Mensuration
Area, Surface Area and Volume of Solids
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Overview
Mensuration is the branch of mathematics dealing with measurement of geometric figures — their lengths, areas and volumes. For CG TET Paper II, this topic carries significant weight as it tests both conceptual understanding and computational accuracy. Questions typically involve finding areas of plane figures, surface areas of 3D solids and volumes of common shapes.
This topic connects directly to real-life applications that teachers must convey to students: calculating land area, paint required for walls, water capacity of tanks, etc. Exam questions range from direct formula application to multi-step problems combining two or more shapes. Mastery requires memorising formulas accurately and knowing when to apply each.
The scope covers plane figures (triangles, quadrilaterals, circles) and solids (cube, cuboid, cylinder, cone, sphere). Understanding the difference between lateral surface area, total surface area and volume is essential for scoring full marks in this section.
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Key Concepts
- **Area** measures the region enclosed by a 2D figure, expressed in square units (cm², m²).
- **Perimeter** is the total length of the boundary of a plane figure, expressed in linear units (cm, m).
- **Surface Area** of a 3D solid is the total area of all its outer faces — divided into Curved/Lateral Surface Area (CSA/LSA) and Total Surface Area (TSA).
- **Volume** measures the space occupied by a 3D object, expressed in cubic units (cm³, m³, litres where 1 litre = 1000 cm³).
- **Lateral Surface Area** excludes the base(s), while **Total Surface Area** includes all faces including base(s).
- **Right solids** have their axis perpendicular to the base — all formulas in the syllabus assume right solids unless stated otherwise.
- **Composite figures** combine two or more basic shapes — solve by adding or subtracting individual areas/volumes.
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Formulas / Key Facts
### Plane Figures (Area and Perimeter)
| Figure | Area | Perimeter | |--------|------|-----------| | Rectangle | l × b | 2(l + b) | | Square | a² | 4a | | Triangle | ½ × base × height | Sum of three sides | | Right Triangle | ½ × base × perpendicular | a + b + c | | Equilateral Triangle | (√3/4) × a² | 3a | | Parallelogram | base × height | 2(a + b) | | Rhombus | ½ × d₁ × d₂ | 4 × side | | Trapezium | ½ × (sum of parallel sides) × height | Sum of all sides | | Circle | πr² | 2πr (circumference) | | Semicircle | ½πr² | πr + 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 Solids (Surface Area and Volume)
| Solid | CSA/LSA | TSA | Volume | |-------|---------|-----|--------| | Cube (side a) | 4a² | 6a² | a³ | | Cuboid (l, b, h) | 2h(l + b) | 2(lb + bh + hl) | l × b × h | | Cylinder (r, h) | 2πrh | 2πr(r + h) | πr²h | | Cone (r, h, l) | πrl | πr(r + l) | ⅓πr²h | | Sphere (r) | 4πr² | 4πr² | (4/3)πr³ | | Hemisphere (r) | 2πr² | 3πr² | (2/3)πr³ |
**Note:** For cone, slant height l = √(r² + h²)
**Useful conversions:**
- 1 m³ = 1000 litres = 10⁶ cm³
- 1 litre = 1000 cm³
- Use π = 22/7 or 3.14 as specified in the question
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Worked Examples
### Example 1: Area of Trapezium **Problem:** A trapezium has parallel sides of 12 cm and 8 cm. The perpendicular distance between them is 5 cm. Find the area.
**Solution:**
- Area of trapezium = ½ × (sum of parallel sides) × height
- Area = ½ × (12 + 8) × 5
- Area = ½ × 20 × 5
- Area = 50 cm²
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### Example 2: Volume and Surface Area of Cylinder **Problem:** A cylindrical water tank has radius 7 m and height 10 m. Find (a) volume of water it can hold, (b) cost of painting the curved surface at ₹15 per m².
**Solution:** (a) Volume = πr²h = (22/7) × 7 × 7 × 10 = 22 × 7 × 10 = 1540 m³
(b) CSA = 2πrh = 2 × (22/7) × 7 × 10 = 2 × 22 × 10 = 440 m² Cost = 440 × 15 = ₹6600
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### Example 3: Composite Solid **Problem:** A solid is in the form of a cone mounted on a hemisphere. Both have radius 3 cm. The height of the cone is 4 cm. Find the total surface area.
**Solution:**
- Slant height of cone, l = √(r² + h²) = √(9 + 16) = √25 = 5 cm
- CSA of cone = πrl = π × 3 × 5 = 15π cm²
- CSA of hemisphere = 2πr² = 2π × 9 = 18π cm²
- Total surface area = 15π + 18π = 33π = 33 × (22/7) = 726/7 ≈ 103.71 cm²
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Common Mistakes
- **Confusing CSA with TSA** → Remember: TSA = CSA + Area of base(s). Read the question carefully — "paint the walls" means CSA; "total material needed" means TSA.
- **Forgetting to calculate slant height for cone** → Students directly use height in πrl formula. Always compute l = √(r² + h²) first.
- **Using diameter instead of radius** → Questions often give diameter. Divide by 2 before applying formulas.
- **Unit conversion errors** → When dimensions are in different units, convert all to the same unit before calculating. Final answer must match the required unit.
- **Adding volumes when shapes are carved out** → If one shape is removed from another, subtract volumes. Adding applies only when shapes are joined.
- **Using wrong formula for hemisphere** → Hemisphere CSA is 2πr² (curved part only), TSA is 3πr² (curved + circular base). A full sphere has no "base."
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Quick Reference
- **Trapezium area:** ½ × (parallel sides sum) × height
- **Cylinder volume:** πr²h; **CSA:** 2πrh; **TSA:** 2πr(r + h)
- **Cone volume:** ⅓πr²h; always find slant height l = √(r² + h²)
- **Sphere volume:** (4/3)πr³; **Surface area:** 4πr² (same as TSA)
- **Hemisphere TSA = 3πr²** (includes the flat circular base)
- **1 m³ = 1000 litres** — essential for tank/container problems