1. Introduction to Mensuration

Mensuration is the branch of mathematics that deals with the measurement of lengths, areas, volumes, and other geometric properties of figures.

Key Concepts

• Perimeter: Distance around a 2D figure
• Area: Amount of space inside a 2D figure
• Surface Area: Total area of all faces of a 3D figure
• Volume: Amount of space inside a 3D figure
• Lateral Surface Area: Area of curved/side surfaces (excluding top and bottom)

2. Two-Dimensional Figures (2D)

Triangles

Basic Triangle

• Perimeter: P = a + b + c
• Area: A = ½ × base × height
• Heron’s Formula: A = √[s(s-a)(s-b)(s-c)] where s = (a+b+c)/2

Special Triangles

• Right Triangle: A = ½ × base × height
• Equilateral Triangle: A = (√3/4) × a² where a = side length
• Isosceles Triangle: Use base and height or Heron’s formula

Quadrilaterals

Rectangle

• Perimeter: P = 2(l + w)
• Area: A = l × w

Square

• Perimeter: P = 4a
• Area: A = a²
• Diagonal: d = a√2

Parallelogram

• Perimeter: P = 2(a + b)
• Area: A = base × height = ab sin θ

Rhombus

• Perimeter: P = 4a
• Area: A = ½ × d₁ × d₂ (diagonals) = a² sin θ

Trapezium (Trapezoid)

• Perimeter: P = a + b + c + d
• Area: A = ½ × (sum of parallel sides) × height = ½(a + b)h

Regular Polygons

General Formula

• Area: A = ½ × perimeter × apothem
• For n-sided regular polygon with side s:
  • Perimeter: P = ns
  • Area: A = (ns²)/(4 tan(π/n))

Common Regular Polygons

• Pentagon: A = (5s²)/(4 tan(36°)) ≈ 1.72s²
• Hexagon: A = (3√3/2)s² ≈ 2.598s²
• Octagon: A = 2(1 + √2)s² ≈ 4.828s²

Circle

• Circumference: C = 2πr = πd
• Area: A = πr²
• Arc length: l = (θ/360°) × 2πr
• Sector area: A = (θ/360°) × πr²
• Segment area: A = (sector area) – (triangle area)

3. Three-Dimensional Figures (3D)

Cube

• Surface Area: SA = 6a²
• Volume: V = a³
• Space diagonal: d = a√3

Cuboid (Rectangular Prism)

• Surface Area: SA = 2(lw + wh + lh)
• Volume: V = l × w × h
• Space diagonal: d = √(l² + w² + h²)

Cylinder

• Curved Surface Area: CSA = 2πrh
• Total Surface Area: TSA = 2πr(r + h)
• Volume: V = πr²h

Cone

• Curved Surface Area: CSA = πrl (where l = slant height)
• Total Surface Area: TSA = πr(r + l)
• Volume: V = ⅓πr²h
• Slant Height: l = √(r² + h²)

Sphere

• Surface Area: SA = 4πr²
• Volume: V = (4/3)πr³

Hemisphere

• Curved Surface Area: CSA = 2πr²
• Total Surface Area: TSA = 3πr²
• Volume: V = (2/3)πr³

Pyramid

Square Pyramid

• Lateral Surface Area: LSA = 2al (where a = base side, l = slant height)
• Total Surface Area: TSA = a² + 2al
• Volume: V = ⅓a²h

General Pyramid

• Volume: V = ⅓ × (base area) × height

Prism

General Prism

• Lateral Surface Area: LSA = perimeter of base × height
• Total Surface Area: TSA = LSA + 2 × (base area)
• Volume: V = (base area) × height

Triangular Prism

• Volume: V = ½ × base × height × length

Frustum of Cone

• Curved Surface Area: CSA = π(r₁ + r₂)l
• Total Surface Area: TSA = π(r₁ + r₂)l + π(r₁² + r₂²)
• Volume: V = ⅓πh(r₁² + r₂² + r₁r₂)

4. Combined Figures

Strategies for Complex Shapes

1. Break into simpler shapes: Divide complex figures into rectangles, triangles, circles, etc.
2. Add or subtract areas/volumes: Combine individual calculations
3. Use appropriate formulas: Apply the correct formula for each component

Common Combined Figures

• Composite rectangles: Add individual areas
• Shapes with holes: Subtract inner area from outer area
• Stepped figures: Break into rectangular sections
• Cylindrical tanks with hemispherical ends

5. Important Relationships and Conversions

Unit Conversions

• Length: 1 m = 100 cm = 1000 mm
• Area: 1 m² = 10,000 cm² = 1,000,000 mm²
• Volume: 1 m³ = 1,000,000 cm³ = 1,000 liters
• 1 hectare = 10,000 m²
• 1 acre ≈ 4047 m²

Key Relationships

• Volume of liquid in cylinder: V = πr²h
• Capacity: 1 liter = 1000 cm³
• Density: Density = Mass/Volume

6. Formula Summary Table

7. Problem-Solving Strategies

Step-by-Step Approach

1. Read carefully: Understand what is given and what to find
2. Draw a diagram: Visualize the problem
3. Identify the shape: Determine which formulas to use
4. List known values: Write down given measurements
5. Apply formulas: Substitute values into appropriate formulas
6. Calculate carefully: Pay attention to units
7. Check reasonableness: Verify your answer makes sense

Common Problem Types

1. Finding missing dimensions: Use given area/volume to find unknown length
2. Cost calculations: Area/volume × rate per unit
3. Material requirements: Surface area for painting, volume for filling
4. Optimization problems: Maximum area for given perimeter
5. Real-world applications: Water tanks, construction, packaging

8. Practical Applications

Real-World Examples

• Construction: Calculating material needed for walls, floors, roofs
• Agriculture: Land area, irrigation, fencing
• Manufacturing: Packaging, containers, storage
• Architecture: Room dimensions, building planning
• Engineering: Tank capacity, pipe volumes

Word Problem Keywords

• Perimeter: Fencing, boundary, border, outline
• Area: Coverage, paint, carpet, land
• Volume: Capacity, filling, storage, liquid
• Surface Area: Painting, wrapping, covering

9. Tips for Success

Calculation Tips

1. Use π ≈ 3.14 or 22/7 as specified in problems
2. Keep track of units throughout calculations
3. Round final answers as requested
4. Show all work step by step
5. Double-check formulas before substituting

Common Mistakes to Avoid

• Confusing perimeter with area
• Using diameter instead of radius (or vice versa)
• Forgetting to include all surfaces in surface area calculations
• Mixing up 2D and 3D formulas
• Unit conversion errors
• Not reading the question carefully

10. Practice Problem Categories

Basic Level

• Direct application of single formulas
• Simple unit conversions
• Basic combined shapes

Intermediate Level

• Multi-step problems
• Combined figures requiring addition/subtraction
• Problems involving ratios and proportions

Advanced Level

• Optimization problems
• Complex combined figures
• Real-world application problems
• Problems requiring multiple formulas and conversions

Mensuration – Worked Examples

1. Two-Dimensional Figures (2D)

Example 1: Rectangle

Problem: A rectangular garden has length 15 m and width 8 m. Find: a) Perimeter b) Area c) Cost of fencing at ₹25 per meter

Solution: Given: l = 15 m, w = 8 m

a) Perimeter: P = 2(l + w) = 2(15 + 8) = 2(23) = 46 m

b) Area: A = l × w = 15 × 8 = 120 m²

c) Cost of fencing: Cost = Perimeter × Rate per meter = 46 × 25 = ₹1,150

Example 2: Triangle (Using Heron’s Formula)

Problem: Find the area of a triangle with sides 13 cm, 14 cm, and 15 cm.

Solution: Given: a = 13 cm, b = 14 cm, c = 15 cm

First, find semi-perimeter: s = (a + b + c)/2 = (13 + 14 + 15)/2 = 42/2 = 21 cm

Using Heron’s formula: A = √[s(s-a)(s-b)(s-c)] A = √[21(21-13)(21-14)(21-15)] A = √[21 × 8 × 7 × 6] A = √[7056] = 84 cm²

Example 3: Circle and Sector

Problem: A circular park has radius 21 m. Find: a) Circumference b) Area c) Area of a sector with central angle 60°

Solution: Given: r = 21 m, θ = 60°

a) Circumference: C = 2πr = 2 × (22/7) × 21 = 132 m

b) Area: A = πr² = (22/7) × 21² = (22/7) × 441 = 1,386 m²

c) Sector Area: Sector Area = (θ/360°) × πr² = (60°/360°) × 1,386 = (1/6) × 1,386 = 231 m²

Example 4: Trapezium

Problem: A trapezium has parallel sides of 12 cm and 8 cm, with height 5 cm. Find its area and the length of the other two sides if they are equal and the trapezium is isosceles.

Solution: Given: Parallel sides a = 12 cm, b = 8 cm, h = 5 cm

Area: A = ½(a + b) × h = ½(12 + 8) × 5 = ½ × 20 × 5 = 50 cm²

For equal non-parallel sides: The difference in parallel sides = 12 – 8 = 4 cm Each side extends (4/2) = 2 cm horizontally Using Pythagorean theorem: Side length = √(h² + 2²) = √(5² + 2²) = √(25 + 4) = √29 ≈ 5.39 cm

2. Three-Dimensional Figures (3D)

Example 5: Cylinder

Problem: A cylindrical water tank has radius 3.5 m and height 6 m. Find: a) Curved surface area b) Total surface area c) Volume d) Capacity in liters

Solution: Given: r = 3.5 m, h = 6 m

a) Curved Surface Area: CSA = 2πrh = 2 × (22/7) × 3.5 × 6 = 132 m²

b) Total Surface Area: TSA = 2πr(r + h) = 2 × (22/7) × 3.5 × (3.5 + 6) = 2 × (22/7) × 3.5 × 9.5 = 209 m²

c) Volume: V = πr²h = (22/7) × (3.5)² × 6 = (22/7) × 12.25 × 6 = 231 m³

d) Capacity in liters: Capacity = 231 m³ = 231 × 1000 = 231,000 liters

Example 6: Cone

Problem: A conical tent has base radius 7 m and height 24 m. Find: a) Slant height b) Curved surface area c) Volume

Solution: Given: r = 7 m, h = 24 m

a) Slant Height: l = √(r² + h²) = √(7² + 24²) = √(49 + 576) = √625 = 25 m

b) Curved Surface Area: CSA = πrl = (22/7) × 7 × 25 = 550 m²

c) Volume: V = ⅓πr²h = ⅓ × (22/7) × 7² × 24 = ⅓ × 22 × 7 × 24 = 1,232 m³

Example 7: Sphere and Hemisphere

Problem: A hemispherical bowl has internal radius 9 cm. Find: a) Curved surface area b) Total surface area c) Volume

Solution: Given: r = 9 cm

a) Curved Surface Area: CSA = 2πr² = 2 × (22/7) × 9² = 2 × (22/7) × 81 = 509.14 cm²

b) Total Surface Area: TSA = 3πr² = 3 × (22/7) × 9² = 3 × (22/7) × 81 = 763.71 cm²

c) Volume: V = (2/3)πr³ = (2/3) × (22/7) × 9³ = (2/3) × (22/7) × 729 = 1,527.43 cm³

Example 8: Cuboid

Problem: A room is 12 m long, 8 m wide, and 3 m high. Find: a) Total surface area b) Cost of painting walls and ceiling at ₹15 per m² c) Volume

Solution: Given: l = 12 m, w = 8 m, h = 3 m

a) Total Surface Area: TSA = 2(lw + wh + lh) = 2(12×8 + 8×3 + 12×3) = 2(96 + 24 + 36) = 2(156) = 312 m²

b) Area to be painted (excluding floor): Area = TSA – floor area = 312 – 96 = 216 m² Cost = 216 × 15 = ₹3,240

c) Volume: V = l × w × h = 12 × 8 × 3 = 288 m³

3. Combined and Complex Figures

Example 9: Combination of Rectangle and Semicircle

Problem: A window consists of a rectangle topped by a semicircle. The rectangle is 80 cm wide and 120 cm high. Find the total area and perimeter.

Solution: Given: Rectangle width = 80 cm, height = 120 cm Semicircle diameter = 80 cm, so radius = 40 cm

Total Area: Rectangle area = 80 × 120 = 9,600 cm² Semicircle area = ½πr² = ½ × (22/7) × 40² = ½ × (22/7) × 1,600 = 2,514.29 cm² Total area = 9,600 + 2,514.29 = 12,114.29 cm²

Perimeter: Rectangle perimeter (without top) = 80 + 120 + 120 = 320 cm Semicircle circumference = πr = (22/7) × 40 = 125.71 cm Total perimeter = 320 + 125.71 = 445.71 cm

Example 10: Cylinder with Hemispherical Ends

Problem: A capsule-shaped container consists of a cylinder with hemispherical ends. The cylinder has radius 3 cm and height 8 cm. Find the total volume and surface area.

Solution: Given: r = 3 cm, cylinder height = 8 cm

Total Volume: Cylinder volume = πr²h = π × 3² × 8 = 72π cm³ Two hemispheres = One sphere volume = (4/3)πr³ = (4/3)π × 3³ = 36π cm³ Total volume = 72π + 36π = 108π = 108 × (22/7) = 339.43 cm³

Total Surface Area: Cylinder curved surface = 2πrh = 2π × 3 × 8 = 48π cm² Two hemispheres = Sphere surface area = 4πr² = 4π × 3² = 36π cm² Total surface area = 48π + 36π = 84π = 84 × (22/7) = 264 cm²

4. Practical Applications

Example 11: Water Tank Problem

Problem: A rectangular water tank 4 m × 3 m × 2 m is ¾ full of water. If water is being pumped out at 500 liters per minute, how long will it take to empty the tank?

Solution: Tank dimensions: 4 m × 3 m × 2 m Tank volume = 4 × 3 × 2 = 24 m³ = 24,000 liters

Water in tank = ¾ × 24,000 = 18,000 liters Pumping rate = 500 liters per minute

Time to empty = 18,000 ÷ 500 = 36 minutes

Example 12: Construction Material

Problem: A cylindrical pillar of radius 35 cm and height 4 m needs to be painted. If 1 liter of paint covers 10 m², how much paint is needed?

Solution: Given: r = 35 cm = 0.35 m, h = 4 m

Surface area to paint = Curved surface area = 2πrh = 2 × (22/7) × 0.35 × 4 = 8.8 m²

Paint needed = 8.8 ÷ 10 = 0.88 liters

Example 13: Frustum of Cone

Problem: A bucket is in the shape of a frustum of a cone. The top radius is 20 cm, bottom radius is 10 cm, and height is 24 cm. Find its volume.

Solution: Given: r₁ = 20 cm, r₂ = 10 cm, h = 24 cm

Volume of frustum = ⅓πh(r₁² + r₂² + r₁r₂) = ⅓ × π × 24 × (20² + 10² + 20×10) = ⅓ × π × 24 × (400 + 100 + 200) = ⅓ × π × 24 × 700 = 8 × π × 700 = 5,600π cm³ = 5,600 × (22/7) = 17,600 cm³ = 17.6 liters

5. Challenge Problems

Example 14: Optimization Problem

Problem: A farmer has 100 m of fencing to enclose a rectangular field. What dimensions give the maximum area?

Solution: Let length = l and width = w Perimeter constraint: 2l + 2w = 100 Therefore: l + w = 50, so w = 50 – l

Area = l × w = l(50 – l) = 50l – l²

For maximum area, take derivative and set to zero: dA/dl = 50 – 2l = 0 Therefore: l = 25 m and w = 25 m

Maximum area = 25 × 25 = 625 m²

Answer: A square with sides 25 m gives maximum area.

Example 15: Complex Combined Figure

Problem: A hall has dimensions 20 m × 15 m × 6 m. It has 8 windows (each 2 m × 1.5 m) and 2 doors (each 3 m × 2 m). Find the cost of painting walls and ceiling at ₹12 per m².

Solution: Room dimensions: 20 m × 15 m × 6 m

Total wall and ceiling area: Walls = 2(20×6) + 2(15×6) = 240 + 180 = 420 m² Ceiling = 20 × 15 = 300 m² Total = 420 + 300 = 720 m²

Area of openings: Windows = 8 × (2 × 1.5) = 24 m² Doors = 2 × (3 × 2) = 12 m² Total openings = 24 + 12 = 36 m²

Area to be painted = 720 – 36 = 684 m²

Cost = 684 × 12 = ₹8,208

6. Tips for Problem Solving

General Strategy

1. Read carefully and identify the shape(s)
2. Draw a diagram when helpful
3. List given information clearly
4. Identify what to find
5. Choose appropriate formulas
6. Substitute values carefully
7. Check units throughout
8. Verify reasonableness of answer

Common Problem Patterns

• Direct application: Use single formula
• Reverse calculation: Given area/volume, find dimension
• Cost problems: Area/volume × rate
• Comparison problems: Find ratios or differences
• Optimization: Maximum/minimum values
• Combined figures: Break into simple shapes