A composite solid is not difficult because it contains many formulae. It is difficult when the learner does not decide which surfaces are exposed, which faces disappear inside a join, and which pieces of volume have been added or removed.
This Secondary 3 Mathematics Learning Guide develops volume and surface area of prisms, cylinders, cones and spheres, composite solids, open containers, hollow solids, hidden faces, volume conservation and conversion between square and cubic units. It extends Arc Length, Sector Area, Radians and Composite Mensuration by concentrating on three-dimensional structure.
Official scope: the current 2027 SEC G3 Mathematics syllabus listing identifies Mathematics as K310. K310 includes volume and surface area of cube, cuboid, prism, cylinder, pyramid, cone and sphere; conversion between cm² and m² and between cm³ and m³; and problems involving composite solids.
Composite-solid routine: split the object into known solids → identify joins and holes → calculate volume by adding or subtracting solids → calculate surface area from exposed surfaces only → convert units at the correct dimensional power → check scale and physical plausibility.
Volume Counts Space; Surface Area Counts Exposed Boundary
Volume measures three-dimensional space and uses cubic units. Surface area measures boundary and uses square units. The same composite object can therefore require different decompositions for the two tasks.
When two solids are joined, their touching faces are usually internal and are not part of the external surface area. Their volumes, however, still contribute unless one solid has been removed from the other.
Worked Example 1: Cylinder With a Hemisphere
A solid consists of a cylinder of radius 3 cm and height 10 cm with a hemisphere of radius 3 cm attached to one circular end. Find its volume and external surface area.
Cylinder volume=πr²h=π(3²)(10)=90π cm³.
Hemisphere volume=2/3πr³=2/3π(27)=18π cm³.
Total volume=108π cm³≈339 cm³.
For external surface area, include the cylinder curved surface 2πrh=60π, the one exposed circular base 9π, and the curved hemisphere 2πr²=18π.
Total external surface area=87π cm²≈273 cm². The joining circle is internal and must not be counted.
Draw a Surface Inventory Before Using Formulae
A surface inventory is a short list: “cylinder curved face, one circular base, hemisphere curved face”. It prevents the common habit of calculating each solid’s total surface area separately and then adding them, which double-counts hidden joining faces.
For volume, use a different inventory: “cylinder volume + hemisphere volume”. Surface and volume inventories answer different questions.
Worked Example 2: Cone on a Cylinder
A cylinder of radius 4 cm and height 6 cm has a cone of the same radius attached on top. The cone has vertical height 3 cm and slant height 5 cm. Find the volume and external surface area.
Cylinder volume=π(4²)(6)=96π cm³. Cone volume=1/3π(4²)(3)=16π cm³.
Total volume=112π cm³≈352 cm³.
External surface area includes cylinder curved area 2π(4)(6)=48π, bottom circle 16π, and cone curved area πrl=π(4)(5)=20π.
Total external surface area=84π cm²≈264 cm². The shared circular face between cone and cylinder is hidden.
Prisms Depend on a Constant Cross-Section
A prism has the same cross-section all along its length. Volume = cross-sectional area × length.
For surface area, the side faces form rectangles whose combined area can be found as perimeter of cross-section × prism length, then add the two end faces.
Worked Example 3: Triangular Prism
A right-triangular prism has a 3-4-5 triangular cross-section and length 10 cm. Find its volume and total surface area.
Cross-sectional area=1/2×3×4=6 cm². Volume=6×10=60 cm³.
Two triangular ends contribute 12 cm². Rectangular side faces contribute (3+4+5)×10=120 cm².
Total surface area=132 cm².
Open Containers Remove Surfaces, Not Volume Capacity
An open-top cylinder still has the same internal volume as a closed cylinder with the same dimensions, but its material surface area excludes the top circular face.
Worked Example 4: Open Cylindrical Container
An open cylindrical container has radius 4 cm and height 10 cm. Find its capacity and the surface area of material required, ignoring thickness.
Volume=π(4²)(10)=160π cm³≈503 cm³.
Material area=curved surface+one base=2π(4)(10)+π(4²)=80π+16π=96π cm²≈302 cm².
Adding a second circular face would model a closed cylinder, not the stated open container.
Hollow Solids Require Subtraction
A hole removes volume and can add new internal surface. Students often remember to subtract the missing material but forget that drilling creates an exposed inner wall.
Worked Example 5: Cylindrical Hole Through a Cylinder
A solid cylinder has radius 5 cm and height 12 cm. A coaxial cylindrical hole of radius 2 cm is drilled completely through it. Find the remaining volume and total exposed surface area.
Remaining volume=π(5²−2²)(12)=π(21)(12)=252π cm³≈792 cm³.
External curved area=2π(5)(12)=120π. Internal curved area=2π(2)(12)=48π.
Each end is an annulus of area π(25−4)=21π, so both ends contribute 42π.
Total exposed surface area=210π cm²≈660 cm².
Square and Cubic Unit Conversion Must Respect Dimension
Since 1 m=100 cm, then 1 m²=100²=10,000 cm² and 1 m³=100³=1,000,000 cm³.
The conversion factor is squared for area and cubed for volume because the unit is repeated across dimensions.
Worked Example 6: Area and Volume Conversion
Convert 45,000 cm² to m² and 2.35 m³ to cm³.
45,000÷10,000=4.5 m².
2.35×1,000,000=2,350,000 cm³.
Dividing the area by 100 or multiplying the volume by 100 would ignore the dimensional power.
Worked Example 7: Convert Before Comparing Capacity
A tank holds 0.42 m³. A second tank holds 390,000 cm³. Which has greater capacity?
0.42 m³=420,000 cm³. Therefore the first tank is larger by 30,000 cm³, or 0.03 m³.
Comparing 0.42 and 390,000 directly is meaningless because the units differ.
Volume Conservation Links Different Shapes
If a school-model problem says a solid is melted and recast with no loss, the volume before and after is equal. Surface area is not conserved because the new shape has a different boundary.
Worked Example 8: Sphere Recast as a Cylinder
A solid sphere of radius 3 cm is melted and recast as a cylinder of radius 1 cm with no loss. Find the cylinder height.
Sphere volume=4/3π(3³)=36π cm³.
Cylinder volume=π(1²)h=πh. Set πh=36π, giving h=36 cm.
The long cylinder is plausible because its radius is much smaller than the sphere radius.
Similar Solids Scale Volume Faster Than Length
If corresponding lengths scale by factor k, surface areas scale by k² and volumes by k³.
This is useful for checking composite-solid results and connects to Similarity, Scale Factors and Mensuration.
Worked Example 9: Scale a Solid
A model solid is enlarged by linear scale factor 2.5. Its surface area is 48 cm² and volume is 32 cm³. Find the enlarged surface area and volume.
Surface-area factor=2.5²=6.25. New surface area=48×6.25=300 cm².
Volume factor=2.5³=15.625. New volume=32×15.625=500 cm³.
A Hemisphere Has Two Different Surface-Area Interpretations
The curved surface area of a hemisphere is 2πr². If the flat circular base is also exposed, total surface area is 3πr².
When the hemisphere is attached to another solid along that base, the circle is usually internal and the curved area alone is counted for the hemisphere.
Worked Example 10: Hemisphere Bowl Versus Solid Dome
A solid hemisphere of radius 5 cm is sitting flat on a table. Find its exposed surface area if the circular base on the table is not exposed.
Only the curved hemisphere is exposed: 2π(5²)=50π cm².
If the object were lifted and the flat circular face were also exposed, total surface area would be 50π+25π=75π cm².
Worked Example 11: Composite Volume With a Removed Cuboid
A cuboid measures 12 cm by 8 cm by 5 cm. A rectangular notch measuring 3 cm by 2 cm by 5 cm is removed completely through its height. Find the remaining volume.
Original volume=12×8×5=480 cm³. Removed volume=3×2×5=30 cm³.
Remaining volume=450 cm³.
Finding surface area after such a cut would require a new exposed-face inventory because the notch creates additional internal faces.
Worked Example 12: Cone Slant Height Before Surface Area
A cone has radius 6 cm and vertical height 8 cm. Find its slant height and total surface area.
Slant height l=√(6²+8²)=10 cm.
Curved area=πrl=60π. Base area=36π. Total surface area=96π cm²≈302 cm².
Using vertical height 8 in the curved-area formula would understate the actual sloping surface.
A Reliable Surface-Area Audit
- Sketch or mentally separate the solid into parts.
- Mark every join between parts.
- Cross out faces hidden inside joins.
- Add new internal faces created by holes or cut-outs.
- Check whether the object is open or closed.
- Use slant height where cone curved surface area requires it.
- Attach square units only after the surface calculation is complete.
A Reliable Volume Audit
- Add volumes of attached solids when they do not overlap.
- Subtract removed holes, cavities or cut-outs.
- Use cross-sectional area × length for prisms.
- Use equal volumes when a problem explicitly states recasting with no loss.
- Convert cubic units with the cube of the linear conversion factor.
Common Errors and Their First Repair
Adding total surface areas of joined solids: remove the two hidden faces at each join.
Subtracting a hole’s volume but ignoring its inner wall: add the new exposed internal surface.
Using cm-to-m conversion directly on cm² or cm³: square or cube the factor.
Using vertical height for cone curved area: find slant height first.
Using surface area when a recasting problem conserves volume: identify what physical quantity is preserved.
Independent Practice
1. A cylinder radius 2 cm, height 8 cm has a hemisphere radius 2 cm attached. Find total volume.
2. For the same solid, find external surface area if the bottom cylinder circle is exposed.
3. A 5-12-13 triangular prism has length 9 cm. Find volume and total surface area.
4. An open cylinder has radius 3 cm and height 7 cm. Find capacity and material surface area.
5. A solid cylinder radius 6 cm, height 10 cm has a central hole radius 2 cm drilled through it. Find remaining volume.
6. Convert 72,000 cm² to m².
7. Convert 0.085 m³ to cm³.
8. A sphere radius 2 cm is recast as a cylinder radius 1 cm. Find the cylinder height.
9. A solid is enlarged by scale factor 3. Its original surface area is 20 cm² and volume 14 cm³. Find the new values.
10. A cone has radius 5 cm and vertical height 12 cm. Find slant height and total surface area.
11. A cuboid 10×7×4 cm has a 2×3×4 cm notch removed. Find remaining volume.
12. Explain why a circular face at the join between a cylinder and a hemisphere is excluded from external surface area.
Explained Answers
1. Cylinder=32π; hemisphere=16π/3. Total=112π/3 cm³≈117 cm³.
2. Cylinder curved=32π; bottom=4π; hemisphere curved=8π. Total=44π cm²≈138 cm².
3. Cross-section area=30 cm²; volume=270 cm³. Surface area=2(30)+(5+12+13)(9)=330 cm².
4. Volume=63π cm³. Area=curved 42π + base 9π=51π cm².
5. Volume=π(36−4)(10)=320π cm³.
6. 72,000/10,000=7.2 m².
7. 0.085×1,000,000=85,000 cm³.
8. Sphere volume=32π/3. Set πh=32π/3, so h=32/3 cm.
9. Surface factor=9, giving 180 cm². Volume factor=27, giving 378 cm³.
10. Slant height=13 cm. Total area=π(5)(13)+25π=90π cm².
11. 280−24=256 cm³.
12. The joining circle lies inside the composite solid, so it is not part of the external boundary.
How to Know the Mensuration Skill Has Transferred
Change which surfaces are exposed. Turn a closed container into an open one, drill a hole, attach a hemisphere, or ask for recast height instead of volume. A student who understands the structure will rebuild the surface or volume inventory instead of applying a memorised total-area formula blindly.
Continue the Secondary 3 Learning Route
Continue with Direct and Inverse Proportion, Constants and Model Choice, Pythagoras, 3D Trigonometry, Elevation and Depression, and Coordinate Geometry, Distance, Gradient, Line Equations and Intersections.
A composite-solid answer is secure when volume, surface inventory, hidden joins and dimensional units all describe the same physical object. Return to the Secondary Mathematics Hub.
