Mensuration
Perimeter, Area, Surface Area and Volume of Solids
Overview
Mensuration is the branch of mathematics dealing with measurement of geometric figures — their lengths, areas and volumes. For UTET Paper II, this topic carries significant weight because it tests both conceptual understanding and computational accuracy. Questions typically involve plane figures (rectangles, triangles, circles) and solid figures (cubes, cuboids, cylinders, cones, spheres).
Mastery requires knowing the standard formulas and understanding when to apply each. Exam problems often present real-life scenarios — fencing a field, painting a room, filling a tank — so you must connect abstract formulas to practical situations. This topic also overlaps with pedagogy questions about how to teach measurement concepts through activities and concrete materials.
Focus areas: memorising formulas with correct units, converting between units, and solving multi-step problems involving composite figures or combined solids.
Key Concepts
- Perimeter is the total length of the boundary of a plane figure. It is measured in linear units (cm, m, km).
- Area is the measure of the surface enclosed within a boundary. It is measured in square units (cm², m²).
- Surface Area of a solid is the total area of all its outer faces. For closed solids, this includes all sides; for open solids (like an open box), exclude the missing face.
- Volume measures the space occupied by a three-dimensional object. It is measured in cubic units (cm³, m³) or capacity units (litres, where 1 litre = 1000 cm³).
- Lateral Surface Area (LSA) or Curved Surface Area (CSA) refers to the area of the sides only, excluding the top and bottom bases.
- Total Surface Area (TSA) includes the lateral surface plus the areas of the base(s).
- Composite figures are shapes formed by combining two or more basic shapes. Find each part's measurement separately, then add or subtract as required.
- Unit conversion is critical: 1 m = 100 cm; 1 m² = 10,000 cm²; 1 m³ = 1,000,000 cm³ = 1000 litres.
Formulas / Key Facts
Plane Figures (Perimeter and Area)
| Figure | Perimeter | Area |
|---|---|---|
| Rectangle (l × b) | 2(l + b) | l × b |
| Square (side a) | 4a | a² |
| Triangle (sides a, b, c; base b; height h) | a + b + c | ½ × b × h |
| Equilateral Triangle (side a) | 3a | (√3/4) × a² |
| Parallelogram (base b, height h) | 2(a + b) | b × h |
| Rhombus (diagonals d₁, d₂) | 4 × side | ½ × d₁ × d₂ |
| Trapezium (parallel sides a, b; height h) | sum of all sides | ½ × (a + b) × h |
| Circle (radius r) | 2πr (circumference) | πr² |
| Semicircle | πr + 2r | ½ × πr² |
Use π = 22/7 or 3.14 as specified in the question.
Solid Figures (Surface Area and Volume)
| Solid | Lateral/Curved SA | Total SA | Volume |
|---|---|---|---|
| Cube (edge a) | 4a² | 6a² | a³ |
| Cuboid (l × b × h) | 2h(l + b) | 2(lb + bh + hl) | l × b × h |
| Cylinder (radius r, height h) | 2πrh | 2πr(r + h) | πr²h |
| Cone (radius r, height h, slant l) | πrl | πr(r + l) | ⅓ × πr²h |
| Sphere (radius r) | — | 4πr² | (4/3)πr³ |
| Hemisphere (radius r) | 2πr² | 3πr² | (2/3)πr³ |
Slant height of cone: l = √(r² + h²)
Worked Examples
Example 1: Area of a Composite Figure
Problem: A rectangular lawn measuring 50 m × 30 m has a circular fountain of radius 7 m at its centre. Find the area of the lawn excluding the fountain.
Solution:
- Area of rectangle = 50 × 30 = 1500 m²
- Area of circle = πr² = (22/7) × 7 × 7 = 154 m²
- Area of lawn excluding fountain = 1500 − 154 = 1346 m²
Example 2: Surface Area of a Cylinder
Problem: A closed cylindrical tank has radius 3.5 m and height 10 m. Find the cost of painting its outer surface at ₹20 per m².
Solution:
- TSA of cylinder = 2πr(r + h) = 2 × (22/7) × 3.5 × (3.5 + 10)
- = 2 × (22/7) × 3.5 × 13.5 = 2 × 22 × 0.5 × 13.5 = 297 m²
- Cost = 297 × 20 = ₹5940
Example 3: Volume and Capacity
Problem: A cuboidal water tank is 2 m long, 1.5 m wide and 1 m deep. How many litres of water can it hold?
Solution:
- Volume = l × b × h = 2 × 1.5 × 1 = 3 m³
- 1 m³ = 1000 litres
- Capacity = 3 × 1000 = 3000 litres
Common Mistakes
- Confusing perimeter with area → Perimeter is a length (one dimension); area is two-dimensional. Check units: m vs m².
- Forgetting to convert units before calculation → Convert all measurements to the same unit first. A mix of cm and m will give absurd answers.
- Using diameter instead of radius (or vice versa) → Read the problem carefully. If diameter is given, divide by 2 before using formulas that require radius.
- Applying LSA formula when TSA is required → Painting "all surfaces" or "completely" means TSA. "Only the curved part" or "walls" means LSA/CSA.
- Incorrect slant height in cone problems → Students often use height directly in πrl. Remember slant height l = √(r² + h²); calculate it first.
- Ignoring open/hollow conditions → An open box has no top; a hollow cylinder has no circular bases. Subtract the missing face(s) from TSA.
Quick Reference
- Rectangle area = l × b; Perimeter = 2(l + b)
- Circle area = πr²; Circumference = 2πr
- Cube: TSA = 6a², Volume = a³
- Cuboid: TSA = 2(lb + bh + hl), Volume = lbh
- Cylinder: CSA = 2πrh, TSA = 2πr(r + h), Volume = πr²h
- Cone: CSA = πrl, Volume = ⅓πr²h, Slant l = √(r² + h²)
- Sphere: SA = 4πr², Volume = (4/3)πr³
- 1 m³ = 1000 litres; 1 litre = 1000 cm³