Grade 10 essential mathematics Measurements and Geometry – Volume and Capacity Notes
Essential Mathematics — 2.0 Measurements & Geometry
2.7 Volume and Capacity (age 15, Kenya)
Specific learning outcomes
- (a) Identify and outline sub-sub-strands: Volume and capacity (cone, pyramid, frustum) and applications.
- (b) Determine volume and capacity of cones and pyramids.
- (c) Compute volume and capacity of a frustum of a cone.
- (d) Calculate volume and capacity of a frustum of a pyramid.
- (e) Apply volumes and capacities in real-life situations (e.g., water tanks, roofs, silos).
- (f) Appreciate the importance of volumes and capacities in everyday life.
Key ideas — definitions
- Volume is the amount of space inside a 3D object (measured in m³, cm³).
- Capacity is how much a container can hold (often measured in litres). Conversion: 1 m³ = 1000 L, 1 L = 1000 cm³.
- General formulas:
- Volume of a cone: V = (1/3) · π · r² · h
- Volume of a pyramid: V = (1/3) · BaseArea · h
- Frustum of a cone (cones with top removed): V = (1/3)·π·h·(R² + Rr + r²)
- Frustum of a pyramid: V = (h/3)·(A1 + A2 + √(A1·A2)), where A1 and A2 are areas of the two parallel bases
Cone
Pyramid (square)
Frustum (cone)
Worked examples
Example 1 — Cone
Find the volume and capacity (in litres) of a right cone with radius r = 5 cm and height h = 12 cm.
V = (1/3)πr²h = (1/3)π(5²)(12) = (1/3)π(25)(12) = 100π cm³ ≈ 314.16 cm³.
Capacity in litres: 314.16 cm³ = 0.314 L (since 1000 cm³ = 1 L).
Example 2 — Pyramid (square base)
A square pyramid has base side 6 m and height 4 m. Find its volume and capacity (in litres).
Base area = 6×6 = 36 m². V = (1/3)·36·4 = 48 m³. Capacity = 48 m³ = 48×1000 = 48 000 L.
Example 3 — Frustum of a cone
A water filter has the shape of a frustum. Top radius r = 3 cm, bottom radius R = 6 cm and height h = 8 cm. Find its volume.
V = (1/3)πh(R² + Rr + r²) = (1/3)π·8(36 + 18 + 9) = (8/3)π·63 = 168π cm³ ≈ 527.79 cm³ ≈ 0.528 L.
Example 4 — Frustum of a pyramid (square)
A square pyramid had base side 10 m. The top is cut off leaving a frustum whose top base is a square of side 6 m and whose vertical height is 4 m. Volume?
A1 = 10² = 100 m², A2 = 6² = 36 m². V = (h/3)(A1 + A2 + √(A1·A2)) = (4/3)(100 + 36 + 60) = (4/3)·196 = 784/3 ≈ 261.33 m³ = 261 330 L.
Notes & tips
- Height (h) is the perpendicular distance between the two bases (not the slant height). Use slant height only for surface area, not volume.
- In many real-life problems you will be given diameters — convert to radius: r = d/2.
- Always check units before converting to litres. Work in m³ for large containers (then multiply by 1000 to get litres).
- Derivations: the frustum formulas come from subtracting similar solid volumes (full cone minus small cone) or using area-sum formula for pyramids.
Real-life applications (Kenyan context)
- Design and capacity calculations for water tanks (often cylindrical or frustum-shaped), jerrycans and household containers.
- Estimating volume of conical roofs (kiosks, traditional granaries) and pyramidal monuments or roof segments.
- Calculating storage space in silos or hoppers (frustums) used in agriculture and dairy.
- Use knowledge when buying materials (soil, sand) by volume and converting to capacity units for transport.
Suggested learning experiences
- Hands-on: Build paper models of cones, pyramids and frustums. Measure dimensions and calculate volumes; compare with water/pebbles to test capacity.
- Field visit: Measure a household water tank (or a kiosk roof). Record dimensions, compute capacity in litres and discuss practical usage.
- Group task: Given plans for a small pyramid monument, compute the volume of stone needed and estimate cost (use local material rates).
- Class experiment: Fill an object whose volume you calculated with water using a measuring jug to check accuracy (converts geometry to real capacity).
- Problem solving: Mix algebraic problems (express volumes in terms of variable r or scale factor) and apply similarity ideas to frustums.
Practice exercises
- A cone has height 15 cm and base diameter 10 cm. Find its volume in cm³ and litres. (Answer: V = (1/3)π·(5²)·15 = 125π ≈ 392.7 cm³ ≈ 0.393 L.)
- A square pyramid has base side 8 m and height 5 m. Find volume in m³ and litres. (Answer: V = (1/3)·64·5 = 106.67 m³ = 106 667 L.)
- A frustum of a cone has top diameter 6 cm, bottom diameter 14 cm and height 10 cm. Find volume. (Hint: convert to radii r=3, R=7.)
- A truncated (frustum) pyramid has top base area 25 m², bottom base area 100 m² and height 6 m. Find the volume.
Solutions for 3 and 4 are left for students; encourage showing all steps and unit conversions.
Appreciation (outcome f)
Understanding volume and capacity helps learners solve everyday problems (how much water a tank holds, how much material is needed for construction, how to pack goods). Recognising units (m³, L, cm³) and converting between them links classroom geometry to real life and local industries.
Reminders: Always label units. Check whether height is perpendicular. Use similar solids to derive frustum formulas.