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.
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.
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.
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
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
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
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
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.
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.