)h → h = 150 × 3/50 = 9 cm

📐 Worked Example 12 — Water Level Problem

A sphere of radius 3 cm is dropped into a cylindrical container of radius 5 cm containing water. By how much does the water level rise?

Volume of sphere = (4/3)πr³ = (4/3)π(27) = 36π cm³

This volume = volume of cylinder slice of height h:
πR²h = 36π
π(25)h = 36π

h = 36/25 = 1.44 cm

Complete Formula Reference Card

📝 Exam Practice Questions

Q1 [3 marks] — A sector of a circle has radius 9 cm and arc length 6π cm. Find (a) the sector angle (b) the area of the sector.

(a) l = (θ/360) × 2πr → 6π = (θ/360) × 18π → 6 = (θ/360) × 18 → θ = (6×360)/18 = 120°

(b) A = (120/360) × π × 81 = ⅓ × 81π = 27π cm² ≈ 84.8 cm²

Q2 [3 marks] — A cone has slant height 15 cm and base radius 9 cm. Find (a) the perpendicular height (b) the volume (c) the curved surface area.

(a) h = √(l²−r²) = √(225−81) = √144 = 12 cm

(b) V = ⅓πr²h = ⅓ × π × 81 × 12 = 324π ≈ 1018 cm³

(c) Curved SA = πrl = π × 9 × 15 = 135π ≈ 424 cm²

Exam Tip: When given slant height and radius, always find perpendicular height using h = √(l²−r²) before calculating volume. The volume formula needs perpendicular height, not slant height.

Q3 [4 marks] — A solid metal sphere of radius 6 cm is melted down and recast into small spheres of radius 1.5 cm. How many small spheres are made? (Assume no metal is lost.)

Volume of large sphere: V = (4/3)π(6)³ = (4/3)π(216) = 288π cm³
Volume of small sphere: V = (4/3)π(1.5)³ = (4/3)π(3.375) = 4.5π cm³
Number of small spheres: 288π ÷ 4.5π = 64 spheres

Q4 [3 marks] — Two similar cones have volumes of 54 cm³ and 128 cm³. The base radius of the smaller cone is 3 cm. Find the base radius of the larger cone.

Volume ratio: 54:128 = 27:64
Linear scale factor: k = ∛(64/27) = 4/3
Radius of larger cone: 3 × (4/3) = 4 cm

Exam Tip: Volume ratio → cube root → linear scale factor. Then multiply any length by the scale factor to find the corresponding length in the other solid.

Q5 [4 marks] — A swimming pool is 25 m long and 10 m wide. One end is 1 m deep and the other end is 3 m deep. The base slopes uniformly. Find (a) the volume of water when full (b) the time to fill it at 500 litres per minute.

(a) Cross-section is a trapezium: a=1m, b=3m, length=25m, width=10m.
Volume = ½(a+b) × length × width = ½(1+3) × 25 × 10
= ½ × 4 × 250 = 500 m³

(b) 500 m³ = 500 × 1000 litres = 500,000 litres
Time = 500,000 ÷ 500 = 1000 minutes = 16 hours 40 minutes

Q6 [4 marks] — A shape is made by removing a cone of radius 3 cm and height 8 cm from inside a cylinder of radius 3 cm and height 8 cm. Find the total surface area of the resulting shape. (Give answer to 3 s.f.)

Slant height of cone: l = √(3²+8²) = √(9+64) = √73 cm

Surfaces present:
— Circular top of cylinder (no cone there): πr² = 9π
— Curved surface of cylinder: 2πrh = 2π(3)(8) = 48π
— Circular base (cylinder base, cone removed — ring shape): π(3²)−π(3²) = 0. Actually the base of cylinder has a hole where cone base was, but cone base IS the same circle, so base = 0 net. Wait — the cone opens downward, so the cone's base circle IS the top opening. Reconsider: cone sits with apex at base, base circle at top. Then:
Top face: annular ring = 0 (cone base fills the top)
— Cylinder curved surface: 48π
— Cone curved surface (inside, now exposed): πrl = π(3)(√73)
— Cylinder base (solid circle): 9π
Total SA = 9π + 48π + 3π√73 = 57π + 3π√73
= π(57 + 3√73) ≈ π(57 + 25.63) ≈ π(82.63) ≈ 259 cm²

Exam Tip: For hollow or composite solid surface area questions, list ALL faces carefully: which are present, which are hidden, which are new exposed surfaces. Draw a sketch and label each face before calculating.

Q7 [3 marks] — A circular running track has inner radius 40 m and outer radius 45 m. Find the area of the track. A groundsman paints the track at a rate of 15 m² per minute. How long does it take to paint the full track?

Area of track = π(45²) − π(40²) = π(2025−1600) = 425π ≈ 1335 m²

Time = 1335 ÷ 15 = 89 minutes (to nearest minute)

← Lesson 5: Geometry Lesson 7: Trigonometry →