Mensuration

Overview

Mensuration is the branch of mathematics concerned with the measurement of geometric figures—their lengths, areas, and volumes. For TN TET, this topic carries significant weightage in the Mathematics section and tests your ability to apply formulas to calculate perimeter, area, surface area, and volume of standard 2D and 3D shapes.

The scope covers figures that students encounter from Classes 1–8: rectangles, squares, triangles, circles, parallelograms, trapeziums, cubes, cuboids, cylinders, cones, and spheres. Questions typically involve direct formula application, unit conversion, or word problems requiring you to identify the correct shape and formula. Mastery here also supports the pedagogy section, as you must understand how to teach these concepts through concrete-pictorial-abstract progression.

Expect 3–5 questions on mensuration in Paper I and Paper II. Accuracy depends on memorising formulas correctly, handling units carefully, and visualising the figure described in word problems.

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Key Concepts

• **Perimeter** is the total length of the boundary of a 2D figure. Think of it as the length of fence needed to enclose a field.

• **Area** is the measure of 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. Lateral (curved) surface area excludes the base(s); Total surface area includes everything.

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

• **Unit Conversion** is critical: 1 m = 100 cm; 1 m² = 10,000 cm²; 1 m³ = 1,000,000 cm³ = 1000 litres.

• **π (pi)** is taken as 22/7 or 3.14 unless specified otherwise. Use 22/7 when dimensions are multiples of 7 for cleaner calculations.

• **Composite Figures**: Many problems combine shapes (e.g., a rectangle with a semicircular end). Break them into standard parts, calculate separately, then add or subtract.

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Formulas / Key Facts

### 2D Figures

| Figure | Perimeter | Area | |--------|-----------|------| | Square (side a) | 4a | a² | | Rectangle (l × b) | 2(l + b) | l × b | | Triangle (sides a, b, c; base b, height h) | a + b + c | ½ × b × h | | Equilateral Triangle (side a) | 3a | (√3/4) × a² | | Parallelogram (base b, height h, sides a, b) | 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 |

### 3D Figures

| Solid | Curved/Lateral Surface Area | Total Surface Area | 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 (total, no base) | 4πr² | (4/3)πr³ | | Hemisphere (radius r) | 2πr² | 3πr² | (2/3)πr³ |

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

**Diagonal of cuboid**: d = √(l² + b² + h²)

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Worked Examples

### Example 1: Area of a Trapezium *A trapezium has parallel sides 12 cm and 8 cm. The perpendicular distance between them is 5 cm. Find its area.*

**Solution**:

• Formula: Area = ½ × (sum of parallel sides) × height
• Area = ½ × (12 + 8) × 5
• Area = ½ × 20 × 5 = 50 cm²

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### Example 2: Volume and Surface Area of a Cylinder *A cylindrical tank has radius 7 m and height 10 m. Find its volume and total surface area. (Use π = 22/7)*

**Solution**:

• Volume = πr²h = (22/7) × 7² × 10 = (22/7) × 49 × 10 = 22 × 70 = 1540 m³
• TSA = 2πr(r + h) = 2 × (22/7) × 7 × (7 + 10) = 2 × 22 × 17 = 748 m²

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### Example 3: Composite Figure *A rectangular field 80 m × 60 m has a circular pond of radius 14 m at its centre. Find the area of the field excluding the pond.*

**Solution**:

• Area of rectangle = 80 × 60 = 4800 m²
• Area of circle = πr² = (22/7) × 14² = (22/7) × 196 = 616 m²
• Remaining area = 4800 − 616 = 4184 m²

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Common Mistakes

1. **Confusing perimeter and area** → Perimeter is a length (single unit like cm); area is in square units (cm²). Always check what the question asks.

2. **Forgetting to halve for triangles and trapeziums** → The formulas include ½. Missing it doubles your answer incorrectly.

3. **Mixing radius and diameter** → Many problems give diameter. Divide by 2 to get radius before substituting into circle/cylinder formulas.

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

5. **Using wrong surface area type** → Read carefully: "curved surface area" excludes bases; "total surface area" includes them. For an open tank, exclude the open face.

6. **Forgetting slant height for cones** → Curved surface area of a cone uses slant height l, not vertical height h. Calculate l = √(r² + h²) if not given.

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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, CSA = 2πrh

• **Triangle**: A = ½ × base × height; **Cone**: V = ⅓πr²h, CSA = πrl

• **Sphere**: V = (4/3)πr³, SA = 4πr²; **Hemisphere**: V = (2/3)πr³, TSA = 3πr²

• **1 litre = 1000 cm³ = 0.001 m³** — essential for tank/container problems