Grade 10 core mathematics Measurements and Geometry – Surface Area and Volume of Solids Notes
Core Mathematics — 2.0 Measurements & Geometry
2.7 Surface Area and Volume of Solids (Age: 15, Kenyan context)
- (a) Identify and outline the sub-sub-strands: Surface Area of Solids and Volume of Solids.
- (b) Determine the surface area of prisms, pyramids, cones, frustums and spheres.
- (c) Calculate the surface area of composite solids.
- (d) Calculate the volume of prisms, pyramids, cones, frustums and spheres.
- (e) Determine the volume of composite solids.
- (f) Explore the use of surface area and volume of solids in real-life situations (e.g., water tanks, silos, packages).
Overview
Surface area (SA) of a solid is the total area of all outer faces. Volume (V) is the space it occupies. Units: use square units for area (cm², m²) and cubic units for volume (cm³, m³). In Kenya use the metric system (cm, m, litre ↔ 1 L = 1 000 cm³). Use π ≈ 3.142 or 22/7 when needed.
Key formulas (with small visuals)
Prism (including cylinder as a circular prism)
Surface area: SA = Lateral area + 2 × (Base area)
Lateral area = (Perimeter of base) × height
Volume: V = (Area of base) × height
Pyramid
Surface area: SA = Base area + 1/2 × (Perimeter of base) × slant height (l)
Volume: V = (1/3) × (Area of base) × height (h)
Cone
Surface area: SA = πr² + πr l = πr(r + l) (l = slant height)
Volume: V = (1/3)πr²h
Frustum (truncated cone or pyramid)
Surface area: SA = π(R² + r²) + π(R + r)l (R = top radius, r = bottom radius, l = slant height)
Volume: V = (1/3)πh (R² + Rr + r²)
Sphere
Surface area: SA = 4πr²
Volume: V = (4/3)πr³
Worked examples
1) Rectangular prism (box)
A box measures 50 cm by 30 cm by 20 cm. Find surface area and volume.
Base areas: 50×30 = 1500 cm², 50×20 = 1000 cm², 30×20 = 600 cm².
SA = 2(1500 + 1000 + 600) = 2(3100) = 6200 cm².
V = 50 × 30 × 20 = 30 000 cm³ = 30 L.
2) Cone
A conical roof has radius r = 1.2 m and vertical height h = 1.6 m.
Slant height l = sqrt(r² + h²) = sqrt(1.44 + 2.56) = sqrt(4.00) = 2.00 m.
SA = πr(r + l) = π × 1.2 × (1.2 + 2.0) = π × 1.2 × 3.2 = 3.84π ≈ 12.06 m².
V = (1/3)πr²h = (1/3)π × (1.44) × 1.6 = 0.768π ≈ 2.41 m³.
3) Frustum of a cone (water trough)
A trough is shaped like a frustum with top radius R = 0.5 m, bottom radius r = 0.2 m, vertical height h = 0.6 m.
Slant height l = sqrt((R − r)² + h²) = sqrt(0.3² + 0.6²) = sqrt(0.09 + 0.36) = sqrt(0.45) ≈ 0.671 m.
Volume: V = (1/3)πh(R² + Rr + r²) = (1/3)π×0.6(0.25 + 0.10 + 0.04)
= 0.2π × 0.39 = 0.078π ≈ 0.245 m³ ≈ 245 L.
Surface area (excluding circular ends if open): lateral SA = π(R + r)l = π(0.5 + 0.2)×0.671 = 0.7×0.671π ≈ 0.4697π ≈ 1.475 m².
4) Sphere (ball)
A metal ball radius r = 10 cm. Find SA and V.
SA = 4πr² = 4π × 100 = 400π ≈ 1256 cm².
V = (4/3)πr³ = (4/3)π × 1000 ≈ 4189 cm³.
Composite solids: example & Kenyan context
Example — Cylindrical water tank with conical roof (common on rural farms)
Cylinder: radius r = 0.9 m, height h_cyl = 1.8 m. Cone roof: same base radius r, cone height h_cone = 0.6 m.
Cylinder SA (side only) = 2πr h_cyl = 2π × 0.9 × 1.8 = 3.24π ≈ 10.18 m².
Cone SA (lateral) = π r l; l = sqrt(r² + h_cone²) = sqrt(0.81 + 0.36) = sqrt(1.17) ≈ 1.082 m.
Cone lateral = π × 0.9 × 1.082 ≈ 0.9738π ≈ 3.06 m².
If top base of cylinder is closed by cone (no circular top area exposed), total external SA (excluding ground base) ≈ cylinder side + cone lateral + cylinder base
= 10.18 + 3.06 + (π × 0.9² ≈ 2.54) ≈ 15.78 m².
Volume: cylinder V_cyl = πr²h_cyl = π×0.81×1.8 = 1.458π ≈ 4.58 m³; cone V = (1/3)πr²h_cone = 0.486π ≈ 1.53 m³.
Total capacity ≈ 6.11 m³ ≈ 6110 L.
Application: estimate paint needed for exterior (use SA) and water capacity (use V).
Notes on composite solids:
- To find SA of a composite solid, sum SA of parts; be careful to exclude internal faces where parts join.
- To find V of a composite solid, add volumes of parts; subtract volumes removed (e.g., a hole).
- Always keep units consistent.
Practice questions (with answers)
- Find SA and V of a right circular cylinder radius 0.6 m and height 2.0 m. (Use π = 3.142)
- A pyramid has square base 4 m by 4 m and vertical height 3 m. Find its volume and total surface area if slant height is 3.5 m.
- A solid is made by joining a hemisphere (r = 0.5 m) to the top of a right circular cylinder (same radius, height 1 m). Find total volume and external surface area (ignore base on the bottom if placed on ground).
- A frustum with radii 0.4 m and 0.2 m and height 0.5 m: calculate its volume.
Answers (click to show)
1) Cylinder: lateral area = 2πrh = 2π×0.6×2 = 2.4π ≈ 7.54 m². Top+bottom = 2πr² = 2π×0.36 = 0.72π ≈ 2.26 m². SA ≈ 9.8 m². V = πr²h = π×0.36×2 = 0.72π ≈ 2.26 m³.
2) Pyramid: Base area = 16 m². Volume = (1/3)×16×3 = 16 m³. Surface area = base + 1/2 × perimeter × slant height. Perimeter = 16 m, so lateral = 0.5×16×3.5 = 28 m². Total SA = 16 + 28 = 44 m².
3) Hemisphere + cylinder: V_cyl = πr²h = π×0.25×1 = 0.25π ≈ 0.785 m³. V_hemi = (1/2)×(4/3)πr³ = (2/3)π×0.125 = 0.08333π ≈ 0.262 m³. Total V ≈ 1.047 m³. External SA: cylinder lateral = 2πrh = 2π×0.5×1 = π ≈ 3.142 m². Hemisphere outer area = 2πr² = 2π×0.25 = 0.5π ≈ 1.571 m². Bottom base (if included) = πr² = 0.25π ≈ 0.785 m². If placed on the ground exclude bottom => SA ≈ 3.142 + 1.571 = 4.713 m².
4) Frustum volume: V = (1/3)πh(R² + Rr + r²) = (1/3)π×0.5(0.16 + 0.08 + 0.04) = (1/6)π×0.28 ≈ 0.04667π ≈ 0.1466 m³.
2) Pyramid: Base area = 16 m². Volume = (1/3)×16×3 = 16 m³. Surface area = base + 1/2 × perimeter × slant height. Perimeter = 16 m, so lateral = 0.5×16×3.5 = 28 m². Total SA = 16 + 28 = 44 m².
3) Hemisphere + cylinder: V_cyl = πr²h = π×0.25×1 = 0.25π ≈ 0.785 m³. V_hemi = (1/2)×(4/3)πr³ = (2/3)π×0.125 = 0.08333π ≈ 0.262 m³. Total V ≈ 1.047 m³. External SA: cylinder lateral = 2πrh = 2π×0.5×1 = π ≈ 3.142 m². Hemisphere outer area = 2πr² = 2π×0.25 = 0.5π ≈ 1.571 m². Bottom base (if included) = πr² = 0.25π ≈ 0.785 m². If placed on the ground exclude bottom => SA ≈ 3.142 + 1.571 = 4.713 m².
4) Frustum volume: V = (1/3)πh(R² + Rr + r²) = (1/3)π×0.5(0.16 + 0.08 + 0.04) = (1/6)π×0.28 ≈ 0.04667π ≈ 0.1466 m³.
Suggested learning experiences (classroom & practical)
- Begin with a discussion: show real objects (box, water drum, cone roof, ball) and ask learners to name shapes and what to measure.
- Group activity: measure a classroom box (length, width, height), compute SA for wrapping paper and volume for storage.
- Field task: visit a farm or water vendor. Measure a cylindrical water tank and estimate water capacity in litres; compare with vendor label if available.
- Practical: build a paper model of a cone and frustum to measure slant height and check SA formula by calculation and direct measurement.
- Problem solving: give composite solids (e.g., a cylinder with a hemisphere on top) — learners must decide which faces to include/exclude for SA and add volumes for capacity.
- Use calculators for square roots and π; practice approximating answers with π = 22/7 or 3.142 and discuss rounding and significance for real-life decisions (e.g., paint purchase).
- Assessment idea: short test with one shape SA, one V, one composite, and a word problem about paint or water.
- Draw a clear diagram, label r, h, l, R before substituting values.
- Watch for which surfaces are exposed — do not count faces that join two parts.
- Keep units consistent and convert when needed (cm ↔ m). For answers include units (m² for area, m³ or litres for volume).
- Use real examples from Kenya: water tanks, grain silos, milk cans, roof cones, boxes for packaging — relate calculations to cost/quantity (paint, water, storage).