O-Level Mathematics · Secondary 3 and 4
Mensuration
Mensuration is the measurement of lengths, areas and volumes. In O-Level Mathematics you find arc lengths, sector areas and segment areas using both degrees and radians, and you calculate the volume and surface area of prisms, cylinders, pyramids, cones and spheres, including composite solids. You need to know which formula fits each shape, how to combine shapes and how to keep units consistent. Mensuration draws on trigonometry and similarity, and radian measure is used throughout Additional Mathematics.
What the syllabus expects
- Convert between degrees and radians, using π radians = 180°
- Find arc length and sector area using degrees, and using s = rθ and A = ½r²θ with θ in radians
- Find the area of a segment as the sector area minus the triangle area
- Find the volume and surface area of cubes, cuboids, prisms and cylinders
- Find the volume and surface area of pyramids, cones and spheres, including hemispheres
- Solve problems involving composite solids and the melting or recasting of solids
Radians
A radian is the angle at the centre of a circle subtended by an arc equal in length to the radius. A full turn is 2π radians, so π radians = 180° and 1 radian is about 57.3°. To change degrees to radians, multiply by π/180, so 60° = π/3 radians and 135° = 3π/4 radians. To change radians to degrees, multiply by 180/π, so 2 radians = 114.6°. Answers in radians are often left in terms of π when the question asks for an exact value. When a question gives an angle in radians, set your calculator to radian mode before using sin, cos or tan, or convert the angle to degrees first. Mixing the two modes produces answers that look reasonable but are wrong, so check the mode every time you switch between a radian question and a degree question.
Arcs, sectors and segments
For a sector of radius r and angle θ at the centre, the arc length is (θ/360°) × 2πr and the area is (θ/360°) × πr² when θ is in degrees. In radians these become much simpler: arc length s = rθ and sector area A = ½r²θ. For example, a sector of radius 6 cm and angle 2.5 radians has arc length 6 × 2.5 = 15 cm and area ½ × 36 × 2.5 = 45 cm². The perimeter of a sector is the arc length plus two radii, here 15 + 12 = 27 cm. A segment is the region between a chord and its arc. Its area is the sector area minus the area of the triangle formed by the two radii and the chord, and that triangle has area ½r² sin θ. When θ is in radians, remember to use radian mode for sin θ.
Prisms, cylinders and pyramids
A prism has the same cross-section all the way through, so its volume is the area of the cross-section multiplied by the length. A cylinder is a prism with a circular cross-section: volume = πr²h, curved surface area = 2πrh, and total surface area = 2πrh + 2πr² for a closed cylinder. A pyramid with any base has volume = ⅓ × base area × perpendicular height. Its surface area is the base plus the areas of the triangular faces. Units matter throughout: lengths in cm give areas in cm² and volumes in cm³, and 1 litre = 1000 cm³. If dimensions are given in different units, convert them all before you start. For example, a cylindrical tank of radius 0.5 m and height 120 cm has volume π × 50² × 120 = 942 000 cm³ to 3 significant figures, which is 942 litres.
Cones, spheres and composite solids
A cone of base radius r, perpendicular height h and slant height l has volume ⅓πr²h and curved surface area πrl, where l² = r² + h² by Pythagoras' theorem. A sphere of radius r has volume (4/3)πr³ and surface area 4πr². A hemisphere has half the volume, (2/3)πr³, and a curved surface area of 2πr², plus a flat circular face of area πr² if it is solid. For a composite solid, such as a cylinder topped by a hemisphere, add the volumes, but for surface area include only the outside faces, so the circle where the two parts join is not counted. When a solid is melted and recast, the volume stays the same, so set the two volume expressions equal and solve. Leave π in the expressions until the end, since it often cancels.
Worked example
A circle has centre O and radius 10 cm. The chord AB subtends an angle of 1.2 radians at O. Find (a) the length of the minor arc AB, (b) the area of the minor segment cut off by AB, (c) the perimeter of that segment.
- Arc length s = rθ = 10 × 1.2 = 12 cm.
- Sector area = ½r²θ = ½ × 100 × 1.2 = 60 cm².
- Area of triangle AOB = ½r² sin θ = ½ × 100 × sin 1.2 = 50 × 0.93204 = 46.602 cm², with the calculator in radian mode.
- Segment area = 60 − 46.602 = 13.398 cm², which is 13.4 cm² to 3 significant figures.
- For the chord, split triangle AOB into two right-angled triangles with angle 0.6 radians at O: AB = 2 × 10 × sin 0.6 = 20 × 0.56464 = 11.293 cm.
- Perimeter of the segment = arc + chord = 12 + 11.293 = 23.293 cm.
Answer: (a) 12 cm (b) 13.4 cm² (c) 23.3 cm
Common mistakes
- Using degree mode with a radian angle. sin 1.2 in degree mode is 0.0209, not 0.932, which makes the triangle almost vanish. Switch to radian mode, or convert 1.2 radians to 68.75° first.
- Using s = rθ with θ in degrees. The formulas s = rθ and A = ½r²θ only work in radians. In degrees, use θ/360° of the circumference or of the area.
- Forgetting the radii in the perimeter of a sector. The perimeter is the arc length plus 2r, not the arc length alone.
- Using the slant height in the cone volume. Volume uses the perpendicular height h, while curved surface area uses the slant height l. Find the one you are missing with l² = r² + h².
- Counting hidden faces in a composite solid. Where a hemisphere sits on a cylinder, the joining circle is inside the solid and is not part of the surface area.
- Mixing units. A radius in metres and a height in centimetres must be converted to the same unit before any calculation, and 1 m³ = 1 000 000 cm³.
Exam tips
- Write down the formula you are using before substituting, so it is clear which height or radius each number stands for.
- Keep π as a symbol until the last line, especially in melting and recasting questions, where π often cancels and leaves a clean answer.
- Check whether the question wants an exact answer in terms of π, an answer to 3 significant figures, or an answer in a different unit such as litres.
Practice questions
1. Convert 135° to radians and 2 radians to degrees.
Show answer
135° = 135 × π/180 = 3π/4 radians, which is 2.36 radians to 3 significant figures. 2 radians = 2 × 180/π = 114.6°.
2. A cone has base radius 6 cm and perpendicular height 8 cm. Find its volume and its curved surface area, leaving your answers in terms of π.
Show answer
Volume = ⅓ × π × 6² × 8 = 96π cm³, about 302 cm³. The slant height is l = √(6² + 8²) = 10 cm, so the curved surface area = π × 6 × 10 = 60π cm², about 188 cm².
3. A solid metal hemisphere of radius 6 cm is melted down and recast into a solid cylinder of radius 4 cm. Find the height of the cylinder.
Show answer
9 cm. Volume of the hemisphere = (2/3)π × 6³ = (2/3)π × 216 = 144π cm³. Volume of the cylinder = π × 4² × h = 16πh. Setting 16πh = 144π gives h = 9 cm.
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