Mensuration
Area, Surface Area and Volume of 2D and 3D Figures
Overview
Mensuration is the branch of mathematics dealing with the measurement of geometric figures — their lengths, areas, and volumes. For GTET Paper-2, this topic carries significant weightage in the Mathematics section and tests your ability to apply formulas accurately to solve practical problems involving 2D shapes (plane figures) and 3D shapes (solid figures).
The syllabus expects you to calculate areas and perimeters of standard 2D figures like triangles, quadrilaterals, and circles, as well as surface areas and volumes of 3D solids such as cubes, cuboids, cylinders, cones, and spheres. Questions typically involve direct formula application, word problems requiring conversion of units, and composite figures combining multiple shapes.
Mastery of this topic requires memorising key formulas, understanding when to apply each, and being careful with unit conversions. Many exam questions test whether candidates can distinguish between surface area (paint needed to cover a solid) and volume (capacity or space occupied).
Key Concepts
- Area measures the region enclosed by a 2D figure, expressed in square units (cm², m²).
- Perimeter is the total length of the boundary of a 2D figure, expressed in linear units (cm, m).
- Surface Area of a 3D solid is the total area of all its outer faces — Curved Surface Area (CSA) excludes bases, while Total Surface Area (TSA) includes all surfaces.
- Volume measures the space occupied by a 3D solid, expressed in cubic units (cm³, m³, litres where 1 litre = 1000 cm³).
- Composite figures are shapes formed by combining two or more basic shapes — solve by adding or subtracting individual areas/volumes.
- Unit conversion is critical: 1 m = 100 cm, 1 m² = 10000 cm², 1 m³ = 1000000 cm³ = 1000 litres.
- Relationship between diameter and radius: d = 2r; always check which is given in the problem.
Formulas / Key Facts
2D Figures — Area and Perimeter
| Figure | Area | Perimeter |
|---|---|---|
| Rectangle | l × b | 2(l + b) |
| Square | a² | 4a |
| Triangle | ½ × base × height | Sum of all sides |
| Right Triangle | ½ × base × perpendicular | a + b + c |
| Equilateral Triangle | (√3/4) × a² | 3a |
| Parallelogram | base × height | 2(a + b) |
| Rhombus | ½ × d₁ × d₂ | 4a |
| Trapezium | ½ × (a + b) × h | Sum of all sides |
| Circle | πr² | 2πr (circumference) |
| Semicircle | ½ × πr² | πr + 2r |
Heron's Formula for triangle with sides a, b, c:
- s = (a + b + c)/2
- Area = √[s(s−a)(s−b)(s−c)]
3D Figures — Surface Area and Volume
| Solid | CSA | TSA | Volume |
|---|---|---|---|
| Cube (side a) | 4a² | 6a² | a³ |
| Cuboid (l, b, h) | 2h(l + b) | 2(lb + bh + hl) | l × b × h |
| Cylinder (r, h) | 2πrh | 2πr(r + h) | πr²h |
| Cone (r, h, l) | πrl | πr(r + l) | ⅓πr²h |
| Sphere (r) | 4πr² | 4πr² | (4/3)πr³ |
| Hemisphere (r) | 2πr² | 3πr² | (2/3)πr³ |
Note: For cone, slant height l = √(r² + h²)
Standard value: π = 22/7 or 3.14 (use as specified in question)
Worked Examples
Example 1: Area of Composite Figure
Problem: A rectangular garden is 20 m long and 15 m wide. A circular fountain of radius 3.5 m is built in the centre. Find the area of the remaining garden. (Use π = 22/7)
Solution:
- Area of rectangle = l × b = 20 × 15 = 300 m²
- Area of circle = πr² = (22/7) × 3.5 × 3.5 = (22/7) × 12.25 = 38.5 m²
- Remaining area = 300 − 38.5 = 261.5 m²
Example 2: Volume and Surface Area of Cylinder
Problem: A cylindrical 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) × 7 × 7 × 10 = 22 × 7 × 10 = 1540 m³
- TSA = 2πr(r + h) = 2 × (22/7) × 7 × (7 + 10) = 2 × 22 × 17 = 748 m²
Example 3: Heron's Formula
Problem: Find the area of a triangle with sides 13 cm, 14 cm, and 15 cm.
Solution:
- s = (13 + 14 + 15)/2 = 42/2 = 21 cm
- Area = √[s(s−a)(s−b)(s−c)] = √[21 × 8 × 7 × 6]
- = √[21 × 8 × 42] = √7056 = 84 cm²
Common Mistakes
- Confusing radius with diameter → Always identify whether the question gives radius or diameter; divide by 2 if diameter is given.
- Using wrong formula for CSA vs TSA → CSA is for lateral/curved surface only (e.g., label on a can); TSA includes top and bottom (e.g., painting entire box).
- Forgetting unit conversion → If length is in metres and another dimension in centimetres, convert to same unit before calculating.
- Applying area formula for volume problems → Area is 2D (square units), volume is 3D (cubic units); check what the question asks.
- Using πr² for circumference instead of 2πr → Remember: πr² is area of circle, 2πr is circumference.
- Ignoring slant height in cone problems → For CSA of cone, you need slant height (l), not vertical height (h); calculate l = √(r² + h²) if not given.
Quick Reference
- Rectangle area: l × b; Square area: a²; Circle area: πr²
- Cylinder volume: πr²h; Cone volume: ⅓πr²h; Sphere volume: (4/3)πr³
- 1 litre = 1000 cm³; 1 m³ = 1000 litres
- Slant height of cone: l = √(r² + h²)
- Heron's formula: Area = √[s(s−a)(s−b)(s−c)] where s = (a+b+c)/2
- TSA includes all faces; CSA excludes bases