GRADE 8 Mathematics GEOMETRY – Common Solids Notes
GEOMETRY: Common Solids
Subject: Mathematics — Age: 13 (Kenya). Short notes on common 3‑D solids, their parts, surface area and volume formulas, simple diagrams and worked examples.
What is a solid (3‑D shape)?
A solid is a figure that has length, width and height. Unlike flat (2‑D) shapes, solids have volume — they take up space.
Common solids and quick facts
Cube
- All sides equal (side = a).
- Faces: 6, Edges: 12, Vertices (corners): 8.
Surface area: SA = 6a²
Volume: V = a³
Volume: V = a³
Cube net (6 equal squares)
Cuboid (rectangular box)
- Sides length l, width w, height h.
- Faces: 6, Edges: 12, Vertices: 8.
Surface area: SA = 2(lw + lh + wh)
Volume: V = l × w × h
Volume: V = l × w × h
Cuboid (front view)
Sphere
- All points same distance r from centre.
- No faces, edges or vertices (smooth surface).
Surface area: SA = 4πr²
Volume: V = (4/3)πr³
Volume: V = (4/3)πr³
Sphere (view)
Cylinder
- Circle base radius r, height h.
- Two circular faces + one curved face.
Surface area: SA = 2πr(h + r) (or 2πr² + 2πrh)
Volume: V = πr²h
Volume: V = πr²h
Cylinder
Cone
- Base radius r, height h, slant height l.
- One circular base and one curved face meeting at a point (vertex).
Surface area: SA = πr(r + l) where l = √(r² + h²)
Volume: V = (1/3)πr²h
Volume: V = (1/3)πr²h
Cone
Square Pyramid
- Square base side s, height h, slant height l.
- Faces: 5 (1 square base + 4 triangular sides), Edges: 8, Vertices: 5.
Surface area: SA = base area + lateral area = s² + 2s l (since 4 triangles each with area 1/2 × s × l)
Volume: V = (1/3) × base area × h = (1/3) s² h
Volume: V = (1/3) × base area × h = (1/3) s² h
Pyramid (side view)
Units & notes
- Volume units are cubic: cm³, m³, mm³. (1 m³ = 1,000,000 cm³)
- Surface area units are square: cm², m², mm².
- Use π ≈ 3.14 or π = 22/7 (choose one and stay consistent).
Worked examples
Example 1 — Cuboid
A box has length l = 8 cm, width w = 5 cm and height h = 3 cm. Find its surface area and volume.
A box has length l = 8 cm, width w = 5 cm and height h = 3 cm. Find its surface area and volume.
Surface area: SA = 2(lw + lh + wh)
= 2(8×5 + 8×3 + 5×3) = 2(40 + 24 + 15) = 2×79 = 158 cm².
Volume: V = lwh = 8×5×3 = 120 cm³.
= 2(8×5 + 8×3 + 5×3) = 2(40 + 24 + 15) = 2×79 = 158 cm².
Volume: V = lwh = 8×5×3 = 120 cm³.
Example 2 — Cylinder
A water tank is a cylinder with radius r = 0.5 m and height h = 2 m. Find its volume (use π = 3.14).
A water tank is a cylinder with radius r = 0.5 m and height h = 2 m. Find its volume (use π = 3.14).
V = π r² h = 3.14 × (0.5)² × 2 = 3.14 × 0.25 × 2 = 3.14 × 0.5 = 1.57 m³.
Example 3 — Cube
A dice has side a = 2 cm. Find its surface area and volume.
A dice has side a = 2 cm. Find its surface area and volume.
SA = 6a² = 6 × 2² = 6 × 4 = 24 cm².
V = a³ = 2³ = 8 cm³.
V = a³ = 2³ = 8 cm³.
Practice questions (try these)
- A cube has side 4 cm. Find SA and V.
- A cylindrical water bottle has radius 3 cm and height 12 cm. Find its volume (use π = 3.14).
- A square pyramid has base side s = 6 cm and height h = 4 cm. Find its volume.
- A sphere has radius 7 cm. Find its volume (use π = 22/7).
Answers (click to reveal)
1) SA = 6×4² = 96 cm²; V = 4³ = 64 cm³.
2) V = πr²h = 3.14×3²×12 = 3.14×9×12 = 3.14×108 = 339.12 cm³.
3) V = (1/3)s²h = (1/3)×36×4 = 48 cm³.
4) V = (4/3)πr³ = (4/3)×(22/7)×7³ = (4/3)×(22/7)×343 = (4×22×49)/3 = (4312)/3 ≈ 1437.33 cm³.
2) V = πr²h = 3.14×3²×12 = 3.14×9×12 = 3.14×108 = 339.12 cm³.
3) V = (1/3)s²h = (1/3)×36×4 = 48 cm³.
4) V = (4/3)πr³ = (4/3)×(22/7)×7³ = (4/3)×(22/7)×343 = (4×22×49)/3 = (4312)/3 ≈ 1437.33 cm³.
- Draw a picture, label all lengths, and write down which formula to use.
- Keep units consistent (convert cm ↔ m first if needed).
- For cones and pyramids, first find slant height l using l = √(h² + (half base)²) when needed.
Prepared for Kenyan syllabus — Age 13