LEARNING OBJECTIVES
What you will be able to do
- convert metric units of length, area, volume, capacity and mass
- calculate perimeter and area of common 2D shapes
- derive and use the area formula ab sin θ
- work with circles, arcs and sectors
- calculate surface areas and volumes of prisms, pyramids and curved solids
- use density, mass and volume relationships
- solve compound-shape and compound-solid problems
AT A GLANCE
INTRODUCTION · THE BIG IDEA
Measure length, area, capacity, surface area and volume with correct units and efficient decomposition.
Mensuration is the mathematics of measurement. A formula produces a number, while the unit tells us whether that number represents a length, area, volume, capacity or mass.
Start by identifying the shape, labelling its perpendicular dimensions and converting all measurements to compatible units. Compound figures become manageable when they are split into familiar, non-overlapping parts.
SECTION 01
Units, capacity and density
Length is measured in units such as mm, cm, m and km. Area uses squared units and volume uses cubed units. If a length conversion factor is k, the matching area factor is k² and the volume factor is k³.
Capacity measures how much a container can hold. The useful links are 1 ml = 1 cm³, 1 litre = 1000 cm³ and 1 m³ = 1000 litres.
Density compares the mass of a material with the volume it occupies. Rearrange the relationship carefully when mass or volume is required.
DETAILED EXPLANATION
- 1 kg = 1000 g and 1 tonne = 1000 kg.
- Convert every measurement to compatible units before substituting into a formula.
Finding the mass of a metal block
- The block has volume 240 cm³ and density 7.8 g/cm³.
- Use mass = density × volume.
- Mass = 7.8 × 240 = 1872 g.
- Convert grams to kilograms by dividing by 1000.
Answer: The mass is 1.872 kg.
SECTION 02
Perimeter and area of polygons
Perimeter is the distance around the outside boundary. Area is the amount of flat surface enclosed. Perimeter therefore uses linear units, while area uses square units.
For a triangle or parallelogram, h is the perpendicular height. It is not a sloping side unless that side is perpendicular to the chosen base. In a trapezium, a and b are the parallel sides.
A kite has two pairs of equal adjacent sides. Its diagonals are perpendicular, so its area is half the product of the diagonal lengths.
DETAILED EXPLANATION
- A general triangle perimeter is a + b + c. The shortcut b + 2s applies only to an isosceles triangle with equal sides s.
- A general trapezium perimeter is the sum of all four sides. The shortcut a + b + 2s applies only when the two non-parallel sides are both s.
- Kite perimeter = 2p + 2q when its two pairs of equal sides have lengths p and q.
Why a parallelogram has area ab sin θ
- Take side a as the base and let side b meet it at angle θ.
- Drop a perpendicular from the upper vertex to the base. This creates a right-angled triangle.
- In that triangle, sin θ = h/b, so h = b sin θ.
- Use parallelogram area = base × perpendicular height: A = a × h.
- Substitute h = b sin θ.
Answer: A = a(b sin θ) = ab sin θ.
SECTION 03
Circles, arcs and sectors
The radius r runs from the centre to the circle, while the diameter is 2r. A sector is the region between two radii and an arc.
A sector with central angle θ is θ/360 of a full circle. Use that same fraction of the full circumference for arc length and of the full area for sector area.
Keep π in the answer when an exact value is requested. For a decimal answer, keep the calculator value of π until the final rounding step.
Major sector with radius 9 cm and minor angle 80°
- Major angle = 360° − 80° = 280°.
- Major sector area = 280/360 × π × 9².
- Simplify 280/360 to 7/9.
Answer: Area = 63π cm², approximately 197.9 cm².
SECTION 04
Prisms, cuboids and cylinders
A prism has the same uniform cross-section along its full length. Its volume is the cross-sectional area A multiplied by length L. Its lateral area is the cross-section perimeter P multiplied by L.
A cylinder is a circular prism. Its curved surface unwraps into a rectangle whose sides are the circumference 2πr and height h.
Surface area totals only the exposed faces. A closed cylinder has two circular ends; an open cylinder may have one or none.
A triangular prism
- The triangular cross-section has base 8 cm and perpendicular height 5 cm.
- Cross-sectional area A = 1/2 × 8 × 5 = 20 cm².
- Its three side lengths are 8 cm, 6 cm and 7 cm, so P = 21 cm.
- The prism length is 12 cm.
- Volume = 20 × 12; total surface area = 21 × 12 + 2(20).
Answer: Volume = 240 cm³ and total surface area = 292 cm².
SECTION 05
Pyramids, cones, spheres and hemispheres
A pyramid or cone narrows from a base to an apex. Its volume is one third of base area times perpendicular height. Surface-area calculations use slant height for the triangular or curved faces.
A square-based pyramid with base side b and equal triangular-face slant height s has base area b² and four triangular faces with total area 2bs.
A hemisphere is half a sphere. Its curved area is half the sphere's surface area, but its total surface area also includes the flat circular base.
DETAILED EXPLANATION
- A hemisphere has curved area 2πr² and total area 3πr².
- For a general pyramid, calculate each triangular face if their slant heights are not equal.
Total surface area of a cone
- The radius is 5 cm and perpendicular height is 12 cm.
- Find the slant height using Pythagoras: l = √(5² + 12²) = 13 cm.
- Curved area = πrl = π × 5 × 13 = 65π cm².
- Base area = πr² = 25π cm².
- Add the exposed curved surface and circular base.
Answer: Total surface area = 90π cm², approximately 282.7 cm².
SECTION 06
Compound 2D shapes
Split a compound region into familiar rectangles, triangles, trapezia, circles or sectors. Add the pieces when they form the required region and subtract holes or cut-outs.
For perimeter, trace the outside boundary once. A cut-out may create new boundary edges, while a shared internal line is not part of the perimeter.
DETAILED EXPLANATION
- Use exact values involving π until the final step.
- Check that the pieces neither overlap nor leave an uncounted gap.
Rectangle with a semicircular end
- A rectangle is 14 cm long and 8 cm wide. A semicircle has diameter 8 cm, so r = 4 cm.
- Area = rectangle area + semicircle area.
- Area = 14 × 8 + 1/2π(4²) = 112 + 8π.
- Perimeter uses two 14 cm sides, one 8 cm side and the semicircular arc πr.
- Perimeter = 28 + 8 + 4π.
Answer: Area = 112 + 8π cm² and perimeter = 36 + 4π cm.
SECTION 07
Surface area of compound solids
When solids are joined, the contact surfaces become internal. Start with the exposed surfaces or subtract hidden areas before adding the new solid's visible surfaces.
A frustum can be handled as a large cone or pyramid with a smaller similar solid removed. Linear scale factor k becomes k² for surface area and k³ for volume.
Cylinder fixed to the top of a cuboid
- A 12 m by 21 m by 10 m cuboid has surface area 2(12 × 21 + 12 × 10 + 21 × 10) = 1164 m².
- A cylinder of radius 4 m and height 7 m covers a circle of area 16π m² on the cuboid, so subtract that hidden circle.
- The cylinder's curved area is 2πrh = 2π(4)(7) = 56π m².
- Its top circle has area 16π m², so add it.
- Total = 1164 − 16π + 56π + 16π.
Answer: The exposed surface area is 1164 + 56π m², approximately 1339.9 m².
QUICK CHAPTER SUMMARY
The ideas to carry forward
- Square or cube a linear conversion factor for area or volume.
- Density is mass divided by volume.
- Perimeter measures the boundary; area measures the enclosed surface.
- Parallelogram area ab sin θ follows from perpendicular height b sin θ.
- Arc and sector calculations use θ/360.
- A prism uses cross-sectional area × length.
- Pyramids and cones use one third of base area × perpendicular height.
- Surface area includes only exposed faces.
- Compound figures are solved by careful addition or subtraction.
QUICK REVISION CHECKLIST
Can you do each of these without your notes?
- convert metric units of length, area, volume, capacity and mass
- calculate perimeter and area of common 2D shapes
- derive and use the area formula ab sin θ
- work with circles, arcs and sectors
- calculate surface areas and volumes of prisms, pyramids and curved solids
- use density, mass and volume relationships
- solve compound-shape and compound-solid problems