After prisms and cylinders in Sec 1, the pyramid, the cone and the sphere complete the Sec 2 mensuration toolkit. A pyramid and a cone start with a flat base but taper to a single point. Each holds exactly one third of the volume of the prism or cylinder that shares its base and its height. A sphere depends only on its radius.
These solids are tested on their own in school exams first. They return at O-Level combined into composite solids, such as an ice-cream cone topped with a hemisphere or a bolt made of a cylinder and a hemisphere. Most marks are lost through small, predictable slips rather than through not knowing the formulas. The commonest are using the slant height instead of the perpendicular height, forgetting to halve a diameter and rounding too early.
What these notes cover
- •Volume of a pyramid
- •Volume of a cone
- •Volume of a sphere and hemisphere
- •Perpendicular height vs slant height
- •Slant height of a cone
- •Surface area of a cone, sphere and hemisphere
- •Surface area of a pyramid
- •Composite solids
- •Melt-and-recast
- •Exact-π answers vs 3 s.f.
- •Reverse problems: finding r or h
- •Common mistakes and summary
Key definitions
- Perpendicular height (h)
- The distance from the apex straight down to the base, meeting the base at right angles. Every h in a volume formula is this vertical height. For a right pyramid, its foot lands at the centre of the base.
- Slant height of a cone (l)
- The length along the sloping surface from the apex down to a point on the rim of the base. It is the hypotenuse of the right-angled triangle formed with r and h, so it is always longer than h. It is used for surface area, not volume.
- Slant height of a pyramid face
- The length from the midpoint of a base edge up the middle of a triangular face to the apex. It is used for surface area.
- Slant edge
- The sharp edge of a pyramid running from a base corner to the apex. It is usually a distractor and is not used in either the volume or the surface-area formula.
- Premature rounding
- Rounding to 3 significant figures in the middle of a calculation. It is penalised because the error compounds, so intermediate values should be kept unrounded or to at least 5 significant figures.
Key formulas
Volume of a pyramid
h is the perpendicular height from the apex to the base. Any polygon base works; find the base area first. This formula is not printed on the O-Level formula sheet, so memorise it.
Volume of a cone
r is the base radius and h is the perpendicular height from the vertex to the centre of the base. Do not use the sloping edge of the cone.
Volume of a sphere
r is the radius. Doubling r multiplies the volume by 2³ = 8. For a hemisphere, halve it: V = (2 ÷ 3)πr³.
Slant height of a cone (Pythagoras' Theorem)
l² = r² + h²
l is the slant height (the hypotenuse), r is the base radius and h is the vertical height. l must be the largest of the three lengths. To find h from l and r, use h² = l² − r².
Surface area of a cone
Curved surface area = πrl; total surface area of a solid cone = πrl + πr²
πrl is the sloping surface only. A solid (closed) cone also has a circular base of area πr². An open cone, such as an ice-cream wafer, has surface area πrl only.
Surface area of a sphere and a solid hemisphere
Sphere = 4πr²; solid hemisphere = 2πr² + πr² = 3πr²
A solid hemisphere has a curved surface of 2πr² plus a flat circular face of πr². Only a hollow, open hemisphere has surface area 2πr².
Worked example
A solid cone has a base diameter of 18 cm and a vertical height of 12 cm. Find (i) the slant height, (ii) the curved surface area, and (iii) the total surface area. Give each area both in terms of π and correct to 3 significant figures.
- 1Halve the diameter first: r = 18 ÷ 2 = 9 cm.
- 2Find the slant height by Pythagoras' Theorem: l² = r² + h² = 9² + 12² = 81 + 144 = 225, so l = √225 = 15 cm.
- 3Check: 15 is larger than both 9 and 12, which is sensible for a hypotenuse.
- 4Curved surface area = πrl = π × 9 × 15 = 135π cm² ≈ 424 cm² (3 s.f.).
- 5Total surface area = πrl + πr² = 135π + π × 9² = 135π + 81π = 216π cm² ≈ 679 cm² (3 s.f.).
Watch out: Do not substitute the diameter 18 cm straight into πrl or πr². Doing so doubles or quadruples the answer. Keep 135π and 216π exact until the last line, and round only once.
Common mistakes
Slant height used in the volume formula
Wrong: Take a cone with radius 6 cm, vertical height 8 cm and slant height 10 cm. Writing V = (1 ÷ 3)π(6)²(10) = 120π cm³ substitutes the slant height for h.
Right: V = (1 ÷ 3)π(6)²(8) = 96π cm³. Volume always takes the vertical height h. The slant height l appears only in the curved surface area, πrl.
Substituting the diameter as the radius
Wrong: Take a cone with base diameter 12 cm and vertical height 7 cm. Writing V = (1 ÷ 3)π(12)²(7) = 336π cm³ uses the diameter as r.
Right: r = 12 ÷ 2 = 6, so V = (1 ÷ 3)π(6)²(7) = 84π cm³. An answer 4 times too big is the classic sign of this error, because (2r)² = 4r².
Forgetting the one-third in pyramid and cone volumes
Wrong: Take a square-based pyramid with base 9 cm by 9 cm and vertical height 10 cm. Writing V = 9 × 9 × 10 = 810 cm³ gives the volume of the cuboid around the pyramid.
Right: V = (1 ÷ 3) × 9 × 9 × 10 = 270 cm³. A pyramid or cone fills exactly one third of the prism or cylinder with the same base and height.
Questions students ask
What is the difference between slant height and vertical height of a cone?
The vertical height h is measured straight up from the centre of the base to the apex. The slant height l is measured along the sloping surface from the apex to the rim. They are linked by l² = r² + h². Use h for volume and l for surface area.
Is the volume of a pyramid formula given in the O-Level formula sheet?
No. The cone and sphere volume formulas are printed on the formula sheet at the front of the O-Level paper, but the pyramid formula is not. Memorise V = (1 ÷ 3) × base area × h. You must also recognise which formula a question needs and which given length goes where.
How do you find the radius of a sphere from its volume?
Substitute into V = (4 ÷ 3)πr³. Multiply both sides by 3 and divide by 4π to isolate r³, then take a cube root: r = ∛(3V ÷ 4π). Do not divide by 3 or take a square root. Keep intermediate values to at least 5 significant figures.
When should I leave my answer in terms of π?
Leave π as a symbol when the question says “leave your answer in terms of π” or asks for the exact value. For example, a sphere of radius 6 cm has volume 288π cm³. With no instruction, use the calculator's π key and round the final answer to 3 significant figures.
In the full PDF
- •Labelled diagrams of the rectangular-base pyramid, the cone with cross-section diameters and the hemisphere-cylinder metal bolt
- •Further worked examples: a rectangular-base pyramid volume, a cone and sphere of equal radius, hemisphere and square-pyramid surface areas, and a cone drilled from a wax cylinder
- •The composite-solids method, covering hemisphere-on-cylinder solids, hidden joined faces, the metal bolt and melt-and-recast questions
- •Reverse problems worked in full: finding a sphere's radius from its volume and a pyramid's height from its volume, plus the SEAB accuracy convention
- •The remaining three common mistakes and the ten-second pre-submission checklist
- •A one-glance formula table and end-of-topic summary