For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.
How this chapter applies
Eclat core: areas, perimeters, volumes, surface areas, composite figures and solids, circle arcs, sectors, segments, and radian conversion form the main route.
School-sensitive extension: frustums, derivations, and unfamiliar composite solids may be school-sensitive depth.
2027 national comparison: K310 Topic G5 includes plane and solid mensuration, unit conversion, composite solids, arc length, sector and segment area, and radian measure.
Check your school: confirm whether formulas are supplied, the expected treatment of frustums, and exact-answer conventions.
Q: What does IP EMaths Notes (Upper Sec, Year 3-4): 11) Mensuration and 3D Geometry cover? A: Surface area, volume, and composite shape strategies with consistent unit handling.
The core idea is simple: Draw the solid, label every length, and keep area and volume units separate.
Use it as a working check: Surface area counts exposed faces; volume counts space filled. For composite solids, split the shape into familiar parts, calculate each part, then add or subtract.
Then go one layer deeper: Use the cylinder, hemisphere, tank, and cuboid examples to practise formula selection, unit handling, and whether the question asks for metal area, capacity, or a hidden diagonal.
Build a formula bank for quick recall and keep units consistent. Draw nets for surface-area questions and label every edge before calculating.
K310 includes arc length, sector area, segment area, and radian measure. Convert degrees to radians before using the compact radian formulas. Arc length equals radius multiplied by the angle in radians. Sector area equals half the radius squared multiplied by the angle in radians.
Worked check: a sector has radius 6 cm and angle 120 degrees, or two-thirds pi radians. Its arc length is 4 pi cm, and its sector area is 12 pi square centimetres.
For a segment, subtract the triangle area from the sector area. Do not stop at the sector when the question asks for the region between the arc and chord.
Area of circle: A=πr2, circumference: C=2πr.
Curved surface area of a cylinder: 2πrh; total surface adds two bases 2πr2.
Volume of a prism: base area × perpendicular height.
Volume of a cone: V=31πr2h.
Mensuration route checkpoint
Before using a formula, decide what the question is actually asking you to count. The same solid can produce different answers depending on whether the task is about exposed material, capacity, or a hidden length.
Question wording
Count this first
Usual formula route
Common trap
"Metal required", "painted area", or "surface area"
Exposed faces only.
Draw a net and add the relevant face areas.
Adding a hidden joined face or a missing lid.
"Capacity", "space inside", or "volume"
Space filled or enclosed.
Split into familiar solids, then add or subtract volumes.
Using surface-area formulae because the object is drawn in 3D.
"Open cylinder", "container", or "tank"
Curved surface plus the base, unless a lid is stated.
Use 2πrh+πr2 for one open end.
Adding both circular ends automatically.
"Longest rod", "internal diagonal", or "corner to corner"
A hidden right triangle in 3D.
Use 3D Pythagoras, d=l2+w2+h2
Common trap: surface area has square units and volume has cubic units. If the unit in the final answer does not match the question wording, the route choice is probably wrong.
Similar solids scale-factor checkpoint
When two solids are similar, do not use the same multiplier for every quantity. Decide whether the question is asking for a length, an area, or a volume before applying the scale factor.
Quantity compared
Multiplier to use
What it means
Common trap
Length, radius, height, or diagonal
k
Every corresponding length is multiplied by the linear scale factor.
Scaling one length but forgetting the matching radius or height.
Surface area or exposed material
k2
Area is built from two length dimensions.
Using k for painted area or metal required.
Volume or capacity
k3
Volume is built from three length dimensions.
Using k2 because the diagram looks like a surface-area problem.
Worked check: two similar containers have corresponding heights in the ratio 2:3. The linear scale factor from the smaller container to the larger container is k=23. If the smaller container has capacity 320cm3, the larger capacity is
320cm3×(23)3=1080cm3.
Misconception check: a scale factor of 23 does not mean the volume becomes 23 times as large. Volume scales with the cube of the linear scale factor.
Worked example - Composite solid volume
A solid consists of a cylinder of radius 3.0cm and height 8.0cm with a hemisphere of the same radius attached on top. Find the volume and give the answer to 3 significant figures.
Base area (one circular base): πr2=π(2.5)2=6.25π≈19.6m2.
Total metal area: 20π+6.25π=26.25π≈82.5m2.
Approximately 82.5m2 of metal is needed.
Worked example - Space diagonal of a cuboid
A storage crate is a cuboid measuring 1.2m by 0.9m by 0.5m. Find the length of the longest rod that fits inside it.
Use the 3D Pythagorean relation: d=l2+w2+h2.
Substitute: d=1.22+0.92+0.52
Compute squares: 1.44+0.81+0.25=2.50.
Square root: d=2.50≈1.58m.
The longest rod that fits is about 1.58m.
Frustum setup checkpoint
For a frustum question, rebuild the original cone before subtracting the small cone. Similar triangles usually give the missing radius or height.
Step
What to find
Why it matters
1. Draw the full cone and the removed small cone
Full height, cut-off height, base radius, and top radius
The frustum is not a standard formula question unless the formula is given.
2. Use similar triangles
Match radius to height with the same scale factor
Cone radius grows in proportion to distance from the vertex.
3. Calculate two cone volumes
Full cone volume and removed cone volume
The frustum volume is the difference between them.
4. Check units
Cubic units such as cm3 or m3
Surface-area units would mean the wrong route was used.
Worked check: a cone has height 15cm and base radius 6cm. If the top 4cm is cut off from the vertex, the small cone has radius r where 6r=154, so r=1.6cm. The frustum volume is then the full cone volume minus the small cone volume.
Misconception check: do not subtract the heights first and then use the original base radius for both cones. The removed cone is similar to the original cone, so its radius is smaller too.
Practice Quiz
Refresh key formulas for cylinders, spheres, cones, and composite solids before tackling exam-style prompts.
Try this
A frustum is formed by cutting the top 4cm of a right cone whose base radius is 6cm and height 15cm. Find the volume of the frustum, giving your answer in cm3.
.
Using only the base diagonal and forgetting the height.
