Mensuration — Study Notes for SSC CHSL

Overview

Mensuration is the mathematical study of measurement — specifically the area, perimeter, surface area and volume of geometric figures. For SSC CHSL, you can expect **3–5 direct questions** from this topic in every Tier-1 exam, making it a high-scoring area if you master the formulas and problem patterns.

The topic divides into **2-D Mensuration** (flat shapes: triangles, rectangles, circles, trapeziums) and **3-D Mensuration** (solid objects: cubes, cuboids, cylinders, cones, spheres). Questions test both direct formula application and mixed-figure problems where you combine multiple shapes. The key to success is formula accuracy and smart substitution — most calculation errors happen in carrying units or squaring/cubing incorrectly.

Mensuration integrates well with **Geometry** (properties of triangles, circles) and **Algebra** (forming equations from given conditions). Practice converting word problems into mathematical expressions quickly — this is where most students lose time in the exam.

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Key Concepts

• **Perimeter vs Area**: Perimeter is the boundary length (measured in units like cm, m); area is the space inside (measured in square units like cm², m²). Never mix units.

• **Surface Area vs Volume**: For 3-D objects, surface area is the total outer covering (square units); volume is the space occupied (cubic units like cm³, m³).

• **Radius–Diameter Relationship**: Diameter = 2 × Radius. In circle formulas, always check whether the given value is r or d.

• **π Approximation**: Use π = 22/7 or 3.14 depending on the question. If the question has 7, 14, 21 in denominators, use 22/7; otherwise use 3.14.

• **Composite Figures**: Many problems combine shapes (e.g. a rectangle with a semicircle on top). Break them into standard shapes, calculate separately, then add/subtract as needed.

• **Unit Conversion**: 1 m = 100 cm; 1 m² = 10,000 cm²; 1 m³ = 1,000,000 cm³. Always convert to the same unit before calculating.

• **Lateral Surface vs Total Surface**: Lateral/Curved surface excludes the top and bottom bases; total surface includes all faces. Distinguish carefully in cylinder and cone problems.

• **Height vs Slant Height**: In cones and pyramids, height (h) is perpendicular to the base; slant height (l) is the diagonal. They're related by Pythagoras: l² = h² + r².

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Formulas / Key Facts

### 2-D Figures

**Rectangle** Perimeter = 2(length + breadth); Area = length × breadth

**Square** Perimeter = 4 × side; Area = side²

**Triangle** Perimeter = sum of all three sides; Area = (1/2) × base × height For equilateral triangle with side a: Area = (√3/4) × a² Heron's formula: Area = √[s(s−a)(s−b)(s−c)] where s = (a+b+c)/2

**Circle** Circumference = 2πr or πd; Area = πr²

**Semicircle** Perimeter = πr + 2r (curved part + diameter); Area = (1/2)πr²

**Trapezium** Area = (1/2) × (sum of parallel sides) × height

**Rhombus** Area = (1/2) × product of diagonals

**Parallelogram** Area = base × height

### 3-D Figures

**Cube** (all edges equal = a) Volume = a³; Total Surface Area = 6a²; Lateral Surface Area = 4a²

**Cuboid** (length l, breadth b, height h) Volume = l × b × h; Total Surface Area = 2(lb + bh + hl); Lateral Surface Area = 2h(l + b)

**Cylinder** (radius r, height h) Volume = πr²h; Curved Surface Area = 2πrh; Total Surface Area = 2πr(r + h)

**Cone** (radius r, height h, slant height l) Volume = (1/3)πr²h; Curved Surface Area = πrl; Total Surface Area = πr(r + l) l² = h² + r²

**Sphere** (radius r) Volume = (4/3)πr³; Surface Area = 4πr²

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

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Worked Examples

**Example 1: Rectangle and Semicircle Composite** A rectangular park 70 m long and 50 m wide has a semicircular garden attached to one of its shorter sides. Find the total perimeter.

**Solution**: Rectangle perimeter (3 sides needed) = 70 + 50 + 70 = 190 m Semicircle (on 50 m side, so diameter = 50 m, radius = 25 m) Curved part = πr = (22/7) × 25 = 550/7 = 78.57 m Total perimeter = 190 + 78.57 = **268.57 m**

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**Example 2: Cube to Cuboid Melting** A cube of side 6 cm is melted and recast into a cuboid of length 9 cm and breadth 4 cm. Find the height.

**Solution**: Volume of cube = 6³ = 216 cm³ Volume of cuboid = 9 × 4 × h = 36h Since volume remains same: 36h = 216 h = 216/36 = **6 cm**

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**Example 3: Cylinder and Cone Height Relationship** A cylinder and a cone have equal base radius and equal volume. If the height of the cylinder is 12 cm, find the height of the cone.

**Solution**: Volume of cylinder = πr²(12) Volume of cone = (1/3)πr²H where H is cone height Given both volumes equal: (1/3)πr²H = πr²(12) (1/3)H = 12 H = **36 cm**

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Common Mistakes

**Using diameter instead of radius**: Many students plug diameter into formulas like Area = πr². Always halve the diameter first. → **Check if given value is d or r before substituting.**

**Forgetting to square/cube in area/volume**: Writing Volume of cube = 3a instead of a³. → **Area formulas have squared terms; volume formulas have cubed terms.**

**Adding areas when perimeter is asked**: For composite shapes, students add areas instead of boundary lengths. → **Perimeter = sum of outer boundary lengths only.**

**Mixing lateral and total surface area**: Using Curved Surface Area when Total Surface Area is asked or vice versa. → **Read the question carefully — does it ask for total or lateral/curved surface?**

**Unit mismatch**: Calculating with length in metres and breadth in centimetres without converting. → **Convert all measurements to the same unit before substituting.**

**Ignoring the "remaining" part**: In problems like "a wire bent into a rectangle then into a circle", students forget perimeter stays constant. → **What's conserved? Length/Perimeter for wire problems; Volume for melting/recasting problems.**

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Quick Reference

• **Circle area–circumference link**: If circumference = 2πr, then area = πr² — both depend on radius.
• **Cube diagonal (space diagonal)**: a√3 where a is the edge.
• **Cylinder to cone volume ratio**: For same base and height, cylinder volume = 3 × cone volume.
• **Sphere to hemisphere volume**: Sphere = 2 × hemisphere (same radius).
• **Melting/Recasting**: Volume before = volume after; shape changes, volume doesn't.
• **Four walls area (room)**: Lateral surface of cuboid = 2h(l + b); useful for painting/wallpaper problems.