Split a composite solid into the basic solids it is built from (cone, cylinder, sphere, hemisphere, pyramid, cuboid), find any shared measurement such as a common radius, then work out each part separately. For volume, simply add the parts (or subtract, if one part has been removed). For surface area, add the parts too, but leave out any face that is hidden where the two solids are joined.
Once the basic formulas for a cone, a cylinder, a sphere and a pyramid are known, most students expect composite solids to be easy: just add the volumes and add the surface areas. The volume part is usually fine. The surface area part is where marks are lost, because a joined solid hides a face that would normally be counted.
A cone glued on top of a cylinder does not have a flat circular top on the cylinder anymore, because the cone sits over it. A hemisphere sitting on a cube does not have a flat circle exposed on the cube's top face either. If a student adds every curved and flat surface from both shapes without removing the hidden join, the surface area comes out too large. This is the single biggest source of lost marks on this topic.
Step 1
Identify the basic solids
Step 2
Find the shared measurement (usually radius)
Step 3
Work out each part separately
Step 4
Combine: add volumes, remove hidden faces from surface area
You will need the standard formulas from volume and surface area of solids: cylinder volume πr²h, cone volume ⅓πr²h, sphere volume 4/3 πr³, cone curved surface area πrl (where l is the slant height, found using l² = r² + h²), sphere surface area 4πr², and cylinder curved surface area 2πrh, which itself builds on the basic perimeter and area of a circle. If you are unsure of any of these, review that guide first, then come back and combine them here.
A solid is made from a cylinder of radius 6 cm and height 10 cm, with a cone of the same radius and height 8 cm fixed on top. Find the total volume, giving your answer to 1 decimal place. Take π = 3.142.
Solution
V = πr²h = 3.142 × 6² × 10 = 3.142 × 36 × 10 = 1131.12 cm³.V = ⅓πr²h = ⅓ × 3.142 × 6² × 8 = ⅓ × 3.142 × 36 × 8 = ⅓ × 904.896 = 301.632 cm³.1131.12 + 301.632 = 1432.752 cm³ ≈ 1432.8 cm³.Quick check: before combining, check that the radius really is shared between both solids. If the question gives two different radii for the cylinder and the cone, the shapes do not join cleanly and you cannot use this shortcut directly. Read the question again to confirm the shared radius.
Using the same solid as above (cylinder radius 6 cm, height 10 cm; cone same radius, height 8 cm), find the total surface area, to 1 decimal place. Take π = 3.142.
Solution
l² = r² + h² = 6² + 8² = 36 + 64 = 100, so l = 10 cm.πr² = 3.142 × 36 = 113.112 cm².2πrh = 2 × 3.142 × 6 × 10 = 377.04 cm².πrl = 3.142 × 6 × 10 = 188.52 cm².113.112 + 377.04 + 188.52 = 678.672 cm² ≈ 678.7 cm².The step students get wrong
Counting the joined circle twice, once as the cylinder's top and again as the cone's base. The same trap appears with hemispheres. A hemisphere's curved surface is half of 4πr², which is 2πr², and its flat circular face is counted only when it is actually exposed. Sitting on top of another solid, that circle is hidden, so you leave it out. Standing on its own as a solid hemisphere, that circle is the base and you must add πr², giving a total surface area of 3πr².
A solid cube of side 8 cm has a hemisphere of radius 4 cm fixed on top, centred on the top face. Find (a) the total volume, and (b) the total exposed surface area, both to 1 decimal place. Take π = 3.142.
Solution
V = 8³ = 512 cm³.4/3 πr³, so V = ½ × 4/3 × 3.142 × 4³ = ½ × 4/3 × 3.142 × 64 = ½ × 268.117… = 134.059 cm³.512 + 134.059 = 646.059 cm³ ≈ 646.1 cm³.8² = 64 cm² each, but the top face has a circle of radius 4 cm removed where the hemisphere sits. Area of that circle: πr² = 3.142 × 16 = 50.272 cm².6 × 64 − 50.272 = 384 − 50.272 = 333.728 cm².2πr² = 2 × 3.142 × 16 = 100.544 cm².333.728 + 100.544 = 434.272 cm² ≈ 434.3 cm².Accuracy: keep the full calculator value for each part and round only at the very end. Rounding each part before you add them can shift the final answer enough to lose the accuracy mark. If you are not sure how many figures to give, check significant figures vs decimal places.
Not every composite solid is made by joining two shapes together. Some are made by removing one shape from another, for example a cylindrical hole drilled through a cuboid, or a cone-shaped hollow scooped out of a cylinder. In these cases the volume of the removed shape is subtracted, not added, and the surface area gains the newly exposed inner surface of the hole while losing the flat area where the hole enters and exits.
Exam tip: read the question carefully to decide whether it describes a solid built up (add the volumes) or a solid with material removed (subtract the volume of the removed shape). The words "solid" versus "hollow", and "on top of" versus "drilled through" or "removed from", are the clues.
Do I always subtract the joined face from the surface area?
Only when two solids are joined so that one face is completely covered, such as a cone on a cylinder or a hemisphere on a cube. If the question describes a solid with a hole or a shape removed, you instead add the newly exposed inner surface and subtract only the small flat area where the hole meets the outer surface, not the whole face.
How do I know which radius to use if the solid has a cone and a hemisphere with different sizes?
Use whichever radius belongs to the part you are currently calculating. Composite solids in O-Level questions almost always share one common radius at the join, since the shapes must fit together, so check the diagram or the given measurements carefully before assuming.
Why does volume always add, but surface area sometimes does not?
Volume measures the total space enclosed, so joining two solids simply combines the space inside each of them, with no double-counting. Surface area measures only the outer skin, and any face hidden inside the join is no longer part of that outer skin, so it must be left out.
What if the question only asks for volume, not surface area?
Then you can ignore the joined-face problem entirely: add the volumes of the basic solids (or subtract, for a removed shape) and you are done. The hidden-face issue only matters for surface area questions.
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