IGCSE Mathematics: Mensuration
Mensuration is the branch of mathematics concerned with measuring lengths, areas, and volumes of geometric figures. It is one of the most practically useful topics in the IGCSE syllabus and appears in almost every examination paper. This chapter covers two-dimensional area, circular measure, surface area and volume of three-dimensional shapes, similar shape scaling, and compound figures built from standard shapes.
Specification Note
Content labelled Extended is required for the Extended tier (Cambridge) or Higher tier (Edexcel) only.
1. Perimeter and Area of 2D Shapes
The perimeter of a shape is the total distance around its boundary. The area is the amount of space enclosed within the boundary. All formulae below assume consistent units; areas are given in square units (e.g. cm², m²).
| Shape | Perimeter | Area |
|---|---|---|
| Rectangle (length $l$, width $w$) | $2(l + w)$ | $lw$ |
| Triangle (base $b$, height $h$) | Sum of three sides | $\dfrac{1}{2}bh$ |
| Parallelogram (base $b$, height $h$) | $2(a + b)$ | $bh$ |
| Trapezium (parallel sides $a$, $b$; height $h$) | Sum of four sides | $\dfrac{1}{2}(a+b)h$ |
| Rhombus/Kite (diagonals $d_1$, $d_2$) | Sum of sides | $\dfrac{1}{2}d_1 d_2$ |
Exam Tip
The height of a parallelogram or trapezium is the perpendicular height, not the slant side. Make sure to identify the correct measurement before substituting into the formula.
Worked Example 1.1 — Area of a Trapezium
A trapezium has parallel sides of length 8 cm and 14 cm, and a perpendicular height of 6 cm. Calculate its area.
Step 1 Identify $a = 8$ cm, $b = 14$ cm, $h = 6$ cm.
Step 2 Apply the formula.
$$A = \frac{1}{2}(a + b)h = \frac{1}{2}(8 + 14)(6) = \frac{1}{2}(22)(6) = 66 \text{ cm}^2$$
Worked Example 1.2 — Perimeter of a Composite Boundary
A field is shaped like a rectangle 20 m by 12 m, with a triangular section of base 20 m and height 9 m attached to one of the longer sides. Find the total perimeter of the field.
Step 1 The rectangle contributes two short sides: $2 \times 12 = 24$ m. The shared long side is internal and does not form part of the perimeter.
Step 2 The triangle's two slant sides each have length $\sqrt{10^2 + 9^2} = \sqrt{181} \approx 13.45$ m.
Step 3 One long side of the rectangle is on the outer boundary: 20 m. The two slant sides: $2 \times 13.45 \approx 26.9$ m.
$$\text{Perimeter} = 24 + 20 + 26.9 \approx 70.9 \text{ m}$$
Practice 1A — Parallelogram
A parallelogram has a base of 11 cm and a perpendicular height of 7 cm. Its slant side is 9 cm. Find (a) its area; (b) its perimeter.
Show Solution
(a) Area $= bh = 11 \times 7 = 77$ cm².
(b) Perimeter $= 2(11 + 9) = 40$ cm.
Practice 1B — Kite
A kite has diagonals of length 10 cm and 16 cm. Calculate its area.
Show Solution
Area of kite $= \dfrac{1}{2}d_1 d_2 = \dfrac{1}{2}(10)(16) = 80$ cm².
2. Circles: Circumference, Arc Length and Sector Area
Circle Formulae
- Circumference: $C = 2\pi r = \pi d$
- Area: $A = \pi r^2$
- Arc length (sector angle $\theta°$): $\ell = \dfrac{\theta}{360} \times 2\pi r$
- Sector area (sector angle $\theta°$): $A_s = \dfrac{\theta}{360} \times \pi r^2$
Figure 5.1 — A circle showing a sector with arc length and area proportional to the central angle.
Worked Example 2.1 — Arc Length and Sector Area
A sector has radius 9 cm and central angle 80°. Find (a) the arc length; (b) the sector area. Give answers to 3 significant figures.
(a) Arc length
$$\ell = \frac{80}{360} \times 2\pi \times 9 = \frac{2}{9} \times 18\pi = 4\pi \approx 12.6 \text{ cm}$$
(b) Sector area
$$A_s = \frac{80}{360} \times \pi \times 9^2 = \frac{2}{9} \times 81\pi = 18\pi \approx 56.5 \text{ cm}^2$$
Worked Example 2.2 — Finding the Radius from Arc Length
An arc of a circle subtends an angle of 150° at the centre. The arc length is 25 cm. Find the radius, giving your answer to 3 significant figures.
Step 1 Use the arc length formula.
$$25 = \frac{150}{360} \times 2\pi r = \frac{5}{12} \times 2\pi r = \frac{5\pi r}{6}$$
Step 2 Solve for $r$.
$$r = \frac{25 \times 6}{5\pi} = \frac{150}{5\pi} = \frac{30}{\pi} \approx 9.55 \text{ cm}$$
Practice 2A — Sector
A sector of radius 12 cm has an arc length of 10 cm. Find (a) the central angle $\theta$ in degrees; (b) the sector area.
Show Solution
(a) $10 = \dfrac{\theta}{360} \times 2\pi \times 12 \implies \theta = \dfrac{10 \times 360}{24\pi} = \dfrac{3600}{24\pi} \approx 47.7°$
(b) $A_s = \dfrac{1}{2} r \ell = \dfrac{1}{2}(12)(10) = 60$ cm² (using the formula $A_s = \dfrac{1}{2}r\ell$).
Practice 2B — Perimeter of a Sector
A sector has radius 7 cm and central angle 120°. Find the perimeter of the sector (two radii plus the arc).
Show Solution
Arc length $= \dfrac{120}{360} \times 2\pi \times 7 = \dfrac{1}{3} \times 14\pi = \dfrac{14\pi}{3} \approx 14.66$ cm.
Perimeter $= 2 \times 7 + \dfrac{14\pi}{3} = 14 + \dfrac{14\pi}{3} \approx 28.7$ cm.
3. Surface Area of 3D Shapes
The surface area of a 3D shape is the total area of all its faces. Think of it as the area of the net of the shape. All answers should include appropriate square units.
| Shape | Surface Area |
|---|---|
| Cube (side $a$) | $6a^2$ |
| Cuboid ($l \times w \times h$) | $2(lw + wh + lh)$ |
| Cylinder (radius $r$, height $h$) | $2\pi r^2 + 2\pi rh$ |
| Cone (radius $r$, slant height $l$) | $\pi r^2 + \pi rl$ |
| Sphere (radius $r$) | $4\pi r^2$ |
| Hemisphere (radius $r$) | $3\pi r^2$ (curved + flat circle) |
Exam Tip — Cone Slant Height
If a question gives the vertical height $h$ of a cone rather than the slant height $l$, you must first calculate $l = \sqrt{r^2 + h^2}$ using Pythagoras' theorem before applying the surface area formula.
Worked Example 3.1 — Surface Area of a Cylinder
A cylinder has radius 4 cm and height 10 cm. Find its total surface area, giving your answer in terms of $\pi$ and also as a decimal (3 s.f.).
Step 1 Identify $r = 4$ and $h = 10$.
Step 2 Total surface area $= 2\pi r^2 + 2\pi rh$.
$$= 2\pi(4)^2 + 2\pi(4)(10) = 32\pi + 80\pi = 112\pi \approx 352 \text{ cm}^2$$
Worked Example 3.2 — Surface Area of a Cone
A cone has base radius 5 cm and vertical height 12 cm. Find its total surface area to 3 significant figures.
Step 1 Find slant height: $l = \sqrt{r^2 + h^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ cm.
Step 2 Surface area $= \pi r^2 + \pi rl = \pi(5^2) + \pi(5)(13) = 25\pi + 65\pi = 90\pi \approx 283$ cm².
Practice 3A — Sphere
A sphere has a surface area of $196\pi$ cm². Find its radius.
Show Solution
$4\pi r^2 = 196\pi \implies r^2 = 49 \implies r = 7$ cm.
Practice 3B — Triangular Prism
A triangular prism has a right-angled triangular cross-section with legs 6 cm and 8 cm. The prism has length 15 cm. Find its total surface area.
Show Solution
Hypotenuse $= \sqrt{6^2 + 8^2} = \sqrt{100} = 10$ cm.
Two triangular faces: $2 \times \dfrac{1}{2}(6)(8) = 48$ cm².
Three rectangular faces: $(6 \times 15) + (8 \times 15) + (10 \times 15) = 90 + 120 + 150 = 360$ cm².
Total surface area $= 48 + 360 = 408$ cm².
4. Volume of 3D Shapes
Volume measures the amount of three-dimensional space enclosed by a shape. Units are cubic (e.g. cm³, m³). The general principle for prisms and cylinders is: Volume = cross-sectional area × length (or height).
| Shape | Volume |
|---|---|
| Cube (side $a$) | $a^3$ |
| Cuboid ($l \times w \times h$) | $lwh$ |
| Prism (cross-section area $A$, length $L$) | $AL$ |
| Cylinder (radius $r$, height $h$) | $\pi r^2 h$ |
| Pyramid (base area $A$, height $h$) | $\dfrac{1}{3}Ah$ |
| Cone (radius $r$, height $h$) | $\dfrac{1}{3}\pi r^2 h$ |
| Sphere (radius $r$) | $\dfrac{4}{3}\pi r^3$ |
Worked Example 4.1 — Volume of a Cone
A cone has base radius 6 cm and height 14 cm. Find its volume to 3 significant figures.
Step 1 Identify $r = 6$, $h = 14$.
$$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(36)(14) = \frac{504\pi}{3} = 168\pi \approx 528 \text{ cm}^3$$
Worked Example 4.2 — Volume of a Sphere
A solid sphere has a volume of $\dfrac{500\pi}{3}$ cm³. Find its radius.
Step 1 Set the volume formula equal to the given value.
$$\frac{4}{3}\pi r^3 = \frac{500\pi}{3}$$
Step 2 Simplify (divide both sides by $\dfrac{\pi}{3}$).
$$4r^3 = 500 \implies r^3 = 125 \implies r = 5 \text{ cm}$$
Worked Example 4.3 — Volume of a Trapezoidal Prism
A prism has a trapezoidal cross-section with parallel sides 5 cm and 9 cm and a perpendicular height of 4 cm. The prism is 12 cm long. Find its volume.
Step 1 Area of trapezium $= \dfrac{1}{2}(5 + 9)(4) = \dfrac{1}{2}(14)(4) = 28$ cm².
Step 2 Volume $= 28 \times 12 = 336$ cm³.
Practice 4A — Cylinder
A cylindrical can has diameter 10 cm and height 18 cm. Find (a) its volume; (b) the volume in litres (1 litre = 1000 cm³).
Show Solution
Radius $r = 5$ cm.
(a) $V = \pi r^2 h = \pi(25)(18) = 450\pi \approx 1414$ cm³.
(b) $450\pi \div 1000 \approx 1.41$ litres.
Practice 4B — Square-Based Pyramid
A square-based pyramid has a base of side 8 cm and a vertical height of 9 cm. Find its volume.
Show Solution
Base area $= 8^2 = 64$ cm².
$V = \dfrac{1}{3}(64)(9) = \dfrac{576}{3} = 192$ cm³.
5. Similar Shapes — Area and Volume Scale Factors
Scale Factor Rules for Similar Shapes
If two similar shapes have a linear scale factor of $k$ (ratio of corresponding lengths), then:
- Area scale factor $= k^2$
- Volume scale factor $= k^3$
These rules apply to any pair of similar shapes, including 2D and 3D figures.
Worked Example 5.1 — Area Scale Factor
Two similar triangles have corresponding sides in the ratio 3 : 5. The area of the smaller triangle is 27 cm². Find the area of the larger triangle.
Step 1 Linear scale factor $k = \dfrac{5}{3}$.
Step 2 Area scale factor $= k^2 = \dfrac{25}{9}$.
Step 3 Area of larger triangle $= 27 \times \dfrac{25}{9} = 75$ cm².
Worked Example 5.2 — Volume Scale Factor
Two similar cones have heights 4 cm and 10 cm. The volume of the smaller cone is 48 cm³. Find the volume of the larger cone.
Step 1 Linear scale factor $k = \dfrac{10}{4} = \dfrac{5}{2}$.
Step 2 Volume scale factor $= k^3 = \dfrac{125}{8}$.
Step 3 Volume of larger cone $= 48 \times \dfrac{125}{8} = 6 \times 125 = 750$ cm³.
Worked Example 5.3 — Finding the Linear Scale Factor from Volume
Two similar spheres have volumes 54 cm³ and 128 cm³. Find the ratio of their surface areas.
Step 1 Volume scale factor $= \dfrac{128}{54} = \dfrac{64}{27}$.
Step 2 Linear scale factor $= k = \sqrt[3]{\dfrac{64}{27}} = \dfrac{4}{3}$.
Step 3 Area scale factor $= k^2 = \dfrac{16}{9}$. So the ratio of surface areas is $9 : 16$.
Practice 5A
Two similar cuboids have corresponding edges in the ratio 2 : 7. The surface area of the smaller cuboid is 48 cm². Find the surface area of the larger cuboid.
Show Solution
Area scale factor $= \left(\dfrac{7}{2}\right)^2 = \dfrac{49}{4}$.
Surface area of larger cuboid $= 48 \times \dfrac{49}{4} = 12 \times 49 = 588$ cm².
Practice 5B
Two similar cylinders have surface areas in the ratio 4 : 9. The volume of the larger cylinder is 540 cm³. Find the volume of the smaller cylinder.
Show Solution
Area scale factor $= \dfrac{4}{9}$, so linear scale factor $k = \sqrt{\dfrac{4}{9}} = \dfrac{2}{3}$.
Volume scale factor $= k^3 = \dfrac{8}{27}$.
Volume of smaller $= 540 \times \dfrac{8}{27} = 20 \times 8 = 160$ cm³.
6. Compound Shapes
A compound shape is formed by combining or subtracting standard shapes. The key strategy is to split the compound shape into recognisable standard shapes, calculate each area or volume separately, then add or subtract as appropriate.
Figure 5.2 — A compound shape formed by a rectangle with a semicircular end. The area is calculated by adding the rectangle area and the semicircle area.
Worked Example 6.1 — Compound 2D Shape
A shape consists of a rectangle 10 cm by 6 cm, with a semicircle attached to one of the shorter sides (diameter 6 cm). Find the total area and perimeter of the shape.
Area
Rectangle: $10 \times 6 = 60$ cm².
Semicircle (radius = 3 cm): $\dfrac{1}{2}\pi r^2 = \dfrac{1}{2}\pi(9) = \dfrac{9\pi}{2}$ cm².
Total area $= 60 + \dfrac{9\pi}{2} \approx 60 + 14.14 = 74.14 \approx 74.1$ cm².
Perimeter
The perimeter consists of: two long sides ($2 \times 10 = 20$ cm), one short side on the opposite end ($6$ cm), and the curved semicircle ($\pi r = 3\pi$ cm). The diameter side is internal and not included.
Total perimeter $= 20 + 6 + 3\pi \approx 26 + 9.42 = 35.4$ cm.
Worked Example 6.2 — Shape with a Hole
A square tile of side 20 cm has a circular hole of radius 5 cm cut from its centre. Find the remaining area.
Step 1 Square area $= 20^2 = 400$ cm².
Step 2 Circle area $= \pi(5)^2 = 25\pi \approx 78.54$ cm².
Step 3 Remaining area $= 400 - 25\pi \approx 400 - 78.5 = 321.5 \approx 322$ cm².
Worked Example 6.3 — Compound 3D Shape
A toy consists of a cylinder of radius 3 cm and height 8 cm, with a hemisphere of radius 3 cm on top. Find its total volume.
Cylinder $V_c = \pi r^2 h = \pi(9)(8) = 72\pi$ cm³.
Hemisphere $V_h = \dfrac{1}{2} \times \dfrac{4}{3}\pi r^3 = \dfrac{2}{3}\pi(27) = 18\pi$ cm³.
Total $V = 72\pi + 18\pi = 90\pi \approx 283$ cm³.
Practice 6A — L-Shaped Room
An L-shaped room can be divided into two rectangles: one measuring 8 m by 5 m, and another measuring 4 m by 3 m. Find the total floor area and the total perimeter of the room.
Show Solution
Total area $= (8 \times 5) + (4 \times 3) = 40 + 12 = 52$ m².
Perimeter: trace the outer boundary carefully. Going around the L-shape: $8 + 5 + 4 + 3 + (8-4) + (5-3) = 8 + 5 + 4 + 3 + 4 + 2 = 26$ m.
7. Practice Problems
Q1 — Area and Perimeter
A triangle has base 14 cm and perpendicular height 9 cm. Find its area.
Show Solution
Area $= \dfrac{1}{2} \times 14 \times 9 = 63$ cm².
Q2 — Circumference
A circle has circumference 30 cm. Find its area, giving your answer to 3 significant figures.
Show Solution
$C = 2\pi r = 30 \implies r = \dfrac{15}{\pi}$.
Area $= \pi r^2 = \pi \left(\dfrac{15}{\pi}\right)^2 = \dfrac{225}{\pi} \approx 71.6$ cm².
Q3 — Sector Area
A sector of a circle has radius 8 cm and sector area $32\pi$ cm². Find the central angle $\theta$ in degrees.
Show Solution
$32\pi = \dfrac{\theta}{360} \times \pi(8)^2 = \dfrac{64\pi\theta}{360}$
$32 = \dfrac{64\theta}{360} \implies \theta = \dfrac{32 \times 360}{64} = 180°$.
Q4 — Surface Area of a Cuboid
A cuboid measures 5 cm by 3 cm by 7 cm. Find its total surface area.
Show Solution
$SA = 2(lw + wh + lh) = 2(5 \times 3 + 3 \times 7 + 5 \times 7) = 2(15 + 21 + 35) = 2(71) = 142$ cm².
Q5 — Volume of a Cone
A cone has diameter 10 cm and slant height 13 cm. Find its volume to 3 significant figures.
Show Solution
Radius $r = 5$ cm. Vertical height: $h = \sqrt{l^2 - r^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ cm.
$V = \dfrac{1}{3}\pi(25)(12) = 100\pi \approx 314$ cm³.
Q6 — Similar Shapes (Area)
Two similar shapes have areas $75$ cm² and $300$ cm². Find the ratio of their corresponding lengths.
Show Solution
Area scale factor $= \dfrac{300}{75} = 4$.
Linear scale factor $= \sqrt{4} = 2$. Ratio of corresponding lengths is $1 : 2$.
Q7 — Compound Shape
A running track consists of a rectangle 100 m by 60 m, with semicircles of diameter 60 m at each end. Find the total area enclosed by the track.
Show Solution
Two semicircles = one full circle, radius $= 30$ m.
Circle area $= \pi(30)^2 = 900\pi$ m².
Rectangle area $= 100 \times 60 = 6000$ m².
Total area $= 6000 + 900\pi \approx 6000 + 2827 = 8827 \approx 8830$ m².
Q8 — Volume and Surface Area of Hemisphere
A solid hemisphere has radius 6 cm. Find (a) its volume; (b) its total surface area.
Show Solution
(a) $V = \dfrac{2}{3}\pi(6)^3 = \dfrac{2}{3}\pi(216) = 144\pi \approx 452$ cm³.
(b) Curved surface $= 2\pi r^2 = 72\pi$. Flat circle $= \pi r^2 = 36\pi$. Total $= 108\pi \approx 339$ cm².
Q9 — Volume Scale Factor
Two similar jugs have heights 12 cm and 20 cm. The smaller jug holds 540 ml. How many millilitres does the larger jug hold?
Show Solution
Linear scale factor $k = \dfrac{20}{12} = \dfrac{5}{3}$.
Volume scale factor $= k^3 = \dfrac{125}{27}$.
Volume of larger jug $= 540 \times \dfrac{125}{27} = 20 \times 125 = 2500$ ml.
Q10 — Arc Length on a Clock
The minute hand of a clock is 15 cm long. Calculate the distance its tip travels in 25 minutes. Give your answer to 3 significant figures.
Show Solution
In 25 minutes, the hand turns through $\dfrac{25}{60} \times 360° = 150°$.
Arc length $= \dfrac{150}{360} \times 2\pi(15) = \dfrac{5}{12} \times 30\pi = \dfrac{25\pi}{2} \approx 39.3$ cm.