Mensuration — SSC MTS Study Notes

Overview

Mensuration deals with measurement of geometric figures — both flat (2-D) and solid (3-D) shapes. In SSC MTS Paper 1, expect 2–4 direct questions testing formulas for area, perimeter, surface area, and volume. Most questions are straightforward formula substitution, though some involve unit conversion or composite figures.

This topic is a score-booster because formulas are fixed and calculation steps are mechanical. Master the 10–12 core formulas, practice unit conversions (cm to m, m² to cm²), and recognize when a problem involves combining multiple shapes. Accuracy in applying the correct formula and careful arithmetic are the keys to full marks.

Common question types: find the area of a triangle given base and height; calculate the cost of painting four walls of a room; determine how much water a cylindrical tank holds; find the diagonal of a rectangle or cuboid. Questions may be direct or wrapped in a one-step word problem.

Key Concepts

• **2-D figures** have only length and width — we measure their **perimeter** (boundary length) and **area** (surface covered). Common shapes: square, rectangle, triangle, circle, trapezium.

• **3-D solids** have length, width, and height — we measure **surface area** (total outer surface) and **volume** (space occupied). Common solids: cube, cuboid, cylinder, cone, sphere, hemisphere.

• **Perimeter** is always in linear units (cm, m) while **area** is in square units (cm², m²). **Volume** is in cubic units (cm³, m³, litres where 1 litre = 1000 cm³).

• **Composite figures**: Some problems combine shapes — e.g. a rectangle with a semicircle on top. Break the figure into known shapes, calculate separately, then add or subtract as needed.

• **Unit conversion**: 1 m = 100 cm, so 1 m² = 10,000 cm² and 1 m³ = 1,000,000 cm³. Always convert to the same unit before calculation.

• **Diagonal** formulas: For rectangle/cuboid, use Pythagoras theorem. Rectangle diagonal = √(l² + b²). Cuboid diagonal = √(l² + b² + h²).

• In **cost problems**, Total Cost = (Area or Volume) × Rate per unit. E.g., cost of flooring = floor area × rate per m².

Formulas / Key Facts

**2-D Figures:**

• **Rectangle**: Perimeter = 2(l + b); Area = l × b; Diagonal = √(l² + b²)

• **Square**: Perimeter = 4a; Area = a²; Diagonal = a√2 (where a = side)

• **Triangle**: Perimeter = a + b + c (sum of sides); Area = ½ × base × height. For equilateral triangle of side a: Area = (√3/4)a²

• **Circle**: Circumference = 2πr or πd; Area = πr² (where r = radius, d = diameter, π ≈ 22/7 or 3.14)

• **Semicircle**: Perimeter = πr + 2r (curved part + diameter); Area = (πr²)/2

• **Trapezium**: Area = ½ × (sum of parallel sides) × height = ½(a + b)h

• **Parallelogram**: Area = base × height

**3-D Solids:**

• **Cube**: Total Surface Area = 6a²; Lateral Surface Area = 4a²; Volume = a³; Diagonal = a√3

• **Cuboid**: Total Surface Area = 2(lb + bh + hl); Lateral Surface Area = 2h(l + b); Volume = l × b × h; Diagonal = √(l² + b² + h²)

• **Cylinder**: Curved Surface Area = 2πrh; Total Surface Area = 2πr(r + h); Volume = πr²h

• **Cone**: Curved Surface Area = πrl (l = slant height = √(r² + h²)); Total Surface Area = πr(r + l); Volume = (1/3)πr²h

• **Sphere**: Surface Area = 4πr²; Volume = (4/3)πr³

• **Hemisphere**: Curved Surface Area = 2πr²; Total Surface Area = 3πr²; Volume = (2/3)πr³

Worked Examples

**Example 1**: A rectangular field is 50 m long and 30 m wide. Find the cost of fencing it at ₹12 per metre.

*Solution*: Perimeter = 2(l + b) = 2(50 + 30) = 2 × 80 = 160 m. Cost = 160 × 12 = ₹1920.

**Example 2**: Find the area of a circle whose circumference is 44 cm. (Use π = 22/7)

*Solution*: Circumference = 2πr = 44. So 2 × (22/7) × r = 44 → (44/7) × r = 44 → r = 7 cm. Area = πr² = (22/7) × 7 × 7 = 154 cm².

**Example 3**: A cylindrical water tank has radius 1.4 m and height 2 m. How many litres of water can it hold? (Use π = 22/7)

*Solution*: Volume = πr²h = (22/7) × 1.4 × 1.4 × 2 = (22/7) × 1.96 × 2 = 22 × 0.28 × 2 = 12.32 m³. Converting to litres: 1 m³ = 1000 litres, so 12.32 m³ = 12,320 litres.

**Example 4**: A room is 5 m long, 4 m wide and 3 m high. Find the cost of painting its four walls at ₹8 per m².

*Solution*: Lateral Surface Area of cuboid = 2h(l + b) = 2 × 3 × (5 + 4) = 6 × 9 = 54 m². Cost = 54 × 8 = ₹432.

Common Mistakes

• **Using perimeter formula for area**: Students confuse 2(l + b) (perimeter) with l × b (area) for rectangles. *Fix*: Remember perimeter measures boundary, area measures surface.

• **Forgetting to square or cube units**: Writing area in cm instead of cm², or volume in m instead of m³. *Fix*: Check dimensionality — area is always squared units, volume is cubed.

• **Wrong unit conversion**: Thinking 1 m² = 100 cm² instead of 10,000 cm². *Fix*: Remember 1 m = 100 cm, so 1 m² = 100 cm × 100 cm = 10,000 cm². Always square or cube the conversion factor.

• **Misapplying π approximation**: Using π = 22/7 when the problem gives decimals or vice versa. *Fix*: Use π = 22/7 when radius or diameter is a multiple of 7; use 3.14 otherwise, or follow the question's instruction.

• **Confusing total and lateral surface area**: Using total surface area when the question asks for curved/lateral (e.g., cost of painting walls excludes floor and ceiling). *Fix*: Read carefully — "four walls" means lateral surface area only.

• **Omitting slant height in cone problems**: Using height h instead of slant height l in curved surface area = πrl. *Fix*: Calculate l = √(r² + h²) first if not given.

Quick Reference

• Rectangle: Area = l × b, Perimeter = 2(l + b), Diagonal = √(l² + b²) • Circle: Area = πr², Circumference = 2πr • Cube: Volume = a³, Surface Area = 6a² • Cylinder: Volume = πr²h, Curved Surface = 2πrh • Cone: Volume = (1/3)πr²h, slant height l = √(r² + h²) • 1 m² = 10,000 cm²; 1 m³ = 1,000,000 cm³; 1 litre = 1000 cm³ • For composite shapes, break into parts and add/subtract areas or volumes.