Mensuration

Overview

Mensuration is the branch of mathematics dealing with the measurement of geometric figures—their lengths, areas, and volumes. For WB TET Paper II (Classes 6–8), this topic carries significant weight because it connects abstract formulas to real-world applications that students encounter daily: calculating the area of a floor to buy tiles, finding the volume of a water tank, or determining how much paint is needed to cover a wall.

In the TET exam, expect questions that test both formula recall and application. You must be comfortable with plane figures (2D shapes like triangles, quadrilaterals, circles) and solid figures (3D shapes like cubes, cuboids, cylinders, cones, spheres). Pedagogy questions may ask how to make mensuration meaningful through hands-on activities or real-life problem contexts.

Mastery requires memorising key formulas, understanding when each applies, and practising unit conversions. The most common errors stem from confusing perimeter with area, mixing up 2D and 3D formulas, or forgetting to square or cube units appropriately.

---

Key Concepts

• **Perimeter** is the total length of the boundary of a plane figure. It is measured in linear units (cm, m).

• **Area** is the amount of surface enclosed within a plane figure. It is measured in square units (cm², m²).

• **Surface Area** of a solid is the total area of all its outer faces. Curved surface area (CSA) excludes the base(s); total surface area (TSA) includes all surfaces.

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

• **Dimensional consistency**: Perimeter → linear (cm), Area → square (cm²), Volume → cubic (cm³). Always check your answer's unit.

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

• **Real-life connections**: Mensuration links to cost calculations—cost = rate × (perimeter/area/volume), depending on context.

---

Formulas / Key Facts

### Plane Figures (2D)

| 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 | | 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) | 2(a + b) | b × h | | Rhombus (diagonals d₁, d₂) | 4 × side | 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 |

**Heron's formula** for triangle area when all three sides are known: Area = √[s(s−a)(s−b)(s−c)], where s = (a + b + c)/2

### Solid Figures (3D)

| 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 (r, h) | 2πrh | 2πr(r + h) | πr²h | | Cone (r, h, slant l) | πrl | πr(r + l) | ⅓ × πr²h | | Sphere (r) | 4πr² | 4πr² | (4/3)πr³ | | Hemisphere (r) | 2πr² | 3πr² | (2/3)πr³ |

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

---

Worked Examples

### Example 1: Area of a Composite Figure *A rectangular garden 20 m × 15 m has a circular fountain of radius 3.5 m in its centre. Find the area of the garden excluding the fountain. (Use π = 22/7)*

**Solution** Area of rectangle = 20 × 15 = 300 m² Area of circle = πr² = (22/7) × 3.5 × 3.5 = (22/7) × 12.25 = 38.5 m² Required area = 300 − 38.5 = **261.5 m²**

---

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

**Solution** TSA = 2πr(r + h) = 2 × (22/7) × 7 × (7 + 10) = 2 × 22 × 17 = **748 cm²** Volume = πr²h = (22/7) × 49 × 10 = 22 × 70 = **1540 cm³**

---

### Example 3: Cost-Based Problem *A room 8 m long, 6 m wide needs flooring with tiles costing ₹45 per m². Find total cost.*

**Solution** Area = 8 × 6 = 48 m² Cost = 48 × 45 = **₹2160**

---

Common Mistakes

1. **Confusing perimeter and area** *Wrong*: Using 2(l + b) when asked for surface to be painted. *Fix*: Perimeter = boundary length (for fencing); Area = surface measure (for painting, tiling).

2. **Forgetting to square/cube the radius** *Wrong*: Writing πr instead of πr² for circle area. *Fix*: Area formulas always have r², volume formulas have r³ or r²h.

3. **Unit mismatch** *Wrong*: Mixing cm and m in the same calculation. *Fix*: Convert all measurements to the same unit before applying formulas.

4. **Using diameter instead of radius** *Wrong*: Plugging diameter directly into πr²h. *Fix*: Always halve the diameter to get r before substituting.

5. **Ignoring curved vs total surface area** *Wrong*: Giving CSA when TSA is asked. *Fix*: Read the question carefully—lateral/curved SA excludes bases; total SA includes everything.

---

Quick Reference

• **Square**: P = 4a, A = a²; **Cube**: TSA = 6a², V = a³
• **Rectangle**: P = 2(l+b), A = lb; **Cuboid**: TSA = 2(lb+bh+hl), V = lbh
• **Circle**: C = 2πr, A = πr²; **Cylinder**: TSA = 2πr(r+h), V = πr²h
• **Triangle (any)**: A = ½ × base × height; use Heron's when heights are unavailable
• **Sphere**: SA = 4πr², V = (4/3)πr³
• **1 litre = 1000 cm³ = 0.001 m³**