Grade 10 essential mathematics Measurements and Geometry – Surface Area of Solids Notes
2.6 Surface Area of Solids
Topic: 2.0 Measurements and Geometry — Subject: Essential Mathematics (Target age: 15, Kenyan context)
Specific Learning Outcomes
- Identify sub‑sub‑strands: surface area of cone, pyramid (square/rectangular base), sphere, frustum, and applications of area of solids.
- Determine the surface area of cones and pyramids from their nets.
- Calculate the surface area of a sphere and a hemisphere.
- Determine the surface area of frustums of cones and pyramids.
- Apply surface area of cones, pyramids and frustums in real life (e.g., roofing, paint, material estimation in Kenya).
- Appreciate applications and importance of surface area in everyday life.
Key formulas (use π ≈ 3.142 or π)
- Cone: Curved surface area (CSA) = π r l, Total surface area (TSA) = π r l + π r² (r = base radius, l = slant height).
- Square/Rectangular pyramid (with equal slant heights on each triangular face): Lateral area = sum of areas of triangular faces. For square base side a and slant height s: Lateral = 2 a s. TSA = base area + lateral area = a² + 2 a s.
- Sphere: TSA = 4 π r².
- Hemisphere: Curved surface = 2 π r²; Total including flat base = 3 π r².
- Frustum of a cone: Lateral area = π (R + r) l where R and r are radii of the two circular ends and l is slant height. TSA = π (R + r) l + π (R² + r²).
- Frustum of a pyramid (square bases): lateral area = sum of 4 trapezoids. If top side a1, bottom side a2 and slant height of each triangular face is s, then Lateral = 2 (a1 + a2) s (since 4 × 0.5 (a1 + a2) s).
Understanding nets (how formulas come from nets)
The curved surface equals the area of the sector with radius = slant height l and arc length = circumference 2πr. That gives CSA = 1/2 × l × (2πr) = π r l.
Each triangular face is ½ × base side × slant height. Sum the four triangles for total lateral area, then add base area.
Worked examples
Given r = 3 cm, slant height l = 5 cm. Find CSA and TSA.
CSA = π r l = π × 3 × 5 = 15π ≈ 47.12 cm².
Base area = π r² = π × 9 = 9π ≈ 28.27 cm².
TSA = 15π + 9π = 24π ≈ 75.40 cm².
Square base side a = 6 cm. Slant height s (triangle height from mid side to apex) = 5 cm. Find TSA.
Area of one triangular face = ½ × base × slant height = 0.5 × 6 × 5 = 15 cm².
Lateral area (4 faces) = 4 × 15 = 60 cm². Base area = 6² = 36 cm².
TSA = 60 + 36 = 96 cm².
Sphere radius r = 7 cm.
Sphere TSA = 4π r² = 4π × 49 = 196π ≈ 615.75 cm².
Hemisphere (r = 7 cm): curved surface = 2π r² = 98π ≈ 307.87 cm². Including flat base (π r²) gives total = 3π r² = 147π ≈ 461.81 cm².
Top radius r = 3 cm, bottom radius R = 6 cm, slant height l = 5 cm.
Lateral area = π (R + r) l = π × (6 + 3) × 5 = 45π ≈ 141.37 cm².
Areas of top and bottom = π (R² + r²) = π (36 + 9) = 45π ≈ 141.37 cm².
TSA = 45π + 45π = 90π ≈ 282.74 cm².
Top square side a1 = 4 cm, bottom side a2 = 8 cm, slant height of each triangular face s = 6 cm.
One face is a trapezoid of area 0.5 × (a1 + a2) × s = 0.5 × (4 + 8) × 6 = 36 cm².
Lateral area (4 faces) = 4 × 36 = 144 cm². Base areas = 4² + 8² = 16 + 64 = 80 cm².
TSA = 144 + 80 = 224 cm².
Applications in Kenyan real-life contexts
- Estimating paint for a conical thatched roof on many traditional houses: use curved surface area of cone to find area to be covered.
- Estimating fabric or material for conical or frustum lampshades, jikos (stove covers) or water-scoop covers.
- Calculating roofing material for truncated cone roofs (frustums) found on water tanks or grain stores.
- Packaging: determining amount of cardboard to make a conical or pyramid-shaped package or lid.
- Sports equipment: calculating surface area for balls (sphere) used in school sports when coating or branding is needed.
Example (practical): A rural community plans to paint the conical thatched roof of a small grain store. Roof base radius 2 m, slant height 3 m. Paint is sold in litres that cover 10 m² per litre. CSA = π r l = π × 2 × 3 = 6π ≈ 18.85 m² → need about 1.9 litres → buy 2 litres of paint (allowing small waste).
Class activities and suggested learning experiences (age 15, Kenyan classrooms)
- Hands‑on: Cut and fold paper nets of cones and square pyramids. Measure r, l and compute areas, then compare to measured surface using grid paper.
- Field task: Survey common roofs, tanks, jikos around school; estimate material needed to cover them using surface area formulas; present findings.
- Group problem solving: Given nets, ask learners to derive formula for cone CSA from the area of the sector (link net to π r l).
- Problem set: Realistic word problems with Kenyan context (paint for roof, cloth for conical hat, lining for hemisphere dome, material for frustum-shaped lamp shades).
Practice questions
- A cone has base radius 4 cm and slant height 6 cm. Find CSA and TSA. (Give answers in terms of π and to 3 s.f.)
- A square pyramid has base side 10 cm and slant height 13 cm. Find the total surface area.
- A sphere of radius 5 m is to be coated with waterproofing. How many square metres must be covered?
- A frustum of a cone has top radius 2 m, bottom radius 5 m and slant height 4 m. Find the total surface area.
- A lampshade is a frustum of a pyramid. Top side length 10 cm, bottom 20 cm, slant height of each face 12 cm. Find material area needed for lateral surface (ignore top/bottom).
Answers (click to reveal)
- CSA = π r l = π × 4 × 6 = 24π cm² ≈ 75.40 cm². TSA = 24π + π(4²) = 24π + 16π = 40π ≈ 125.66 cm².
- Lateral area = 4 × (½ × 10 × 13) = 4 × 65 = 260 cm². Base = 10² = 100 cm². TSA = 360 cm².
- Sphere area = 4π r² = 4π × 25 = 100π ≈ 314.16 m².
- Lateral = π (5 + 2) × 4 = 28π ≈ 87.96 m². Areas of ends = π(25 + 4) = 29π ≈ 91.11 m². TSA = 57π ≈ 179.07 m².
- Area of one trapezoid = 0.5 × (10 + 20) × 12 = 180 cm². Four faces → lateral = 4 × 180 = 720 cm².
Assessment pointers and appreciation
- Check that students can identify parts (radius, slant height, heights, top & bottom radii for frustums) on diagrams and nets.
- Expect stepwise working: identify formula, substitute values with units, compute and give final answer with appropriate unit and rounding.
- Discuss why surface area matters: cost, material conservation, safety and design — encourage learners to link formulas to concrete tasks in their communities.