Mensuration (Class 6–8)

Overview

Mensuration is the branch of mathematics dealing with the measurement of geometric figures—their lengths, areas, and volumes. For UPTET Paper II, this topic bridges arithmetic skills with spatial reasoning, testing whether candidates can apply formulas to everyday objects like fields, tanks, rooms, and containers. Questions typically involve calculating perimeter and area of plane figures (rectangles, triangles, circles, parallelograms, trapeziums) and surface area and volume of solids (cubes, cuboids, cylinders, cones, spheres).

This is a high-scoring section because formulas, once memorised, can be directly applied. However, examiners often set traps by mixing units (cm and m), giving diameter instead of radius, or asking for total vs curved surface area. Mastery here requires knowing which formula applies to which shape, careful unit conversion, and quick mental checks to verify reasonableness of answers.

Key Concepts

• **Perimeter** is the total length of the boundary of a 2D figure; **area** is the measure of the region enclosed within that boundary.

• **Surface area** of a 3D solid is the total area of all its outer faces; it can be **curved surface area (CSA)** (excluding flat ends) or **total surface area (TSA)** (including all faces).

• **Volume** is the amount of 3D space a solid occupies, measured in cubic units (cm³, m³, litres where 1 litre = 1000 cm³).

• For composite figures, break them into standard shapes, compute separately, then add or subtract as required.

• Unit consistency is critical: convert all measurements to the same unit before applying any formula.

• The value of π is taken as 22/7 or 3.14 unless otherwise specified; use whichever simplifies calculation.

• Doubling linear dimensions quadruples area and multiplies volume by eight—useful for ratio-based questions.

Formulas / Key Facts

### 2D Figures — Perimeter (P) and Area (A)

| Figure | Perimeter | Area | |--------|-----------|------| | Rectangle (l × b) | 2(l + b) | l × b | | Square (side a) | 4a | a² | | 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, 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 |

### 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, slant height l, height h) | πrl | πr(r + l) | ⅓ × πr²h | | Sphere (radius r) | 4πr² (TSA = CSA) | — | (4/3)πr³ | | Hemisphere (radius r) | 2πr² | 3πr² | (2/3)πr³ |

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

Worked Examples

**Example 1 — Area of a trapezium-shaped field**

A field is in the shape of a trapezium with parallel sides 25 m and 15 m and perpendicular distance 8 m. Find its area.

*Solution:* Area = ½ × (sum of parallel sides) × height = ½ × (25 + 15) × 8 = ½ × 40 × 8 = 160 m²

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**Example 2 — Volume and TSA of a cylinder**

A cylindrical water tank has diameter 1.4 m and height 2 m. Find its volume and total surface area. (Use π = 22/7)

*Solution:* Radius r = 1.4/2 = 0.7 m, h = 2 m

Volume = πr²h = (22/7) × (0.7)² × 2 = (22/7) × 0.49 × 2 = (22 × 0.49 × 2)/7 = 21.56/7 × 2 (simplify: 0.49 × 22 = 10.78; 10.78 × 2 = 21.56) = 3.08 m³ = 3080 litres (since 1 m³ = 1000 litres)

TSA = 2πr(r + h) = 2 × (22/7) × 0.7 × (0.7 + 2) = 2 × (22/7) × 0.7 × 2.7 = (44/7) × 1.89 = 44 × 0.27 = 11.88 m² ≈ 11.88 m²

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**Example 3 — Composite solid**

A toy is made by placing a cone of height 4 cm on a hemisphere of radius 3 cm. Find the total surface area. (Use π = 3.14)

*Solution:* Hemisphere CSA = 2πr² = 2 × 3.14 × 9 = 56.52 cm²

Cone slant height l = √(r² + h²) = √(9 + 16) = √25 = 5 cm Cone CSA = πrl = 3.14 × 3 × 5 = 47.1 cm²

Total surface area = 56.52 + 47.1 = 103.62 cm²

(Note: The circular base of the cone sits on the hemisphere, so we do not add it.)

Common Mistakes

• **Using diameter as radius:** Many problems give diameter; always halve it before applying πr² or 2πr formulas.

• **Confusing CSA with TSA:** Read the question carefully—"curved surface area" excludes flat ends; "total surface area" includes them.

• **Ignoring unit conversion:** Mixing cm and m in the same calculation leads to answers off by factors of 100 or 1000. Convert first.

• **Forgetting to square or cube units:** Area is in cm² or m²; volume is in cm³ or m³. Omitting exponents costs marks.

• **Wrong formula for trapezium vs parallelogram:** Trapezium has two different parallel sides, so use ½(a + b)h; parallelogram has equal opposite sides, so use base × height directly.

Quick Reference

• Rectangle area = l × b; perimeter = 2(l + b)
• Circle area = πr²; circumference = 2πr
• Cylinder volume = πr²h; TSA = 2πr(r + h)
• Cone volume = ⅓πr²h; slant height = √(r² + h²)
• Sphere volume = (4/3)πr³; surface area = 4πr²
• 1 m³ = 1000 litres; always unify units before calculation