Area and Volume — Study Notes for TNPSC Group II/IIA
Overview
Area and Volume (Mensuration) is a high-weightage topic in TNPSC Group II/IIA Aptitude section. Questions typically test your ability to calculate surface areas, volumes, and perimeters of standard 2D and 3D shapes.
Mastery requires memorizing formulas and knowing when to apply each. The exam favors questions involving combined figures (hemisphere on cylinder, cone inside cube), unit conversions, and finding one dimension when others are given. Speed matters—so formula recall must be instant.
Focus areas: triangles (especially right-angled and equilateral), circles, cubes, cuboids, cylinders, cones, and spheres. Composite solids appear frequently in recent papers.
Key Concepts
- Area measures the surface covered by a 2D shape (in square units); Volume measures the space occupied by a 3D object (in cubic units).
- Perimeter/Circumference is the total boundary length of a 2D figure—often needed as an intermediate step.
- Lateral/Curved Surface Area (CSA) excludes the base(s); Total Surface Area (TSA) includes all surfaces.
- For 3D shapes, distinguish between hollow and solid objects—hollow objects have thickness affecting volume calculations.
- Unit consistency is critical: convert all measurements to the same unit before calculating. 1 m = 100 cm; 1 m³ = 1000 litres.
- Slant height (l) for cones and pyramids differs from vertical height (h). Use Pythagoras: l² = h² + r².
- When a shape is inscribed or circumscribed, identify the geometric relationship (e.g., diagonal of cube = diameter of circumscribed sphere).
Formulas / Key Facts
2D Figures
| Shape | Area | Perimeter |
|---|---|---|
| Rectangle | l × b | 2(l + b) |
| Square | a² | 4a |
| Triangle (general) | ½ × base × height | Sum of three sides |
| Right-angled Triangle | ½ × base × perpendicular | a + b + hypotenuse |
| Equilateral Triangle | (√3/4) × a² | 3a |
| Circle | πr² | 2πr (circumference) |
| Semicircle | ½πr² | πr + 2r |
| Sector (angle θ°) | (θ/360) × πr² | (θ/360) × 2πr + 2r |
Heron's Formula for triangle with sides a, b, c:
- s = (a + b + c)/2
- Area = √[s(s−a)(s−b)(s−c)]
3D Figures
| Shape | Volume | CSA/LSA | TSA |
|---|---|---|---|
| Cube (side a) | a³ | 4a² | 6a² |
| Cuboid (l, b, h) | l × b × h | 2h(l + b) | 2(lb + bh + hl) |
| Cylinder (r, h) | πr²h | 2πrh | 2πr(r + h) |
| Cone (r, h, slant l) | ⅓πr²h | πrl | πr(r + l) |
| Sphere (radius r) | (4/3)πr³ | — | 4πr² |
| Hemisphere (r) | (2/3)πr³ | 2πr² | 3πr² |
Slant height of cone: l = √(h² + r²)
Diagonal of cuboid: √(l² + b² + h²)
Diagonal of cube: a√3
Worked Examples
Example 1: Cylinder Volume
A cylindrical tank has diameter 14 m and height 5 m. Find its capacity in litres.
Solution:
- Radius r = 14/2 = 7 m
- Volume = πr²h = (22/7) × 7 × 7 × 5 = 770 m³
- Capacity = 770 × 1000 = 7,70,000 litres
Example 2: Cone Surface Area
Find the total surface area of a cone with radius 6 cm and height 8 cm.
Solution:
- Slant height l = √(8² + 6²) = √(64 + 36) = √100 = 10 cm
- TSA = πr(r + l) = (22/7) × 6 × (6 + 10)
- TSA = (22/7) × 6 × 16 = 2112/7 = 301.71 cm²
Example 3: Combined Solid
A hemisphere is mounted on a cylinder of same radius. If radius = 7 cm and cylinder height = 10 cm, find total volume.
Solution:
- Volume of cylinder = πr²h = (22/7) × 49 × 10 = 1540 cm³
- Volume of hemisphere = (2/3)πr³ = (2/3) × (22/7) × 343 = 718.67 cm³
- Total volume = 1540 + 718.67 = 2258.67 cm³
Common Mistakes
- Confusing radius and diameter → Always check whether the question gives radius or diameter. Halve the diameter before using formulas.
- Using height instead of slant height for cone CSA → Cone's curved surface needs slant height (l), not vertical height (h). Calculate l using Pythagoras.
- Forgetting to square or cube properly → Area involves r², volume involves r³. A small error in exponent creates large errors in answers.
- Wrong unit conversion → Students multiply instead of cube when converting volume units. Remember: 1 m³ = 10⁶ cm³, not 100 cm³.
- Adding CSA instead of TSA for closed figures → Read whether the question asks for curved/lateral surface or total surface. Closed containers need TSA.
- Using 3.14 when 22/7 simplifies better → For TNPSC, if radius or diameter is a multiple of 7, use π = 22/7 for cleaner calculations.
Quick Reference
- Equilateral triangle area: (√3/4)a² — memorize √3 ≈ 1.732
- Cylinder volume: πr²h — "Area of base × height"
- Cone volume: ⅓ × cylinder volume of same base and height
- Sphere volume: (4/3)πr³ — hemisphere is exactly half
- Always find slant height first for cone problems: l = √(h² + r²)
- 1 litre = 1000 cm³ = 0.001 m³ — critical for tank/cistern problems