Mensuration

Overview

Mensuration is the branch of mathematics dealing with the measurement of geometric figures—their lengths, areas, and volumes. For AP TET Paper I (Classes 1-5) and Paper II (Classes 6-8), this topic forms a substantial portion of the mathematics section, typically contributing 3-5 questions per paper.

At the primary level, students encounter perimeter and area of basic 2D shapes (rectangles, squares, triangles, circles). At the upper primary level, the scope expands to include surface area and volume of 3D solids (cubes, cuboids, cylinders, cones, spheres). Mastery requires memorising formulas and—equally important—understanding when and how to apply them in word problems.

Questions often combine mensuration with real-life contexts: finding the cost of fencing a field, the amount of paint needed for walls, or the capacity of a tank. Expect both direct formula application and multi-step reasoning problems.

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Key Concepts

• **Perimeter** is the total length of the boundary of a 2D shape. Think of it as the distance you would walk if you traced the outline.

• **Area** measures the surface enclosed within a 2D boundary, expressed in square units (cm², m²).

• **Surface Area** of a 3D solid is the total area of all its outer faces. Distinguish between **Curved Surface Area (CSA)**—the curved part only—and **Total Surface Area (TSA)**—curved plus flat faces.

• **Volume** measures the space occupied by a 3D object, expressed in cubic units (cm³, m³). For liquids, 1000 cm³ = 1 litre.

• **Unit Consistency**: Always convert measurements to the same unit before calculating. Common conversions: 1 m = 100 cm; 1 km = 1000 m; 1 m² = 10,000 cm².

• **Composite Figures**: Many problems involve shapes made by combining or removing standard figures. Break them into simpler parts, calculate separately, then add or subtract.

• **Practical Applications**: Fencing (perimeter), tiling/painting (area), constructing tanks/boxes (surface area/volume), filling containers (volume in litres).

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Formulas / Key Facts

### 2D Figures — Perimeter and Area

| Figure | Perimeter | Area | |--------|-----------|------| | Square (side a) | 4a | a² | | Rectangle (length l, breadth b) | 2(l + b) | l × b | | Triangle (sides a, b, c; base b, height h) | a + b + c | ½ × b × h | | Right Triangle (legs a, b) | a + b + √(a² + b²) | ½ × a × b | | Equilateral Triangle (side a) | 3a | (√3/4) × a² | | Parallelogram (base b, height h, side a) | 2(a + b) | b × h | | Rhombus (diagonals d₁, d₂; side a) | 4a | ½ × d₁ × d₂ | | Trapezium (parallel sides a, b; height h) | Sum of all sides | ½ × (a + b) × h | | Circle (radius r) | 2πr (circumference) | πr² | | Semicircle (radius r) | πr + 2r | ½ × πr² |

*Use π = 22/7 or 3.14 as specified in the problem.*

### 3D Figures — Surface Area and Volume

| Solid | Curved/Lateral SA | Total SA | Volume | |-------|-------------------|----------|--------| | Cube (edge a) | 4a² | 6a² | a³ | | Cuboid (l, b, h) | 2h(l + b) | 2(lb + bh + hl) | l × b × h | | Cylinder (radius r, height h) | 2πrh | 2πr(r + h) | πr²h | | Cone (radius r, height h, slant l) | πrl | πr(r + l) | ⅓ × πr²h | | Sphere (radius r) | 4πr² | 4πr² | (4/3)πr³ | | Hemisphere (radius r) | 2πr² | 3πr² | (2/3)πr³ |

*Slant height of cone: l = √(r² + h²)*

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Worked Examples

### Example 1: Area and Perimeter (Rectangle) **Problem**: A rectangular garden is 25 m long and 18 m wide. Find the cost of fencing it at ₹45 per metre and the cost of laying grass at ₹12 per m².

**Solution**:

• Perimeter = 2(l + b) = 2(25 + 18) = 2 × 43 = 86 m
• Cost of fencing = 86 × 45 = ₹3,870
• Area = l × b = 25 × 18 = 450 m²
• Cost of grass = 450 × 12 = ₹5,400

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### Example 2: Volume and Capacity (Cylinder) **Problem**: A cylindrical tank has radius 7 m and height 10 m. Find its volume in litres. (Use π = 22/7)

**Solution**:

• Volume = πr²h = (22/7) × 7² × 10 = (22/7) × 49 × 10 = 22 × 70 = 1,540 m³
• Convert to litres: 1 m³ = 1,000 litres
• Capacity = 1,540 × 1,000 = 15,40,000 litres (or 15.4 lakh litres)

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### Example 3: Surface Area (Cone) **Problem**: A cone has base radius 6 cm and height 8 cm. Find its Total Surface Area.

**Solution**:

• First find slant height: l = √(r² + h²) = √(36 + 64) = √100 = 10 cm
• CSA = πrl = (22/7) × 6 × 10 = 1320/7 cm²
• Base area = πr² = (22/7) × 36 = 792/7 cm²
• TSA = CSA + Base = 1320/7 + 792/7 = 2112/7 = 301.71 cm² (approx. 302 cm²)

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Common Mistakes

• **Confusing perimeter with area** → Perimeter is a length (single unit like cm); area is in square units (cm²). If asked for fencing, use perimeter; for painting a floor, use area.

• **Mixing radius and diameter** → Many problems give diameter. Always halve it to get radius before using formulas. Forgetting this doubles or halves your answer incorrectly.

• **Ignoring unit conversions** → If length is in metres and breadth in centimetres, convert to the same unit first. Mixing units gives absurd answers.

• **Using wrong surface area formula** → TSA includes all faces; CSA excludes flat circular/rectangular bases. Read whether the question asks for "curved" or "total" surface area.

• **Forgetting the ⅓ factor for cone/pyramid volumes** → A cone's volume is one-third that of a cylinder with the same base and height. Students often forget this fraction.

• **Misapplying Heron's formula** → For triangles with three sides (no height given), use Heron's formula: Area = √[s(s−a)(s−b)(s−c)] where s = (a+b+c)/2. Don't confuse s with side length.

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Quick Reference

• **Square**: P = 4a, A = a²; **Cube**: V = a³, TSA = 6a²

• **Rectangle**: P = 2(l+b), A = lb; **Cuboid**: V = lbh, TSA = 2(lb+bh+hl)

• **Circle**: C = 2πr, A = πr²; **Cylinder**: V = πr²h, TSA = 2πr(r+h)

• **Triangle**: A = ½ × base × height; use Heron's formula when only sides are given

• **Cone**: V = ⅓πr²h, slant height l = √(r²+h²)

• **Sphere**: V = (4/3)πr³, SA = 4πr²

• **1 m³ = 1,000 litres** — crucial for capacity problems