Mensuration
Area, Surface Area and Volume of 2D and 3D Figures
Overview
Mensuration is the branch of mathematics dealing with measurement of geometric figures — their lengths, areas and volumes. For TS TET Paper II (classes 6-8), this topic carries significant weightage in the Mathematics section and tests your ability to apply formulas correctly to practical problems.
The topic divides naturally into two parts: 2D figures (plane figures where we calculate perimeter and area) and 3D figures (solid shapes where we calculate surface area and volume). Exam questions typically present word problems requiring you to identify the correct formula, substitute values carefully and compute the answer. Many questions involve unit conversions or combine multiple shapes.
Mastery requires memorising all standard formulas and understanding when each applies. The pedagogy aspect may ask about activity-based methods for teaching mensuration concepts to upper primary students.
Key Concepts
- Perimeter is the total length of the boundary of a 2D figure; area is the region enclosed within that boundary.
- Surface area of a 3D solid is the total area of all its outer faces — think of it as the amount of paper needed to wrap the object completely.
- Volume measures the space occupied by a 3D object — the capacity it can hold.
- Lateral (curved) surface area excludes the top and bottom faces; total surface area includes all faces.
- Units matter: area is always in square units (cm², m²), volume in cubic units (cm³, m³). Converting between units requires squaring or cubing the conversion factor.
- Composite figures are solved by breaking them into standard shapes, calculating separately, then adding or subtracting as needed.
- π (pi) is taken as 22/7 or 3.14 unless specified otherwise in the problem.
Formulas / Key Facts
2D Figures — Perimeter and Area
| Figure | Perimeter | Area |
|---|---|---|
| Rectangle | 2(l + b) | l × b |
| Square | 4a | a² |
| Triangle | a + b + c | ½ × base × height |
| Right Triangle | a + b + c | ½ × base × perpendicular |
| Equilateral Triangle | 3a | (√3/4) × a² |
| Parallelogram | 2(a + b) | base × height |
| Rhombus | 4a | ½ × d₁ × d₂ |
| Trapezium | sum of all sides | ½ × (a + b) × h |
| Circle | 2πr (circumference) | πr² |
| Semicircle | πr + 2r | ½ × πr² |
Here: l = length, b = breadth, a = side, r = radius, h = height, d₁ and d₂ = diagonals
3D Figures — Surface Area and Volume
| Solid | Lateral/Curved SA | Total SA | Volume |
|---|---|---|---|
| Cuboid | 2h(l + b) | 2(lb + bh + hl) | l × b × h |
| Cube | 4a² | 6a² | a³ |
| Cylinder | 2πrh | 2πr(r + h) | πr²h |
| Cone | πrl | πr(r + l) | ⅓ × πr²h |
| Sphere | — | 4πr² | ⁴⁄₃ × πr³ |
| Hemisphere | 2πr² | 3πr² | ⅔ × πr³ |
Here: l = slant height for cone, calculated as l = √(r² + h²)
Useful Conversions
- 1 m = 100 cm → 1 m² = 10,000 cm² → 1 m³ = 1,000,000 cm³
- 1 litre = 1000 cm³ = 0.001 m³
Worked Examples
Example 1: Area of a Trapezium
A field is in the shape of a trapezium with parallel sides 25 m and 15 m. The perpendicular distance between them is 8 m. Find its area.
Solution: Area of trapezium = ½ × (sum of parallel sides) × height = ½ × (25 + 15) × 8 = ½ × 40 × 8 = 160 m²
Example 2: Volume and Surface Area of a Cylinder
A cylindrical water tank has radius 1.4 m and height 2 m. Find its volume and total surface area. (Use π = 22/7)
Solution: Volume = πr²h = (22/7) × (1.4)² × 2 = (22/7) × 1.96 × 2 = (22/7) × 3.92 = 12.32 m³
Total SA = 2πr(r + h) = 2 × (22/7) × 1.4 × (1.4 + 2) = 2 × (22/7) × 1.4 × 3.4 = 2 × 22 × 0.2 × 3.4 = 29.92 m²
Example 3: Composite Figure
A rectangular sheet of paper 44 cm × 20 cm is rolled along its length to form a cylinder. Find the volume of the cylinder.
Solution: When rolled along length (44 cm), this becomes the circumference. Circumference = 2πr = 44 r = 44 × 7 / (2 × 22) = 7 cm
Height of cylinder = breadth of sheet = 20 cm
Volume = πr²h = (22/7) × 7 × 7 × 20 = 22 × 7 × 20 = 3080 cm³
Common Mistakes
- Confusing radius and diameter: Many problems give diameter; students forget to halve it before applying formulas. → Always check: is the given measurement radius or diameter?
- Mixing lateral and total surface area: Questions may ask specifically for curved surface area, but students add the base areas. → Read the question carefully; "curved" or "lateral" excludes bases.
- Ignoring unit conversions: If length is in metres and breadth in centimetres, computing directly gives wrong answers. → Convert all measurements to the same unit before calculation.
- Using wrong height in cone/cylinder problems: Students sometimes use slant height instead of vertical height for volume. → Volume formulas always use perpendicular height; slant height is only for lateral surface area of cone.
- Forgetting the fraction in volume formulas: Cone volume has ⅓, sphere has ⁴⁄₃, hemisphere has ⅔. → Memorise these fractions as part of the formula, not as an afterthought.
Quick Reference
- Rectangle area = l × b; Perimeter = 2(l + b)
- Circle area = πr²; Circumference = 2πr
- Cuboid volume = l × b × h; TSA = 2(lb + bh + hl)
- Cylinder volume = πr²h; CSA = 2πrh; TSA = 2πr(r + h)
- Cone volume = ⅓πr²h; Slant height l = √(r² + h²)
- Sphere volume = ⁴⁄₃πr³; SA = 4πr²