Mensuration
Area, Surface Area and Volume of Solids
Overview
Mensuration is the branch of mathematics dealing with the measurement of geometric figures—their lengths, areas and volumes. For JKTET Paper II, this topic bridges arithmetic computation with spatial reasoning, testing whether candidates can apply formulas to real-world and exam-style problems involving two-dimensional shapes and three-dimensional solids.
This topic carries significant weight because questions are straightforward once formulas are memorised, yet careless errors in unit conversion or formula selection cause avoidable mark loss. Mastery requires knowing the standard formulas for plane figures (triangles, quadrilaterals, circles) and solids (cuboid, cube, cylinder, cone, sphere), understanding when to use lateral versus total surface area, and being comfortable converting between cm², m², cm³ and litres.
Expect 2–4 direct application questions in the mathematics section. Speed and accuracy here can boost your score reliably.
Key Concepts
- Area measures the extent of a two-dimensional surface; expressed in square units (cm², m²).
- Perimeter is the total length of the boundary of a plane figure; expressed in linear units (cm, m).
- Surface area of a solid is the total area of all its outer faces; for solids with a base and top, distinguish between lateral (curved) surface area (LSA/CSA) and total surface area (TSA).
- Volume measures the space enclosed by a three-dimensional object; expressed in cubic units (cm³, m³) or litres (1 litre = 1000 cm³).
- For composite figures, break them into standard shapes, compute individually, then add or subtract as required.
- Unit consistency is critical: convert all measurements to the same unit before substituting into formulas.
- The value of π is typically taken as 22/7 or 3.14 unless otherwise specified.
Formulas / Key Facts
Plane Figures (Area and Perimeter)
| Figure | Area | Perimeter |
|---|---|---|
| Rectangle | l × b | 2(l + b) |
| Square | a² | 4a |
| Triangle (general) | ½ × base × height | sum of three sides |
| Right triangle | ½ × base × height | a + b + hypotenuse |
| Equilateral triangle | (√3/4) × a² | 3a |
| Parallelogram | base × height | 2(a + b) |
| Rhombus | ½ × d₁ × d₂ | 4a |
| Trapezium | ½ × (sum of parallel sides) × height | sum of all sides |
| Circle | πr² | 2πr (circumference) |
| Semicircle | ½ πr² | πr + 2r |
Three-Dimensional Solids
| Solid | Volume | CSA / LSA | TSA |
|---|---|---|---|
| Cuboid | l × b × h | 2h(l + b) | 2(lb + bh + hl) |
| Cube | a³ | 4a² | 6a² |
| Cylinder | πr²h | 2πrh | 2πr(r + h) |
| Cone | ⅓ πr²h | πrl (l = slant height) | πr(r + l) |
| Sphere | (4/3)πr³ | 4πr² | 4πr² |
| Hemisphere | (2/3)πr³ | 2πr² | 3πr² |
Slant height of cone: l = √(r² + h²)
Diagonal of cuboid: √(l² + b² + h²)
Diagonal of cube: a√3
Worked Examples
Example 1 — Volume and Surface Area of a Cylinder
A cylindrical water tank has radius 7 m and height 10 m. Find its volume and total surface area. (Use π = 22/7)
Solution:
Volume = πr²h = (22/7) × 7² × 10 = (22/7) × 49 × 10 = 22 × 70 = 1540 m³
TSA = 2πr(r + h) = 2 × (22/7) × 7 × (7 + 10) = 2 × 22 × 17 = 748 m²
Example 2 — Cone Problem
A cone has base radius 6 cm and height 8 cm. Find its slant height, curved surface area and volume. (Use π = 3.14)
Solution:
Slant height l = √(r² + h²) = √(36 + 64) = √100 = 10 cm
CSA = πrl = 3.14 × 6 × 10 = 188.4 cm²
Volume = ⅓ πr²h = ⅓ × 3.14 × 36 × 8 = ⅓ × 904.32 = 301.44 cm³
Example 3 — Composite Solid
A solid is made by placing a hemisphere of radius 3 cm on top of a cylinder of the same radius and height 5 cm. Find the total surface area.
Solution:
The top circular face of the cylinder is covered by the hemisphere, so we exclude it.
TSA = CSA of cylinder + CSA of hemisphere + base of cylinder
CSA of cylinder = 2πrh = 2 × (22/7) × 3 × 5 = 660/7 cm²
CSA of hemisphere = 2πr² = 2 × (22/7) × 9 = 396/7 cm²
Base of cylinder = πr² = (22/7) × 9 = 198/7 cm²
Total = (660 + 396 + 198)/7 = 1254/7 ≈ 179.14 cm²
Common Mistakes
| Wrong Thinking | Correct Fix |
|---|---|
| Using diameter instead of radius in formulas | Always halve the diameter first; r = d/2 |
| Confusing CSA with TSA | CSA excludes bases; TSA includes all faces—read the question carefully |
| Forgetting to compute slant height for cones | Calculate l = √(r² + h²) before finding CSA |
| Mixing units (e.g., cm and m in the same problem) | Convert all measurements to a single unit before substitution |
| Using 2πr²h for cylinder volume | Correct formula is πr²h; avoid doubling unnecessarily |
| Adding areas when a shape sits on another (composite solids) | Subtract the common interface area that is no longer exposed |
Quick Reference
- Rectangle area: l × b; Square area: a²
- Circle area: πr²; Circumference: 2πr
- Cylinder volume: πr²h; TSA: 2πr(r + h)
- Cone volume: ⅓ πr²h; Slant height: √(r² + h²); CSA: πrl
- Sphere volume: (4/3)πr³; Surface area: 4πr²
- 1 litre = 1000 cm³ = 0.001 m³