Mensuration

Perimeter, Area and Volume of Plane Figures and Solid Shapes

---

Overview

Mensuration is the branch of mathematics dealing with measurement of geometric figures — their lengths, areas and volumes. For UPTET, this topic carries significant weight in the Mathematics section of both Paper I (Classes 1–5) and Paper II (Classes 6–8). Questions typically test your ability to apply formulas quickly and accurately rather than derive them.

At the primary level, expect questions on perimeter and area of basic 2D shapes (rectangles, squares, triangles, circles). At the upper-primary level, the scope expands to surface area and volume of 3D solids (cuboids, cylinders, cones, spheres). Many questions are word problems requiring you to identify the correct formula, substitute values carefully and handle unit conversions.

Mastery here demands two things: memorising the standard formulas and practising enough problems to recognise which formula applies in a given context. This is a high-scoring area if you are systematic.

---

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 plot.

• **Area** measures the surface enclosed by a 2D figure. It tells you how much paint is needed to cover a wall or how much carpet to buy for a floor.

• **Volume** measures the space occupied by a 3D object. It tells you the capacity of a container or the amount of material in a solid block.

• **Surface Area** is the total area of all outer faces of a 3D solid. Curved Surface Area (CSA) excludes the flat bases; Total Surface Area (TSA) includes them.

• **Units matter**: Perimeter is in linear units (cm, m), area in square units (cm², m²), volume in cubic units (cm³, m³). Converting between units (e.g., m to cm) requires squaring or cubing the conversion factor for area and volume respectively.

• **Composite figures** are shapes formed by combining basic shapes. Break them into simpler parts, compute separately, then add or subtract as needed.

• **Practical applications** include finding the cost of fencing (perimeter), tiling (area), or filling a tank (volume).

---

Formulas / Key Facts

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

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

**Heron's Formula** (when only three sides are known):

• s = (a + b + c)/2
• A = √[s(s − a)(s − b)(s − c)]

### 3D Solids — Surface Area and Volume

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

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

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

---

Worked Examples

### Example 1 — Area of a Triangle (Heron's Formula) **Problem**: Find the area of a triangle with sides 13 cm, 14 cm and 15 cm.

**Solution**: 1. Semi-perimeter s = (13 + 14 + 15)/2 = 21 cm 2. s − a = 21 − 13 = 8; s − b = 21 − 14 = 7; s − c = 21 − 15 = 6 3. Area = √(21 × 8 × 7 × 6) = √7056 = 84 cm²

**Answer**: 84 cm²

---

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

**Solution**: 1. Volume = πr²h = (22/7) × 7² × 10 = (22/7) × 49 × 10 = 22 × 70 = 1540 m³ 2. TSA = 2πr(r + h) = 2 × (22/7) × 7 × (7 + 10) = 2 × 22 × 17 = 748 m²

**Answer**: Volume = 1540 m³; TSA = 748 m²

---

### Example 3 — Composite Figure **Problem**: A rectangular park 60 m × 40 m has a circular fountain of diameter 14 m at its centre. Find the area of the park excluding the fountain. (π = 22/7)

**Solution**: 1. Area of rectangle = 60 × 40 = 2400 m² 2. Radius of fountain = 14/2 = 7 m 3. Area of fountain = πr² = (22/7) × 49 = 154 m² 4. Required area = 2400 − 154 = 2246 m²

**Answer**: 2246 m²

---

Common Mistakes

• **Confusing radius and diameter**: Many errors occur because the problem gives diameter but the formula uses radius. Always check and halve if necessary.

• **Ignoring unit conversion**: If length is in metres and another dimension in centimetres, convert to the same unit before applying the formula. For area, 1 m² = 10 000 cm²; for volume, 1 m³ = 10⁶ cm³.

• **Using wrong height**: In cones and triangles, height must be perpendicular to the base. Slant height is different from vertical height.

• **Forgetting ½ or ⅓ factors**: Triangle area needs ½; cone and pyramid volumes need ⅓. Missing these factors doubles or triples your answer incorrectly.

• **Adding when you should subtract (composite figures)**: For a shaded region between two shapes, subtract the inner area from the outer. Read the question carefully.

---

Quick Reference

1. Rectangle: P = 2(l + b); A = lb 2. Circle: C = 2πr; A = πr² 3. Cylinder: V = πr²h; TSA = 2πr(r + h) 4. Cone: V = ⅓πr²h; CSA = πrl 5. Sphere: V = (4/3)πr³; SA = 4πr² 6. Heron's formula: A = √[s(s−a)(s−b)(s−c)] where s = (a+b+c)/2