Mensuration
Overview
Mensuration is the branch of mathematics dealing with measurement of geometric figures — their lengths, areas and volumes. For Bihar TET Paper II, this topic carries significant weight as it tests both conceptual understanding and computational ability. Questions typically involve plane figures (2D shapes like triangles, quadrilaterals, circles) and solids (3D objects like cubes, cylinders, cones, spheres).
This topic bridges pure geometry with real-world applications — calculating land area, tank capacity, material required for construction. Upper-primary students encounter mensuration extensively in Classes 6–8, so TET aspirants must master not just the formulas but also the pedagogical approach to teach these concepts effectively. Expect 3–5 direct questions requiring formula application, unit conversion and multi-step problem solving.
Success demands memorising key formulas, understanding when to apply each, and avoiding common unit-conversion errors. The exam favours questions combining two or more concepts — for example, finding the cost of painting a room (surface area) or the time to fill a tank (volume and rate).
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Key Concepts
- **Perimeter** is the total length of the boundary of a plane figure; measured in linear units (cm, m).
- **Area** measures the region enclosed by a plane figure; measured in square units (cm², m²).
- **Surface area** of a solid is the total area of all its outer faces; includes lateral (curved) surface area and total surface area.
- **Volume** measures the space occupied by a solid; measured in cubic units (cm³, m³) or capacity units (litres, where 1 litre = 1000 cm³).
- **Lateral Surface Area (LSA)** excludes the base and top; **Total Surface Area (TSA)** includes all surfaces.
- **Right prisms and cylinders** have uniform cross-section; their volume = Base Area × Height.
- **Pyramids and cones** have volume equal to one-third of the corresponding prism/cylinder with same base and height.
- **Unit conversion** is critical: 1 m = 100 cm; 1 m² = 10000 cm²; 1 m³ = 1000000 cm³ = 1000 litres.
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Formulas / Key Facts
### Plane Figures (2D)
| Figure | Perimeter | Area | |--------|-----------|------| | Rectangle | 2(l + b) | l × b | | Square | 4a | a² | | Triangle | a + b + c | ½ × base × height | | Right Triangle | a + b + c | ½ × leg₁ × leg₂ | | Equilateral Triangle | 3a | (√3/4) × a² | | Parallelogram | 2(a + b) | base × height | | Rhombus | 4a | ½ × d₁ × d₂ | | Trapezium | sum of all sides | ½ × (a + b) × h, where a, b are parallel sides | | Circle | 2πr (circumference) | πr² | | Semicircle | πr + 2r | ½ × πr² |
**Heron's Formula** for triangle with sides a, b, c:
- s = (a + b + c)/2
- Area = √[s(s−a)(s−b)(s−c)]
### Solids (3D)
| Solid | Lateral/Curved SA | Total SA | 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, slant l) | πrl | πr(r + l) | ⅓πr²h | | Sphere (r) | — | 4πr² | (4/3)πr³ | | Hemisphere (r) | 2πr² | 3πr² | (2/3)πr³ |
**Slant height of cone**: l = √(r² + h²)
**Diagonal of cuboid**: d = √(l² + b² + h²)
**Diagonal of cube**: d = a√3
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Worked Examples
### Example 1: Area of Trapezium *A field is in the shape of a trapezium with parallel sides 25 m and 15 m. The perpendicular distance between them is 8 m. Find the area.*
**Solution:**
- Area = ½ × (sum of parallel sides) × height
- Area = ½ × (25 + 15) × 8
- Area = ½ × 40 × 8 = 160 m²
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### Example 2: Volume and Surface Area of Cylinder *A cylindrical water tank has radius 3.5 m and height 5 m. Find its capacity in litres and the cost of painting its curved surface at ₹20 per m². (Use π = 22/7)*
**Solution:**
- Volume = πr²h = (22/7) × 3.5 × 3.5 × 5 = (22/7) × 12.25 × 5 = 192.5 m³
- Capacity = 192.5 × 1000 = 1,92,500 litres
- Curved Surface Area = 2πrh = 2 × (22/7) × 3.5 × 5 = 110 m²
- Cost of painting = 110 × 20 = ₹2200
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### Example 3: Combined Solid *A solid is made by placing a cone on top of a hemisphere. Both have radius 7 cm. The total height of the solid is 17 cm. Find the total surface area. (π = 22/7)*
**Solution:**
- Hemisphere radius r = 7 cm
- Height of cone h = 17 − 7 = 10 cm
- Slant height l = √(r² + h²) = √(49 + 100) = √149 ≈ 12.2 cm
- Curved SA of hemisphere = 2πr² = 2 × (22/7) × 49 = 308 cm²
- Curved SA of cone = πrl = (22/7) × 7 × 12.2 = 268.4 cm²
- Total Surface Area = 308 + 268.4 = 576.4 cm²
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Common Mistakes
- **Confusing LSA and TSA** → LSA excludes bases; TSA includes all faces. Read the question carefully — "painting the walls" needs LSA; "painting a closed box" needs TSA.
- **Forgetting to square or cube during unit conversion** → 1 m² = 10000 cm² (not 100). Always convert dimensions first, then calculate, or convert the final answer using the correct power.
- **Using diameter instead of radius** → Formulas use radius. If diameter is given, divide by 2 before substituting.
- **Wrong formula for cone/pyramid volume** → These are ⅓ of the prism/cylinder formula, not ½. Remember: "pointed solids get one-third."
- **Ignoring slant height in cones** → Curved surface area uses slant height l, not vertical height h. Calculate l = √(r² + h²) if not given directly.
- **Mixing up perimeter and area** → Perimeter is for fencing (linear); area is for covering (square). Match the unit to the context.
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Quick Reference
- Rectangle area = l × b; Perimeter = 2(l + b)
- Circle area = πr²; Circumference = 2πr
- Volume of cylinder = πr²h; TSA = 2πr(r + h)
- Volume of cone = ⅓πr²h; CSA = πrl
- Volume of sphere = (4/3)πr³; SA = 4πr²
- 1 m³ = 1000 litres = 10⁶ cm³