Secondary Mathematics · Worked Repair Guide 26
Mensuration becomes unreliable when a learner sees a three-dimensional object and immediately searches memory for a formula. The safer approach is structural: identify the solid, identify which measurement is requested, decide whether the object is complete or composite, and track units through every stage.
This guide develops one central habit: separate boundary from capacity. Surface area measures exposed two-dimensional faces. Volume measures three-dimensional space occupied. A hollow or composite object may require adding or subtracting volumes while its surface area requires a completely different boundary count.
All containers, materials and dimensions below are invented teaching examples. The Units, Scale and Measurement guide remains the prerequisite repair route for unit conversion and scale.
1. Volume counts three-dimensional space
A cuboid of length 8 cm, width 5 cm and height 3 cm has volume 8×5×3=120 cm³.
The cubic unit matters. Volume is not measured in cm or cm².
Entry check: doubling only the length of that cuboid doubles its volume to 240 cm³. Doubling all three dimensions multiplies volume by 2³=8.
This scaling distinction returns repeatedly in similar solids.
2. Surface area counts the exposed boundary
For a closed cuboid with dimensions 8,5,3, the three distinct face areas are 40,24 and15 cm². Each occurs twice.
Total surface area=2(40+24+15)=158 cm².
Do not multiply all three dimensions for surface area. That gives volume.
A useful check is dimensional: every surface-area term must have square units.
3. Prism volume is cross-sectional area times length
A prism has a constant cross-section along its length.
If the cross-sectional area is 18 cm² and the prism length is 7 cm, volume=18×7=126 cm³.
The cross-section can be triangular, trapezoidal or another fixed shape. Calculate its area first, then multiply by the perpendicular prism length.
Do not use a slanted edge as the prism length unless it is the direction in which the cross-section is repeated.
4. Prism surface area can be organised as two ends plus a wrap
For a prism, total surface area can often be organised as:
2×cross-sectional area + perimeter of cross-section × prism length.
This works because the lateral faces unwrap into rectangles whose combined width equals the cross-section perimeter.
A net is not merely a drawing aid; it shows exactly which faces belong to the boundary.
5. Cylinder volume uses circular cross-sectional area
A cylinder of radius r and perpendicular height h has volume V=πr²h.
For r=3 cm and h=10 cm, V=90π cm³.
If a diameter is given, halve it before substituting as r.
A cylinder with diameter 12 cm and height 5 cm has radius 6 cm and volume 180π cm³.
6. Cylinder surface area separates curved and circular parts
The curved surface unwraps into a rectangle of width equal to the circumference 2πr and height h. Curved area=2πrh.
A closed cylinder also has two circular ends, total area 2πr².
Thus total surface area=2πrh+2πr².
For r=3 and h=10, curved area=60π and total area=78π cm².
7. Cone volume contains the one-third factor
A right circular cone of radius r and perpendicular height h has volume V=(1/3)πr²h.
For r=3 cm and h=4 cm, volume=12π cm³.
The height in the volume formula is perpendicular to the base. It is not the slant height.
If a cone has the same base radius and perpendicular height as a cylinder, its volume is one third of that cylinder’s volume.
8. Cone surface area uses slant height
For a right cone with radius r and slant height l, curved surface area=πrl.
Total surface area including the circular base=πrl+πr².
If r=3 and perpendicular height=4, Pythagoras gives slant height l=5. Curved area=15π and total surface area=24π cm².
This is a classic example of two different “heights” doing different jobs in different formulas.
9. Sphere formulas use radius, not diameter
A sphere of radius r has volume (4/3)πr³ and surface area 4πr².
For r=3 cm, volume=36π cm³ and surface area=36π cm².
The numerical coefficient happens to match for r=3, but the units and meanings are different.
Always inspect the power of r and the requested unit before trusting a copied number.
10. A hemisphere has curved area plus a possible circular base
Half a sphere has volume (2/3)πr³. Its curved surface area is 2πr².
If the flat circular face is part of the exposed surface, add πr², giving total 3πr².
For r=3, volume=18π cm³, curved area=18π cm² and total area including the base=27π cm².
The phrase “surface area of a hemisphere” can therefore be ambiguous unless the question or diagram makes clear whether the circular base is exposed.
11. Composite volume usually adds or subtracts simple solids
Suppose a cylindrical hole of radius 2 cm is drilled completely through the axis of a cylinder radius 4 cm and height 10 cm.
Remaining volume=π(4²)(10)−π(2²)(10)=160π−40π=120π cm³.
This subtraction is valid because both cylinders occupy the same height and the smaller solid is removed from the larger.
Surface area would require more care because drilling exposes an additional inner curved surface.
12. Composite surface area is not found by blindly adding whole-solid areas
If two solids are joined, the contact surfaces become internal and should not be counted as exposed area.
For example, placing a hemisphere exactly on the circular top of a matching cylinder removes that circular interface from the exposed boundary.
The exposed area would include the cylinder’s curved surface, its bottom circle and the hemisphere’s curved surface.
A boundary-tracing habit is safer than adding two total surface-area formulas and subtracting later from memory.
13. Capacity connects cubic centimetres to millilitres
1 cm³=1 mL, and 1000 cm³=1 L.
An invented rectangular tank 50 cm by 40 cm by 30 cm has volume 60,000 cm³, so its full mathematical capacity is 60 L.
A real container may not be filled to the brim, and wall thickness can reduce internal capacity. School models normally use the stated internal dimensions unless told otherwise.
Do not convert cm² into litres; capacity comes from volume.
14. Similar solids scale with powers of the length factor
If corresponding lengths scale by factor k, corresponding surface areas scale by k² and volumes by k³.
For k=2.5, area factor=6.25 and volume factor=15.625.
If two similar solids have volume ratio 27:1, their length ratio is 3:1 because ∛27=3.
This is why a small increase in linear size can produce a much larger change in capacity.
15. Density adds a rate between mass and volume
Density ρ=mass/volume, so mass=density×volume.
If an invented material has density 2.7 g/cm³ and an object made from it has volume 50 cm³, mass=135 g.
The units verify the structure: (g/cm³)(cm³)=g.
Do not use external material densities unless supplied or sourced appropriately; this example’s 2.7 value is simply stipulated for the calculation.
16. Open containers change the surface-area count
An open-top cuboid of base 6 cm by 4 cm and height 3 cm has material area:
base 24 + two 6×3 sides 36 + two 4×3 sides 24 =84 cm².
A closed cuboid would add another 24 cm² top face, giving 108 cm².
Words such as open, hollow, capped, covered and removed directly affect which boundaries count.
17. Nets help distinguish edge length from face area
A net lays a solid’s faces flat without changing their dimensions.
For a cylinder, the curved surface becomes a rectangle whose width is 2πr, not πr².
For a prism, the side rectangles use the side lengths of the cross-section as their widths.
Use the net to check that every exposed face is included once and every internal interface is excluded.
18. A capsule-shaped solid combines a cylinder and a sphere
Imagine a cylinder of radius 2 cm and length 10 cm with a hemisphere attached at each end. The two hemispheres together form one sphere of radius 2.
Cylinder volume=π(2²)(10)=40π.
Sphere volume=(4/3)π(2³)=32π/3.
Total volume=40π+32π/3=152π/3 cm³.
The circular interfaces are internal, so they would not appear in the external surface-area calculation.
19. Capstone: reverse a formula and check feasibility
A cone has radius 5 cm and volume 100π cm³. Find its perpendicular height.
100π=(1/3)π(25)h. Cancel π and multiply by 3: 300=25h, so h=12 cm.
The slant height is then √(5²+12²)=13 cm.
Total surface area including base=π(5)(13)+25π=90π cm².
One set of dimensions now supports volume, Pythagoras and surface-area checks.
20. Independent practice
- Find the volume of a cuboid 8 cm by 5 cm by 3 cm.
- Find its total surface area.
- A prism has cross-sectional area 18 cm² and length 7 cm. Find its volume.
- Find the volume of a cylinder radius 3 cm, height 10 cm.
- Find the total surface area of that closed cylinder.
- Find only its curved surface area.
- A cone has radius 3 cm and perpendicular height 4 cm. Find its volume.
- Find the cone’s slant height and total surface area including its base.
- A sphere has radius 3 cm. Find its volume and surface area.
- A hemisphere has radius 3 cm. Find its volume and total surface area including its base.
- A cylinder radius 4 cm, height 10 cm has an axial cylindrical hole radius 2 cm drilled through it. Find remaining volume.
- A 10×8×6 cm cuboid has a 4×3×6 cm rectangular tunnel removed. Find remaining volume.
- Similar solids have length scale factor 2.5. Find area and volume scale factors.
- Similar solids have volume ratio 27:1. Find their positive length ratio.
- Find the volume of a cylinder diameter 12 cm and height 5 cm.
- A cone has radius 5 cm and volume 100π cm³. Find its height.
- A sphere has surface area 100π cm². Find its radius.
- An object has volume 50 cm³ and stipulated density 2.7 g/cm³. Find its mass.
- Find the material area of an open-top cuboid with base 6×4 cm and height 3 cm.
- Find the volume of a capsule consisting of a cylinder radius 2 cm, length 10 cm, capped by two hemispheres radius 2 cm.
21. Worked answers
1. 120 cm³.
2. 158 cm². 2(40+24+15).
3. 126 cm³.
4. 90π cm³.
5. 78π cm².
6. 60π cm².
7. 12π cm³.
8. Slant height 5 cm; total area 24π cm².
9. Volume 36π cm³; surface area 36π cm².
10. Volume 18π cm³; total area 27π cm².
11. 120π cm³. Subtract 40π from 160π.
12. 408 cm³. 480−72.
13. Area factor 6.25; volume factor 15.625.
14. 3:1.
15. 180π cm³. Radius is 6 cm.
16. 12 cm.
17. 5 cm. 4πr²=100π.
18. 135 g.
19. 84 cm².
20. 152π/3 cm³. Cylinder 40π plus sphere 32π/3.
22. Diagnose mensuration errors by the boundary being measured
Common failures include using diameter as radius, using slant height in a cone-volume formula, counting internal contact faces as exposed surface, confusing square and cubic units, or scaling volume by a length factor instead of its cube.
A useful repair note says “volume → space inside”, “surface area → exposed boundary”, “cone volume uses perpendicular height”, “curved cone area uses slant height”, or “similar volume factor=k³”.
Then redraw the object as a net or split it into named solids. A formula becomes safer when the learner can point to the physical region each term represents.
23. Continue through the BTT learning routes
Return to the BTT Mathematics Hub or BTT Mathematical Lab. Use Units, Scale and Measurement for unit control, Angles, Similarity and Geometric Reasoning for scale, and Trigonometry, Pythagoras, Bearings and Elevation when slant lengths need right-triangle calculations.
Within Batch 07, continue to Sampling, Scatter Plots, Correlation and Lines of Best Fit or Functions, Domain, Range and Graph Behaviour.
24. Sources and scope
The solids, dimensions, materials and practice questions are original teaching material. Standard prism, cylinder, cone and sphere formulas are used as mathematical definitions or standard mensuration results.
For the current Singapore Secondary curriculum doorway, see MOE: Curriculum for secondary schools. Match sphere, composite-solid and density extensions to the learner’s actual subject level and school programme.
Complete Secondary Mathematics repair route: browse all 48 worked repair guides or return to the Secondary Mathematics Learning Hub.