The O-Level formula sheet gives you three volume/surface area formulas: cone (V = ⅓πr²h, curved SA = πrl), sphere (V = 4/3πr³, SA = 4πr²), and pyramid (V = ⅓ × base area × height). Everything else, cylinder, prism, and cuboid, you must recall from memory. When a question gives you radius and height but asks for slant height or curved surface area, find the slant height first using Pythagoras.
Volume and surface area questions look like pure substitution, so students expect them to be easy marks. They lose marks anyway, for two reasons. First, they confuse which formulas are given on the formula sheet and which they need to know by heart, so they either waste time hunting for a formula that was never provided, or misremember one that was. Second, cone and pyramid questions often give you the base radius and vertical height, then ask for the slant height or curved surface area, which needs an extra Pythagoras step the student did not expect.
Fix both problems by memorising exactly what the formula sheet gives you (below), and by always checking whether a question needs slant height before you start substituting.
Cone (given)
V = ⅓πr²h
Curved SA = πrl
Sphere (given)
V = 4/3πr³
SA = 4πr²
Pyramid (given)
V = ⅓ × base area × height
Cylinder (not given)
V = πr²h
Total SA = 2πrh + 2πr²
Prism (not given)
V = cross-section area × length
SA = sum of all faces
Cuboid (not given)
V = l × w × h
SA = 2(lw + lh + wh)
For a cone, the slant height l is the distance from the tip to a point on the base circle's edge. It forms the hypotenuse of a right-angled triangle with the vertical height h and the radius r as the two shorter sides:
l² = r² + h², so l = √(r² + h²)
A cone has base radius 6 cm and vertical height 8 cm. Find its volume and curved surface area, taking π = 3.142.
Solution
V = ⅓πr²h = ⅓ × π × 6² × 8 = ⅓ × π × 36 × 8 = ⅓ × 288π = 96π.π = 3.142: V = 96 × 3.142 = 301.6 cm³ (1 decimal place).l, which is not given. Find it with Pythagoras: l = √(r² + h²) = √(6² + 8²) = √(36 + 64) = √100 = 10 cm.Curved SA = πrl = π × 6 × 10 = 60π = 60 × 3.142 = 188.5 cm² (1 decimal place).Quick check: the slant height must always be longer than both the radius and the vertical height, since it is the hypotenuse. Here l = 10, which is greater than r = 6 and h = 8, so the value is sensible.
A solid sphere has radius 5 cm. Find its total surface area, then find the total surface area of a solid hemisphere with the same radius, taking π = 3.142.
Solution
SA = 4πr² = 4 × π × 5² = 4 × π × 25 = 100π = 100 × 3.142 = 314.2 cm².½ × 4πr² = 2πr².πr².2πr² + πr² = 2 × π × 5² + π × 5² = 50π + 25π = 75π = 75 × 3.142 = 235.7 cm² (1 decimal place).The step students get wrong
Students find the curved part of the hemisphere correctly, then forget the flat circular face entirely, because it does not look like part of a "sphere" formula. Any solid hemisphere problem asking for total surface area needs both the curved half and the flat circle. If the question only asks for curved surface area, then 2πr² alone is correct and the flat face is left out.
Before starting any mensuration question, glance at the formula sheet and note which of the three solids (cone, sphere, pyramid) it covers. If your question is about a cylinder, prism, or cuboid, you will not find a formula there. Do not waste exam time searching for it: recall it directly, or derive it from the cross-section area times length rule for prisms.
Do I need to memorise the cone and sphere formulas if they're on the formula sheet?
You do not need to memorise them to write them down, but you do need to recognise when to use each one, and to know that curved surface area for a cone needs the slant height, not the vertical height. Mixing up l and h in πrl is a common error even when the formula is provided.
Why is the pyramid volume formula the same shape as the cone volume formula?
Both are one third of a "prism-like" volume: ⅓ × base area × height for a pyramid, and ⅓πr²h for a cone, which is the same idea with a circular base area of πr². Any solid that comes to a single point (apex) above a base uses this one-third rule.
What if the cone or pyramid is a frustum, not a full cone?
A frustum (the shape left when the top of a cone or pyramid is sliced off) is not directly on the O-Level formula sheet. The usual approach is to find the volume or surface area of the full solid, then subtract the volume or surface area of the smaller cone or pyramid that was removed. Because the smaller solid is similar to the full one, the area and volume scale factor rules are often the fastest route to its measurements. The same add-or-subtract thinking carries over to composite solids, where a shape is built from two or more of these basic solids joined together.
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