Mensuration
Overview
Mensuration is one of the most formula-intensive yet scoring topics in the Maharashtra Police Bharti Mathematics section. It deals with the measurement of geometric shapes—calculating perimeter, area for 2-D figures and surface area, volume for 3-D figures. Questions typically involve direct formula application or multi-step problems combining two or more shapes.
This topic carries significant weightage because it connects to real-world scenarios—land measurement, tank capacity, wire bending, painting costs—making it a favourite for examiners. Mastery requires memorising formulas accurately and understanding when to apply each. Most questions are straightforward once you identify the correct formula; the challenge lies in unit conversion and visualising composite figures.
Students who invest time in learning formulas systematically and practising 15–20 varied problems can expect to solve mensuration questions quickly and accurately in the exam.
Key Concepts
- Perimeter is the total length of the boundary of a 2-D shape. Think of it as the length of fencing needed to enclose a plot.
- Area is the surface enclosed within the boundary of a 2-D figure. It tells you how much paint or carpet is needed.
- Surface Area of a 3-D object is the total area of all its outer faces—relevant for painting, wrapping, or coating problems.
- Volume measures the space inside a 3-D object. It answers "how much water/sand/air can this container hold?"
- Lateral/Curved Surface Area (LSA/CSA) excludes the top and bottom faces—used when only the sides are painted or wrapped.
- Unit consistency is critical: if dimensions are in cm, area is in cm², volume in cm³. Convert before calculating, not after.
- Composite figures combine basic shapes. Break them into standard parts, calculate separately, then add or subtract as needed.
Formulas / Key Facts
2-D Figures
| Figure | Perimeter | Area |
|---|---|---|
| Square (side a) | 4a | a² |
| Rectangle (l × b) | 2(l + b) | l × b |
| Triangle (sides a, b, c; base b, height h) | a + b + c | ½ × b × h |
| Equilateral Triangle (side a) | 3a | (√3/4) × a² |
| Right Triangle (legs a, b) | a + b + √(a² + b²) | ½ × a × b |
| Circle (radius r) | 2πr (circumference) | πr² |
| Semicircle (radius r) | πr + 2r | ½ × πr² |
| Ring/Annulus (outer R, inner r) | — | π(R² − r²) |
| 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 |
Heron's Formula for triangle area when only sides are known:
- s = (a + b + c)/2
- Area = √[s(s−a)(s−b)(s−c)]
3-D Figures
| Figure | 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, slant l, height h) | 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)
Diagonal of cuboid: √(l² + b² + h²)
Diagonal of cube: a√3
Worked Examples
Example 1: Area of a Rectangular Field with Path
A rectangular park is 50 m × 40 m. A 5 m wide path runs inside along the boundary. Find the area of the path.
Solution:
- Outer dimensions: 50 × 40 m
- Inner dimensions (subtracting path width from both sides): (50 − 10) × (40 − 10) = 40 × 30 m
- Area of path = Outer area − Inner area = 2000 − 1200 = 800 m²
Example 2: Volume of a Cylinder
A cylindrical tank has diameter 14 m and height 5 m. Find its capacity in litres. (Use π = 22/7)
Solution:
- Radius r = 14/2 = 7 m
- Volume = πr²h = (22/7) × 7 × 7 × 5 = 22 × 7 × 5 = 770 m³
- 1 m³ = 1000 litres
- Capacity = 770 × 1000 = 7,70,000 litres
Example 3: Surface Area of a Hemisphere
A hemispherical dome has radius 21 cm. Find the cost of painting its outer surface at ₹5 per cm². (Use π = 22/7)
Solution:
- Curved Surface Area = 2πr² = 2 × (22/7) × 21 × 21 = 2 × 22 × 3 × 21 = 2772 cm²
- Cost = 2772 × 5 = ₹13,860
Common Mistakes
- Confusing radius with diameter → Always check if the problem gives diameter; divide by 2 to get radius before applying formulas.
- Mixing up LSA and TSA → Read carefully: "curved surface" or "sides only" means LSA; "total surface" or "all faces" means TSA.
- Ignoring unit conversion → If length is in metres and you need area in cm², first convert all dimensions to cm, then calculate.
- Using wrong formula for irregular triangles → When only sides are given (no height), use Heron's formula, not ½ × b × h.
- Forgetting to add base area for open/closed containers → Open cylinder (no top) has TSA = CSA + one base = 2πrh + πr². Read if "open" or "closed."
- Path problems: inside vs outside → Inside path reduces inner dimensions; outside path increases outer dimensions. Visualise before calculating.
Quick Reference
- Square: Area = a², Perimeter = 4a
- Rectangle: Area = lb, Perimeter = 2(l+b)
- Circle: Area = πr², Circumference = 2πr
- Cube: Volume = a³, TSA = 6a²
- Cylinder: Volume = πr²h, CSA = 2πrh
- Cone: Volume = ⅓πr²h, Slant height l = √(r² + h²)
- Sphere: Volume = (4/3)πr³, Surface = 4πr²
- 1 m³ = 1000 litres; 1 litre = 1000 cm³