Mensuration — RRB NTPC Study Notes

Overview

Mensuration is the quantitative study of geometric figures — computing area, perimeter, surface area and volume. For RRB NTPC, this topic consistently delivers 2–4 direct questions each year, making it a high-yield area. Questions test both formula recall and the ability to apply formulas to word problems involving real-world objects like water tanks, metal sheets, ground plots, and packaging.

Mastery means knowing the right formula instantly and executing arithmetic cleanly. Most errors come from unit confusion (mixing cm and m), incorrect formula selection (lateral vs. total surface area), or algebraic slips when solving for an unknown dimension. Since mensuration questions are computation-heavy, speed and accuracy in multiplication, squaring, and handling π are critical.

The syllabus explicitly covers 2-D shapes (triangle, rectangle, square, circle, trapezium) and 3-D solids (cube, cuboid, cylinder, cone, sphere, hemisphere). You won't see complex composite solids, but do expect problems combining two shapes or converting between volume and capacity (1 litre = 1000 cm³).

Key Concepts

• **Area** measures the space inside a 2-D boundary; units are always square (cm², m², etc.). Perimeter measures the boundary length itself.
• **Surface area** for 3-D solids comes in two flavors: **lateral (curved) surface area** excludes the top/bottom bases; **total surface area** includes all faces.
• **Volume** measures the space a 3-D object occupies; units are cubic (cm³, m³). One cubic metre = 1000 litres; one litre = 1000 cm³.
• The number **π (pi)** appears in all circular formulas. Use π = 22/7 or 3.14 as instructed; RRB typically accepts 22/7 unless specified otherwise.
• For composite figures, break them into standard shapes, compute each separately, then add or subtract areas/volumes as needed.
• Always convert units to a common system before calculating. Mixing metres and centimetres is the #1 source of wrong answers.
• The diagonal of a rectangle with sides a and b is √(a² + b²); for a cube of side a, space diagonal = a√3.
• Heron's formula for triangle area works when three sides are known but no height is given: use the semi-perimeter method.

Formulas / Key Facts

**2-D Figures**

• **Rectangle**: Area = length × breadth; Perimeter = 2(length + breadth)
• **Square**: Area = side²; Perimeter = 4 × side; Diagonal = side√2
• **Triangle**: Area = ½ × base × height; Heron's formula: Area = √[s(s−a)(s−b)(s−c)] where s = (a+b+c)/2
• **Circle**: Area = πr²; Circumference = 2πr; where r = radius
• **Semicircle**: Area = πr²/2; Perimeter = πr + 2r (curved part + diameter)
• **Trapezium**: Area = ½ × (sum of parallel sides) × height
• **Rhombus**: Area = ½ × d₁ × d₂ (d₁, d₂ are diagonals)
• **Parallelogram**: Area = base × height
• **Equilateral Triangle**: Area = (√3/4) × side²; Height = (√3/2) × side

**3-D Solids**

• **Cube** (side a): Volume = a³; Total Surface Area = 6a²; Lateral Surface Area = 4a²
• **Cuboid** (length l, breadth b, height h): Volume = l × b × h; TSA = 2(lb + bh + hl); LSA = 2h(l + b)
• **Cylinder** (radius r, height h): Volume = πr²h; Curved Surface Area = 2πrh; TSA = 2πrh + 2πr² = 2πr(h + r)
• **Cone** (radius r, height h, slant height l = √(r² + h²)): Volume = (1/3)πr²h; Curved Surface Area = πrl; TSA = πrl + πr² = πr(l + r)
• **Sphere** (radius r): Volume = (4/3)πr³; Surface Area = 4πr²
• **Hemisphere** (radius r): Volume = (2/3)πr³; Curved Surface Area = 2πr²; TSA = 2πr² + πr² = 3πr²

**Conversion**: 1 m³ = 1000 litres; 1 litre = 1000 cm³

Worked Examples

**Example 1: Circle and square combination** A circular park of radius 14 m is surrounded by a path 3.5 m wide. Find the area of the path. (Use π = 22/7)

*Solution*: Outer radius = 14 + 3.5 = 17.5 m Area of path = πR² − πr² = π(R² − r²) = (22/7)(17.5² − 14²) = (22/7)(306.25 − 196) = (22/7) × 110.25 = 22 × 15.75 = 346.5 m²

**Example 2: Cylinder volume and capacity** A cylindrical water tank has radius 70 cm and height 1.5 m. How many litres of water can it hold?

*Solution*: Convert to common units: radius = 70 cm, height = 150 cm Volume = πr²h = (22/7) × 70 × 70 × 150 = 22 × 10 × 70 × 150 = 2,310,000 cm³ Capacity in litres = 2,310,000 ÷ 1000 = 2310 litres

**Example 3: Cuboid surface area** A room is 8 m long, 6 m wide, 4 m high. Find the cost of painting its four walls at ₹50 per m².

*Solution*: Lateral Surface Area (four walls) = 2h(l + b) = 2 × 4 × (8 + 6) = 8 × 14 = 112 m² Cost = 112 × 50 = ₹5,600

Common Mistakes

• **Confusing radius and diameter** → Always confirm whether the problem gives r or d. If diameter = 14, radius = 7. Many students plug d directly into r formulas and get a 4× error.
• **Using lateral SA when total SA is asked** → Read carefully: "curved surface area" excludes bases; "total surface area" includes them. For painting a closed tank, use TSA; for a tube open at both ends, use CSA only.
• **Forgetting unit conversions** → Mixing metres and centimetres inflates or deflates answers by factors of 100 or 1000. Convert everything to one unit before computing.
• **Misapplying volume formulas** → Volume of cone and pyramid are (1/3) × base area × height, not just base × height. Missing the 1/3 factor triples your answer incorrectly.
• **Incorrect slant height in cone problems** → If h and r are given, you must calculate l = √(r² + h²) before finding curved surface area. Don't assume l = h.

Quick Reference

• **Square diagonal**: side√2
• **Cube space diagonal**: side√3
• **Cylinder**: Volume = πr²h; Curved SA = 2πrh
• **Cone**: Volume = (1/3)πr²h; slant height l = √(r² + h²)
• **Sphere**: Volume = (4/3)πr³; Surface Area = 4πr²
• **1 m³ = 1000 litres; 1 litre = 1000 cm³**